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TRANSISTOR CIRCUIT ANALYSIS SIMON md SCHUSTER TECH OUTLINES ALFRED D. GRONNER theory and step-by-step solutions to 235 problems TRANSISTOR CIRCUIT ANALYSIS ALFRED D. GRONNER Singer-General Precision, Inc. REVISED EDITION SIMON ad SCHUSTER TECH OUTLINES SIMON AND SCHUSTER, NEW YORK "X ! HA r 35 COLLEGE J I PRESTON /I HjQoSS. y Copyright©1966, 1970 by Simon & Schuster, Inc. All rights reserved. No part of this material may be reproduced in any form without permission in writing from the publisher. Published by Simon and Schuster Technical and Reference Book Division 1 West 39th Street New York, N.Y. 10018 Published simultaneously in Canada Printed in the United States of America PREFACE This book combines the advantages of both the textbook and the so-called review book. As a textbook it can stand alone, because it contains enough descrip- tive material to make additional references unnecessary. And in the direct manner characteristic of the review book, it has hundreds of completely solved problems that amplify and distill basic theory and methods. It is my intention that this book serve equally well as a basic text for an introductory course, and as a collateral problem- solving manual for the electrical engineering student at the junior- or senior-level, who has had a course in circuit theory. It is also a useful supplement for the student taking advanced courses in related areas that require a knowledge of transistors. The analysis and design problems should benefit professional engineers encountering transistors for the first time. Although the principles of transistor circuit design and analysis are developed in an academic manner, a practical emphasis is maintained throughout; i.e., the stu- dent is shown how to "size up" a problem physically, and to estimate the approximate magnitudes of such parameters as quiescent operating point, impedances, gain, etc. Moreover, a scrupulous effort is made in the solved problems to keep sight of under- lying analytical and physical principles, thereby establishing a strong background for the practical problems that arise in the analysis and design of circuits. New concepts, definitions, and important results are tinted in grey throughout the text. The solved problems are generally comprehensive, and incorporate numerous applications. Supplementary problems are included not only for exercise but also to strengthen the skill and insight necessary for the analysis and design of circuits. After a preliminary discussion of semiconductor principles in Chap. 1, a com- plete chapter is devoted to graphical analysis of semiconductor circuits. Thus the foundation is laid for succeeding chapters on small- and large-signal parameters. Nonlinearities, in particular, are easily investigated by means of the graph- ical methods described. Chapter 3 provides a thorough coverage of the small-signal equivalent circuit, with emphasis on the tee-equivalent and hybrid configurations. The hybrid-* circuit is introduced in connection with the high-frequency limitations of transistor behavior. Chapter 4 presents a variety of bias circuit configurations, including leakage effects, stability factors, temperature errors, and methods of bias stabilization. Chapter 5 establishes the basic formulae for the small-signal amplifier. Multi- stage amplifiers, together with various feedback circuits, are considered in Chap. 6. Power amplifiers, both single-ended and push-pull, are covered in Chap. 7. Chapter 8 rounds out much of the material on feedback developed in earlier chapters, and investigates the operational amplifier and the stability of high-gain feedback amplifiers by Nyquist and Bode techniques. The appendices provide a convenient reference to transistor characteristics, important formulae, asymptotic plotting, and distortion calculations. I am deeply grateful to Mr. Sidney Davis, who made important contributions to both the first and second editions in organizing the problems, unifying the nota- tion, and commenting on the contents as a whole. I also wish to acknowledge the editorial efforts of Raj Mehra of Simon & Schuster, Inc., towards the revision of the first edition. Alfred D. Gronner White Plains, New York TABLE OF CONTENTS 1 Page SEMICONDUCTOR PHYSICS AND DEVICES 1 .1 Basic Semiconductor Theory 1 1 .2 Effects of Impurities 3 1 .3 The p-n Junction 5 1 .4 The Transistor 1 2 1 .5 The Ebers-Moll Model of the Transistor 15 1 .6 Basic Transistor Amplifier Circuits 17 1 .7 Transistor Leakage Currents 18 1 .8 Transistor Breakdown 19 1.9 D-C Models 20 1.10 The Hybrid-Jt Equivalent Circuit 21 1.11 Supplementary Problems 22 TRANSISTOR CIRCUIT ANALYSIS 2.1 Characteristic Curves 24 2.2 The Operating Point 26 2.3 The Load Line 27 2.4 Small- and Large-Signal A-C Circuits 30 2.5 Supplementary Problems 37 SMALL-SIGNAL EQUIVALENT CIRCUITS 3.1 Introduction 38 3.2 Hybrid Equivalent Circuit 38 3.3 Tee-Equivalent Circuit 43 3.4 Common-Base Parameters 47 3.5 Derivation of Common-Base Parameters 48 3.6 Calculation of Amplifier Performance 52 3.7 Hybrid-jr Equivalent Circuit 62 3.8 Supplementary Problems 66 8 BIAS CIRCUITS AND STABILITY 4.1 Introduction e _ 4.2 Leakage Current 67 4.3 Tee-Equivalent Circuit Representation of Leakage 68 4.4 Constant Base Voltage Biasing Techniques 71 4.5 Stability Factors 73 4.6 Emitter Bias Circuit 83 4.7 Bias Compensation 8 c 4.8 Self-Heating gg 4.9 Thermal Runaway 8g 4.10 Approximation Techniques 92 4.1 1 Supplementary Problems 96 SINGLE-STAGE AMPLIFIERS 5.1 Introduction g 5.2 Common-Emitter Circuit g7 5.3 Common-Base Circuit 107 5.4 Common-Collector Circuit (Emitter-Follower) 109 5.5 High-Frequency Performance -, 1 o 5.6 Hybrid-* Circuit 1 ^ 5.7 Supplementary Problems 120 MULTI-STAGE AMPLIFIERS 6.1 Introduction 12 1 6.2 Capacitor Coupling 1 24 6.3 Transformer Coupling 1 38 6.4 Direct Coupling 144 6.5 Complementary Transistors 155 6.6 Supplementary Problems 159 POWER AMPLIFIERS 7.1 Introduction 1 6 q 7.2 Distortion 1 66 7.3 Power Amplifier Design Equations '.'.'.'.'.'. 1 71 7.4 Common-Base Connection j 1 73 7.5 Common-Collector Power Amplifier Stage \ 1 76 7.6 Push-Pull Amplifiers _" "i 78 7.6a Class A Push-Pull Amplifier 178 7.6b Class B Push-Pull Amplifier ................ ...^80 7.7 Supplementary Problems 185 FEEDBACK 8.1 Basic Concepts of Feedback \ 1 86 8.2 Types of Feedback ! 1Qn 8.3 Stability. . . , ""'.'.}'.'.'.'.'.'.'.'.['.'.'. ]g 6 8.4 The Bode Diagram 1Qg 8.5 Operational Amplifiers .'.' I.' .'.'!.'.' .' 200 8.6 Supplementary Problems '.'.'.'.['.'.'.'.'. 202 f\ TRANSISTOR CHARACTERISTICS A.1 Types 2N929, 2N930 n-p-n Planar Silicon Transistors 203 A.1a Typical Characteristics 205 A.2 Types 2N1 162 thru 2N1 167 Transistors 209 A.2a Peak Power Derating 211 A.3 Types 2N1302, 2N1 304, 2Jsl 1306, and 2N1308 n-p-n Alloy-Junction Germanium Transistors .x. 21 2 A.3a Typical Characteristics. . . .\ 213 A.4 Types 2N1529A thru 2N1532A, 2N1534Athru 2N1537Aand 2N1529thru 2N1 538 Transistors 216 A.4a Collector Characteristics at 25°C: Types 2N1529A thru 2N1532A and 2N1 529 thru 2N1 533 Transistors. . . \. 217 A.4b Determination of Allowable Peak Power . 220 ft SUMMARY CHARTS 221 APPENDIX C FREQUENCY RESPONSE PLOTTING C.1 Introduction 227 C.2 The Asymptotic Plot 227 C.3 More Complex Frequency-Response Functions 233 D DISTORTION CALCULATION D.1 Distortion 239 E LIST OF SYMBOLS 241 APPENDIX INDEX 243 SEMICONDUCTOR PHYSICS AND DEVICES 1 CHAPTER 1 .1 Basic Semiconductor Theory Solid-state devices such as the junction diode and transistor are fabricated from semiconductor materials. These materials have electrical resistivities which lie between conductors and insulators. The princi- pal semiconductors used are the elements germanium and silicon, which in a pure state occur in crystalline form; namely, where the atoms are arranged uni- formly in a periodic pattern. To fully appreciate the operation of solid-state devices, a familiarity with atomic physics is needed. Refer to Fig. 1.1, which shows the atomic models of germanium and silicon. The nuclei of the atoms have 32 and 14 units of positive charge or protons, respectively, while around the nuclei orbit an identical number of units of negative charge or electrons. This equalization of charges results in the atoms possessing a total effective charge which is neutral. The electron orbits are arranged in shells designated by the letters K, L, M, N, ... . According to quantum mechanics, the maximum allowable number of electrons in shell K is 2, in L, 8, in M, 18, and in N, 32. A filled shell has very little influence on chemical processes involving a particular atom. The electrons in their individual orbits around the nucleus exhibit specific energy values, called discrete energy levels. These are determined by the momentum of the electrons and their distance from the nucleus. The bond between the electron and the nucleus is inversely proportional to the distance between them. The closer they are, the greater the energy required to free the electron from the atom. Subsequently, electrons which are remote from the nucleus require less energy to free themselves from the atom. Valence band Valence band Va lence band Valence band (a) (b) Fig. 1.1 Models of (a) germanium and (b) silicon atoms, and their simplified representations. Transistor Circuit Analysis Conduction band Conduction band Valence band (b) Valence electrons are those in the outer orbit which can break away more freely from the atom. The inner orbit electrons can be combined with the nucleus; in effect, simplified to a central core or kernel (Fig. 1.1), which may then be considered a modified nucleus. The valence band electrons or outer orbit elec- trons determine the chemical and crystalline properties of the elements. Valence electrons exist at excitation levels if energy is supplied from some external source. When the energy source is removed, the electrons normally fall back into the valence band. The most common source of energy that moves valence electrons into excitation levels is beat. At absolute zero, electrons do not exist at excitation levels. Valence electrons at excitation levels are called free electrons. They are so loosely held by the nucleus that they will move relatively freely through a semiconductor in response to applied electrical fields as well as other forces. Now consider a semiconductor crystal wherein the atoms are arranged uni- formly in a periodic pattern. The proximity of neighboring atoms leads to modi- fications in the energies of the valence electrons. The energies are distributed in an energy band that represents the range of energies of the valence electrons in the crystal. Although the energies of the specific electrons have discrete values, the energy bands corresponding to the valence electrons in the crystal appear almost as a continuous band of energy distribution. There is also a corresponding energy band for every shell within each atom of the crystal. The bands are separated by energy gaps, which represent the energy required to move electrons between bands. Energy is generally expressed in electron volts (1 ev = 1.6 x 10" 1 ' joules). Quantum mechanics demonstrates that electrons can only exist at energy levels within the bands and not at levels within the forbidden gaps. Electron motion within an energy band can only occur if the band is not filled, such as in the case of the valence band. If sufficient energy is applied to an electron, it can move from its band to a higher band. Heat or energy supplied by an external electric field can move an electron from the valence band to the con- duction band, where it may travel with relative ease through the crystal. If all valence bands in a crystal are filled, conduction can only occur if electrons are first moved to the conduction band. The vacant sites left in the valence band are called holes. Conduction band Fig. 1.2 Comparison of energy gaps between valence and conduction bands for (a) conductors, (b) semi- conductors, and (c) insulators. PROBLEM 1.1 What distinguishes conductors, semiconductors, and insulators in terms of the forbidden energy gap? Solution: In a conductor, the forbidden gap between conduction and valence bands is zero (they actually overlap in most conductors.) Therefore, no energy is needed to move electrons into the conduction band and electron flow is large for small applied voltage. In a semiconductor, the forbidden gap is on the order of 1 v. Temperature will excite some electrons across it, but the number so excited is small. In an insulator, the forbidden gap is very wide and almost no electrons are available for conduction. Therefore a large amount of energy is required to cause conduction. Figure 1.2 shows the above energy levels in a convenient, diagrammatic form. The valence bands of both germanium and silicon atoms have 4 electrons each (Fig. 1.1), and in crystals form covalent bonds; i.e., adjacent atoms share pairs of valence electrons. At absolute zero temperature the valence band is filled, and there are no electrons available for conduction. The semiconductor is then said to have infinite resistivity. As temperature increases, the valence electrons absorb energy and a certain number break their covalent bonds Semiconductor Physics and Devices ^=7 ?sr/ \/ Electron \/ Hole /V, Of / \^ Electron / \ Fig. 1.3 The generation of mobi le electron-hole pairs due to thermal agitation in a germanium crystal shown two-dimensionally. (Fig. 1.3). The broken bonds move electrons into the conduction band, leaving holes in the valence band. This makes conduction possible in both bands. In the conduction band, the free electrons move in response to an applied electric field, while in the valence band, electrons move by shifting from one hole to the next. The latter process is most easily visualized by regarding the holes as positive particles, moving under the influence of an electric field. When the holes reach an electrode, they neutralize electrons at the electrode, so that the resultant current cannot be distinguished outside the semiconductor from the more familiar conduction band current (Fig. 1.4). The valence electrons of common semiconductors require relatively large amounts of energy to break their covalent bonds, and thus exhibit a characteris- tic poor conductivity. The valence electrons of silicon and germanium need, respectively, 1.1 ev and 0.72 ev to excite them out of their covalent bonds. The greater energy needed for the silicon electrons indicates that pure silicon has higher ohmic resistance than pure germanium. The resistivity of the pure semi- conductor is its intrinsic resistivity. PROBLEM 1.2 Why does the conductivity of a semiconductor increase, rather than decrease with temperature, as does the conductivity of a metal? Solution: As temperature increases in a semiconductor, the number of electron- bole pairs generated by thermal agitation increases. The liberated electrons and holes are current carriers, and thus provide increasing conductivity. But at very high temperatures, when sufficiently large numbers of free elec- trons and holes are generated, collisions tend to increase resistance by reducing the average speed of the current carriers. 1.2 Effects of Impurities When an electron moving through a semiconductor crystal encounters a hole, recombination occurs. We may think of the electron as "en- tering" the hole, and the electron-hole pair thereby ceasing to exist. At any given temperature, equilibrium exists where the rate of thermal generation of electron-hole pairs equals the recombination rate. It may thus be inferred that in a pure semiconductor crystal, the number of electrons equals the number of holes. The crystal is, of course, electrically neutral. •- © e-© © •— © ©— ■ I (+) Positive ions (3) Negative ions Q Free holes O Free electrons Fig. 1.4 Movement of electron s and holes in two types of semi- conductors. Transistor Circuit Analysis To create a useful semiconductor device, a small amount of a specific im- purity element is added to the pure semiconductor crystal. The technique is called doping. The most common impurity elements are atoms of approximately the same volume as the atoms of the crystal or host, in order to minimize dis- location of the crystal structure. However, the impurity atoms have either one electron more (pentavalent) or one electron less (trivalent) in their valence bands than the host. When the impurity atoms are introduced into the crystal structure to form covalent bonds with the host atoms, there will be - depending on the type of impurity - either an extra electron or extra hole in the vicinity of each impurity atom. Impurities that contribute extra electrons are called donor or n-type (n for negative) impurities, and the crystal thus treated becomes an n-type semiconduc- tor. Analogously, impurities that contribute extra holes are called acceptor or p-type (p for positive) impurities, and the crystal thus treated becomes a p-type semiconductor. Figure 1.5 shows how the type of impurity determines whether a semiconductor becomes either an n-type or p-type. (a) Fig. 1.5 Effect of impurities on pure germanium crystals, (a) Donor impurity provides mobile electrons. The positively-charged atoms are not free to move, (b) Acceptor impurity provides mobile holes. The negatively -charged atoms are not free to move. Typical numbers showing impurity effects are of interest. Pure silicon, for example, has approximately 10*° charge carriers (electrons and holes) per cubic centimeter at room temperature, and an intrinsic resistivity of 240,000 Q-cm. Typically, a crystal of silicon might be doped by one donor atom per 10' host atoms with a corresponding reduction in resistivity. PROBLEM 1.3 What effect do added impurities have on semiconductor conductivity? Solution: Added impurities contribute electrons or holes which are not rigidly held in covalent bonds. Thus electrons may move freely through n-type material, thereby creating an electric current. Similarly, the principal current in p-type material is that of boles moving through the crystal in the opposite direction to the movement of electrons. Electron and hole motion constitute components of current flow. The charge carriers contributed by the impurity atoms lead to substantially increased conductivity. In n-type material, electrons are called majority carriers and holes are called minority carriers. In p-type material, the holes are majority carriers and the electrons minority carriers. Both p-type and n-type materials are normally elec- trically neutral even though free holes and electrons are present. Semiconductor Physics and Devices PROBLEM 1.4 Would you expect minority carrier current flow in response to an applied voltage? Solution: Minority carrier flow occurs in response to an applied voltage since it is a current carrier. Majority carrier flow, however, is predominant, except at high temperatures where thermally-generated electron-hole pairs lead to a higher proportion of minority carriers. 1 .3 The p-n Junction If p-type and n-type materials are mechanically joined together to form a single crystal, and they thereby create a junction in which the continuity of the crystalline structure is preserved - such a junction is called a p-n junction or junction diode. Since both the p-type and n-type materials exist at different charge levels be- cause of natural and impurity differences, they seek equilibrium between one another and an energy exchange occurs. Thus electrons and holes migrate across the p-n junction by the fundamental process of diffusion; i.e., the spread of charge carriers from regions of high concentration to regions of low concentra- tion, ultimately tending toward uniform distribution. By diffusion, the holes mi- grate from p-type to n-type material, while the electrons move in the opposite direction. Figure 1.6a shows the p-n junction. During diffusion, the excited or ionized areas on either side of the junction become relatively free of charge carriers due to the annihilation of electrons and holes by recombination, and are called the depletion layer or region. An electric field also builds up, generated by the newly created positive and negative ions located in the opposing materials, and conduction decreases. A potential difference or barrier is thus created in the depletion region (Fig. 1.6b) which inhibits further electron and hole migration. This potential difference is called the potential barrier voltage or contact po- tential, and is about 0.3 v for germanium and 0.7 v for silicon at room temperature. An equilibrium condition or barrier balance in which conduction is limited by the potential difference exists between the p-type and n-type materials. How- ever, if electron-hole pairs are formed by thermal agitation in the p-type mate- rial, electrons will flow across the p-n junction aided by the electric field. Similarly, holes in the n-type material will also migrate. Therefore minority carriers continue to flow despite the barrier balance, assisted by the potential difference established by the diffusion of majority carriers. Of course any net movement of minority carriers due to increasing temperature will be balanced by further diffusion of majority carriers, and a resultant widening of the depletion Now suppose an external potential is applied to the p-n junction of Fig. 1.7a. With the polarity shown in Fig. 1.7b, the junction is forward-biased and the field of the applied potential difference opposes the internal field across the depletion layer. Majority carriers therefore will flow freely across the barrier. When the polarity of the externally applied voltage is reverse- or back-biased (Fig. L7c), die internal field across the junction is increased, and majority carriers cannot flow. However, minority carriers generated by thermal agitation continue to flow freely. This property of conducting essentially in one direction makes the p-n junction a rectifier. Note that the depletion region gets wider as applied reverse voltage is in- creased. Since the depletion layer does not contain many current carriers, it acts as an insulator, and the depletion region can be regarded as a capacitor whose plate distance varies with the reverse voltage. Depletion layer Electric field III"' -vS>.\\ _O + 0,0: 0^0 ! O0] 0©i© + © : + T — 1 + + O]GK} 0_© O_0 ©Oi©© :©©ib.©_ ;@©i©© + Junction (a) — -4 + — 4- OOI00 00:00 OO|0 OOI0O + O0jOO 0O1OO ©@!0 ©©!oo (b) Q Donor atoms \~^ Acceptor atoms + Holes — Electrons • Imperfections, etc. Fig. 1.6 A p-n junction . (a) loni zed regions on each side of the junction form a depletion layer, (b) As a re- sult of the depletion layer, a contact potential, represented symbolically by a battery, is established across the junction . © + ©♦ © + ©♦©+ © + .© .© .© .© .© .© (a) p n ©♦J3*J3* -©_..© .© _©__©__© H 1 — (b) p n £♦ 2 + §♦ & £+ 9. -§. -0. -§. -©, -©. -@. H h (c) Fig. 1.7 (a) Unenergized p-n junc- tion, (b) The effect on thep-n junc- tion of application of forward bias, (c) When the battery connections are reversed, the electrons and holes are drawn away from the p-n junction . 6 Transistor Circuit Analysis A plausible expression for current flow across a p-n junction as a function of applied voltage may be developed by using relationships from semiconductor physics. Referring to Figs. 1.7a-c, consider first the case where no external bias is applied. There are four current components flowing simultaneously across the junction: 1. A diffusion current l d „ due to electron flow from the n-type material with its relatively high concentration of mobile electrons. 2. A diffusion current I dp due to hole flow from the p-type material with its relatively high concentration of mobile holes. These two current components are majority carrier currents since they are due to electrons in the n-region and holes in the p-region. As a result of the flow of these current components, at the junction the n-type material develops a net positive charge and the p-type material a net negative charge, which leads to a potential barrier across it. This barrier limits further diffusion except for the ef- fect of thermally-generated electron-hole pairs on both sides of the junction. Consequently the remaining two current components are: 3. A current /„„ due to thermally-generated free electrons in the p-region, which are accelerated across the junction by the barrier voltage. 4. A current l ep due to thermally-generated free holes in the n-region, which are accelerated across the junction by the barrier voltage. These last current components are minority carrier currents, since they are due to electrons in the p-region and holes in the n-region. Since holes flowing in one direction across the junction, and electrons flow- ing in the opposite direction correspond to the same current direction: Total diffusion current l d = l dB + / dp 'dm Total thermally-generated current /„ = l a + I 'en< (1.1) (1.2) This latter current component, /,, is called the saturation current, and is dis- cussed later. In the absence of applied voltage, under open-circuit or short-circuit condi- tions, the net current flow must be zero. Therefore, The barrier potential adjusts itself until an equilibrium exists between the diffu- sion and thermally-generated current components. An externally applied voltage modifies this equilibrium, and leads to a con- dition of net current flow across the junction. Depending on the polarity of the applied voltage, the potential barrier is either raised or reduced. To arrive at a quantitative picture of current flow with applied voltage, it is necessary to relate /„ to barrier potential. Although the detailed background physics is beyond the scope of this book, it is intuitively clear that the rates of diffusion of electrons or holes depend on the concentrations, n and p, of elec- trons and holes, respectively, in the regions from which they diffuse. In the typical materials used in making common semiconductor devices, the concen- trations of majority carriers are affected to only a slight extent by the relatively low concentrations of thermally- generated electron-hole pairs. From semiconductor physics, I dp = pe* T (1.3) and l dn = ne' fcr < V B~V) ,?*..* (1.4) Semiconductor Physics and Devices where V B = barrier potential at equilibrium, volts, k = Boltzmann's constant » 1.38 x 10"" joules/°K, T = absolute temperature, °K, q = electron charge = 1.6 x 10 -1 ' coulomb. In consistent units, for T = 300 °K (approximately room temperature), — =0.026. (1.5) Note that I dp and / dn are proportional to the p and n concentrations, respec- tively, as noted earlier. Note further that diffusion currents decrease exponentially as V B increases, which is intuitively plausible, and consistent with previous reasoning. The total diffusion current is, therefore, Id = 'dp + '; 1 dm a or, substituting and simplifying, /^Ae*' (1.6) where A is a constant. I Since the total junction current is the difference between diffusion current r and saturation current (/,), \ l-ld-l., i Now, at V = and / = 0, ;, Hence, or /. / = i4e* T -/.. / = /,(e* T -1), (1.7) f' This is the expression for the diode current, and is often referred to as the recti- fier equation. Since kT/q = 0.026 v at room temperature, the expression for / simplifies for V » 0.026 v; that is, i^/.e* 1 " for V» 0.026 v. (1.8) ..Similarly, for V negative by more than about 0.1 v, The diode characteristic in the forward and reverse regions are plotted in i'.Fig. 1.8. The rapid increase in reverse current at a large reverse voltage is -discussed later. PROBLEM 1.5 Explain the flat character of the saturation current of the p-n junction. Solution: The reverse current is dependent on the number of minority carriers in the semiconductor crystal. When a few tenths of a volt reverse voltage is applied, all the thermally-generated minority carriers are swept across the junction. Re- verse current is limited by the number of available minority carriers. The saturation or reverse leakage current increases sharply with temperature (Fig. 1.9). As a very approximate rule of thumb, this leakage component doubles for every 10 °C increase in temperature. Crystal I breakdown 7=/.(e Minority _ carrier conduction Majority carrier conduction V I I 0.6 v I-" for Si, 0.3 v for Ge -1) (a) y^ (b) Fig. 1.8 Current vs. voltage across a p-n junction, (a) The magnitudes of the leakage current I s and forward voltage V F are greatly exaggerated, (b) A scaled drawing of a diode characteri sti c. Transistor Circuit Analysis 100 I, amp Fig. 1.10 Comparing silicon and germanium diode characteristics. Note the modified scales for plus and minus ordinates. Voltages VpQ and Vps are forward volt- ages caused primarily by the po- tential barriers. 10 u o 0.1 Germanium / Si licon 25° 50° 75° 100° 125 c Junction temperature, °C — 150° 175° Fig. 1.9 Typical variation of I s with temperature for germanium and si licon diodes. PROBLEM 1.6 Would you expect germanium or silicon to have the greater value of reverse leakage? Why? Solution: Since reverse leakage is predominantly due to minority earners gen- erated by thermal agitation, it is evident that germanium, with its lower energy gap (0.72 ev for Ge compared to 1. 1 ev for Si) between valence and excited elec- tron states, would have the greater leakage. The greater energy gap of silicon allows it to be used at much higher temperatures (to 150 °C approximately) than germanium, whose maximum junction temperature is slightly above 100 °C. PROBLEM 1.7 Sketch a curve similar to Fig. 1.8a, roughly comparing a silicon and a germanium diode. How does the width of the energy gap affect diode characteristics? Solution: Figure 1.10 provides the required sketch. In the forward direction, more voltage is required to overcome the barrier potential (related to the energy gap) for silicon. In the reverse direction, the tighter covalent bonds mean greater crystal breakdown voltages. The saturation current, as previously discussed, is higher for germanium. Note the different scales in the forward and reverse directions. PROBLEM 1.8 Derive a formula for the diode incremental forward resistance or dynamic resistance where r, = dV/cB for currents in the forward direction. Note that this is the resistance to small changes in current and voltage about a par- ticular operating current, /. Solution: The rectifier equation for the diode current is / = / 8 (e* r Differentiating (1.7) with respect to V, dV kT -1) kT [1.7] (1.9) Semiconductor Physics and Devices For large positive voltage, e<* /M "> v » 1, and I?I B e<«" [T5v from (1.7). Sub- stitute this in the expression (1.9) for dl/dV: /, and _ dV_ *T f/ ~ d/ ~ qr/' (1.10) Note that the incremental diode forward resistance r, varies inversely with the diode forward current. PROBLEM 1.9 At room temperature, find the incremental resistance r { as a function of /. Introduce numerical values for k, T, and q. Solution: Room temperature, 25°C, corresponds to 273 + 25°C = 298°K. Substi- tute in (1.10) using numerical values for k and q: 26 Q, (1.11) where / is in milliamperes. PROBLEM 1.10 What is the incremental resistance of a forward conducting p-n junction with 2 ma current? Solution: Substitute in (1.11): ff ^26 = 26 =13Q / 2 PROBLEM 1.11 If forward diode resistance r f is 13 fl at 25°C and 2 ma, what is the resistance at 125 °C and 2 ma? Solution: Refer to (1.10) in which r t is directly proportional to absolute ♦ temperature: t tt 273+125 (13)=^(13)=17.4fl. ' 125 ° c 273 + 25 V 298 The effect of temperature on the diode forward characteristics may now be estimated. From (1.10), the incremental diode forward resistance r, is inversely proportional to current and directly proportional to temperature, °K. Since °K = 273 f °C, relatively small changes in °C about a room temperature ambient lead to much smaller percentage changes in °K. Consequently r, is relatively in- dependent of temperature. However, diode obmic resistance decreases with in- creasing temperature due to increased thermal agitation. A family of diode forward characteristics as a function of temperature has the general appearance of Fig. 1.11. Points at the same current level have the same incremental resistance (except, of course, at very low voltages where e Vq/kT is not much greater than unity), so that the curves are essentially parallel. It turns out that for a fixed forward current, the forward voltage drop decreases by about 1.5 to 3 mv/°C. As a rule of thumb, a good average figure for temperature sensitivity is - 2.5 mv/°C. This determines the separation of the individual curves of the family shown in Fig. 1.11. PROBLEM 1.12 A diode has a forward drop of 0.6 v at 10 ma where the tem- perature is 25 °C. If current is held constant, what is the forward drop at 125 °C? 75° 50° 25° 0° Fig. 1.11 Family of forward diode characteristics for different tem- peratures. At / = ^F l i a constant forward current, the curves are ap- proximately uniformly spaced for equal temperature increments. 10 Transistor Circuit Analysis Approximately what additional voltage is required at 125°C to increase current to 12 ma? Solution: At 125 °C and constant current, the diode forward drop decreases by (2.5 mv/°C) x 100 = 250 mv. The forward drop at high temperature is now only 0.350v. Using the method of Prob. 1.11, r, = 3.5 0. The additional voltage corre- sponding to a current increment of 2 ma is AV = r t M = 3.5 x 2 x 10~ 3 = 0.007 v. There is a 7 mv increase in voltage. •V (Reverse bias) Fig. 1.12 Leakage components in a semiconductor: / s = saturation com- ponent due to minority carriers, lj_ = surface leakage component, 1]_t = total leakage. An additional component of leakage current not previously described corre- sponds to surface leakage along the semiconductor surface between terminals. Humidity and surface impurities contribute to this leakage component. The magnitude of the leakage is proportional to the reverse voltage applied across the junction, in contrast to the constant /. component. At low reverse voltages, the surface leakage component is negligible. Figure 1.12 shows the effect of reverse leakage on the diode characteristic curve. At low temperatures, the surface leakage components usually predomi- nate. At high temperatures, leakage resulting from thermal agitation becomes in- creasingly important. Because silicon requires more thermal agitation to gen- erate electron-hole pairs than does germanium, the saturation leakage of silicon is much less than for germanium. Surface leakage is therefore more important in silicon p-n junctions. PROBLEM 1.13 A germanium diode has a saturation leakage of 200 //a at 25°C. Find the corresponding leakage component at 75 °C. Solution: From Fig. 1.9, it is estimated that the leakage increases over the temperature range by a ratio of 20: 1. The high temperature leakage is therefore 4 ma. PROBLEM 1.14 A silicon diode has a saturation leakage of 10 //a at 25°C. Find the corresponding leakage components at 75°C and 125°C. Solution: Again refer to Fig. 1.9. The leakage increase ratio is about 6:1, leading to a 60 ^aleakage at 75°C. At 125°C, leakage becomes 40x 10=400 //a. PROBLEM 1.15 A silicon diode operates at a reverse voltage of 10 v and has a total leakage of 50 /za. At 40 v, the leakage is 80 ^a. Find the leakage re- sistance R L and the leakage currents. Solution: The total leakage consists of a voltage-independent component / s , and a surface leakage component l L : Ilt = I S +I L , 10 50 fia = I s + R, (1.12a) 80 tia = I a+ i°, Rt (1.12b) where R L equals the equivalent resistance corresponding to surface leakage. Now subtract (1.12a) from (1.12b) and solve for R L : 30 M a=|°, i? L = — M a=lMn. Rr 30 Semiconductor Physics and Devices 11 Thus l LT = I„ + V/l Mft where V is the applied reverse voltage. We then solve for / s as follows: 10 50 fia = / s + 1 I a = 40 fia. The final expression is Ilt =(40+ V)(ia. The 40 n a component varies sharply with temperature, as previously discussed. When sufficiently large reverse voltages are applied, the potential gradient (electric field) across the p-n junction may measure in the hundreds of thousands of volts per inch. Such a gradient will impart a very high kinetic energy to the minority carriers normally flowing across the junction. Thus the minority car- riers, as a result of increased momenta, will collide with the atoms of the crystal with such a force as to release additional carriers, which in turn, are : accelerated by the gradient. An avalanche breakdown therefore occurs, and there is a very rapid increase in current for slight increases in reverse voltage ^(Fig. 1.8). But as long as the allowable junction power dissipation is not ex- . ceeded, the diode can operate in the avalanche breakdown mode without damage. : This characteristic makes the avalanche or Zener diode suitable for voltage regulating circuits. True Zener breakdown refers to the disruption of covalent bonds because of : the presence of a high electric field. In practice, however, the diode generally breaks down because of the avalanche effect. PROBLEM 1.16 A diode breaks down under a reverse voltage of 40 v. A 60 v battery is applied to the diode through a 1000 ft resistor. Find the power dissi- pation at the junction. Solution: The diode drop is 40 v, leaving a 20 v drop across the resistor. The current, by Ohm's law, is 20/1000 = 20 ma. The p-n junction dissipation is 40 v x 20 ma = 0.8 w. PROBLEM 1.17 A diode is in series with a 100 ft resistor and a 2 v battery (Fig. 1.13a). Find the circuit current and show, qualitatively, the effect of in- creased temperature.* Solution: The problem is easily solved by superimposing a load line corre- sponding to the 100 ft resistor on the diode forward characteristic. The graphi- cal solution of Fig. 1.13a, as shown in Fig. 1.13b, is much easier to obtain than an analytical solution based on (1.7). The load line is drawn with a slope of magnitude 2v/20 ma = 100 ft. The load line intersects the diode characteristic at point P, which is the operating point. From the curve it is estimated that the diode voltage = 0.6 v and diode current = 14 ma. At higher temperatures, the char- acteristic curve shifts to the left and the operating point is located at P '. The diode may be conveniently represented for analytical purposes by the Jbtraight-iine piecewise linear approximation of Fig. 1.14. The slope of the gitraight-line approximation corresponds to r t , the average forward resistance in die vicinity of the operating current This straight line intersects the horizontal pods -at E, the voltage corresponding to the battery of the equivalent model. For flatious temperatures, the straight line portions are parallel. (a) Increased temperature Forward characteristic (b) Fig. 1.13 Graphical solution of Prob. 1.17. f 1 V (forward bias) Fig. 1.14 Forward characteristi c of diode approximated by two-line seg- ments (piecewise linear approxima- tion). The equivalent model is a bat- tery in series with Vp , as long as there is substantial forward conduction. *In this and the following problems, the diode is operated as a forward-biased device. 12 Transistor Circuit Analysis 0.184 0.210 K, volt — »»- 26 mv (a) 10Q WAr 0.5v J (b) io 12 -W\r 0.184v .26 mv .0.5 v • 10ma = 2.6fl (c) Fig. 1.15 Solution to Prob. 1.18. (a) Diode characteristics, (b) cir- cuit, and (c) equivalent model. 10 0.134 0.184 0.210 V, volt — »» Fig. 1.16 Idealized diode charac- teristic showing effect of 20°C rise in temperature. PROBLEM 1.18 A diode having the characteristics of Fig. 1.15a is energized as shown in Fig. 1.15b. Calculate the forward current. Solution: The diode equivalent circuit used to calculate the current is given in Fig. 1.15c. Current is (0.5 - 0.184)/(10 + 2.6) = 25 ma. This analysis maybe compared with the graphical method of Fig. 1.13. PROBLEM 1.19 In Prob. 1.18, show the effect of a 20 °C rise in temperature. Solution: A 20 °C rise in temperature means a 50 mv (20 °C x 2.5 mv/°C) shift to the left in the diode forward characteristic. This is shown in Fig. 1.16 as a shift in the idealized characteristic. The high temperature current is 0.5-0.134 10 + 2.6 = 29 ma. Actually, r t is somewhat reduced at this higher current, but the error in neglect- ing this is small. PROBLEM 1.20 A germanium diode, for which saturation current / s = 10 //a, is conducting 2 ma at room temperature. What is the forward voltage drop? Solution: Use the diode equation (1.7) and solve for voltage drop. We have l=I s (e kT -1). [1.7] Substituting numerical values, 2x 10~ 3 = 10- 5 [e 0026 _1] or Taking the natural logarithm of both sides, V 199 x 10 -5 = e 0026 . 0.026 5.31, V = 5.31x0.026 = 0.14 v. PROBLEM 1.21 A diode with 10 /xa saturation current is in series with a 100 fi resistor. What current is developed with an applied voltage of 0.220 v? Solution: The problem is solved by trial and error. Assume a series of values for V, the voltage across the diode itself, typically between 0.1 v and 0.2 v. Let kT/q= 0.026v. Compute / using (1.7). To the IR drop across the 100 Q, resistor, add the assumed V to establish a figure for applied voltage. Interpolate (perhaps graphically) to find where the calculated applied voltage equals the actual ap- plied 220 mv. This occurs for / = 1 ma, with a 120 mv drop across the diode. (Problems of this kind are very well suited for solution on a digital computer.) PROBLEM 1.22 22 ma current? In the preceding problem, what is the applied voltage for a Solution: The solution to this problem is direct. The IR drop is 2.2 v. Since / and / s are known, the diode equation is readily solved for diode voltage, 0.2 v, corresponding to a total applied voltage of 2.4 v. 1.4 The Transistor The previous section describes the behavior of the p-n junction diode as a rectifying device. Consider now two p-n junctions, /, and /,, as in Fig. 1.17. Junction / t is forward-biased so that majority carriers (in this case, electrons) flow from n- to p-material. Junction /, is reverse-biased, so that only /, is flowing. If we now combine these junctions and make the p-region very Semiconductor Physics and Devices 13 thin, so that majority carriers do not recombine with holes to any appreciable ex- tent, then these majority carriers are almost all accelerated to the right by the barrier potential of /,. If we call a the ratio of current going through /, to the " current going through / |( The holes that do recombine with the few electrons in the p-region are sup- ■ plied by I B . Since a is close to but less then unity, Ib is almost entirely trans- ferred to the right-hand junction. Because these carriers are accelerated by the i/, barrier potential, which can be high, this current can flow through a high ex- eternal resistor to produce voltage amplification. I The three-layer device described is called a function transistor. The equilib- |SiUm conditions previously mentioned occur automatically at each junction. Fig- lire 1.18 shows the two junctions biased so as to produce transistor behavior for Poth the n-p-n and p-n-p cases. The emitter-base junction is forward-biased and phe base-collector junction is reverse-biased for ordinary amplifier operation. For |J|mi-P transistors, holes flow from emitter to base, and electrons from base to fitter. This is the typical flow of majority carriers characteristic of forward- ised diodes. In all cases, the emitter current l B is equal to the sum of base and collector If; In practical junction transistor design, very few impurities are introduced into flae base, so that the base has relatively few electron charge carriers compared m> the large number of hole charge carriers of the emitter (p-n-p device). In addi- pon, the base is made quite thin. Both of these factors tend to minimize recom- &nation in the base. Because the current flow from base to collector occurs across M relatively high reverse bias, the power obtained in the collector-base circuit is ponsiderably higher than the power in the base-emitter circuit. This power gain, properly exploited, is the basis for the amplifying properties of the transistor. m Figure 1. 19 shows a p-n-p transistor in the so-called common-base connection; p,e., the base is common to both the emitter and collector circuits. vvv II 9 — < n P > (a) R 2 -WV 1 s) < p J 2 (b) I B= I E ~ l C R l 1+ 1+ Rl r-wv — 'rT^r — v ^~i Fig. 1.17 Juxtaposition of two p-n junctions to make a transistor, (a) Junction /, is forward-biased, (b) junction J 2 is reverse-biased, and (c) an n-p-n junction transistor. Emitter Collector -o + + o- Emitter Collector -o- (a) n-p-n Fig. 1.18 A junction transistor showing flow of currents in normal amplifier operation. vii r\j V E E 'cc Fig. 1.19 Common-base connection of a p-n-p transistor. Lower case letters represent small-signal components of voltage and current superimposed on steady d-c bias components. 14 Transistor Circuit Analysis 0.99 0.98 0.97 0.96 0.95 Fig. 1.20 Variation of /S with a. Note that large changes in j8 cor- respond to small changes in oc. Voltage gain = PROBLEM 1.23 The transistor in the common-base circuit of Fig. 1.19 has a current gain of a. Determine voltage gain and power gain with respect to small- signal input variations. Solution: The transistor input resistance R, is related to the low forward re- sistance from emitter to base. The load resistance R L in the reverse-biased col- lector circuit can be quite high. Thus, Input voltage = v, = i a R, , Output voltage = v e = i e R Li Power gain =» current gain x voltage gain = a 1 _ R,' While the common-base circuit offers a less than unity current gain, it can provide high voltage and power gain. Other transistor connections, discussed later, can also provide large current gains. The current gain p may be defined as the ratio of collector current to base current: PROBLEM 1.24 Derive a formula for /3 as a function of a. Solution: Start with the defining relationships: Ie=Ib + Ic, Ic = <xl E , = ^-. Substituting l c /ai for l R , and I B » / c //3, we obtain for l B - l B + I c , — =— +/ c or CC=- P Therefore, ^ (1.15) l + j 8 l-a (1.16) PROBLEM 1.25 A transistor has an a of 0.98. For an emitter current of 2 ma calculate the base current I B . Also calculate /3 = / C // B . Solution: For the stated conditions, collector current equals 0.98 x 2 = 1.96 ma. The difference between emitter and collector is necessarily the base current of 40 pta. Therefore, fl = £c = L96 /„ 0.04 = 49. PROBLEM 1.26 Sketch the curve of /3 vs. a for a between 0.95 and 1. Solution: Calculate points using the formula ft = <x/(l - a): _a jS 0.95 0.95/0.05 = 19 0.96 0.96/0.04 = 24 0.97 0.97/0.03=32 0.98 0.98/0.02 = 49 0.99 0.99/0.01 = 99 1.00 1.00/0 = oo The curve is plotted in Fig. 1.20. Note the sharply increasing current gain as a = 1. A transistor with an a of 0.99 has three times the ft of a transistor with an a of 0.97. Semiconductor Physics and Devices 15 PROBLEM 1.27 A transistor has a jS of 60. Find a. Solution: Start with the equation /3 = a/(l - a). Solve for a: a = /3- 0tj8, 1 + /3 (1.17) Substitute j8 = 60, and solve for a: a = ^ = 0.984. 61 The point may be estimated from the curve of Fig. 1.20 as a a 0.98. 1 .5 The Ebers-Moll Model of the Transistor We are now in a position to develop a general model of the transistor applicable to forward- and reversed-bias conditions, and adaptable to a-c as well as d-c conditions. The Ebers-Moll model, in effect, presents the transistor static characteristics derived from previously developed concepts in convenient form. Although the Ebers-Moll model will later be expanded to cover high-frequency characteristics, for the moment its use will be confined to low frequency where capacitive effects are negligible. The low-frequency model is the starting point for the development of a variety of equivalent circuits and their analyses in succeeding chapters. The following relationships have already been derived: -I. (e*"4 / = /. (e tT -ll, [1-7] It is obvious that since the transistor is essentially a symmetrical device, a cor- rect model must likewise exhibit this symmetry. From the point of view of the model, either outer terminal relative to the base could be the emitter, with the remaining outer terminal the collector. Whether a terminal is forward- or reverse- biased depends on whether it is the emitter or collector. The following equations cover the symmetry conditions described above: / qVbb \ Ief = Ibs Ier = - Q-r Icr> IcF = - CCjf l|!Fi *vbc 'cs ■ Ics I qV B C \ (1.18a) (1.18b) (1.18c) (1.18d) Using the notation of Fig. 1.21, the conditions defined by these equations are depicted on the Ebers-Moll model of Fig. 1.22, which also defines the symbols. The representation of the polarities of the separate components of current and voltage is particularly important. As indicated, the model defines the transistor static characteristics, and applies to all polarities of bias conditions. Since the transistor is usually used with a forward-biased emitter junction, and a reverse-biased collector junction, the Ebers-Moll model may be simplified for this condition. This simplified model is particularly well-suited to the analy- sis of amplifier circuits. Veb Base, B -Q — ' B \ V C B Collector, C — -Q (a) (b) Fig. 1.21 (a) Representation of n-p-n transistor showing stan- dard polarities of voltages and currents, (b) Alternate represen- tation of n-p-n transistor. 16 Transistor Circuit Analysis ~lER= a RlcR a F I EF=~ICF EO •■ H4- * » >f- 'CR -oc Fig. 1.22 The Ebers-Moll model of the transistor. Figure 1.23 shows such a model or equivalent circuit, adapted only to the specific set of bias conditions. For these conditions, a R 'ck « IefI therefore we neglect (X R / CR , and A small-signal equivalent circuit is one which depicts the response of the transistor to small-signal inputs appearing as small variations about the bias or operating points. The small-signal model is basic to the design of small-signal a-c amplifiers. The most useful forms of the equivalent circuit are the tee-, the hybrid, and the hybrid-7r models. The tee- and hybrid models provide a simple and direct representation of transistor behavior, and are readily suited to circuit calculation at audio frequencies. The hybrid-^ is useful for analyzing the performance of transistor amplifiers at high frequencies. I CS=IcBO EO OC EO- -WAr- e -oc a/ E Fig. 1.23 Ebers-Moll model with forward-biased emitter junction and reverse-biased collector junction. Fig. 1.24 Simplified tee-model with equivalent repre- sentation of emitter diode. 'e EO- 'b B Fig. 1.25 Equivalent tee-circuit with leakage current omitted. For the present, the tee-equivalent circuit will be developed. Referring to Fig. 1. 23, the input diode of the Ebers-Moll model can be replaced by the diode equivalent circuit of Fig. 1.24. This consists of an equivalent battery E, and an incremental resistance r,. Since the equivalent battery is, in effect, part of the d-c bias, it can be neglected in the incremental a-c equivalent circuit, leading to the further simplified model of Fig. 1.25. The leakage current component is also deleted. This equivalent circuit is a good practical representation of the small-signal behavior of the transistor biased for amplifier application. However, it can be refined somewhat to improve accuracy. As V C b is increased, more of the carriers injected into the base from the emitter reach the collector. Recombination in the base region is then reduced. Semiconductor Physics and Devices 17 Thus, collector current increases with V C b for constant Ie- This is equivalent to connecting a resistor across the current source, Ot /g. This collector resist- ance, r e , is defined by the equation AAA* — i d Vr. (1.19) 'C r , . li£ B constant An increase in F CB causes an increase in emitter current for constant V EB . This feedback effect can be represented by a resistance r fa ' in the base lead. In addition, the ohmic or base spreading resistance r bb ' of the thin base region is also included in the base lead. Then the total base resistance r b = t b ' + r bb ' as in Fig. 1.26. Resistance r t is labeled r e , the emitter resistance. 1 .6 Basic Transistor Amplifier Circuits There are three basic transistor circuits, namely, the common-base, the common-emitter, and the common-collector connections. Their configurations derive from the choice of the input and output terminals and the terminal common to both. 'In each of them, the basic bias conditions are satis- fied; i.e., the collector-base junction is reverse-biased, and the emitter-base junc- tion is forward-biased. (Of course in an actual circuit, the polarities of the bias- ing voltage depend on whether the transistor is p-n-p or n-p-n.) Figures 1.27-28 show the configurations of p-n-p and n-p-n transistor circuits. As will be illustrated by numerous examples throughout the book, each configura- tion has its own area of superiority in specific applications. Fig. 1.26 More accurate equivalent tee-circuit for the common-base connection. fr "t L 7¥ T T H T Common-base connection l c =al E (a) Common-emitter connection Common-collector or emitter- follower connection /c=/3/ B » a 1-a (b) 1-a (c) Fig. 1.27 Common circuit configurations for the p-n-p transistor. For the above, I E =I B + l c , a - Iq/Iei P ~ l c/ l B- The circuits are simplified. Bias and load resistors are not shown. IE h€5R' Vcc V B B - Common-base connection (a) Vbb Common-emitter connection (b) Vbb ~ v cc Common -col lector or emitter- follower connection (c) Fig. 1.28 Basic transistor circuits using an n-p-n transistor. 18 Transistor Circuit Analysis Characteristic curves relating transistor currents and voltages may be used to describe the behavior of each circuit. Typical characteristic curves for the^! eZ io:*:™** 10 " are Pr ° Vided by Fi8 ' U9 ' Corresponding sets ofcu^s exist for the common-emitter and common-collector connections. Note that^he common-base connection, collector current flows with v„ T V CB =0 (b) Fig. 1.29 Common-base (a) output and (b) input characteristic (a) 7wi>/ a) (c) Fig. 1.30 Leakage components for 1.7 Transistor Leakage Currents nt , . , In transi stor circuits a problem arises from the variation of leakage currents with temperature in the collector-base junction. This leaka~ component ^analogous to diode leakage (Fig. 1.9). Since the emitter-base junc- tion is normally forward-biased, leakage current here is not significant. Figure 1.30 a-c shows the basic leakage current components. Current I co (as i is usually called) is the common-base leakage comment, and is more pre- crrcutted. Analogously. I EO rI EBO applies to the emitter-base junction when it is reverse-biased (not a normal condition) and the collector is open-circuited Now consider 1 cbo . This current varies as noted in Fig. 1.9 with temperature It is much more significant in relatively high leakage germaniu. "SoSST than m silicon transistors. As temperature increases, l CBO riees , and the junction wanner due to the increased current component. This, in turn, further increases «?£? ^T ^ t0 ^ UnStaWe C00diti0n known as th '™«l runaway (dis- cussed in detail in Chap. 6). Leakage current variation leads to shifts in the d-c operating or b ias point of the transistor, and can result in nonlinear operation. 17"'; ^ SC COnnection ^Pen-circuited, is an important parameter, which can be expressed in terms of / CBO and fl. ■=• ~~ PROBLEM 1.28 Find / tfie principal transistor configura- cbo a s a function of I CBn and B usins th» *».,.,;..*«. equation defined in Fig. 1.24, but including the l4age cutnt 7 C Solution: The basic equation 1 cbo ■ tions. Since the emitter circuit is not normally reverse-biased, I EO is not signifi cant. IS tc = a l E + / CBO i l B = lc + h Therefore, 'c=a(/ c +/ B ) + / Cfl o or f c (l- a ) = a/ fl +/ c BO- Semiconductor Physics and Devices 19 Solving for l c , lc = - — — Ib + I; ) Icbo- 1 - a \l-a/ Since /3 = a/(l - a) and 1 - a = 1/(1 + j8), The first component is the output current resulting from l B , and the second com- ponent is IcEOt f ot the common-emitter configuration: There is an interesting graphical interpretation of the above expression. Re- fer to Fig. 1.31. Note that when l B = 0, l c = Icsoi when collector and base cur- rents are equal, l B = -I c = l CB o ( the emitter being open). PROBLEM 1.29 A transistor has an 1 C bo = 50 /ia when measured in the grounded base configuration. If /3 = 100, find / ceo . Solution: I CEO = (j8+ 1)/cbo = 101 x 50^5000 ^a, or 5 ma. CBO —tcBO ° 1 B — »- Fig. 1.31 Leakage components on an Iq vs. Ig curve. For /g = 0, only the IcEO leakage current flows. 1 .8 Transistor Breakdown In a transistor the conditions for breakdown, when reverse voltage is applied from collector to base, correspond roughly to the breakdown of a diode under reverse bias conditions. Usually, breakdown is an avalanche ef- fect as previously described. However, when the base-emitter circuit is involved, the breakdown mechanism becomes more complex. Figure 1.32 shows l c under reverse voltage conditions which occur in tran- sistor applications. The figure defines the most significant breakdown condi- BVcbo BV CB0 BV, CEO BV C ES Reverse voltage- Fig. 1.32 Breakdown voltages for different transistor connections; BVqer corresponds to breakdown voltage for a resistor i?j connected from base to emitter. This curve must lie between BV CE0 (R = <x>) and BV CES {R = 0). tions. Note that the reverse voltage breakdown across the collector-base junction is the highest of the several breakdown voltages, and may not be used safely as a voltage limit in most transistor circuits. The correct breakdown voltage rating depends on the circuit, and cannot usually exceed BV CBS , the breakdown voltage from collector to emitter with the base connected to the emitter. Voltage break- down is usually not harmful if current is limited, so that the junction does not overheat. Many voltage regulating devices depend on the voltage breakdown phe- nomenon for their behavior. Voltage breakdown of transistors will be covered in more detail in Chap. 7, as it is most important in power amplifiers. 20 Transistor Circuit Analysis Et> © C l CBO HEH «-* (tlr e -w- • o fB *B Fig. 1.33 Simplified d-c transistor equivalent circui t, common-base con fi gu ra ti on . (|8+l)/ C BO Fig. 1.34 Simplified d-c transistor equivalent circui t, common -emitter confi guration. 1.9 D-C Models Let us briefly relate the tee-equivalent circuit (Figs. 1.22- 25) derived from the Ebers-Moll model to the n-p-n transistor under static (d-c) conditions as shown in Fig. 1.33. The tee-equivalent circuit suggests, loosely speaking, the representation of the transistor as two back-to-back junction diodes. Output current includes two components, leakage current and amplified input a l E , as was previously explained. The forward-biased emitter circuit impedance varies substantially with input voltage, as shown in Fig. 1.29b. This impedance, because it is of such low value, is usually swamped by external resistances. The emitter is most conveniently dealt with analytically by assuming a constant volt- age drop of several tenths of a volt from emitter to base, and perhaps adding a small resistive component and working from input currents rather than input volt- ages. This is exactly analogous to the earlier study of diode circuits. For all except low collector voltages, the emitter-base characteristic is independent of collector voltages. The collector is always reverse-biased for normal amplifier operation. Its leakage current varies a great deal from transistor to transistor, and also with temperature, much as does the leakage current of the diode. At very low collector voltages and emitter currents, decidedly nonlinear transistor behavior occurs. The equivalent circuit parameters vary over a wide range. Notwithstanding this vari- ability, the tee-equivalent circuit is an invaluable aid in preliminary design, in visualizing the effects of changing external circuitry, and in the establish- ment of optimum circuit performance. Furthermore, the equivalent circuit itself provides a firm basis for the evaluation of the effects of these variations on over- all circuit behavior. The entire subject of transistor circuit biasing is closely tied to the study of the effects of changes in transistor characteristics and methods to minimize the effects of these changes on the operating point. Shifts in the collector current with changes in transistor leakage, with forward current gain a, or with bias voltage, constitute significant problems in the maintenance of a stable operating point. An orderly study of this topic is presented in Chap. 4. Figure 1.34 shows the d-c equivalent circuit for the common-emitter configura- tion. Note the leakage component (fi + l)/ CBO , which is the previously derived result. The base-collector current gain is /8. PROBLEM 1.30 The common base d-c equivalent circuit of an n-p-n transistor is shown in Fig. 1.35 driving a resistance load R L . Using the equivalent circuit, calculate the d-c input and output resistances, R t and R ol respectively. Neglect leakage current 1 C bo, and the small resistor r E shown in the emitter circuit. Solution: The basic equations required to find R t are Ve = -(.V B e +IbRb)> l=aI E , I B = (1 - a)I E . Referring to Fig. 1.35b and combining equations, V E _[(1- 0O/ £ R b + V bb ] R, -U -I, Ri = (1 - a) R B + U For this idealized configuration, output resistance R is infinite as long as Vcb is an effective reverse bias. Output current is, of course, a I E , regardless of R L . Semiconductor Physics and Devices 21 cc (a) «») Fig. 1.35 Common-base amplifier, (a) D-c circuit, and (b) equivalent model. PROBLEM 1.31 Analyze the circuit of Fig. 1.36 for input and output impedances (resistances), R t and R ot respectively. Neglect leakage components and the rela- tively small ohmic resistance in the base circuit. B O *\rW R< VbiTZT Re, (j3+i) I C BO -©- e |8/ B ic I E Rt b v cc (a) (b) Fig. 1.36 Common-emitter amplifier, (a) D-c circuit, and (b) equivalent model. Solution: From the equivalent circuit the basic equations are V B =V RB +R E l E *E i Combining equations, V c = Vcc-RlIc, / c = /3/b, / b = (j8 + 1)/b. V b = Veb + Re(P + 1)Ib- The input impedance R, is the sum of Veb/Ib and R E (1 + /3). It is characteristic B of this type of circuit which features moderately high input resistance. The output resistance, R =<», since the output current is supplied by an ideal current source. The output impedance is very high as long as the circuit biasing is correct. Ib *\ f ICBO a F / E 1.10 TheHybrid-7T Equivalent Circuit The hybrid-n is an equivalent circuit configuration par- ticularly suited to high-frequency calculations. It may be derived from the Ebers- Moll model of Fig. 1.22 as shown in Fig. 1.37, and some physical considerations. Fig. 1.37 Simplified tee-equivalent circuit used as a basis for deriva- tion of hybrid-?/ equivalent circuit. 22 Transistor Circuit Analysis B Tbb ' O—VW 6m v b'e I The hybrid-iT circuit is derived for the common-emitter configuration in Fig. 1.38. Only the small-signal model is considered. The pertinent small-signal parameters are v b. I Sn ~ dv BE (1.21a) U m Fig. 1.38 Simplified representation of hybrid-77 equivalent circuit for the common -emitter connection. (1.21b) Note that lower case letters and subscripts are used for small-signal parameters. These small-signal relationships lead to the simplified equivalent circuit of Fig. 1.38, in which l CBO and the d-c diode drops are not pertinent. In Fig. 1.38, B' represents a theoretical base point within the transistor, while B is the base terminal. The bulk resistance, r bb > , intervenes. This resistance component be- haves as a pttte ohmic resistance. The simplified equivalent circuit of Fig. 1.38 may be improved as shown in Fig. 1.39. The added resistors r b > and r c . are related to the increasing width of the collector-base junction depletion layer with increasing collector-base reverse bias. As the reverse bias voltage increases, the width of the base is effectively reduced so that a increases, while recombination in the base de- creases. The increase in collector current with voltage is represented by r co in Fig. 1.39. The reduced base current is achieved in the equivalent circuit by the feedback from C to B' through r b i e . Cft'c oc r bb bo — WV Fig. 1.39 More accurate hybrid-TT equivalent circuit forthe common- emitter connection. Oc O E Fig. 1.40 Hybrid-77 equivalent circuit incorporating capaci- tances for a more accurate representation of high-frequency characteristics. High-frequency features may now be incorporated simply by including capac- itances between collector and base, and emitter and base (Fig. 1.40). The ca- pacitances are excellent representations of effective capacitances actually appearing across the junctions. This high-frequency model is relatively unaf- fected by frequency over a wide and useful range. The capacitances appear primarily because the depletion layers themselves appear as capacitors. The widening of the collector-base depletion layer, for example, behaves like a reduction in capacitance.. This leads to transient cur- rents whose effects are represented very satisfactorily by the capacitors of Fig. 1.40. 1.11 Supplementary Problems PROBLEM 1.32 Explain (a) kernel, (b) valence electron, (c) excitation level, and (d) free electrons. PROBLEM 1.33 Explain the meaning of holes in semiconductors. Semiconductor Physics and Devices 23 PROBLEM 1.34 What is the significance of electron-hole pair generation? What are its main causes? PROBLEM 1.35 Define majority carrier and minority carrier current components. PROBLEM 1.36 For kT/q= 0.026 v and a saturation current of 10 fia, plot / from v = -5 to.v = +l. Use (1.7) and suitable increments to obtain a smooth curve. PROBLEM 1.37 For a germanium diode, / s = 0.0005 a at 25°C, determine the leakage current at 75°C. PROBLEM 1.38 Find the incremental resistance of a diode at room temperature (25°C) with a forward current of 1 ma. PROBLEM 1.39 Explain the behavior of a p-n junction which is (a) back-biased and (b) forward-biased. PROBLEM 1.40 In a transistor in the common-base connection, i b = 0.1 ma and i c = 5 ma. Determine Ot and fi. PROBLEM 1.41 Why is the current gain of a transistor in the common-base con- nection always less than unity? PROBLEM 1.42 Why is the voltage gain of the common-collector circuit always less than unity? PROBLEM 1.43 For/ CBO = 10 ^ain the grounded base configuration and £= 50, find I ceo- PROBLEM 1.44 Show the directions of all currents in a p-n-p transistor for the three common connections. 2 CHAPTER TRANSISTOR CIRCUIT ANALYSIS V CE BO o c Fig. 2.1 Voltoges and currents in a p-n-p transistor. Fig. 2.2 Comm sistor circu i-base n-p-n /?L is a loa tor. CC tran- d 2.1 Characteristic Curves The static characteristic curves of the transistor define the steady-state relationships among its input and output currents and voltages. The curves are easily obtained by means of d-c measurements, and provide: 1. The basis of graphical design procedures. 2. A description of transistor performance under nonlinear conditions. 3. The starting point and final reference in the development of analytical pro- cedures. The characteristic curves thus constitute the basis for understanding transistor operation. The transistor is a three-terminal device that in general has six variables comprised of three currents and three voltages as shown in Fig. 2.1. Since any two currents or any two voltages determine the third respective quantity, the actual number of variables is reduced to four. If any two of the four variables are speci- fied, the remaining two are automatically determined. Now in general terms, let x, and x t be a pair of specified independent varia- bles, and y, and y, be the automatically determined dependent variables. Mathe- matically, (2.1) (2.2) where /, and /, are the functional relationships between the independent and de- pendent variables. To graphically describe y, and y 2 , we require a separate family of curves for each. Thus, two families or sets of curves are necessary for a complete steady- state description of the transistor. From the two sets of curves, all other possible curves may be derived. The choice of the two independent and two dependent variables from the six possible transistor variables is a matter of convenience. It may depend, for ex- ample, on which sets of curves are easier to use than others, or on the particular transistor under investigation. In general, it is most convenient to ascribe one set of curves to the transistor input, and the other to the transistor output. Thus for the common-base configuration of Fig. 2.2, V EB is plotted versus the inde- pendent variable l B for various values of the second independent variable V CB as shown in Fig. 2.3a. The output collector current / c is plotted versus the same independent variables l E and V CB in Fig. 2.3b. This figure constitutes the col- lector family of curves. The preferred choices of variables have become somewhat standardized by manufacturers who wish to present their data in the manner most suitable for use. Characteristic curves such as the ones illustrated by Fig. 2.3a are especially 24 Transistor Circuit Analysis 25 1.0 0.8 -o 0.6 IB ^ 0.4 0.2 V CB = p V CB >U 2 4 6 8 10 12 14 1 E , ma ►- (a) 10 y 1 1 10ma 8m a 6 ma 6 r 1 4ma ** 2 2ma »• 7 E = 10 15 20 25 30 35 V CB , volt fr- (b) Fig. 2.3 Common-base (a) input and (b) output characteristi cs of the 2N929 transistor at room temperature. convenient, because the independent variable is essentially a function of only one dependent variable, in this case, the emitter current l E . Curves such as those of Fig. 2.3b, which are uniformly spaced parallel straight lines over most of the useful transistor operating range, provide a basis for linearizing transistor param- eters and simplifying circuit analysis. Static characteristics are directly obtainable by means of elementary methods. Figure 2.4 shows how common-base parameters are measured. The collector-base voltage and emitter current are the independent variables; the collector current and base-emitter voltage are the dependent variables (see Fig. 2.3). Analogous to the common-base configuration of Fig. 2.2 and its curves of Fig. 2.3 are families of curves particularly descriptive of the behavior of the com- mon-emitter and common-collector configurations. As with the common-base con- nection, the circuit characteristics of the latter are conveniently described by static characteristic curves as in Figs. 2.5-6, respectively. Note that the char- acteristics - although shown to different scales - are essentially identical be- Fig. 2.4 Simple d-c circuit for ob- taining the curves of Fig. 2.3. Shown is a common-base connec- tion for a p-n-p transistor. With an n-p-n transistor, the polarities of all voltages and currents are reversed. 1.0 0.8 ^ 0.4 0.2 P_^~ . v CE - lv v CE =o 100 200 300 400 500 600 700 I B , fta ► (a) (b) Fig. 2.5 Common-emitter (a) input and (b) output characteristics of the 2N929 transistor at room temperature. 26 Transistor Circuit Analysis 1.0 0.8 i0.6 o > w 5?0.4 0.2 V CE >lv P-^ V r CE = _ 20 40 , pa- rt 60 80 cause I E = l c , and that the curves completely define transistor circuit behavior under d-c or low-frequency conditions. Figure 2.7 shows how the curves of Fig. 2.5 may be determined experimentally. 2.2 The Operating Point In normal operation of a transistor circuit, as for example, in a linear amplifier, currents and voltages are applied to the transistor to estab- lish a bias or quiescent operating point in the linear region of the output charac- teristics (e.g., point P, Fig. 2.5b). In this region variations in input (base current) lead to proportional changes in output (collector current). The proportionality constant represents a current gain - comparable to /3 in Chap. 1 - that can be de- termined from the characteristic curves. By means of this constant as well as other similar ones applicable in the linear regions, the transistor may be analyzed using a linear equivalent circuit or model, just as in the case of a vacuum-tube device. <fp, lg=4.Slla 20/te -15fia 10/Ua S/ia /n = 5 10 V EC , volt (b) IS 20 PROBLEM 2.1 The tabulation below lists sets of values of given transistor cur- rents and/or voltages. Verify by inspection of the curves of Figs. 2.3 and 2.5 that the currents and/or voltages are consistent.* V CB = 3 v l E = 3 ma l B = 10 ma I E = 10 ma /« = I B = 30 fia l B = 10 pa V CB = 20 v V CB = 5 v V CB = 20 v V CE = 20 v ' CE = 30 v V EB = 0.57 v Fig. 2.3a l c = 10 ma Fig. 2.3b I c = 10 ma Fig. 2.3b / c = Fig. 2.3b l c = 8.5 ma Fig. 2.5a l c = 3 ma Fig. 2.5a PROBLEM 2.2 Given the common-emitter characteristics of Fig. 2.5, and I B = 20 fxa, and V CE = 20 v, find the collector current and the base-emitter voltage drop. Fig. 2.6 Common -co I lector (a) input and (b) output characteristics at low ba " CU a7roo°m 11™™^°' * , ." t '™ , *** P * ** 2 * ^ ^ base-emitter voltage drop V BE = 0.58 v. Point P of Fig. 2.5b shows the collector current I c = 5.4 ma. PROBLEM 2.3 For a common-collector configuration using the 2N929 transistor, V EC = 2 v and l E = 0.6 ma. Find the collector current l c and the base-emitter voltage drop V BE . Solution: Use the common-collector characteristic curves of Fig. 2.6. (Note that if common-collector curves are unavailable, common-emitter curves are almost identical for most transistors.) In Fig. 2.6b, operating point P is located. By interpolation, l B = 4.5 fia (between the l B = and 5 pa curves). Thus, l c = 0.6 ma. In Fig. 2.6a, point P is located corresponding to l B = 4.5 n a, and V CE > 1 v. The base-emitter drop is found to be V BB = 0.53 v. PROBLEM 2.4 The operating point of a common-emitter circuit using the 2N929 transistor is I c = 5.6 ma, and V CE = 20 v. Find l B , V BE , V CB , and l E . Protective resistor Fig. 2.7 Simplified d-c circuit for obtaining the curves of Fig. 2.5. "The characteristic curves of the Jesas Instrument 2N929 n^p-n silicon transistor are used for most or tne problems in this chapter. Transistor Circuit Analysis 27 Solution: Use the characteristic curves of Fig. 2.5b to locate point P l at V CE = 20 v, / c = 5.6 ma. By interpolation, I B = 21.5 fja. From Fig. 2.5a, V BE = 0.585 v. By summing voltages, V CB = V CE - Vbe = 20 - 0.585 = 19.42 v. Since the algebraic sum of the three transistor currents must equal zero, 1e = Ic + Ib = 6.02 ma. This analysis gives all six parameters of the operating point. PROBLEM 2.5 For a common-base connection using the 2N929 transistor, l c = 5 ma and V CB = 5 v. Determine the remaining four voltage and current parameters for the operating point. (Note that in most data sheets only common-emitter curves are provided by the transistor manufacturer. Therefore use only the common-emitter curves of Fig. 2.5 to solve this problem.) Solution: Start by estimating that for the conditions of this problem, V BE lies between 0.5 v and 0.6 v. Therefore, for V CB = 5 v, Vce = 5.6 v. For V CE = 5.6 v and lc = 5 ma (given), from Fig. 2.5b, l B is estimated at 20 /za (point P 2 ). For l B = 20 fua and V CE > 1, V BE = 0.58 v in Fig. 2.5a. More accurately: V CE = V B e + V CB = 5.58 v, l E =lc +'b = 5.0 ma + 20 /xa = 5.02 ma. All parameters are known. The preceding examples illustrate how calculations may be made using any available family of curves. Often, however, a particular set of curves leads to greater accuracy or convenience in a specific problem. In carrying out the above and other calculations, the following points are worth remembering: 1. The emitter and collector have nearly the same current. Typically, 99% of the emitter current flows in the collector circuit. For this reason, it is usually important to specify base current, and either collector or emitter current. 2. Base-emitter current increases very rapidly with increasing base-emitter voltage. The base-emitter current is, in effect, the forward current of a diode. Excessive base-emitter voltage overheats the base-emitter junction and may burn out the unit. It is important to control base current and not apply low impedance voltage sources to the base-emitter circuit. 2.3 The Load Line Transistors, of course, are not used as isolated elements. They are usually operated in essentially resistive networks where resistors are employed in biasing circuits to establish an operating point for the transistor, and as load elements. In a-c applications, capacitors are generally used to isolate d-c signals while permitting a-c signals to pass. The combining of transistors and resistors presents no special problem, because the current flow through the re- sistors determines their corresponding voltage drops, as well as the voltages ap- pearing on the transistor terminals. The graphical treatment of transistor circuits makes use of the concept of the load line. Consider, for example, the output or collector characteristics of a typi- cal common-emitter transistor circuit shown in Fig. 2.8. Assume a collector volt- age supply Vcc in series with a load resistor Rl- A straight line may be super- imposed on the collector characteristics corresponding to the volt-ampere charac- 28 Transistor Circuit Analysis teristic of the battery-resistor combination. The slope of the load line represents the voltage-current characteristic of resistance R L . The equation for the load line is Vgb**V cc -IcRi,< <2.3) Setting l c = in (2.3). V CB = V cc ; setting V CB - 0, l c . V CC /R L . These two points or intercepts define the load line whose slope equals -1/R,. If /. . / at point P, the quiescent operating point is defined. Figure 2.8 also shows load* lines (dashed lines) for increased V cc , and increased V cc combined with reduced R L . The quiescent point corresponds to the given base current. Figure 2.9 provides a numerical illustration of the use of the load line. The supply voltage V cc - 30 v and R t = 5000 fi. The load line is superimposed on the collector characteristics. The horizontal intercept occurs at l c - and equals 30 v; the vertical intercept occurs at V CB = and equals 30 v/5000 Q = r>ma. The figure also shows the load line for V cc = 10 v and R L =.- 2000 fl. Increased V cc , Reduced R L Fig. 2.8 Load lines superimposed on transistor collec- tor characteristics for the common -emitter connection. Fig. 2.9 Graphical analysis using the load line. PROBLEM 2.6 A common-emitter circuit using the 2N929 transistor has a load resistance R L = 5000 Q in the collector circuit; V cc = 30 v, 7 C = 3.7 ma. Find Ib> V ce , l E , and V BE . Solution: The set of curves in Fig. 2.9 shows a superimposed load line repre- senting the voltage-current characteristic of the collector circuit resistor. At l c = 3.7 ma, l B = 15 ^a, and V CE = 11.6 v. Hence, I E = / c + I B = 3.7 + 0.015 = 3.715 ma. From Fig. 2.6a, which is applicable to the 2N929 transistor, V BE = 0.57 v for V CE > 1 v, and I B = 15 /za. Note that both voltage across the load resistor and the collector current can be much larger than V BE and I B , indicating the possibility of large current, volt- age, and power gains for the common-emitter circuit. The load line approach is simpler to use at the transistor input whose charac- teristics are nearly independent of collector voltage. The load line establishes a simple intersection with the significant input characteristic to determine the operating point. Figure 2.10 shows the manner in which base current is deter- mined from a given voltage and resistance in the base circuit. The parameters in question are indicated. Transistor Circuit Analysis 29 1.0 0.8 W (I) &. 0.4 0.2 V CE >U V CE =0 2N929 I B - R,= !»0. \ \Load \line V, = l V - k 1 ~ Vbe 1 V^CE >lv (b) 50 100 150 200 250 300 350 1 B , (Ja p* Fig. 2.10 Base circuit load line. 15 20 25 V CB , volt »- 30 35 Fig. 2.11 Common-base characteristics with superim- posed load line. PROBLEM 2.7 For a common-base connection using the 2N929 transistor, V cc = 25 v, I c = 3 ma, and R L = 5000 Q. Find l B , V BE , and V CB . Solution: Refer to Fig. 2.11. Draw a load line corresponding to the 5000 fi load resistance. For l c = 3 ma, V CB = 10 v corresponding to point P. Emitter current I E = 3 ma. From Fig. 2.3a, which is applicable to the present problem, V BE = 0.57 v. It is worth noting that since 1 B and / c are almost equal, current amplification does not occur in the common-base configuration. Because of the higher impedance level of the output collector circuit, voltage gain can be realized. PROBLEM 2.8 For the common-collector configuration of Fig. 2.12a using the 2N929 transistor, V EE = 30 v, l E = 3.7 ma, and R L = 5000 fi. Find l B , l c , V BE , V, , and V . 10 E W **3Stla 1 30; la 35, 1SL 2 Op 15fia P^ 10/la / B = o (o) | o r i i i i 5 10 15 20 25 V ECl volt »- (b) Fig. 2.12 (a) Elementary common-collector amplifier, (b) Common-collector output char- acteristics with superimposed load line. 30 35 30 Transistor Circuit Analysis Solution: Refer to Fig. 2.12b for the load line construction. From the load line, V EC = U.6 V , Ib = 15 fja , V BE = 0.57 v (from Fig. 2.6a), V = 3.7 ma x 5000 = 18.5 v, V, = 3.7 ma x 5000 + V BE = 19.07 v. Note that for the common-collector (emitter-follower) circuit, V is always somewhat lower than V,. There is no voltage gain. The input current l B is much smaller, however, than the output current l E . This configuration is essentially a current amplifier with less than unity voltage gain. IN "ig. 2.13 Elementary common- emitter amplifier. 2.4 Small- and Large-Signal A-C Circuits Refer to the basic common-emitter circuit used as an a-c amplifier in Fig. 2.13. The input coupling capacitor blocks d-c signals but per- mits transmission of a-c signals. In effect, relatively small a-c signals con- stitute a perturbation or small modification to the bias point. Figure 2.14 shows the load line superimposed on the collector characteristics of the 2N929 transistor. When a bias point is chosen in the center of the transis- tor's linear region, the output-to-input ratio with moderate signal amplitudes is constant and may be expressed as a gain. To determine this gain, we need only to vary the input by a small amount about the bias point and determine the cor- responding variation in output. The transistor parameters associated with small- signal excursions about the bias point, usually in the linear region, are referred to as small-signal parameters. Small-signal parameters actually vary somewhat with the bias point, even in the linear region. PROBLEM 2.9 Given the common-emitter transistor amplifier of Fig. 2.13, ca- pacitor C presents negligible impedance to a-c. Resistance R B is adjusted so that l B = 15 /za. An input a-c current having a 5 /xa peak-to-peak (p-p) amplitude is im- pressed at the input terminals. Using point-by-point graphical construction, plot i B (f). Also repeat for a 30 ^a p-p input current amplitude. Solution: The required graphical construction is developed from Fig. 2.14. The intersections of the load line with the collector family of curves leads to the l c vs. l B transfer characteristic of Fig. 2.15. The small and large sinusoidal base currents are shown superimposed on the 15 fi a quiescent current. Observe the clipping of the output, resulting from the large amplitude input current, which drives the transistor into the nonlinear region. Largersigaai parameters are based on the static d-c characteristics of the transistor over the fall operating range, including nonlinear regions. As a result, they vary substantially with signal amplitude. Large-signal behavior is important in d-c amplifier design, in the design of biasing circuitry, in switching applica- tions where the transistor is intentionally driven into nonlinear regions, and in power amplifiers where large signals are permitted, with distortion maintained within tolerable limits. Transistor Circuit Analysis 31 £ » 4 o r ^ — 25 ^-a y~~ ^ 20 Ma 15 /ia r < ^ PS, 10/ia 5/xa _ J B =0 k 10 15 'CE> 20 /oh 25 30 40 Fig. 2.14 Common-emitter output characteristics with superimposed load line. PROBLEM 2.10 For the common-emitter circuit using the 2N929 transistor with a 5000 Q load and V cc = 30 v, find: (a) I B needed to operate at /c = 5 ma. (b) The power P c dissipated in the collector junction. (c) The d-c voltage V L across the load, and the power P L dissipated in the load resistor. (d) The input d-c power P B to the base. (e) The variation in the parameters l c , V CE , and v l if 1 b is decreased by 5 fia. (f) The variation in the parameters V BE , P B , and P L if l B is decreased by Fig. 2.15 Collector current variation -L 40 i B , 30^(a Peak-to-peak Time 5 iia. Solution: (a) Refer to Fig. 2.16. Draw the load line corresponding to 5000 Q. At l c = 5 ma, I B = 20 /*a. (b) D-c power is current multiplied by voltage. From Fig. 2.16, V CE = 5.3 v, I c = 5 ma, and P c = 26 mw. (c) The power P L =IcRl = 125 mw. The voltage V L = 30 v - 5.3 v = 24.7 v. (d) The power P B = V BE l B . From the applicable curve of Fig. 2.17, V B = 0.58 v. Hence, P B = 0.58 x 20 x 10 -6 = 11.6 jzw. (e) Since l c must remain on the load line of Fig. 2.16, the new operating point must be at the intersection of the load line and the l B = (20 - 5) or 15 pa characteristic. From this new operating point, it is found that with sinusoidal base current drive. Note the large-signal distortion re- sulting from saturation. 0.7 0.6 0.5 0.4 0.3 0.2 0.1 Slope | 2.2mv/f l 3^*^ Fce-!v v CE = o 20 40 Ib , P-°- 60 80 Fig. 2.17 Low level input character- istics in common -emitter connection. Fig. 2.16 Common-emitter output characteristics with superimposed load line. 32 Transistor Circuit Analysis V CE = 11.6 v, lc = 3.7 ma, V L = 30 v - 11.6 v = 18.4 v. Use A notation to designate changes in these parameters with the shift in l B : AF CE = 11.6-5.3 = 6.3 v, A/ c =3.7- 5 ma = -1.3 ma, AV L = 18.4 -24.7 = -6.3 v. The negative signs indicate that the changes are reductions. (f) The reduction is A/ B = -5 pa. From Fig. 2.17, V BE = 0.569. Hence, V BE is 0.011 v lower at the reduced base current. This is most accurately estimated from the tangent at the original operating point as the rate of change of V BE with I B . Thus, A7 B£ =0.011 v, AP B = V BE2 I Bi - V BEi l Bi = [(0.569 x 15) - (0.58 x 20)] x 10* = -3 x 10" 6 w . Now find A P L =P L2 -P Li ; p l 2 = Ic 2 2 Rl = (3.7 x 10- 3 ) 2 x 5000 = 58 mw at I Bi = 15 ,*, p i, = 1 c\Rl = (5 x 10- 3 ) 2 x 5000 = 125 mw at / Bj = 20 ^a, AP L = 58-125=-67mw. PROBLEM 2.11 From Prob. 2.10, determine the d-c current gain, i.e., the ratio of collector to base current. Also find the incremental current gain, i.e., the ratio of a change in I c to a change in I B . Solution: From Prob. 2.10(a), I c = 5 ma, and 7 B = 0.02 ma. Thus, D-c current gain = -£ = = 250 1 B 0.02 From Prob. 2.10(e), A/ c = -1.3 ma for A/ B = -0.005 ma. Thus, Incremental current gain = c = ~ = 260 A/ B -0.005 PROBLEM 2.12 Using the conditions of Prob. 2.10, determine the input resist- ance (static d-c value), and the incremental resistance to small input changes. Solution: Static input resistance is V BE 0.58 v I B 20xl0- 6 a Incremental input resistance is A V BE -0.011 29,000 n. A/ B -5xl0' 6 a 2200 fi. Note the very large difference between static and incremental resistances. The input resistance is decidedly nonlinear. Transistor Circuit Analysis 33 PROBLEM 2.13 Referring to Prob. 2.10, find the ratio of output power (in R L ) to input power (to the base of the transistor) for static and incremental conditions. p Solution: Static (d-c) power gain is — — 0.125 w 11.6 x 10" 6 w 10,800. AP L -6.7 xlO -3 From Prob. 2.10(f), incremental power gain is AP t -3x10' = 22,300. PROBLEM 2.14 For the conditions of Prob. 2.10, find the incremental voltage gain, A V L /A V BE . Solution: From Prob. 2.10(e), -6.3 AV, =-6.3 v, AV BE =-0.011 v, Voltage gain = -0.011 570. PROBLEM 2.15 For the 2N929 transistor in the common-emitter circuit of Fig. 2.13, calculate the static and incremental output impedance for V CE = 10 v and I B = 15 fia from the transistor characteristics. Solution: Incremental output impedance is defined as A V CE /A lc for constant l B . To obtain it, use A V CE = +5 v (arbitrarily), so that V CE ranges from 10 v to 15 v. Since A V CB is noncritical in the linear region of the characteristics, it is selected for convenience. Referring to Fig. 2.18a-b, Point P t : Point P 2 : V CE = 10 V CE = 15 v, lc = 3.7 ma 1 > l B = 15 n a (given), v, lc = 3.8 ma J Incremental output impedance = AV CE Static output impedance = Air CE — — =50,ooon = — 0.0001 h ol 10 = 2700 fl = — . 0.0037 h OE The much higher incremental output impedance as compared with the static value is, of course, due to transistor nonlinearity. Note that h OE and h oe are the static and incremental output conductances for the common-emitter connection by definition. (a) (b) Fig. 2.18 (a) Determining transistor small-signal out- put impedance from the common-emitter output charac- teristics, (b) Enlarged view of (a) showing transistor characteristics in region of interest. 34 Transistor Circuit Analysis PROBLEM 2.16 For the transistor circuit of Prob. 2.15, using the characteristic curves of Fig. 2.17, calculate the static and incremental input impedances. Use a + 5 p increment for l B , Solution: From Fig. 2.17, for V CE > 1 v, / B = 15pa, l B =20fia, A/ B =5pa, V BE = 0.569, V BE = 0.580, A V BE =0.011 v, Static input impedance = ' = 38,000 Q = h IE , 15 x 10 -6 Incremental input impedance = ' * _ = 2200 fi = h ie . Parameters h JB and h ie are static and incremental input impedances, respec- tively, in the common-emitter circuit. Note that the values of input impedance are the same as would be deduced from the calculations of Prob. 2.10, in which V CE is not constant, due to the load resistance R L . The reason is that the V CE vs. l B curve is insensitive to collector voltage except when this voltage is very low. Since the calculations are carried out with respect to / B = 15 pa rather than l B = 20 pa as in Prob. 2.13, h IE is much higher, due to the nonlinearity of the base-emitter junction. PROBLEM 2.17 Using the common-emitter circuit of Prob. 2.15, calculate the static and incremental ratios of collector current to base current for V CE constant. Set l B = 15 pa as operating point, and A/ B = +5 pa as increment. Refer to Fie 2.18. B Solution: Choosing V CE = 10 v (Fig. 2.18a), I B = 15 pa, l c = 3.65 ma, Ib = 20 pa, I c = 5.1 ma. Static current ratio is Incremental current ratio is ^L = i^_=246=/, FE . 1 B 0.015 FE A/ c 1-45x10-' A/*" 5x10- - 290 = ^ The static current ratio with V CE constant in the common-emitter connection is designated as i FB . This is known as the forward current gain. The incremen- tal forward current gain is designated h la . These current gains may be compared with the values of Prob. 2.11. Differences are due to differences in V CE for dis- parate operating conditions. PROBLEM 2.18 A commonly-used two-transistor circuit is shown in Fig. 2.19. Find the quiescent operating point and the over-all incremental current gain A 'c 2 /A I Bi . Solution: Since / Bi is undefined, assume / Cj = 5 ma, which corresponds to a 5.3 v collector voltage V C e 2 on the 5 KQ load line of the applicable curve of Fig. 2.16. From Fig. 2.16, / Bj = 20 pa, and from Fig. 2.19, I B =l E . Transistor Circuit Analysis 35 O V C c = 30v 50 40 30 o ^ 20 2N929 Fig. 2.19 Simplified schematic diagram for Prob. 2.18. 10 0.3fia 0.2Ma 0.166 0.1 Ma IB 0.0983 10 15 20 V CE , volt- 25 30 35 Fig. 2.20 Common-emitter output characteristics in the very low current region of the 2N929 transistor. Consider Fig. 2.20, the common-emitter curves for the 2N929 transistor ap- plicable to the low current region. For / Ej « / Cl = 20 /xa and V CEl = 30 v, l B = 0.166 /xa = 166 ma, scaled from the figure. To determine incremental gain, change I Bl to 0.166/2 = 0.083 /xa, which yields (from the curves used above) / Bj = 10 pa and / Cj = 2.4 ma. v !'• Hence, A/ f 83 ma, A/ c = 4.9 - 2.4 = 2.5 ma, ' rt" SwV < -L4 R L = 5Kfi -Wr-o : V cc = 30v 2N929 100 fi Incremental current gain =—^ r^r = 30,000 83 x 10" for the two amplifier stages. PROBLEM 2.19 In the circuit of Fig. 2.21, switch Sw is closed. Assume an op- erating point at l B = 10f*a. Determine from the characteristic curves the incre- mental voltage gain for a change of + 10 mv in V B , the voltage from base to ground. Solution: Refer to the curves of Figs. 2.16 and 2.22. From the slope on Fig. 2.22, 10 mv (A V BE ) corresponds to A/ B = 3 /xa at the specified operating point. From Fig. 2.16, points P l and P 2 locate the operating points for I B = 10 /xa and l B - 13 /xa, respectively. At P,: I B = 10 /xa, l c = 2.4 ma, V CE = 18 v. At P 2 : l B = 13 tx a, l c = 2.9 ma, V CE = 15.6 v. And therefore, A I B = 3 /xa, A / c Thus, Voltage gain A v Fig. 2.21 Circuit for Probs. 2.19-20. 0.7 0.6 0.5 I 0.4 0.5 ma, A7 CE =-2.4v. w 0.3 .°5 0.2 0.1 -2400 mv 10 mv = -240. Slope at iB=10ii = 3.33mv a y VCE>}v v CE =o 20 Note that a small increase in V B leads to increased collector current and a reduced collector voltage. This is represented mathematically by the negative voltage gain. Because of the exceptional sensitivity to base-emitter voltage, the base is usually fed a current input. The gain calculated above is in effect the a-c gain to small-signal voltages about the quiescent operating (bias) point. 40 j b, H<> 60 80 Fig. 2.22 Calculation of input im- pedance at an operating point from the common-emitter input characteristi cs. 36 Transistor Circuit Analysis PROBLEM 2.20 Repeat Ptob. 2.19 with switch Sw open. Solution: Because R L and R E are in series and the collector current is nearly equal to the emitter current, the load line of Fig. 2.16 is applicable to the present problem. The most direct approach is to find the base to ground voltages which lead to the same operating points as P t and P 2 in Fig. 2.16, and then calculate voltage gain from these values. At P lt from Prob. 2.19, l B = 10 (ia, I c = 2.4 ma, I c Re = 2.4 ma x 100 Q, = 0.240 v. AtP 2 , AV BE = 10 mv, l B --, 13 fia, I c = 2.9 ma, l c R B = 2.9 ma x 100 fl = 0.290 v. The total change in base to ground (input) voltage is the increase of 10 mv in base to emitter voltage plus the change in drop across the emitter resistor. Hence, Change in base-ground voltage = (0.29 - 0.24) + 0.01 = 0.06 v. As in Prob. 2.19, the change in output = A V CE = -2.4 v : Effective voltage gain A v = -2.40 0.06 = -40. ■Afn Fig. 2.23 Circuit and input current for Prob. 2.21. ., The calculated voltage gain is substantially reduced by the introduction of irO = 5v *-^ e ^e resistor. Note that the 10 mv contribution, which is the increase in base- ^ A A*. q emitter voltage, is a relatively minor factor in establishing gain. The drop across R E is a negative feedback voltage which stabilizes gain. As V B is increased, the increased emitter drop tends to reduce the base-emitter voltage. If the base- emitter voltage is relatively small, the base-ground voltage approximately equals the emitter resistor drop. Thus, A ~ ^ L K E as long as the voltage gain is much higher with R E short-circuited. 2N929 10 .? 4 35jxa^ 30/ia 25/Ua K~^ 20/Xa lSua lOfte 5jUa ' — \ p ? / s =o 10 volt- 15 20 Fig. 2.24 Solution to Prob. 2.21. PROBLEM 2.21 Refer to Fig. 2.23. If I B varies as shown between and 25 /xa, what is the variation in output voltage V ? What is the variation if / B varies be- tween and 35 \i a? Solution: Figure 2.24 shows the common-emitter characteristics applicable to the present problem. A load line is drawn corresponding to 7 cc =5v and a 1000 fi load resistor. From the load line, it is apparent that as long as l B > 20 ft a, the operating point remains at P,. The transistor is saturated and is said to be on. The collector-emitter voltage drop cannot be substantially reduced by further in- creases in l B , Similarly, if I B = or less (reversed in polarity), the operating point moves to P 2 where the collector-emitter drop equals V cc , and the transistor is said to be off. Therefore, Vq varies between 5 v and about 0.7 v. The mode of operation described here is called switching, since the output is either on or off, with output voltages independent of l B in the extreme nonlinear regions. PROBLEM 2.22 Using the circuit of Fig. 2.23 but with R L = 0.25 MQ, find the variation in V as I B varies between and 10 i&. Transistor Circuit Analysis 37 Solution: Refer to Fig. 2.25, which is the common-emitter characteristic for low values of collector current. Draw the load line for a 0.25 Mil resistor. The verti- cal axis intercept of the load line corresponds to a collector current of = 20 jxa . 250,000 n Collector-emitter voltage varies between 0.25 v at P t Q c = 19 /za) and 5 v at Pi Oc = °)- The on collector current is determined by the circuit and not by the value of l B , as long as l B is greater than the value needed to sustain l c , in this case, about 0.2 y. a. 2.5 Supplementary Problems PROBLEM 2.23 From the curves of Fig. 2.5 for the 2N929 transistor, deter- mine the operating points (a) l c when V CE = 30 v and l B = 0.01 ma, (b) / B when V CE = 15 v and / c = 5 ma, and (c) V CE when l B = 30fta and / c = 8 ma. PROBLEM 2.24 Using the characteristics of the 2N929 transistor of Fig. 2.5, draw a load line for V cc = 30 v and R L = 10,000 O. Find I c and V CE for I B = 0.01 ma. PROBLEM 2.25 Repeat Prob. 2.24 with R L = 4000 12. PROBLEM 2.26 A transistor with a very high j8 is connected in the common- base mode. Draw a load line for V cc = 20 v and R L = 5000 fl, and find V CB and l c for l E = 1 ma. PROBLEM 2.27 For the common-emitter circuit using the 2N929 transistor with a 6000 Q load and V cc = 30 v, find (a) l B needed to operate at l c = 5 ma, (b) the power P c dissipated in the collector junction, (c) the d-c voltage V L across the load and the power P L dissipated in the load resistor, (d) the input d-c power P B to the base, (e) the variation in the parameters l c , Vce, and v l if 1 b is de- creased by 5 jxa, and (f) the changes in V BE , P B , and P L if l B is decreased by 5/ia. PROBLEM 2.28 From Prob. 2.27, determine the d-c current gain, i.e., the ratio of collector to base current. Also, find the ratio of a change in l c to a change in l B (incremental current gain). PROBLEM 2.29 Using the conditions of Prob. 2.27, determine the input resis- tance (static d-c value), and the incremental resistance to small input changes. PROBLEM 2.30 Referring to Prob. 2.27, find the ratio of output power (in R L ) to input power (to the base of the transistor) for static and incremental conditions. PROBLEM 2.31 For the conditions of Prob. 2.27, find the incremental voltage gain Ar L /AV BE . 50 40 30 10 ' 0.3fia ' 0.2(la \ O.lfta ■\ , p > '?.= ?. S 10 15 20 V CE , volt-*> Fig. 2.25 Solution to Prob. 2.22. 3 CHAPTER SMALL- SIGNAL EQUIVALENT CIRCUITS 3.1 Introduction 1.0 0.8 0.6 - 0.4 0.2 Although the tee-equivalent circuit introduced in Chap. 1 provides an easily visualized model of transistor behavior, there are other equiv- alent circuit configurations that offer characteristic advantages. Alternate models are now presented here on a small-signal basis, where essentially linear rela- tionships hold for small-signal excursions about the operating (Q) point on the characteristic curves. Figures 3.1a-b show typical transistor input and output characteristics with small-signal excursions about the operating point. Note that the assumption of linearity is more valid for the output characteristics - which are well approxi- mated by parallel straight lines - than for the highly-curved input characteristics. **\Kv rjA's v CE=10v W t / B =30//a ) operatin point 9 ! 100 200 300 400 I Bl fla — (a) 500 600 700 (b) (c) I B =32— f'v 9 ' 3-1 T/Pe 2N929 common-emitter characteristic curves, (a) Input characteristics, £^^ZL r < b > output characteristics, and (c) enlarged view of critical region of output character- istics. Note thatA = reference point; C= final point; A/ Ci = 0.25 ma for A V CE = 5 v, where A/ Ci is the change in I c due only to the change in V CE ; A/ c = 1.4 ma for A/ B = 5 Ha, where A/ C2 is the change in I c due only to the change in I B . 3.2 Hybrid Equivalent Circuit The hybrid equivalent circuit is the most widely used for describing the characteristics of the transistor. It is termed hybrid because 38 Small-Signal Equivalent Circuits 39 it combines both impedance and admittance parameters, known as the b-parameters. The ease of measurement of the A-parameters has contributed to its widespread .a^pttoa. A set of A-parameters can be derived for any black box having linear ele- ments and two input and two output terminals. Each of the three basic circuit configurations of the transistor, that is, the common-base, common-emitter, and common-collector, has a corresponding set of A-parameters, both for small- and large-signal operation. The development of the hybrid equivalent circuit is illustrated by the fol- lowing problem. PROBLEM 3.1 Derive the equivalent common-emitter circuit equations from the following functional relationships that characterize the families of curves shown $y Pigs. 3.1a-b: 'c = IcWcE' 'b)> Solution: Both (3.1) and (3.2) may be expanded into differential forms: dl c = fa dV BB = dVcE *V BB dV, CB r CE VcE = 'B> "CE dV, BE a/ B V C E (3.1) (3.2) (3.3) (3.4) Assuming small-signal linear conditions, the partial derivatives, dtr dV t CB dV, Mi dV CE dl c \ d, B I V CE BVt BE dl B 'CE become constants whose values are determined from the characteristic curves. Hence substitution of the appropriate constants leads to the required equations. The above constants are given a special nomenclature because of their im- portance: *v 9Vmb dV cs die dl B dV BE = A oe , output admittance (mhos) , = A,., reverse voltage ratio (a numeric) , = h f9 , forward current gain (a numeric), dl K CB : r GE A is> input resistance (ohms). (3.5) (3.6) (3.7) (3.8) VBE = hjE/s + h RE V CE Ir=h FE'B (a) *OE v CE Sow using lower case letters for small-signal operation, (3.3) and (3.4) become ': Note the mixed or hybrid nature of the A-parameters in (3.5) through (3.8). Tim second subscript e is applied to the individual A-parameters, since in this Vbe =hie>b + J>re" C e 'e = hfe'b + h oe v ce (b) Fig. 3.2 Block diagram representa- tion of the hybrid equivalent circuit for the common-emitter connection, (a) Large-signal parameter (d-c) and (b) small-signal parameter. 40 Transistor Circuit Analysis v be = 'fih/e + h re v ce h te'b PC 'c = h oe v 'ce + h fe'k ce I (b) Fig. 3.3 Equivalent circuit (model) representation of the common-emitter configuration, (a) Input side and (b) output side. instance, it signifies the common-emitter connection. For the common-base and common-collector connections, the subscripts b and c apply, respectively. Figures 3.2a-b illustrate the character of this black-box approach by black- box representations of the common-emitter circuit for small- and large-signal parameters. As was explained in Chap. 2, the large-signal parameters exhibit decidedly nonlinear characteristics. PROBLEM 3.2 Illustrate the physical significance of (3.9) and (3.10) by ref- erence to Figs. 3.1a-c. Also establish numerical values for the parameters at the operating points on the input and output characteristics. Solution: Consider Figs. 3.1b-c in relation to the expression *'c =^oe^ce +f>fei, [3.9] and remember that h oe and h te are assumed constant for small-signal operation. Now A is the reference point, and C, a new point that shows the shift due to changes, v ce and i b . On the I B = 30 M a curve, l B is constant, so that i b = 0; hence i c = h oe v ce . At point B, V CE = 15 v and AV CE = v ce = 5 v. The change A/ C7 = i c in I c is due only to a change in V CE . The slope of the characteristic curve is Mr AF, K e = CE 0.25 ma 5 v 50 x 10 -6 mhos. Now consider the component change in l c due to a change in /„, V r * = con- stant (v ce = 0): Mr. KM B , Mr M* = h. 1.4 x 10 _ From Fig. 3.1c, M c = 1.4 ma, A/ B = 5 /ia, A 5x10- With parameter values substituted in the expression for i c , 280. = A„ + A,„i„ = 50 x 10~ 6 v ce + 280 i b . A similar procedure can be followed with respect to the input characteristics of Fig. 3.1a, whose defining equation (3.10) is repeated here: v be =K e v ce +h ie i b , [3.10] The input characteristic curves, for all but very low values of V CE , are al- most independent of V CE . Thus, for practical operating points, h re may be set equal to zero, so that v be = h le i b . From Fig. 3.1a, at I B = 30 na, M B = 100 fia, AF BE =0.13v, A7, BE v be 0.13 A/ B 100 x 10- = 1300 fi, h ie = 1300 fl , and (3.10) reduces to v be = 1300 i b . As already mentioned, an analogous set of A-parameters can be obtained for both the common-base and common-collector connections, since there is nothing in the preceding analysis which depends on the transistor configuration. All that is necessary for each connection is the analogous set of characteristic curves with the operating point identified. Small-Signal Equivalent Circuits 41 PROBLEM 3.3 Show how (3.9) and (3.10) may be represented by equivalent Solution: Figures 3.3-4 show the set of equations and the equivalent circuit representations. It is seen that the circuit equations are identical with (3.9) and (3.10). The equivalent circuit or model provides an exceptionally simple basis for calculation. PROBLEM 3.4 For the 2N929 transistor whose characteristic curves and opera- ting points are defined in Figs. 3.5a-b, compute the /i-parameters for the common- emitter connection, and draw the equivalent circuit. Solution: The output operating point, A, is defined in Fig. 3.5b as h re v ce hfe'fa i c AAA^-^-CV-»— £>-) * " P( l B = 15 u&, Proceeding as before in Probs. 3.1-2, 'CE = 12 v be = h ie'b + h te v ce >c = h fe'b + h oe v ce Fig. 3.4 Complete hybrid parameter model for the common-emitter con- nection. This circuit applies to small-signal operating conditions. Mr bV CE . A/ c b " ~~ bT B 0.3 x 10" V C E h.„ = bI B bV BE 10 1.4 xlO" 3 5 xl0~ 6 0.22 = 30 x 10 -6 mhos , 290, V CE bv, CE Ib 2,200 n, 100 x 10- 6 (essentially), for V CE > 1 v. The equivalent circuit corresponding to these parameters is shown in Fig. 3.6. i.o . • . . 1 1 1 io 0.8 u m 0.4 CO 0.2 V CE >lv ■ ^1 i L— — ■ J^*" — r^ {Av BE = 0.22v V C B = /&1b = 100(Ja 35p 30fia j 25/Jfl ^S 20/to D C _{_ 15/te -+(V<~i? = A/ Cj ! i nt)* | '* - A Av CE =10\ , B t = A/c t = 0.3 ma lOfla. I B = ° 100 200 300 400 500 600 700 5 I B , fia fc- (a) Fig 3.5 Computing h-parameters for Prob. 3.4. (a) Input characteristics and (b) output characteristics. 10 15 20 VcEl volt — (b) BO- h ie = ■ 2200fl -WAr— t 25 290i„ e 30 35 -OC -ArW = 30 X 10~ 6 mhos (= 33,000Q) The equivalent circuit based upon small-signal variations about an operating point is, of course, immediately adaptable to the analysis of small-signal a-c Fig . 3.6 Equivalent circuit corre amplifiers. The calculation of amplifier performance is merely the determination sp0 nding to the ft -parameters Of Output Signals With given input Signals. derived in Prob. 3.4. «£ 42 Transistor Circuit Analysis PROBLEM 3.5 We are given the circuit of Fig. 3.7 whose operating conditions and parameters correspond to those of Fig. 3.6. The input capacitor is assumed to have zero a-c impedance. (a) For an input signal v g = 10 mv rms, calculate the currents i b and i c , and the voltage across and power in R L . (b) Calculate the current gain A, = iji b . (This is not the same as h te , since the external resistance R L enters into the calculations. Parameter h le is defined for short-circuit conditions; i.e., v ce = 0.) (c) Calculate the voltage gain A v = v ce /v be . (d) Calculate the power gain, i.e., the ratio of a-c load power to transistor a-c input power. (e) Calculate the input impedance Z t . (f) Calculate the output impedance Z . It is assumed that h re = 0. This is generally a valid or realistic assumption for small-signal operation which results in simplified calculations. R B . :!'* looo J2 c b /y^\ ' 2N929 10 mv rms 5000 0, -o v. cc ;30v Fig. 3.7 Circuit for Prob. 3.5. The bias point is set at a base current of 15 fla supplied through R B . Ca- pacitor Cg blocks the d-c bias current to prevent its flowing in the low impedance generator circuit, com- ponents Rg and Cjg are neglected in calculations, and Vg and Rg are a-c generator voltage and internal resistance, respectively. R g = 1000 fl A/W- 220012 -Wr- 10 mv rms 290 i b — vw e =30x 10" 6 mhos (=33,000 Q) ,R L = y L ' 5000 fi Fig. 3.8 The a-c hybrid model corresponding to the circuit of Fig. 3.7. Solution: The first step is to draw an a-c model for the circuit of Fig. 3.7, as shown by Fig. 3.8. Calculations are then made in a straightforward manner using ordinary circuit theory, (a) To calculate i b : 10" R g + h ie 1000 + 2200 = 3.1x10-* a. The current generator develops h le i b , or 290 x 3.1 x 10~ 6 = 0.90 ma. The current source output divides between h oe and R L in accordance with Ohm's law: L = h f „ i Ife'b ^ 1= 0.90 I \ ma = 0.90 ( + J_ \R L h oe + l) { 5000x30xl0- 6 + l, ~ 1.15 = 0.78 ma. Small-Signal Equivalent Circuits 43 The voltage across R L is v L = 5000x(-0.78xl0" 3 ) = -3.9 v rms. The voltage is negative because of the assumed current and voltage polarities of Fig. 3.8. The power dissipated in R L = 0.78 x 3.9 v = 3.04 mw. (b) The calculation of current amplification is 'b 0.78 x 10" 3.1 x 10" = 252. (c) To calculate voltage amplification, input voltage is taken as the input voltage at the transistor base: 4„ = v be -3.9 i h h ia -3.9 -3.9 3.1 x 10- 6 x 2.2 x 10 3 6.8 x 10" ■574. Note that input voltage = i b Z t where Z, = h ie . The minus sign in the voltage gain calculation arises because i c is flowing away from the assumed positive side of of R L . In the common-emitter circuit at low frequencies, the output voltage is 180° out of phase with the input signal. (d) Calculation of power gain: Load power, previously calculated, is 3.04 mw, Power input = i b x v be = (3.1 x 10~ 6 )(6.8 x 10~ 3 ) = 21 x 10~ 9 w, Power gain 3.04 x 10" 145,000 , 21 x 10- 9 or calculated somewhat differently, Power gain = 1-4,4,1 = 252 x 574 = 145,000. (e) Input impedance = h ie = 2200 fl. (f) Output impedance = l/h oe = 33,000 Q. This is the a-c impedance seen looking toward the transistor from R L . 3.3 Tee-Equivalent Circuit The tee-equivalent circuit (Fig. 3.9) provides a close ap- proximation of transistor behavior. Within the scope of the linearity assumptions, it is easy to relate its circuit parameters to physical ones. However, it is difficult to measure tee-parameters directly with high accuracy, in contrast to the ease with which h-parameters may be measured. The best way therefore to determine tee-parameters is to convert from known A-parameters. The basis for converting between A- and tee-parameters depends on the nec- essary identity of behavior of each configuration fpr different input and output conditions. The conditions that will be used in the subsequent analysis are as 1. Input impedance i» measured with output short-circuited. 2. Output impedance is measured with input open-circuited. 3. Reverse voltage ratio is measured with input open-circuited. 4. Foward current gain is measured with output sbort-circeited. For the above conditions, all valid equivalent circuits must yield the same nu- PROBLEM 3.6 Using the common-emitter hybrid and tee-equivalent circuits of Figs. 3.9-10, calculate the four quantities listed above. Also, develop comparable equations for ^e two circuit coaf^uratioaa. Pit T n Vbe r b -vw- — vw- ^T Fig. 3.9 The tee-equivalent circuit for the common -emitter connection. The parameters r bl r e , rj, and p are constant only for small-signal oper- ation. The fixed bias currents ond voltages are not shown. Bo Wv ± V V 7" Vbe h fe'b -€>■ "oe -VW 1- Fig. 3.10 Common-emitter hybrid model. 44 Transistor Circuit Analysis ' Solution* (a\ Tft f*a\r*ii\ata tkft inrait tmtiaflannA £«• «4> A >_,. ^al^l-^A iJ.^.*! _ > i •*wfi#iiw«. \»/ tu uaicuittic uic input lrnpcuancv tor ine lOe^OQulVaURlt ClfCuit, r©~ draw Fig. 3.9 as shown in Fig. 3.11, with the output short-circuited. The current entering node A is (/S + l)i b . The voltage across the parallel shunting resistors Impedance i * 1 iTW is theref ° re . ' - : "- ; / : .'. . , ;, - -^ ' ^ Fig. 3.11 Calculation of input im- The input voltage equals pedance of the tee-configuration. Since the input voltage = Z,i b where <Z, is the input impedance, The input impedance of the hybrid equivalent circuit is. determined by in- spection of Fig. 3.10. Note that h„ v e , * 0, since the output (y„,) is short- circuited. Therefore, (b) To determine the output impedance from the tee-equivalent circuit of Fig. 3.9, the input is open-circuited. Since i b = 0, /3i 6 * 0, and the output im- pedance Z is Analogously, referring to Fig. 3.10, with i b = and therefore h,,if, « 0, (c) The reverse voltage ratio is also calculated for i b = 0. From Fig. 3.9, *?>» = *e» and for (he hybrid equivalent circuit of Fig. 3.10. (d) Referring to Fig. 3.11 to calculate the forward current gain in the tee- equivalent circuit, the current in r d is '6 08 + 1) -^ ' o + T d\ Solving for i c , i''o-/3ib-08 + l)i fc ~^ Since r g « r d , the following approximation is valid: For the hybrid circuit of Fig. 3.10, since the output is short-circuited for the forward current gain calculation, : 'c = "le'bi ~T~ = "t»' Small-Signal Equivalent Circuits 45 Notice the simplicity with which results are obtained from the hybrid equiva- lent circuit. The simple relationships between easily measureable circuit char- acteristics and hybrid parameters are a prime reason for using h-parameters. PROBLEM 3.7 Using the results of Prob. 3.6, derive formulae for conversion between tee- and A-parameters. Solution: The equations derived in Prob. 3.6 lead naturally to expressions for the h-parameters in terms of the tee-parameters. These are listed below: Z, - h t , » r b + (1 + /8)-^- . (3.13) Z = J_ « f . + r„, A .= — (3.14) A„. '- + «■ !'oe » + r d The above formulae may be solved for tee-parameters in terms of A-parameters. Obviously, Dividing (3.15) by (3.14), ** From (3.14) add (3.17), LL_ r>tt X-^L = izA2.. (3.i8) K0 Km K» K» Solving for x b in (3.13), r b «A,.-(l + 0>i r . + r d However, A. K oe But /3 = A, e and Since h„ « 1, r i = A *. -(1 + */•)** (4# r» = fti. -a + **>!=-. (3.i9) PROBLEM 3.8 Using the results of Prob. 3.7, convert the hybrid equivalent circuit of Fig. 3.6 to the corresponding tee-equivalent circuit for the common- emitter connection. 46 Transistor Circuit Analysis r b = 2200fi B o VW EO- 290; €> r d = 33,000li — — wv -oc -OE Fig. 3.12 Equivalent tee-circuit, identical to the hybrid circuit of Fig. 3.6 because h re is assumed to be zero. "fe'b Solution: From Fig. 3.6, Applying the formulae of Prob. 3.7, the parameters of the tee-configuration are - 33,000 Q, ic-e^ 2200 n. With these parameters, the tee-equivalent circuit of Fig. 3.12 is identical to the hybrid circuit of Fig. 3.6. This is so because A,, was assumed to be zero, a perfectly valid approximation. Because A,, is so small, it is generally not feasible to determine it from the transistor static characteristics. Instead, it may be obtained by making small- signal a-c measurements on the transistor. With the transistor properly biased, a small a-c voltage is applied to the output with the input circuit open (to a-c),' as in Fig. 3.13. The ratio of the open-circuit voltage to the applied voltage is h„. This technique is also used for determining h rb and A rc . For example, using the above procedure, the value of A,, for the 2N929 tran- sistor at V CB = 12 v and I B - 15 fia was found to be 2 x 1<T 4 . This means that a 1 v change in V CB produces a 0.2 mv change in the coupled base-to-emitter volt- age. This magnitude of change cannot be seen on the characteristic curves, which explains why the previously derived h„ equals zero. PROBLEM 3.9 For the equivalent circuit of Fig. 3.8, determine the errors in voltage gain (A v ) and current gain (A,) resulting from the assumption that A,. = 0. (In this problem, A re = 2 x 10" 4 .) Solution: The equivalent circuit with the same operating point as in Prob. 3.5 but with the h„ v ce voltage source inserted, is shown in Fig. 3.14. It will be recalled that v ce = -3.9 v, yielding an h te v ce = 2 x 10" 4 x -3.9 = -0.8 mv rms. If i b is held constant, v g must decrease by 0.8 mv to 9.2 mv. For this condition the output current i e is unchanged. The input voltage v be becomes v be = 9.2 x h,. + R. = 9.2 2200 2200 + 1000 6.3 mv, -3.9 6.3 x 10 - = -618 (vs. -574, neglecting h re ). Fig. 3.13 showi Simplified circuit diagram The a PP roxi ™te gain is 7% lower than the "exact" value. This approxi- ng how to measure h re . mation error is almost always acceptable, since the variability in transistor parameters is much greater than 7%. The current gain is unchanged, because it depends primarily on h te . PROBLEM 3.10 Obtain the tee-equivalent circuit from the hybrid model of Fig. 3.14. Small-Signal Equivalent Circuits 47 Solution: The A-parameters are h ie =2200fi, A oe = 30 x 10 -6 mhos, A re = 2 x 10- 4 , h„ = 290. Substitute in the conversion formulae: j8 = A /# = 290, r d =-— = 33,000 0, 2x10- 30 x 10" = 6.67 O, r b = h ie - (1 + A, e ) ^- = 2200 - 291(6.67) = 260 O. A„„ h oe = 30 X 10" 6 mhos lOOOfl =220011 J3; b =290i b r b =260fl BO VW — -WAr- 'b r e =6.67fl r d = 33,000fl -oc Fig. 3.14 Hybrid equivalent circuit, common-emitter connection. The tee-equivalent circuit is shown in Fig. 3.15. Note the substantial change in t b in contrast to Fig. 3.12. The coupled voltage (i e through r e ) in- troduces a large effective resistance value, equivalent in over-all effect to the previous 2200 12 base resistance. \>. Fig. 3.15 The tee-model derived from the hybrid circuit of Fig. 3.14. 3.4 Common-Base Parameters Since many manufacturers' data sheets list only the common-emitter characteristics, it is important to be able to determine the common- base parameters by calculation. The parameters under discussion are ftf 6 , ho b , ha, and A,*. The defining equations for the common-base circuit are v*. » A«> »'• + A* v «b. ( 3 - 2 °) PROBLEM 3.11 Determine the common-base A-parameters for the 2N929 tran- sistor at an operating point l c = 4 ma, V C b = 12 v. Solution: Refer to (3.20) and (3.21). These expressions are most easily in- vestigated by letting i e = A/ E = 0, and in turn, v cb = AV CB = 0, then graphically determining the relationships among the remaining variables. Consider Fig. 3.16a: 48 Transistor Circuit Analysis 10 8 t« o B ^4 h ob = Ale AT, CB A/ Cl AB = 0, '/b A/ c A/ f CB BC ^ 2 x IP' 3 A/ E ~ 2x 10" 3 It is clear that for a high-quality transistor with low leakage current and high- current gain, the collector family curves are almost useless in establishing the output circuit parameters in the common base configuration. .-- — 10 ma 8 ma C 6 ma r JA/ C 4 ma A [s >V CB B 2 ma / E =o 1.0 0.8 |0„ 0.4 s 0.2 10 15 20 VcBi volt— (a) 25 30 35 01 A Av BE i_L Ai E B T 10 12 14 "ig. 2 4 6 8 Is , mo— ► (b) 3.16 Type 2N929 common-base characteristic curves, (a) Output characteristics and (b) input characteristics. The input characteristics are more amenable to calculation. Referring to Fig. 3.16b, A,6 = AV, BE A/* CB 0.06 v 8 x 10- 3 a = 7.5 il. Parameter h Tb = 0, since V BE is almost independent of V CB . The h ib parameter can be established with fair accuracy from the characteristic curves; the remaining hybrid parameters cannot. The parameters can still be measured by a-c techniques, as previously explained, but it is usually more con- venient to compute them from the generally available common-emitter parameters. 3.5 Derivation of Common-Base Parameters Common-base parameters may be derived from common- emitter parameters by the following procedure: 1. Redraw the common-emitter hybrid equivalent circuit, taking the transistor base as the common terminal between the emitter and collector sides. 2. For the redrawn circuit, calculate the four quantities listed in Sec. 3.3 from which the hybrid parameters are derived. 3. Equate the results obtained in Step 2 to the hybrid parameters of the common-base circuit. PROBLEM 3.12 Using the procedure given above, calculate the common-base hybrid parameters of a transistor from known values of the common-emitter hybrid parameters. Solution: Refer to Figs. 3.17a-b which show the hybrid common-emitter circuit, and its redrawing, in which the base B is made common to the input and output. Small-Signal Equivalent Circuits 49 The four quantities to be calculated are repeated below: 1. Input impedance, measured with output short-circuited. 2. Output impedance, measured with input open. 3. Reverse voltage ratio, measured with input open. 4. Forward current gain measured with output short-circuited. Calculate the input impedance of Fig. 3.17b with the collector short-circuited to the base, as shown in Fig. 3.18a. The circuit may be simplified by replacing i K„v the active sources, A^Vce and n /e i b , by equivalent resistances. The base leg E can be simplified as follows: hfe'l t k =>: (3.22) Because the output is short-circuited, v ce = vj, a , and (3.22) becomes i* = v b .O.-K.) ™le r^h (a) /ife'h r^h Solving for the equivalent resistance of this leg, "6» ";• Consider the current generator, h t9 i b . Using the value of i b from (3.22), v b ,h, e (l-h n ) hmh - »ie Since the voltage across the current generator is v be , the current generator can be replaced by an equivalent resistor: Veb=~v b (b) Fig. 3.17 Deriving the common-base parameters, (a) Original common- emitter circuit, and (b) redrawn so the base is now the common terminal. "le With the above simplifications, the equivalent circuit takes the form shown in Fig. 3. 18b, with three resistors in parallel and the active sources eliminated. The circuit can be further simplified by the following approximations: Ke h to (1 ~ A re) h le With these approximations, the equivalent circuit reduces to h le in parallel with Ai./A,„. The common-base input impedance is therefore, hf.it h* x l>U I 'e i h, '/« Me This may be simplified and equated to the hybrid common-base input impedance parameter: Vbe \\'b , /i/ e (l - h re ) Aj HI i + fi" (3.23) fe The forward current gain h a is determined in a similar manner. Referring again to Fig. 3. 18b and using the above approximations, -7- C <c (b) Fig. 3.18 Determining hjb ond ''/b- (a) Circuit of Fig. 3.17 (b) redrawn with C short-circuited to B. (b) Elimination of active sources. ftfb = (3.24) 50 Transistor Circuit Analysis and >c = v b. 21*. Substituting the last two expressions in (3.24), h tb = - l + ht. (3.25) Consider now the calculation of the hybrid output admittance parameter in the common-base configuration. Referring to Fig. 3. 17b, let i. = and calculate the admittance from collector to base. By Kirchhoff's laws, 'b + »'c = 0, Vb c = v be + v ec , -«' b = A /e i b - v ec A Solving this last expression for v ec . Also, ib d + hte) Vbe = ibh, e + A re v co (3.26a) (3.26b) (3.26c) (3.27) (3.28a) h ib o— w\, 1 h (b'b h ob (i /WV • h rh v, rbV cb ^ v b» = 'h hi e - h ta V ec . Substitute (3.27) and (3.28b) in (3.26b), Vbc = i* a,. - *±- ib (1 + hie)+ (bSLthiA, Simplifying, (1 + *,. ) 'b Ke This is the output impedance. In terms of common-base parameters, V " c = A le+ ii-f-^si(i_A F#)ii 1 - 1+ ^(l-A ro ). 'ob " OB = f>le + Using the previous approximations, > c and substituting, 'cb ",. « 1, A,. « l±JU*_, A*- . A b = 1 + h, (3.28b) (3.28c) (3.29) (3.30) Fig. 3.19 Hybrid model for the common-base configuration. ' B It remains now to determine h tb in terms of the common-emitter hybrid pa- rameters. Refer to the common-base hybrid model of Fig. 3. 19. This equivalent circuit should be compared with Fig. 3.17b, where Veb = v eb h tb , v be = v bc h rb . Substituting (3.27) in tt.28b) for v be and using G.28c) for v br . Small-Signal Equivalent Circuits 51 i& h le - ^re 'b (1 + h te ) u Vbe "rb = ' " hi e h oe v bc . - h te d + ftf e) ft le* »b "le ~ 'b U + «/«)+ »b 7 "oe "oe 1 + A/e Making the usual approximations, Then l>,5 becomes 1+h le Ke »hi 1 + 6*. <3.31) Vbe (3.32) 3 'b r? 'b rd AAA/ o — AAA •- Fig. 3.20 Common-emitter tee- equivalent circuit. ;,;. The above problem concludes the development of formulae for the conversion W-eaminon-etnittef hybrid parameters to coauaea-base hybrid parameters. The Simplified conversion formulae are summarized below: 'ib J ie hg, «*- h ob = 1 +.*,.' [3.23] 1 + h te 1 + h [3.25] v °" A- [3.30] lit /l.K = h&hsS. _ /,; ««* 1 + h lm L3.32] ^'b ■AAAr- -AAA * Vcb Fig. 3.21 Circuit of Fig. 3.20 re- drawn so that the base is now the common terminal. PROBLEM 3. 13 Convert the parameters of the common-emitter tee-equivalent ^Circuit to corresponding common-base parameters. (a) AAAr-^ j- P'fd (b) ^v-O-O^ r d fi'e r d fii c r d (c) o P'etd (d) P'd ■A/W- »Iution: The common-emitter tee-equivalent circuit is shown in Fig. 3.20. It redrawn in Fig. 3.21 for the common-base configuration. For a true common- ue model, the network consisting of {H b in parallel with r d must be developed . terms of input current i a , rather than input current i b . Therefore convert the current source /Si b in parallel with r d shown in Fig. ^22a to an equivalent voltage source in series with a resistor, as in Fig. 3.22b. equivalency of the two networks is obvious from the figures, where they ibibit equal open-circuit voltages and equal output impedances. The network of Fig. 3.22b must be expressed in terms of i e instead of i b . following fundamental equations have been developed in Chap. 1: ».-ju -i> /Sib* -AAAr- r c =(l+j8)r d (e) Fig. 3.22 Steps in the network con- version of Prob. 3.13. For (a) and (b): output resistance = r^, open cir- cuit voltage = pi b r d . 52 Transistor Circuit Analysis Using these relationships. Fig. 3.22b takes the modified form of Pig. 3.22c, in which two voltage generators are shown. Note that the generator Bi e r d has a voltage drop opposing i' c exactly equivalent to the drop across a resistor, Br,,. This suggests ^m^^s^Mit^muk-W^g^-M^^^ ■"' . ' "\ ■■■ . : -.- " .' The above network is converted to a current source in parallel with a re- sistor to obtain the common-base tee-equivalent circuit. The output impedance of the network is r d (l + B). The parallel current source, multiplied by r d (l + 8), must equal the open-circuit voltage, Bi u r d . Thus, the current source is [B/(l + B)]i e , or ai.. Figure 3.22e shows this parallel configuration. The re- sultant common-base tee-equivalent configuration is given in Fig. 3.23. i Figures 3.24-7 summarize approximate conversion formulae between hybrid B and tee-models for the common-base, common-emitter, and common-collector con- c . , , r . figurations. Those formulae not derived here may be verified using the methods rig. 3.23 Common-base tee- - .. . ,. ...... „ -»•.«. .........vo equivalent circuit. of the P^vious problems. A table of "exact" formulae is given in Appendix B, but these are rarely used in practice. 3.6 Calculation of Amplifier Performance A prime application of transistor models is in the calcu- lation of small-signal amplifier performance. This includes the determination of voltage, current, and power gains, and input and output impedances. r€T] 'be f , i ltd (a) Common-emitter configuration. (b) Hybrid equivalent circuit. Hybrid 'ie h le h oe Common- base hn 1 + A/b 1+A h t b tb hfb 1 + A, 1+A« Common- collector 1-A,c ■(1 + A/c) Tee- equivalent Tb + r e 1-a r e (1 -a)r c a 1 - a 1 (1 - a)r c (c) Approximate parameter conversion formulae. A,„ = 2200 n A re = 2 x 10" 4 A fe = 290 A oe = 30 x 10" 6 mhos (d) Typical values for type 2N929 transistor at Iq = 4 ma, Vce = 12 v. Fig. 3.24 Conversion to common-emitter h-parameters. Small-Signal Equivalent Circuits 53 v eb v cb i )^J> (a) Common-base configuration. (b) Hybrid equivalent circuit. Hybrid Common- emitter 1 + h le h i e " o e j. 1 hi. l + *«. ft oe 1 + fcf. Common- collector fc,«-l- ft/c^oc 1 + ftfc A/e _ A OC /l /c Tee- equivalent r e +(1 -Cl)r b 'b (c) Approximate parameter conversion formulae. h, b = 7.57 Q ).268 x -0.996 ft rb = 0.268 x lO" 4 /i o6 = 0.103 x 10" 6 mhos (d) Typical values for type 2N929 transistor. Fig. 3.25 Conversion to common -base h-parameters. I > 1 (a) Common -co I lector configuration. h ■ l b 11 1 r. —zL r~r tc'b « '•oe v ec (b) Hybrid equivalent circuit. Hybrid h ic hie Common- emitter !-*,. = ! -a+A,.) Aoe Common- base Ait 1-ft, 1 + h, b * 1 + h tb ~ 1 1 + Afb Aob 1+A /6 Tee- equivalent " 1-0 1 r -± — = 1 (l-a)r e -1 1- a 1 (l-a)fe (c) Approximate parameter conversion formulae. h lc = 2200 n h rc = 0.9999^1.0 h lc = - 291 h oc = 30 x lO" 6 mhos (d) Typical values for type 2N929 transistor. Fig. 3.26 Conversion to common-collector h-parameters. 54 Transistor Circuit Analysis ie eo 'VAA/ f • — *\AiA* — • — oc (a) Tee -equivalent ci rcuit, common-base. l-a -■b=-p' b BO VW f * VW-1 — oc <■<*=«•<; (1-<X)=. i+)8 (b) Tee-equiva lent circuit. common -em i tter . Tee- param- eter P Common- emitter 1 + A/e h le + 1 Ke_ Ke h h re (l + h le ) "ie Ke Common- base - h ib i~Kb K b h ib ~a + h tb )^l Kb Kb Kb hfb Common- collector 1 + A, _ his. Ke "lc + 1 f h lb (c) Approximate parameter conversion formula a+A/c) a = +0.996 r c = 9.7 Mfi r e = 6.667 fi r„ = 260 [l (d) Typical values for type 2N929 transistor Fig. 3.27 Conversion to tee-paramete CC 2N929 30v Fig. 3.28 The a-c emitter-fol lower amplifier for Prob. 3.14. Choke I. and capacitors C l and C 2 isolate a-c and d-c circuits. PROBLEM 3.14 Figure 3.28 shows a single-stage, a-c transistor amplifier. The collector characteristics for the type 2N929 transistor are given in Fig. 3.29 Determine: (a) A-c voltage gain, for a 10 mv input signal. (b) Input impedance. (c) Output impedance. Solution: (a) This is a common-collector circuit, thus requiring common-collector parameters to be used in the analysis. Let us first establish the d-c operating point at which the small-signal parameters are to be determined. Since I E = /,! for the very low base current of this example, a load line can be superimposed on Frg. 3.29, even though the 5000 ft resistor is in the emitter circuit. The op- erating (Q) point is determined as V CE = 11.6 v, I c = 3.7 ma, I B = 15/za (as given). This is close to the operating point of Fig. 3.6, in which the /.-parameters were determined for the common-emitter connection. These parameters are repeated below tor convenience: h ie = 2200 ft, Ke = 2 x 10- 4 , h le = 290, h oe = 30 x 10- 6 mhos. The above parameters are converted to common-collector parameters using the conversion formulae of Fig. 3.26c: Small-Signal Equivalent Circuits 55 h lc =h le = 220012, A™ = 1, h k = -(l + fi fe ) = -291, h nr = h nB = 30 x 10- 6 mhos. •oc — "oe The common-collector hybrid equivalent circuit therefore takes the form of Fig. 3.30. Using this equivalent circuit, »ec - v o" Fig. 3.29 Determining the operating point in Prob. 3.14. B f 1)c =2.2KQ -o wv— h tc 'b = -291 /„ e U j «'o| Z a v Fig. 3.30 Common-collector equivalent circuit corresponding to the amplifier of Fig. 3.28. The basic equations are 'i = '6 v 6 - A rc v ec 'ic h, c i fc 'b h (c i '« = l B = fc 'b Rl.+ 1 h oe R L + 1 v =-i a R L- Since v ec = v 0) we may combine the above equations to solve for v in terms of i b : 1 + h oc R L h ic Simplifying,. d _ A tc *z. x Ajc.\ = _ A fc R L /M (3i 3 3) \ l+h oc R L h,J l + h oc R L \h ic J Substituting numerical values, l + h oc R L =1.15, h ^= ^ r =4.54xl0- 4 , 'fc 291 | Thus, h lc 2200 (-253) (5000) 1 + /VP, 1.15 = - 253. 1- 2200 (-253) (5000) vg 2200 Amplifier voltage gain is now determined: &> Transistor Circuit Analysis v 6 1 + 575 575 = 1 - 0.00174 ^1. The voltage gain is slightly less than unity. (b) If v g = 10 mv, then v ec = v ^ 10 mv. Now substitute in (3.22) and cal- culate i b in order to determine the input impedance Z,- : h lc h.„ • f 3 - 22 ] i b = Vbc ~ h ™ V ec = V g ~ h rc V Q Since the voltage gain is only slightly less than unity, let v = v e where Av /v is the per unit deviation of output from that corresponding to ex- actly unity gain. Substituting this expression in (3.22), (3.34) ib = v 6 h ic ~ v e(l- Av Let h rc = 1, and simplify: ib h ic ra Calculate the input impedance > Zi- z t ib hic Av Substitute numerical values: h le - 2200, ^l = 000174; z = 2200 = L2fi4 x l0 . m im m v o U.U0174 (c) The output impedance is obtained by inspection of Fig. 3.30 as 1 Z„ = — +h c Substituting numerical values, z ° = T/sooo + aoTiF = 435oa The common-collector amplifier studied in the preceding problem exhibits the following characteristic features: 1. The voltage gain is very close to unity. 2. Input impedance is relatively high. 3. Output impedance is relatively low. This circuit is called an emitter. follower, analogous to the vacuum tube cathode- follower. While the calculations of Prob. 3.14 usually provide satisfactory accuracy it is often important to know the deviation of voltage gain from unity to a high degree of precision. The approximation h rc = 1 is not sufficiently accurate, and a more precise figure is required. Small-Signal Equivalent Circuits 57 PROBLEM 3.15 For the emitter-follower of Fig. 3.28, using the "exact" value of h rc , calculate the percent deviation from unity gain and the input impedance. Use the equivalent circuit of Fig. 3.30. Solution: Rewrite (3.33) to obtain the voltage gain v /v g : -h tc Rl 1 + Kc Rl [hi J Vg i _ ftfc r l 1 + h nr . R oc "L Equation (3.35) may be expressed as (3.35) ~l + h oc R L (h lc \ . h tc R L \h TC ) The quantity in brackets is very much less than unity. Therefore, dividing and keeping the first two terms of the quotient, 1 1 + (L±J1ocRl\ (hjc. h fc R L j \h rc For h r 1, 1 + h oc R L \ Ihu 1 575' ■M1--V This is the value that was obtained in Prob. 3. 14. More accurately, h tc = 1 - h re = 1 - 2 x 10- 4 , J- = i s 1 + 2 x 10- 4 , h rc 1-2x10-* 4=(l+2x 10" 4 )(1 - 17.4 x 10-*), A v e 1 - 15.4 x 10- 4 . The deviation from unity voltage gain is 0. 154%. The input impedance may be calculated from (3.34): ■•^K^)} z - Y±- hie 1-h. (-^) From the previous gain calculation, 1 - ^Se. = 1 _ 15.4 x Hr 4 , Substituting in (3.37), (3.36) [3.34] (3.37) i„ = l-2x 10- 4 , = 1 - (1 - 2 x 10- J, )(1 - 15.4 x 10- 4 ) = 17.4 x lO" 4 . Z = 2200 1Q , l2M Mfl . 17.4 58 Transistor Circuit Analysis R L = 500012 Fig. 3.31 The a-c amplifier ci rcui t for Prob. 3.16. The input impedance is not particularly susceptible to the error introduced by using the approximate value of h rc . PROBLEM 3.16 Calculate the input impedance Z t and the voltage gain v /v e for the circuit of Fig. 3.31. Solution: Figure 3.31 is treated as a common-emitter circuit where R E is a cur- rent feedback resistor. Since I E and l c are nearly equal, the effective load line resistor is R L + R B , as shown in Fig. 3.32. The Q point is approximately 3.7 ma at 11.2 v. This is close enough to the operating point of Prob. 3.4; therefore the same h e parameters may be used: h ie = 2200 0, A re =2x 10- 4 , Afo = 290, Ke = 30 x 10- 6 mhos. h f „i le'b r€h (b) Fig. 3.33 Model for the ampli- fier of Fig. 3.31. (a) Exact equivalent circuit and (b) simplified output circuit. IS 20 V CE • v olt ► Fig. 3.32 Finding the operating point for the circuit of Fig. 3.31. The equivalent circuit takes the form of Fig. 3.33a. The emitter resistor acts as a coupling element between the base and collector circuits, and is best dealt with on an approximation basis. This practice is almost always legitimate in transistor circuitry where parameters are rarely known to a high degree of ac- curacy. The following two approximations are very helpful. 1. Let h te £ 0, since v ce h re is very small for A v h re « 1. (This may be checked after the voltage gain has been approximately determined, as explained below.) 2. Let i e = i c , an excellent assumption for high current gain transistors. Using the above approximations, the output equivalent circuit can be simpli- fied as shown in Fig. 3.33b and analyzed by conventional methods: ™ie 'b 1 ^e't From Fig. 3.33a, Rr + R, tbh ie + l + (R E +R L )h oe (3.38) Re hie l + (R B +R L )h c Small-Signal Equivalent Circuits 59 Solving for Z t = v g/i b , Z, = h ie + R ^ . (3.39) Substituting numerical values, ?9 000 Z, = 2200 + *»>}™>. = 27,3000. ' l + (30x 10" 6 )(5100) The output voltage v is 1 +h oe (R E + R L ) Substituting V g/Z { for i b , The voltage gain is y -**< h f X ^ ° l + h oe (R E + R L ) Z { A Is = Z^Jlle x _L. (3.40) v & l + h oe (R E + R L ) Z i Substituting for Z f and simplifying, A v = =^^ t (3>41) A ie + /i ie h oe {R E + R L ) + h, e R E Before substituting numerical values, observe that if h fe is very large, the voltage gain reduces to A r = =^. (3.42) K E This approximate formula is very valuable in estimating the approximate behavior of circuits having the configuration of Fig. 3.31. The numerical results of this problem confirm the validity of (3.42). Us- ing (3.41), -5000x290 A„ = 2200 + 2200(30 x 10" 6 )(5100) + (290) (100) - 5000 x 290 = -46. 2200 + 337 + 29,000 This compares well with a value of - 50 determined from the approximation of (3.42). Check the validity of our assumption that h Te v ce is negligible. Assume v g = 10 mv. Then v = A v v g = 460 mv. Thus, h re v ce = 2 x 10 -4 x -460 = - 0.092 mv. This voltage aiding the input signal is less than 1% of v e , confirming the sound- ness of the earlier assumption. The characteristics of the above circuit can be further clarified by solving for input impedance and voltage gain using the common-emitter tee-model. The pa- rameters were determined earlier in Prob. 3.8: 291 r b =260O, r a = 6.67 0, r c = ^-^ = 9.7 MO, fl = 290, r d = -^- = 33,0000. 1+ P 60 Transistor Circuit Analysis PROBLEM 3.17 Using the above parameters, solve the tee-equivalent circuit of Fig. 3.34 for Z { and A v . Solution: The tee-circuit is easily solved using conventional two-mesh analy- sis. Since most engineers are more adept in using voltage generators, these are substituted for the current generators of Fig. 3.34, yielding the modified circuit of Fig. 3.35. The mesh equations are v 6 = (r b +r e + R E )i b + (r e + R E )i c , Ph B r b +1 4 — wv— -4 — • f^p h = (Re + r e )i b + (r e + R E + ^S- + R L \ if (3.43) The second of these equations may be rewritten as 1+0 Using determinants, solve (3.43) and (3.44) for i b : (3.44) 1 Fig. 3.34 Tee -equivalent circuit for common -emitter connection showing resistor R E . - V t- r „ + R K r e +R E + —£- + R L 1 + + R B+T f^ + R L where A is the determinant of the system equations. Solving for Z, = v 6 /i b , A Z. r e + R B + (3.45) 1 + + Rr The determinant A is 1 + B 'b 1 + B r b -• — VW- &l A = 7 t b + r e + R E r B +Rr R E +r e - + Rl +1 /~\ ?° ' c (~) ^ '" 'M Je (u) R L ^ v =r b ^r e+ R B+ -i- + R L \ + (r e + R E )( fc + R L ). Substituting (3.46) in (3.45), Fig. 3.35 Equivalent circuit solved by using mesh currents. Z l = r b + (r. + R E )(r c + R L ) (r e + R E ) + (3.46) (3.47) Insert numerical values: 1 + /3 + Rl Z, - 260 + a06-67)(9.7xl0' + 5000) _ ^^ (106.67) + 9J x 1Q6 + 5000 291 This checks very closely with the value of input impedance calculated for the hybrid circuit of the previous example. Simplifying (3.47) by making the following approximations, r e + R E « -^ +R L , r. <v (r e +R E )(r c +RL) 1 + /3 T h « 1 + /3 + Rr Small-Signal Equivalent Circuits 61 we obtain Z,^ (t e +R E )(t c +R L ) + Rl (3.48) 1 + A further level of approximation with consequent simplification is introduced by assuming that » Ri, l + i8 which leads to Z, = (1 + /3)(r e + R E ). (3.49) Check this approximate Z t with the more accurate value of 27,10012 by sub- stituting numerical values: Z, = (1 + 290) (106.7)= 31,000 fl. This approximation is perfectly satisfactory for many purposes. Now calculate voltage gain. Determine i c by the use of determinants, t b + t e + R E £lc 1_ A 1 + 18 Since v = -i c R L , and A v = v /v g , We now substitute the expression for the determinant: A = t b It. +R E + y±- + R^j + (r e + R E )(r c + R L ). Simplifying (3.50) and (3.46) by the following approximations, (3.50) [3.46] r c » t e + R E , the equation reduces to 1 + »r e +R E , r c »R L p. ll^ 1 + •(iV*) (3.51) + r c (r e + R E ) Substituting numerical values, A„ = 290 -(5000) — 9.7 x 10« 291 260 f 9,7 x 1Q6 + 5000J + 9.7 x 10 6 (106.7) = -46.3. r i i 'E, ^ Transistor Circuit Analysis This confirms the previously calculated value of - 46. The expression for gain (3.51) can be further simplified. The second term in the denominator of (3.51) is often much greater than the first term when R B is present in the circuit. Therefore, A.* - Rl (3.52) r e +R B or simply, A v ^ ~ R ^ , r e +R E As a check, substitute numerical values from the above problem: ^^?— • This is obviously an excellent check of the approximate expression. The significance of the simplified gain formula -^.//fe is readily apparent from a physical viewpoint. The base-emitter voltage is small, so that changes in base voltage must be approximately duplicated by changes in emitter voltage. The presence of an emitter resistor means that emitter (and therefore collector) current must follow changes in base voltage. If the collector current follows base voltage changes, then the drop across R L must do likewise. The drop across R L is equal to R L /R E times the emitter resistor drop, which approximately equals the base voltage. Hence, the voltage gain approximates -R L /R E . If R B is substantially larger than the variations of r e due to temperature, long term drift in the point and voltage gain will be relatively insensitive to these factors. Voltage gain is also relatively insensitive to changes in /3. In the above cir- cuit, if doubles, the voltage gain will increase by about 0.2%. Although this use of R B — an example of current feedback —reduces voltage gain, it provides a simple method of introducing negative feedback around a single stage for in- creased stability and input impedance. 3.7 Hybrid-7T Equivalent Circuit h ie = 2200fi h le = 290 h re = 2 X 1CT 4 h oe = 30 X 10- 6 mho This equivalent circuit was originally derived by means of fundamental considerations in Chap. 1. Its value rests primarily on its use at (°> "*& frequencies. However, in this chapter we will concentrate on its low-fre- quency characteristics, in particular, the development of conversion formulae be- tween h- and hybrid-* parameters. Accordingly, the capacitors which represent high-frequency effects are momentarily ignored. Figures 3.36a-b show the h- and hybrid-* models for comparison purposes. Note that the hybrid-* circuit has five independent low-frequency parameters; i.e. one more parameter than the other equivalent circuits we have thus far studied. One parameter may be specified in an arbitrary manner. Usually it is convenient to select r bb , as this arbitrary parameter, although at high frequencies the actual value can be measured. No problems arise from this arbitrary selection; however _. , ,, , . u L ltlS P referable to choose r bb , so that all hybrid-* parameters are then positive. rig. j.Jo (a) Hybrid equivalent cir- cuit for common -emitter connection. PDnni CM ■» in ,-■ ,. (b) Hybrid-77 equivalent circuit for ™ UB L-CM J.18 Given the fi-parameters of Fig. 3.36a, derive conversion for- common -emitter connection. Capaci- ">ulae for the hybrid-* parameters. tive components are not shown in C»l„»!„. tl l this low-frequency model. Solution. The basic approach is to compare the behavior of the circuits for the conditions of the output short-circuited and the input open-circuited; i.e., short- (b) Small-Signal Equivalent Circuits 63 circuiting the collector to the emitter so that v ce = 0. Then i c /i b is calculated for both configurations and equated. For the hybrid-??, (3.53) 'c = SmVb'e = >bgm r b'c + Tb'e For the /i-parameter circuit J 'e = hfe 'b $ Equating, Bm = » f eq feq = rb'c Tb'c Tb'e A r b ' e (3.54) Vc ■•- r b ' e Note that r eq = r b ' B since normally r b ' c » r b ' e . In a similar manner, the short-circuit input impedance is calculated. For the hybrid-;7, Z l = T bb ' + r.q, and for the h-parameter circuit, Z, = h le . Equating, fcfe = Tbb' + fau- lt we determine r bb ' either by measurement at high frequency or arbitrarily, we can determine r eq : feq =h le -i bb '. (3-55) From (3.54), g m is determined: *. = — ^— . (3.56) h le ~ 'bb' A simple approximate expression for the large-signal value of g m may be de- rived. From (3.53), { e = &m V b 'e- Dividing both sides by I B , b_ =gm Y^S.. (3.57) /a 'a However, based on (1.11) at room temperature (300°K), V b , e _ 0.026 Substituting in (3.57), i '* oil 0.026 For commercial silicon transistors, a somewhat more accurate expression for £m is is f ' (3.58) Sm 0.043 If no measured value of r bb ' is available, this expression can be used to estimate g m and r bb '. 64 Transistor Circuit Analysis To continue our conversion of parameters, v c ./v b8 is determined with the in- put circuit open, and v c , applied at the collector. For the hybrid-*, ;V».-v c . ■ *>'> , y , '■ For the A-parameter circuit, Equating these expressions. r b'« + r 6 ', v be = /»r« v c (3.59) *W -A* TABLE 3.1 Conversion from hybrid to hybrid-TT parameters. &m z A /e [3.56] - A ie - r b b r b'c- A ie - r bb > Ke [3.60] fb'e = hie - r bb > 1-A,. = ft /e ~ ffcb' [3.61] -!--*-- ,, 1+A/e ""re- -tl t e r ce "fe - Tbb' = ftoe [3.63] **t - *** At* ~ *** A re "re For these same conditions, Substituting (3.60) for t b - c in (3.59) to determine r b '„ * * A*»~«'tri» > (3.60) [3.59] (3.61) Again, for these same conditions, the base circuit open and v c . applied to the collector circuit, v b'» = v c . «V. f 6'» + rv e All currents at node C of the hybrid-jr circuit are now summed: [3.59] *'. - vc U m tb '° + — + — * — \ (362V This may be modified by substituting the hybrid parameters already determined: But by definition, so that v ee h l9-T bt) ' h lm -r bb ' r eo r- A oe , Ac - A M * + *'■-*" = A oe . (3.63) r c« "le-r bb i This completes the required conversion. The results are summarized in Table 3.1. PROBLEM 3.19 Using the parameters of Fig. 3.36a, derive the corresponding hybrid-7r parameters for r 66 / = 0. Solution: The required parameters are found by direct substitution. Referring to Table 3.1, 290 * m= 2200 = <U32mh0 ' 2200 Tb ' c = 2710= r b ' e = 2200 0, = llx 10 6 Q, [3.56] [3.60] [3.61] Small-Signal Equivalent Circuits 65 1 -= 30 x 10- 6 - 2 x 10- 4 | J?L ] = 3.4 x lO" 6 mhos, \2200 ) [3.63] so that r ce = 286,000 £1. PROBLEM 3.20 Repeat the previous problem using r bb ' = 255 Q. Solution: Proceed as before: 290 1945 1945 0.149 mho, -=9.73x 10 6 fl, oc "2xl0- r b >. = 1945 fl, — = 30 x 10- 6 - 2 x 10- 4 -££-= mho, t.. 1945 [3.56] [3.60] [3.61] [3.63] so that r ce = ■». For r bb > > 255 Q, r ce would be negative, making it inconvenient for calcula- tions. PROBLEM 3.21 Figure 3.37 shows the hybrid-77 amplifier model. Using the hy- brid-w values found in Prob. 3.20, calculate load power, and current and voltage gains. R R e = K r bb ' = r b'c - n lKO b 25511 B ' 9 - 73x lo6fl « Wr-f WV f- g - rbb' lKfl B 225fi -W\» • W^ — f 9.73 X 10" fl -vs/v- 5Kft K L =0.149v b ' e R L = 745v t -._.,,., .... . , n ■ o on Fig. 3.38 Hybnd-77 amplifier circuit Fiq. 3.37 Hybrid-TT amplifier circuit for Prob. 3.20. M „ ' . , s ' of Fig. 3.37, with current source re- placed by voltage source for easier calculation. Solution: Start by replacing the current source and R L of Fig. 3.37 with the equivalent voltage source shown in Fig. 3.38. The basic equations are v g = {R £ + r bb ' + r b ' e ) i t - t b > e i 2 , V L = - r b ' e ii + (ffc'e + *Vc + Rl) V Substituting numerical values and solving by determinants, I i l = 3.38 fj. a, \ i 2 = 0.44 n a. Since v b > e = r b ' e (i, - i 2 ), j v L = 745 r b ' e (»', - i a ) = 4.12 v, ( and continuing, - v ce = v L + i 2 R L = 4.12 + 0.44 x 10" 6 x 5 x 10 s 2 4.12 v, i„ = - 4.12 R L ~ 5000 = 0.824 ma, p l =i c v ce = 3.12 mw. 66 Transistor Circuit Analysis Thus, A Lm 824x10-' i, 3.38 x 10- 6 A v = Zls± = ~ v " = - 625. v be v b'e + Tb'b 'b 3.8 Supplementary Problems PROBLEM 3.22 Give the generalized definitions of static and incremental h- parameters. PROBLEM 3.23 Define mathematically the static and incremental fc-parameters for the common-base connection. PROBLEM 3.24 For the characteristics shown in Fig. 3.5, find the /i-parameters for a 2N929 transistor when the CE connection is at V CE = 10 v and I c = 2 ma. PROBLEM 3.25 Determine r e , r b , and r d for the A-parameters of Prob. 3.24. PROBLEM 3.26 In the circuit of Fig. 3.7, let R L = lOKfl, R B = «, R g = 2KQ, C B = oo,and I B = 20/xa. Determine (a) the operating point, (b) the incremental A-parameters at the operating point, and (c) r e , r b , and r d . Draw this circuit, replacing the transistor by its tee-equivalent network, and determine the input and voltage gain using standard circuit analysis and assuming r d infinite as an approximation. PROBLEM 3.27 How are r e and r d measured? PROBLEM 3.28 Design a simple (3 measuring circuit. PROBLEM 3.29 In the circuit of Fig. 3.28, if l B = 5 pa, find (a) the maximum rms output voltage without distortion, (b) the v^ that generates the maximum voltage, and (c) the approximate input impedance. PROBLEM 3.30 If R L = 10 6 fl in Fig. 3.31, the approximate formula (3.42) is no longer valid. Why? PROBLEM 3.31 If h ie = 5000 Q, /, fe = 300fl, h re = 10~ 4 fl, and h oe = 10~ 6 O for a transistor, find (a) the tee-equivalent circuit parameters and (b) the input and output impedances. PROBLEM 3.32 For Fig. 3.2b, define the small signal /i-parameter in physical terms for the common-emitter circuit. BIAS CIRCUITS AND STABILITY 4 CHAPTER 4.1 Introduction The key to correct transistor operation is the establish- ment of a. quiescent operating (Q) point. This corresponds to the steady current condition that occurs in the absence of an input signal. Setting up a bias point for transistors is more difficult than for vacuum-tube circuits because of transis- tor leakage currents, which are extremely sensitive to temperature. A transistor which is correctly biased at room temperature may be incorrectly hiasfid at high temperature. This drift in operating point is far more characteristic of high-leakage ger- manium transistors than low-leakage silicon ones. However, leakage currents are Important even for silicon transistors. Therefore, this chapter not only considers rSow to set the correct bias point but also how to compare circuit configurations for sensitivity to leakage changes. It will be found that some configurations are for more stable than others with changing temperature. 4.2 Leakage Current Diode leakage has been described in Chap. 1. The re- verse-biased collector-base junction of a transistor (emitter open) exhibits simi- lar leakage characteristics. Transistor collector-base leakage l CBO varies with temperature typically as shown in Fig. 4.1. Since leakage currents in silicon transistors are far lower than in germanium devices, Fig. 4.1 tells only part of the story. PROBLEM 4.1 A germanium transistor has a leakage current of 5 pa at room temperature (25°C). If Fig. 4.1 applies, find the leakage at 75°C. Solution: The room temperature value is 5 pa. At 75°C, this is multiplied by a factor of 30, for a leakage current of 150 /xa. PROBLEM 4.2 A 2N929 silicon transistor has a maximum 25°C leakage current of 0.01 pa. Find the maximum leakage from Fig. 4.1 at 125°C. Solution: From Fig. 4.1, the leakage is multiplied by a factor of 45, for a leak- age of 0.45 pa. Because germanium transistors have much greater leakage currents than sili- con ones, a germanium 2N1308 n-p-n type has been selected for examples in this chapter. Figures 4.2-3 show its common-emitter characteristic curves, which will provide the necessary data for subsequent calculations. Although this transistor 67 68 Transistor Circuit Analysis 100 10 u « 1.0 0.1 0.01 -50° Silic on / German um i r f- / / / / Germanium (multiply scale by 10 . ■ i i »v f -25 1 " 25 L 50 1 - 75 u 100° 125° Junction temperature, °C — ► Fig. 4.1 Variation of l CBO with temperature relative to 25°C. 0.6 0.4 0.2 V CE =2v -20°C 25°C r r 70°C r 0.1 0.2 0.3 0.4 0.5 0.6 0.7 l B , ma-* Fig. 4.2 Input characteristics of 2N1308 n-p-n germa- nium transistor vs. junction temperature for the com- mon-emitter connection. Fig. 4.3 Collector characteristics of 2N1308 n-p-n germanium transistor vs. junction temperature for the common-emitter connection. 'CBO Fig. 4.4 Common-base tee-equivalent circuit. The directions of current flow correspond to an n-p-n transistor. has a 5 pa maximum leakage at room temperature, 1 ^a is more typical and assumed here. 4.3 Tee-Equivalent Circuit Representation of Leakage* The tee-equivalent circuit, discussed in Chaps. 1 and 3, is particularly convenient for studying the effects of leakage in transistor cir- cuits. Figure 4.4 gives the equivalent circuit for d-c or bias conditions. If the emit- ter circuit is open, l B « 0, and the output current from collector to base is / CBO (sometimes abbreviated to l co ). This circuit, in effect, provides a definition of collector-base leakage. A modified tee-equivalent circuit for the common-emitter connection is pro- vided in Fig. 4.5. Here, the leakage component is between collector and emitter. The value of leakage current, l CBO , in terms of I CBO , may be determined from simple transistor equations. * As we are analyzing only d-c signals in this chapter, /? and a correspond to static characteristics. Currents and voltages are shown with their normal polarities. Bias Circuits and Stability 69 PROBLEM 4.3 Derive a formula for l CEO in terms of current gain /3 and / cbo . Solution: The basic transistor current equations are We combine the two expressions to eliminate l E and then solve for / c : f^ V . 1+a ■ (4.1) o Now we substitute the common- emitter current gain /3 as a new variable to re- place a in (4.1), where j8 = a/(l - a). Therefore, h = Icbo (/3 + 1) + h P- (4-2) Since the second term in (4.2) is the normal transistor output current, the first term must be the required leakage component, /cec- Therefore, Fig. 4.5 Tee-equivalent circuit in the common-emitter connection. Icmo m l&solX + /8X (4.3) The current gain factor j8 leads to a relatively high leakage current in the com- mon-emitter circuit. Since the leakage current occurs with the base open, the leakage is identical in the common-collector circuit. The multiplied leakage currents in the common- emitter and common-collector connections, and their high temperature sensitivity, lead to bias point instability. The drift in bias point due to temperature or inter- changing transistors of the same type is the central problem of biasing. The es- sential requirement is that l c be maintained constant over all operating condi- tions, because this fixes the quiescent point on the load line. Figure 4.6 provides additional clarification on the effect of leakage. The figure shows / c vs. l B in the common-emitter connection. Note that at l B = 0, the collector current must equal I C bo- Furthermore, when base current becomes negative and equal to the collector current, emitter current must be zero, so that the coordinates of point P are as shown. PROBLEM 4.4 For the circuit of Fig. 4.7, determine R B at 25°C, such that I c = 19 ma at the operating point. Solution: Refer to Figs. 4.2 and 4.8a. On the latter, draw a load line corre- sponding to R L = 100. The l c intercept is 5 v/100 = 50 ma. Thus, l c = 19 ma at l B = 0.1 ma (point P,). , From Fig. 4.2, for I B = 0.1 ma, V BE = 0.22 v. The voltage across R B is | npu therefore 5 - 0.22 = 4.78 v. For this voltage, and a base current of 0.1 ma, *-/« Fig. 4.6 Leakage in a common- emitter circuit. Rr 4.78 0.1 x 10 3 = 47,800 Q. PROBLEM 4.5 For the conditions of Prob. 4.4, find I c at 70°C. Solution: From Fig. 4.2, V BE has decreased to 0.12 v. Now I B is 4.88 v Fig. 4.7 Common-emitter amplifier with bias set by adjusting R B . 47,800 Q, = 0.102 ma. In Fig. 4.8a, the operating point has moved from P t (/ c = 19 ma) to P 2 , where l c = 23 ma. This is a considerable change. 70 Transistor Circuit Analysis 60 50 40 t □ E 30 U 20 10 — 70 °c °c i ■— "**"* _ r 0.5 "" c Oma r — .20 ma if *** 1 s \t ^€ ^q 0.1 5 ma 1/^- — " •^ 0.10 ma ys" p, 0.10 ma Y 0.05 ma 1 '" fy 0.05 ma Y .\.. / B =o ' -x r B =o 70°C 0.005 ma / B =o 0. 005 ma I B =0 3 4 F CE ,volt- (a) 1 1.5 2 2.5 3 V CE / vo 1 1 *• (b) Fig. 4.8 (a) Collector characteristics of common-emitter connection with superimposed load line, (b) Expanded region showing load lines. The germanium transistor of the preceding examples has a relatively low Icbo- Considerably higher leakages are common, leading to correspondingly in- creased drift of the operating point. PROBLEM 4.6 Referring to Fig. 4.9, determine R B so that V CE = 1.25 v at 25°C. With this same R B , find V CB at 70°C. Solution / Draw a new load line on Fig. 4.8b. At point P 3 , V CE = 1.25 v and B = 0.005 ma. From Fig. 4.2, V BE = 0.22 v. The drop across R B is 1.5 - 0.22 = 256,000fi. 1.28 v. With a base current of 0.005 ma, s„- L28 Input CE 5x 10- 6 At 70°C, V BE decreases to 0.12 v (approximately - 2.2 mv change per de- gree C). Base current, at 70° C can be calculated: Fig. 4.9 Resi for V CE stor R B is adjusted 1.25 vat 25°C. 1.5-0.12 _ 1.38 256,000 ~ 256 ;10~ 3 =■ 0.0054 ma. This results in an operating point (P 4 ) of V CE = 0.2 v. At. such low voltage, the transistor is almost inoperative, showing the possible critical effects of leakage change with temperature. PROBLEM 4.7 In Prob. 4.6, find the quiescent collector current I c at 25°C and at 70°C. Solution: From Fig. 4.8b, / c = 1 ma at 25°C (P 3 ) and 4.5 ma at 70°C (P„). This is an enormous percentage change and is due to the fact that at low oper- ating levels, l c contains a particularly high leakage component. In the previous examples, a large bias resistor is in series with the base, essentially presenting a current source characteristic. Although base current is relatively independent of temperature, collector current may change significantly, perhaps drastically, when the transistor is operating at low current levels, It is thus worth comparing this simple bias circuit with other bias circuits to see whether or not the operating point stability with temperature can be improved. Bias Circuits and Stability 71 4.4 Constant Base Voltage Biasing Techniques The following problems illustrate the stability of the transistor operating point (collector current) with the base voltage held reason- ably constant with temperature, in contrast to constant base current conditions. PROBLEM 4.8 Referring to Fig. 4.10, determine R B so that V C e Find l B and l c - Calculate I B at 70° C. 2 v at 25° C. 2.5 eq 1000 + Rr 0.22 R. 1000 x R B 1000 + R B = 5 x 10- 6 . Solving, R B = 97 O. At 70°C, V BE is reduced to 0.12 v, and the base current becomes 2.5 x 97 1097, 0.12 1000 x 97 1097 1. 13 ma. 1KQ Solution: Draw a load line on Fig. 4.8b. The operating point for V CE = 2 v is shown at P 5 . At this point, I c = 1.25 ma and I B = 5 fia. To determine R B , use Thevenin's theorem which states (with reference to this problem) that the source resistance R eq driving the transistor base equals the parallel resistance of R B and 1 Kfl, and that the equivalent source potential 7 eq equals the open base circuit voltage at the junction of the two resistors. Figure 4.11 shows how the circuit is simplified for analysis by the application of Thevenin's theorem. The base-to-emitter drop is represented by a battery of 0.22 v at room temperature (see Fig. 4.2). Resistance R B is now calculated: 2N1308 Fig. 4.10 Voltage divider for adjust- ing base bias voltage. o 1000 xR B 1000 + R B — W\r 22v J- V eq =2.5v 1000 +R, (b) Fig. 4.11 Simplification of the cir- cuit of Fig. 4.10 using Thevenin s theorem. This is over two hundred times the base current at room temperature. Obviously we may conclude that constant voltage base-emitter bias is impractical. The previous considerations may be examined from a somewhat simpler view- point. The change in base-emitter voltage is 0. 1 v. The effective resistance is 97 x 1000/1097 = 88.5 fi. The change in current is 0.1/88.5 = 1.13 ma. The result might have been expected. Figure 4.2 shows that small changes in base-emitter voltage can lead to very large current changes when the effective external resistance in the base circuit is small. A large base resistance is nec- essary to achieve relative insensitivity to changes in V BE . PROBLEM 4.9 For the circuit of Fig. 4.12 and the transistor characteristics of Fig. 4.13: (a) Determine I B and /c for V C e = 2 v at 25° C. (b) Using the above value of I B , find Vce and l c at 70° C. (c) Determine a new l' B at 70°C, so that l c is restored to its 25°C value. /? L = 50on 2N1308 Fig. 4.12 Transistor in common- emitter circuit with constant base current drive. 72 Transistor Circuit Analysis Solution: From Fig. 4.13: (a) Point P, at I B = 0.005 ma; l c = 1 ma at 25°C. (b) At 70°C, for l B = 0.005 ma, l c = 4.7 ma and V CE ^ 0.2 v at point P 2 . (c) Since l B = -0.025 ma gives V CE = 2 v, l c = 1 ma at point P t at 70°C. - 0.005 ma 0.025ma 0.005 ma Fig. 4.13 Low level transistor characteristics of common-emitter connection with superim- posed load line. V CE , volt — *- Fig. 4.14 Common-emitter collector characteristics of the 2N1308 transistor at 25 C. The two values of base current for each curve correspond to two different tran- sistors of the same type. Aside from the effects of temperature on transistor stability due to changes in V BE and Iqbo • there is a considerable spread in characteristics among tran- sistors of the same type. Replacing a transistor in a given circuit can lead to a major shift in operating point. Figure 4.14 displays a family of collector characteristics for a 2N1308 ger- manium transistor. The values of l B in parentheses correspond to a low current gain unit. The unbracketed base currents refer to a medium j8 transistor. The three-to-one variation in ft is not uncommon; similar variations occur over mili- tary temperature ranges such as -55 c to -t85°C. Fig. 4.15 Common-emitter circuit with approximately constant base current drive for Prob. 4.10. PROBLEM 4. 10 The transistor used in Fig. 4.15 has the characteristics of Fig. 4. 14. Calculate the following: (a) R B , such that I c = 9 ma with the low /3 transistor. Also determine V C b- (b) With R B fixed, change to the higher /3 unit. Obtain a new operating point, and discuss its usability. (c) Readjust R B to achieve I c = 9 ma for the high (3 transistor. Discuss the usability of the new operating point. Solution: (a) Referring to Fig. 4.14, at l c = 9 ma (point P,), l B = 0.15 ma and V CB = 2.7 v. The operating point is centrally located in the usable area of the characteristics. Since the base-emitter drop is 0.22 v (see Fig. 4.2), the drop across R B is 5 v - 0.22 v = 4.78 v: R* 4.78 0.15 x 10- 3 31,800 0. Bias Circuits and Stability 73 (b) The base current is essentially unchanged with the high (3 transistor, be- cause l B is almost entirely determined by the values of R B and V C c- Assuming l B = 0.15 ma on the same load line as before, the new operating point is located at P 2 , where l c = 19 ma and V C e = 0.2 v. This point is not in the useful oper- ating region of the transistor (the transistor is in saturation). (c) To restore I c = 9 ma using the high /3 transistor, reduce I B to 0.05 ma, thus returning to the original operating point in the center of the linear region. A new R B is now required: 5 - 22 R B = — — ^~ = 95,600 0. 0.05 x 10- 3 If we now use the low /S transistor with the new R B , the operating point moves to near cut-off, point P 3 . The foregoing example illustrates the difficulty in maintaining a stable oper- ating point as transistor parameters vary. For stable operation, it is important to bold l c reasonably constant as V BE , Icbo, "rid fi vary with temperature, aging, and from transistor to transistor. The next section develops a quantitative ap- proach to stability, so that different circuits may readily be compared. 4.5 Stability Factors For the purpose of comparing the relative stabilities of different transistor circuits, a stability factor S is defined as where 5 is a measure of the sensitivity of the collector current I c to changes in leakage current Icbo, and varies with the circuit configuration, such that the lower the value of S, the more stable the circuit. This stability factor can be calculated using convenient formulae applicable to the specific circuit con- figurations. The permissible value of S depends on both the transistor material and the requirements of the application. Generally speaking, low-leakage silicon tran- sistor circuits tolerate a much higher S than relatively high-leakage germanium transistor circuits. For silicon, leakage current may typically be 0.01 /na, while a comparable germanium transistor may have an Icbo ot 5 //a. For an S of 25 and similar bias circuitry, l c changes by 0.25 //a in the silicon circuit, and by 125 fia in the germanium one. The following discussion investigates common bias circuits, establishes de- sign procedures, and evaluates stability. The common-base circuit is not con- sidered because a constant bias current Is leads to a practically constant Ic- This follows since I c = I E . The common-emitter circuits considered will lead to results directly applicable to the common-collector circuits, by setting the collector circuit 'resistance R L equal to zero. Figure 4.16a shows the general form of the most commonly-used bias circuit. A single battery source and current feedback {Re) characterize the circuit. A simplified equivalent circuit (r c = <*, r E = 0) is shown in Fig. 4.16b. If R t in parallel with R 2 constitutes a very low equivalent resistance /? eq in the base circuit, the base voltage V B is essentially constant. The drop across R E is significantly higher than V BE . The emitter current adjusts to satisfy the relationship: 74 Transistor Circuit Analysis Re Re if V B » Vbe, which is a valid approximation. Since lc = Is, the collector cur- rent remains about as constant as the voltage at the base, notwithstanding changes in V BE , 1 C bo', and /3. The behavior of the circuit depends on a low value of the combination of R in parallel with R 2 . The current in these resistors should be substantially more than the base current. However, the resistances must be high compared to the reactance of the blocking capacitor at the minimum a-c input signal frequency. Note that in practical a-c amplifiers, R B is by-passed to avoid gain reduction due to a-c negative feedback. The negative feedback is, of course, the basis for d-c stabilization, on which the capacitor has no effect. PROBLEM 4.11 Derive formulae for / c and the stability factor S in terms of circuit parameters for the common-emitter amplifier of Fig. 4.16a. Input © (a) R„ o /WV , VccR 2 R,+R 2 Fig. 4.16 (a) General bias circuit for common-emitter connection, (b) Simplified equivalent circuit. Solution: The collector current is lc - fiI B + Q8+1)/cbo.' From Fig. 4.16b, (b) | )(P + l)l C BO [4.2] ' cc R, + R 2 -V B E=*-&r l B + R E l E . R x + R 2 Also (4.5) The above equations may be combined, eliminating / B and l E , and solved for l c in terms of the parameters of the circuit. The basic bias equation becomes Re + /_!_} JL* (4.6) Equation (4.6) is a fundamental one, readily modified and adapted to the solu- tion of a variety of problems. Now differentiating (4.6) with respect to l CBO yields Bias Circuits and Stability 75 He Ut + aJ \1 + P) R l + R, Re S = -^ — ^— ^ LJ ^ / • (4.7) R* + Since 1/(1 + |8) is very small, minimum S occurs when R B » R l R a /(R l + /?,). As this condition is approached. 5 approaches unity. It is appropriate at this point to introduce the additional stability factors w = JZc_ and N ^ilc t (4>8) which are measures of the sensitivity of collector current to changes in V BB and /3, respectively. Applying these definitions to (4.6), If.-ife-.. l±l « Zl . (4.9) dv BE „ / 1 \iA. p funu R > R > RB + \Trfi)R-^t RBil+fi)+ R^t Solving for N = ril c /(ifi is tedious but ab&ululth direct. The mathematics can be simplified by differentiating with respect to a: remembering that jS = a/(l - a) and 1/(1 + /3) = 1 - a. Let R,R 2 /(/?i + R 2 ) =/? eq . an equivalent base circuit resistance, and k = /?,/(/?, + R 2 ), an attenuation factor. Substituting in (4.6), (* V cc - V BB ) a + /cso («b + R. q ) / c _ ■ . (4. 10) Differentiate with respect to a: 81 c „. (Ru + *.,)(* ^cc - ^be + /cboR«») „ tiX = JV*= :: . (4.11) The expression (4.11) gives the variation of /c with a, and is a valuable meas- ure of stability in its own right. The symbol for d lc /dOL is N*. Substitute /3 for a in the expression for JV*: 3a = WW 76 Transistor Circuit Analysis The factor N* is often more convenient to use than N since it changes very little An additional, sometimes convenient relationship, is established by compar- ing the expressions for S and N*, i.e., (4.7) and (4.11): r^.Sx^il^i. (4 . 13 ) There are many practical approximations which increase the utility of the above formulae. Consider a room temperature bias condition where leakage is a negligible component of collector current. The approximate collector current is easily obtained fro® (4.6): A. -■*,*£** {kV cc -V BB ) **$$. R«q Comparing with (4»13)» where / e is the quiescent collector current, neglecting leakage current. This formula shows how S may be used as a convenient figure of merit, even for dis- covering sensitivity to changes in /3. Since N* => dl c /dai, the following approximate relations hold: A/ = iV*Aa = S/ ^Aa, — 0- = per unit change in quiescent collector current = S . (4.15) fa . -a.- . •*' The percentage change in collector current equals S times the percentage change A further realistic simplification in the above formulae assumes that R B » R« q /(1 + /3). To minimize the influence of variations in j8, it is important to ad- just the circuit resistors so that this inequality is valid. When this is not the case, it is necessary to return to the more exact formulae as originally derived. The three sensitivity formulae simplify as follows: «■+*- S1+ ^L, (4>16) W= ^£- = - — (for /3 > 10) , (4.17) R s + R»q Rb 1 + /8- N* = S 'tVce- - -Vbb + 1 cm &*£ a R B + 3s_' 1 + /S . (4.18) Bias Circuits and Stability 77 The last approximation assumes that Icbo is negligible under nominal room tem- perature conditions. PROBLEM 4.12 Refer to the circuit of Fig. 4.17. Assume I C bo = 3 /za at room temperature. (a) Calculate the current I c in R L using (4.6), after first determining the ap- proximate operating point from the collector characteristics and the load line. (b) Calculate V C e at the operating point. (c) Calculate S, M, and N*. (d) If Icbo = 3 n& at 25°C, what change occurs in I c due to the change in Icbo at 30° C? (e) If V B e changes by -2.2 mv per degree centigrade, and is 0.22 v at 25° C, find the change in Ic resulting from the change in Vbe if the temperature in- creases from 25° C to 30° C. (f ) If /3 is reduced to 0.9 of its nominal value, what is the corresponding change in l c ? (g) For the conditions of (c/), what is the change in I c between 25° C and 75° C? (Note: For parts (c-f) above, use the approximate expressions for S, M, and N*, which apply especially well to small changes.) Solution: Refer to the basic formula of (4.6) and to Fig. 4.18 which shows the transistor collector characteristics. The formula is repeated here: /3 1 + (* V cc -V BE )+ Icbo (.Re + R eq ) [4.6] Rf.+ i + y s Now determine the numerical values of the parameters: k = R, 1380 = 0.274, R, + R 2 3650 + 1380 R E =50, R E + R eq = 1050 0, R, + R 2 1380 x 3650 5030 = 1000 fl , ' cc 5v, k V CC = 1-370 v. =3650fi R 2 = 1380fi I CBO = 3/Ua at 25 C "=■ Fig. 4.17 Common-emitter ampli fier with voltage divider providing base bias voltage. 60 50 40 E . 30 o 20 10 1 7 o°c 5°C ^ u o.^^r , -"■""" _ — —• """~"c .20tna 1 / ._ — — ---" Q.lSjj va ^f^ 2 1/ /c" If : = = : = =.= ~6~025 0. 00b ma "1)75 05 ma" V CE , volt- -0.025 ma Fig. 4.18 Common -emitter output characteristics at 25 C and 70 C with superimposed load line. 78 Transistor Circuit Analysis (a) On Fig. 4.18, draw a load line. The resistance which determines the slope of the load line is the sum of R L and R E (since l c = I E ). Estimate the approximate voltage at point A of Fig. 4.17 as t/ e 1380 V A = 5 x = 1. 37 v. 3650 + 1380 From Fig. 4.2, P BE = 0.22 v; therefore the voltage at point B is 1.37- 0.22 = 1.15 v. For R E = 50 Q, I E = 1.15/50= 23 ma. This establishes the op- erating point at P, (Fig. 4.18) where I B = 0.125 ma (by interpolation) and Vce = 1.6 v, and 23 £dc = -j^J^ = 184 ( the d "C value h FE , not h te ). Having determined the approximate operating point, using (4.6), , 184 (1.37 - 0.22) + (3 x 10~ 6 )(1050) nn „ 'c = — — ■ = 20.8 ma. 185 „ 1000 50+ 185 This corresponds to point P 2 where I B = 0.12 ma. (b) At P 2 , V CE = 1.85 v. (c) The sensitivity formulae are „ , Req 1000 S^l+— i = i+ -— = 21, [4.16] K E 5U M= ^-=^=-0-02 ma/mv, [4. 17] K E 5U »;* SI „„ 20.8 N = ^ = 21x o^ = a44a > [4 - 18] since «--£-. ^ = 0.9946. 1 + B 185 (d) At 25°C, l CBO = 3fta. From Fig. 4.1, for a germanium transistor, 1 C bo in- creases to 4.4 /za at 30° C. Therefore M C bo = 1.4 jia and A/ c =SM CBO = 21 x 1.4 = 29 fja. (e) Since, from 25°C to 30°C, &V BE = -11 mv, A/ c = _ _ A V BE = - 0.02 x A V BE = + 0.22 ma. K E (f) The effect of a small reduction in /S is easily estimated from N* deter- mined above: 184 Nominal B = 184, a = — = 0.9946, 185 1fi4 Reduced B = 0.9 x 184 = 164, a = — = 0.9939, 165 A/ c = W*Aa = -0.0007x0.44 =-0.31 ma. (g) From Fig. 4.1, l CBO is 22 times greater at 70°C than at 25°C: A/cso = 3(22- l)=63/za, A l c = S A I co = 21 x 63 = 1. 32 ma. Bias Circuits and Stability 79 Note that a more accurate calculation can be obtained by using (4.6). An example comparing the use of the fundamental bias equation with the simple approxima- tion above is provided in Prob. 4.14. PROBLEM 4.13 Figure 4.19a shows a very general configuration of a bias cir- cuit, in which the collector-base feedback is incorporated for improved stability. Show that this circuit can be reduced in special cases to a simpler configuration by the following procedure: (a) Draw the equivalent tee-circuit for Fig. 4. 19a. (b) Derive an expression for Ic including leakage current. (c) Derive expressions for Output N* = dl c da (a) (d) Develop simple approximations to the above expressions. Solution: (a) Figure 4.19b shows the equivalent tee-circuit sketched in accor- dance with the principles previously developed in Chaps. 1 and 3. The collector resistance r D is assumed to be infinite. (b) Write the basic circuit equations for Fig. 4.19b: Vcc = Ie Re » Vbe +I d Ri+ Ud + Ic) Rl, V B e+IeRe = (!d -I B )R i , Ic =/3/ B + (0 4 I) Icbo. Ic = Ie - Ib- Combining (4.2) and (4.20b), solve for I B : / B = f. /8+1. -I CBO- (4.19) (4.20a) Rl S P'b( \ [4.2] (4.20b) (4.21) (/3+i)/ C bo Substitute (4.21) into (4.20a): 1b Re + Vbe - Id Ri = [ - — - + Icbo ) Now substitute (4.21) into (4.20b) and simplify: Ic = Ie - Ib (4.22) (b) Fig. 4.19 (a) Generalized bias cir- cuit incorporating feedback from col- lector to base for increased stabi- lity . (b) Simp li fied equivalent circuit. = /. m +/, CBO- (4.23) Then solve (4.23) for l E , and (4.22) for l D R 2 : Ie = dc - Icbo) Id R 2 = Ib Ub + -^p-) + Vbe - Icbo R t (4.24) Substitute the expression for l D from (4.24) and for l c from (4.23) into (4.19) to obtain an equation in terms oi Ie- Kcc - Vbe = Ie Re + Ie Rl cbo^l -u VBE I R, }■ 80. Transistor Circuit Analysis Simplify by separating out terms in I E : Vcc-VgE + IcBoRi-^— (R l + R L ) = I E [-fcH^-^)] Solve the above expression for l E and substitute in (4.23). This leads to an ex- j pression for I c : : 'M I C = 'CBO J up Vcc - V BE + I cbo R, - ^ CR, + R L ) r e U + X^) +Rl (JL.\_±\ + J± /8' (4.25) Combine terms (4.25) and simplify to obtain a final general expression for l c : This is the required expression for l c . (c) By partial differentiation of (4.26), S = dI c /dl CBO is obtained: 26) 5 - J!° Similarly, M = R..±Rl+ Re (l + ^tBA eicso ~ Re L r, + rA ' jg^; \ R, I 1 + /8 _JL_( l+ R 1 + RA l + P\ R> I (4.27) \ R* J 1-/9 Remembering that a = /3/(l + /S) and 1 + /9 = 1/(1 - a), substitute in (4.26) a[v cc -v BE {i^ R ^- R ^ + i Ci «■ (l^ & ^ 56 ) + Hl + (1 - a) J? x Differentiating with respect to a , (4.28) R 1 + Rl+Re (l + ^^\ (4.29) N* = ^Is. Vcc - V BE fl.^L. \ R2 + I CBO Ri R* 1- R, + R L By comparing (4.27) and (4.30), the following simplification is obtained: (4.30) N* M l+ ~Kf~) +/?r ' + * i(i ~ co + lcBoRi j T . R F (i,x-,±XL.y._ Ri , Riil _ a) (4.31) (d) Examine the expression for l c in (4.29) for the case where l CBO at room temperature is small compared with I c . For this condition, Bias Circuits and Stability 81 a h = - / R. -t RA Vcc- V BE 1+ - l *" A k,_..L (4 . 32) « E (i + -^-^Uk l + k 1 (1-cO Comparing this expression with the value of N* in (4.31) for Icbo = 0> N*=~1^-. (4.33) .-""' '',■' ■'. •:■/."' '■'.•'' '■.':■■','■ " -'"-' y ■.:■'' '-'''■ ^7.V:;jJ);iS^'^ -:- This last expression again demonstrates that S is a good measure of quiescent point stability, even with respect to changes in a. Further approximations may be introduced: Substituting in the expressions for 5, M, and N , S? 1 t 7 ^ r (4.34) R «( 1+51 lf) +R - ,W^ - , (4.35) R E + ^ 1 , R i + ^ N **S1SL. (4.36) Now let us apply the above formulae to a numerical example. PROBLEM 4.14 Solve the circuit of Fig. 4.20, for the following quantities: (a) The operating point (/ c , V C e\ (b) The sensitivity formulae S, M, and N*. (c) Using the results of (b), compute lc at 70° C. Assume /3 increases 1.5 times, Icbo goes from 3/za to 66 /za, and Vbe changes from 0.22 v to 0.12 v. (d) Repeat (c) using the exact bias formula. Use the output characteristics of Fig. 4.21. Solution: (a) As a first approximation, assume I L and I E are equal (a perfectly realistic assumption), and draw a load line on the characteristics curve of Fig. 4.21. Assume further that I B «/ D , I D «I C > and therefore, I c = Is- Then, (V CC -I E R L ) * 2 =/ E K E+ 0.22. i?j + R 2 Substituting numerical values, (5 - 100 I E ) (0.455) = 50 l E + 0.22. Solve for I E : l E = 21.5 ma = lc- This operating point is shown as P^ on Fig. 4.21. Note l B = 0.12 ma and V CE = 1.8 v. 82 Transistor Circuit Analysis d± Vr.r. =5v Input R 1 =3200Q , . r— ^r^V f If O Output J? 2 = 2670n R, = 100Q 2N1308 R E =50Q Fig. 4.20 Bias circuit incorporating collector-base feedback. 1.8 2 3 4 V CE , volt-*- Fig. 4.21 Collector characteristics with superim- posed load line for Prob. 4.14. Hence, /3dc = d-c current gain = 21.5 0.12 179. This preliminary calculation has given us an approximate result, in particular, an approximate value for /3 DC . This value, together with the circuit parameters of Fig. 4.20, permit a more accurate calculation of / c . For convenience, (4.26) is repeated here: A. lr.= 1+/3 Vcc-V B e 1 + Ri+Ri R, +/, CBO R^Rl+Re 1 + Ii+RlS R* I. \ R R l\ +r , + ^l. [4.26] -a / 1 + /3 The numerical values to be substituted are /3 = 179, R, = 3200 fl, R 2 = 2670 Q, 50 0, / CBO 3 x 10' 6 a, V BE = 0.22 v, cc = 100 fl, = 5 v. Now make the substitutions: 179 180 5-0.22 1 + / 1+ 3300\ \ 2670/ + (3x 10- 6 )(3200+ 100+ 112) 112 + 100 + 3200 180 = 19.7 ma. This gives point P 2 on Fig. 4.21. Since this is in the close vicinity of P, in an essentially linear region, the value of /3 may be assumed as unchanged. At P 2 , V CE = 2.02 v. (b) The sensitivity formulae are 1 + *i (-^) = 1 + + Rr 3200 112 + 100 = 16, [4.34] Bias Circuits and Stability 83 M* -1 R E + Rr R, 50 + 100 = - 0.0105. [4.35] 1 + 3300 2670 From (4.33), assuming low I C bo at room temperature, OV c N*=-I Q = 16 19.7 x 10" 3 = 0.316. a ' 0.9944 (c) By definition, A/ c = SA/ CBO + MAV BE + N*Aa. The given data is A/ CBO = +63xl0' 6 , AV BE = -0.lv, Act =0.0019. Note that at /S = 1.5 x 179 = 269, a = 0.9963. For j6 = 179. a = 0.9944. Therefore A a = 0.0019. Now substituting numerical values, A/ c = 16 x 63 x 10- 6 + (-0.0105) (-0.1)+ 0.316 x 0.0019 = 1.01 x 10- 3 + 1.05 x 10- 3 + 0.60 x 10- 3 = 2.66 ma. At room temperature, I c was 19.7 ma. At 70° C, I c = 19.7 + 2.66 = 22.4 ma. . (d) The collector current is ■O Output Input -V, EE M lr. = 269 270 •\ 2670/ 5-0.12 1 + + (66 x 10- 6 )(3200+ 100+ 112) 112+ 100 3200 270 = 22.1 x 10- 3 a = 22. 1 ma. The excellent correlation between approximate and exact results demonstrates . — VS/V — —j| ° Ue the validity of the approximation. *■ » 4.6 Emitter Bias Circuit : The emtper Mm, <dtanjit..af Pg. 422 is -«& esfeeietly useful configuration that may be analyzed by straight-forward circuit methods as illustrated in the following examples. P ROBLEM 4. 15 Referring to Fig. 4.22, determine the following: I c , S, M, and N*. Solution: Calculate / c , using the previously developed fundamental relations: V EE - V BE = I B R B * l B R, - lgR B + fefii - 'c ^i> /c = /8/s-/3fc+0 + l)'cso, Veb ~ Vbe + 'c^ V- i B " v BE \ H/3+i)i CB o -v EE (b) Fig. 4.22 (o) Bias circuit using a separate source of emitter bias voltage, (b) Simplified tee- equivalent circuit. [4.2] -*.*• Rk + &i \ 84 Transistor Circuit Analysis Combining equations and solving, /3/ c fi, + /3 V EE - /3 V BB + QS + l)/ CBO (R £ + «,) (/3 + l)/ c = *E + *l The last expression may be simplified and solved for I c : 'c- 7; — r (^ee ~ Vbe) + Icbo (Re + Ri) p + 1 R* + *» (4.37) 1 + /3 Equation (4.37) for the emitter bias configuration is the same as (4.6) with R l R 2 /(R l + R 2 ) = R and V cc [R 2 /(R t + «,)] = P„. Therefore all formulae dealing with the circuit of Fig. 4.16 can be applied to Fig. 4.22 as shown below: s = Ml.1 R E +Ri dl CO .**+...: *» M di c 1 + /9 £+1 B 0! *?■ BE Re + ,*>. i + 8 K* (1 + /3) + R, Vvtr — Var _ 'CBO Ri R B + 1 + X B + Ri (4.38) (4.39) (4.40) As before, if l CBO at the quiescent point is essentially negligible , a r, =100 n« K — o v cc = PROBLEM 4.16 For the circuit of Fig. 4.23, calculate the values of I c and S. Solution: Examine the collector characteristics of Fig. 4.18. Observe that v ee + Vcc = 5 v is applied to R E and R L in essentially a series circuit. As 0u, P ut previously described (Prob. 4.12), we get an estimated I c = 23 ma, and a pre- 2N1308 liminary value of /3 of 184. More accurately, using (4.37), J~ (Vbb - V BE )+l CBO (R E+ R 1 ) ' i|l (1.37 - 0.22) + 3 x 10" 6 (1050) ^ + P 185 Rp=50fl EE =-1.37v 'o = 'c R* + 1 + )S 50 + 1000 185 = 20.8 ma, Fig. 4.23 Emitter bias circuit for From (4.38), Prob. 4.16. S = R E +Rl R* + 1 + 1050 55.4 = 19. The emitter bias circuit is particularly advantageous when the base is driven by an input transformer. Bias is, of course, adjusted by choosing V BB and R El while the base is essentially at ground potential with respect to d-c. With R B = 0, the stability factor is unity, a theoretically optimum condition: Bias Circuits and Stability 85 This leads to a desirable low value of N*. There is no particular improvement inltf. 4.7 Bias Compensation When a particular circuit configuration is selected, it is possible to improve stability by using the nonlinear and temperature-sensitive characteristics of auxiliary diodes and transistors. Some of these compensation methods are now illustrated in the following examples. PROBLEM 4.17 For a common-emitter circuit with an n-p-n transistor, show how to use a diode to compensate for the effects of temperature change on V BE . Solution: Figure 4.24 shows a circuit using diode compensation. The current / is adjusted so that V D (the forward diode drop) equals V BE (thus cancelling one another). The values of R l and R 2 are adjusted for the required bias. The cancellation occurs over a wide temperature range because the diode and transistor junctions follow identical laws. The circuit becomes equivalent to that of Fig. 4.16 but with V BE = over the whole temperature range. PROBLEM 4.18 The circuit of Fig. 4.25 shows a method of compensating for the effects of temperature on l CBO . Analyze the circuit's performance. Solution: Leakage current l CBO flows in transistors Q 1 and Q 2 . If the tran- Fig. sistors are matched, the leakage currents should be equal over the temperature range. The lc B o drawn from the base circuit of Q t by Q 3 results in a reduction of /S/ CBO in the collector current of Q,. As the component of / Ci corresponding to leakage current is /cbo(1 + fi)> the effective collector leakage is reduced from (£ + 1)/ C bo t0 IcBO* thereby providing the required compensation. Both compensation techniques described above can be used simultaneously, but are not required very often. Circuits are generally designed for good bias stability with passive elements, and relatively complex compensation methods are thus avoided. It is both difficult to match transistors and to hold junctions at equal temperatures. Compensation with diodes and transistors is used only in special cases. 4.8 Self-Heating Transistor parameters must correspond to actual junc- tion temperatures for accurate analysis of transistor performance. The junction temperature is the sum of ambient temperature T a plus a temperature rise re- sulting from power dissipation at the junction. For small-signal amplifiers, junction dissipation is almost entirely due to bias currents. Since I c = l B , the total power dissipation at the two junctions of a transistor is essentially V CB l c . (Situations in which this is not the case will be discussed elsewhere.) The junction temperature is 4.24 Circuit with diode bias compensation. CC -VL Fig. 4.25 Method of compensation for the effect of temperature on IqbO- T, = T. :+(0,-.)'c»W (4.41) where 6, a is the thermal resistance from the junctions to the ambient environ- ment expressed in °C/watt. Thermal resistance fy_« is normally given on tran- sistor data sheets for specific recommended mountings (heat sinks) in free air. 86 Transistor Circuit Analysis The problem of including temperature rise in transistor calculations results from the fact that l c must be evaluated at the final junction temperature, which is unknown at the start of calculations. Iterative procedures are suggested, but are rarely warranted. PROBLEM 4.19 For the circuit of Fig. 4.26 at an ambient temperature of 70°C, calculate I c . Include the effect of junction temperature rise due to power dis- sipation. Assume y _ a = 200°C/w, and that /3 is independent of temperature. Solution: The circuit is identical to that of Fig. 4.17 in Prob. 4.12. We there- fore use the results of that problem as an initial approximation: l c at 25°C = 20.8 ma, A/ Ci (due to change in I CBO ) = 1.32 ma (for 70°C), Af = —0.02 ma/mv. Since we are evaluating operation at 70°C, Ar = 70°C-25°C = 45°C, &Vbe =-45x2.2 = -99mv, M Cl =Mt± V BE = /-0.02 —\ (-99 mv) = 2 ma. The increased collector current at 70°C can be estimated as / C(70°c) =/ C(25°c) + A/ C! + A/ c 2 = 20.8 + 1.3 + 2 2T24 ma. R, = 36500 Input O If — !► f? 2 =1380fl OF cc =5v R L = 100Q, i » 1 C O Output 2N1308 R E =son icBO = 3/*a at 25°C Fig. 4.26 Circuit for self-heating calcu lotion. 60 50 40 30 20 10 1 70°C ■^25 tna — * •""" ^r: _ — •■ 0.20 m a tf «'" f / 3.15 ma 4/*'"' SC- a y bma . _ —j ,_ |^^— - — 3 4 V CE , volt- Fig. 4.27 Collector characteristics of the 2N1308 transistor with superimposed load line. From the load line (Fig. 4.27), the operating point P t is at I c = 24 ma, V CE = 1.4 v. Therefore, T, = T a + e,_ e <J c V CE ) = 70 + 200(0.024 x 1.4) = 76.7°C. Since it is assumed that /3 does not vary with temperature, a value which does not include an I CBO component is required. This is obtained, for all practical Bias Circuits and Stability 87 purposes, from the room temperature characteristics in which l CB o is a very small percentage of l c . From point P 2 on the characteristic curves, I c = 17.5 ma, I B = 0.1 ma, = ^=175. Is Using the high temperature value of Iqboi ^cbo^ 76.7°C) = 35 x/ CB0 at25°C (see Fig. 4.1). Now substituting the numerical values in (4.6), IZi fs x — - 0.12| + (3 x 35 x 10- 6 )(50 + 1000) /r _ 176 ^ 5030 / = 24.3 ma. 50 + 5.7 This value is close enough to the previously calculated 24 ma so as not to re- quire an improved approximation. PROBLEM 4.20 In the circuit of Fig. 4.28, calculate I c at an ambient tem- perature of 45°C. Assume that Iqbo = 3 //a at 25°C is the specified maximum leakage, and fy_ a = 100°C/w. Use the characteristic curves of Figs. 4.2-3. The diode is adjusted for the same voltage drop as V BE at its operating point. Solution: As a first approximation, the voltage at A is V A = 5 (— ) +V D = 0.108 + V D . \930j Since V D = V BE , the voltage across R E is V E = V A - V BE = 0.108 + V D - V BE = 0.108, , V E 0.108 R* = 54 ma = I c . -OV cc =5v C Primary d-c resistor: 111 2N1308 :2fl Fig. 4.28 Amplifier circuit with bias compensation for temperature variation of V BE . 60 50 40 30 20 10 1 70°C _,. p.. oT2Sm a ^ -* """ ^ *■» m ___ — ~~ p > . 3.20wa 1 ^ f / »^— '" ■^r - — 0.15 ms f y / /> — rX r s 0.10m a / 0.05ti na r>- — .r-_=-— =="== ==-- 12 3 4 5 6 V CE , volt—*- Fig. 4.29 Collector characteristics of the 2N1308 transistor with superimposed load line. Transistor Circuit Analysis Now consider junction temperature effects. Power dissipation = 0.054 V CE . On Fig. 4.29, draw a load line for R L + R E = 3(2. At 54 ma or point P u V CB = 4.84 v (V CE = V cc -R L l c - R E l E ). This gives Power dissipation = P, = 0.054 x 4.84 = 0.262 w, Junction temperature = 7) = T B + fy_ a (/ c V CE ) = 45 + (0.262 x 100) = 71°C. Use this estimated operating temperature for a more accurate calculation of I c by means of (4.6). Since V BE and the diode forward voltage drops are always equal, they may both be ignored with no sacrifice of accuracy. In the region of interest, /3 can be obtained from Fig. 4.29. It is the d-c |6, excluding any l CBO component, which may therefore be taken from the 25°C curves (as before). At point P 2 , I c = 50.5 ma, l B = 0.25 ma, = 50.5/0.25 = 202. From Fig. 4.1, l CBO = 1 CB0 (25°C) x 22 = 66 ^a. Substituting in (4.6), 202 / 20 \ ^5x^+66 xl0-(2 + 19.6) 'c = — — = 52 ma, 2 + 0.097 T, = 45 + (100 x 4.84 x 52 x 10~ 3 ) = 70.2°C. This is close enough to the first approximation so as not to warrant an additional computation. PROBLEM 4.21 If the diode D of Fig. 4.28 is omitted and R 2 is increased to 65fi, calculate I c at an ambient temperature T B = 45°C. Assume l CBO is negli- gible at room temperature. Also assume, as before, a thermal resistance of 100°C/w junction dissipation. Solution: Calculate V A : V A = — x 5 = 0.33 v 975 (approximately, neglecting base current drawn from the voltage divider). From Fig. 4.2, V BE = 0.22; l CB0 = 3/xa (assumed negligible). Hence, V B = V A - V BE = 0.33 - 0.22 = 0.11 v, = 55 ma. V E 0.11 R E *■ For a more accurate value of / c , using /3 = 202, from Prob. 4.20, and sub- stituting in (4.6), || (0.33- 0.22) + C " 2T0l 48m "- Using this value of collector current, estimate the junction temperature 7}. Let Pj = 7 C V CE = approximate junction dissipation. Then, 48 P i = 7 c V CE = 1555" [5 ~ 3 (°- 048 )] = 0.233 w. Tj = T a + d^Pj = 45 + 100(0.233) = 68.3°C. At this high junction temperature, leakage current increases markedly, and must be included in a more accurate temperature estimate. From Fig. 4.1, at 68.3°C, l CBO = 63 x 10" 6 a. From Fig. 4.2, V BE = 0.12 v. Substituting in (4.6), Bias Circuits and Stability 89 la =■ 202 203 (0.33 - 0.12) + 63 x 10" 6 (2 + 60.5) = 93 Therefore, and 60.5 ~203~ VcK = 5-^ = 4.72 1000 7} = 45 + 100 (^-1 (4.72) = 89°C. 93 \ ,1000/ The temperature at the junction has increased sufficiently above the previous estimates to warrent a third approximation. Assume now a junction temperature of 89°C. At this temperature, V BE = 0.12 - (19 x 2.2 x 10" 3 ) = 0.078 v. (Note that V BE = 0.12 at 70°C, and changes -2.2 mv/°C.) From Fig. 4.1, l CBO increases one hundred fold over the value at 25°C. Again using (4.6), it is found that / c = 117 ma and "CE 5 -0.117(3) = 4.65 v. Hence, T, = 45 + (100 x 0.117 x 4.65) = 99.5°C. Note that once again the previous calculation was inaccurate, and a try at a closer approximation is indicated. At 99.5°C, V BE = 0.12 - (29.5 x 2.2 x 10~ 3 ) = 0.055 v; l CBO = 180 times the room temperature value, or 0.540 ma, and I c = 130 ma. This successive approximation process could be continued until the series of values of / c converges, if it ever does. Because a germanium transistor can- not operate above a junction temperature of about 100°C, the calculations be- come academic; the transistor will eventually be destroyed. This process, re- sulting from the reduction in V BE with increasing temperature, can be avoided by bias compensation. 4.9 Thermal Runaway There is another type of thermal destruction generally called thermal runaway. This is caused by a regenerative increase in I C bo. Increasing temperature leads to increasing l CBO with its associated increase in dissipation, in turn leading to further heating, and a continuation of the process. Leakage current / CBO increases until it is limited by the external circuit, or until the transistor is destroyed. This section presents an approximate analysis of thermal runaway. There are three basic equations required for the analysis of thermal runaway in the circuit of Fig. 4.30: (4.42) (4.43) Vce= V CC -I C (R E +R L ). (4.44) These expressions may be combined and differentiated to arrive at the thermal runaway condition in which the increase in I CBO due to an increase in tempera- r i «r. + i _./» J>: 'C r CE> A/W— O v cc Fig. 4.30 Simplified circuit for analysis of thermal runaway. 90 Transistor Circuit Analysis ture leads in turn to a further increase in Icbo an ^ corresponding futher in- crease in temperature, etc., until runaway occurs. PROBLEM 4.22 Using the above equations and the circuit of Fig. 4.30 as a starting point, establish the condition for thermal runaway. Solution: Combining (4.42) and (4.43), T,-T a + 0,_ a V CB I c . [4.41] Differentiating (4.41) with respect to T t , dT L= \dlc_ V + dV " I VcE + ~dfT Ic e Ha = 1. (4.45) Also, From (4.44), dI c = d/ c ^ dlcBO dT) dl CBO dT, ^CB fD , D \ dl c ., dlcBO = -(R E + R L ) -fi£- x dT, dl C BO $ T i Substituting in (4.45), 1 dl c dI rBn r„ . ,„ „o ....._ ._. dlr- dl. e i-a dlcBO dT) dl CBO dT) Simplifying, and recalling that dl c /dl CB0 = S, 1 = s d_l^± [Vcc _ 2 J iR + R) i (4-46) This expression represents an equilibrium condition, wherein the increased l CBO and dissipation at high temperature are compared to the associated temperature rise for the increased l C BO' Thermal runaway occurs when either S or 6> a is increased, upsetting the equality of (4.46). Thus, the condition for stability is 1 > dJzzo_ [Vcc _ 2Ic (Re + RDl (4i47) e Ha .s dT, PROBLEM 4.23 Remembering that l CBO for germanium approximately doubles for every 10°C rise in temperature, modify (4.47) by substituting an appropriate expression for dI CBO /dT). Solution: Let Icbo = leakage current at a reference operating point and tem- perature. Let temperature increase from Tjq to 7). Then, T i- T iQ ICBO = 'CBOQ x 2 , In I CB0 = In l CBOQ + [ Tl ~ Tl9 ) In 2. Differentiating, dlcBO _ 1" 2 ,_ Icbo ~ 10 " Bias Circuits and Stability 91 or dlcB0 = 0.0695 l CBO ~ 0.07 l CBO . (4-48) dT, Substitute (4.48) into (4.47): > 0.07 I CBO [V cc - 2/ c (Re + Rz.)] • (4-49) This expression applies to germanium transistors. Silicon transistors almost never exhibit thermal runaway due to their low leakage. The values of l c and l CB0 must correspond to the highest design value of junction temperature. Be- cause of the approximate nature of the analysis, large safety factors are sug- gested in design to avoid thermal runaway. PROBLEM 4.24 (a) For the circuit of Fig. 4.28, calculate the stability factor S at 70°C. (b) Determine whether thermal runaway occurs at 70°C. Use the col- lector characteristics of Fig. 4.3. For this particular transistor, 0,_ a is 100°C/w. Assume that l CBO = 200 ^ia at 70°C, the poorest case condition. At 70°C, V BE = 0.12 v, but is compensated by diode D. Thus, variation of V BE with temperature does not aggravate the thermal stability problem. Solution: (a) The stability factor is given by the expression R E + s= R t + R 2 [ 4>16 ] K E R 1+ R 2 1 + j8 Calculations are to be carried out at 70°C. First, calculate the approximate emitter current: V A = 5 x 2 ° = 0.108 v 20 + 910 (since V BE = V D ). Therefore, the drop across the 2Q emitter resistor equals the drop across the R 2 = 20 Q resistor: r _ YE. . 0O°l = 0.054 a. B R E 2 Refer to Fig. 4.31. Draw the load line, locate point P, and determine /3 in this region. This has been done in Prob. 4.20, in which j8 = 202, so that we will use this value to determine S: 20x910 + 930 mc S = = 10.3. 20 x 910 1 930 203 From (4.49), J_ > 0.0695 (200 x 10- 6 ) [5 - (0.108) (3)] = 65 x 10"* , se,_ a 970 xlO" 6 > 65x10"*. S6,_ a 10.3x100 92 Transistor Circuit Analysis sn -- --70^ 25°C __„. ---> oSs *a 40 f --- "" — * .--■ 0.20 »a in O.lSma Y'' a ?n ¥' m 0.05 na Is 1740Q ©V cc =20v OV Q 2N929 260 Fig. 4.32 Evaluating V by an ap- proximate analysis. 12 3 4 5 6 7 VcE< volt— *• Fig. 4.31 Collector characteristics of the 2N1308 transistor with superimposed load line. This represents a stable system and thermal runaway will not occur. Again, due to the approximate knowledge of l CBO , d t _ a , and therefore T,, it is important to calculate for the worst possible conditions using ample safety factors. Experimentally, 7} is determined from V BB . Either the manufacturers' data on V BB are used, or, if greater accuracy is required, V CB vs. temperature for a fixed value of I E may be determined experimentally. 4.10 Approximation Techniques The analytical techniques developed above provide ac- curate circuit solutions. However, it is important to start out with approximate ! operating points, so that preliminary calculations can be accomplished quickly. The examples below illustrate methods for rapidly approximating the operating point of transistor circuits. These methods are also of value in setting up bias conditions in the initial stages of circuit design. The following simplifications apply: 2N929 Vbb " 1 0. 'c 2 for germanium 6 v for silicon ' a = 0.996 — Fig. 4.33 Evaluating Iq by an ap- proximate analysis. r e = oa > fe =* Of and r b = in the equivalent tee-circuit. Except where specifically called out in the problem, I CBO is neglected. PROBLEM 4.25 In the circuit of Fig. 4.32, find V . Solution: The potential at point A is (260 x 20)/(1740 + 260) * 2.6 v. For a sili- con transistor, V BE = 0.6 v, so that the drop across the lKfi resistor is 2 v Since 1 E 2T2ma £7 C , V = V cc - I c R L =20 - 0.002(5000) = 10 v. PROBLEM 4.26 Referring to Fig. 4.33, estimate I c . Solution: Assume I rT , n = 0: Bias Circuits and Stability 93 V EE - 0.6 = (10,000 + 2,000) I E - 10,000 I c , I c =CLl E = 0.996 l E , 3.4 = 12,000 I E - 9960 I E = 2040 l B , 3.4 /* = 1.67 ma, 2040 l c = 1.67 x 0.996= 1.66 ma. PROBLEM 4.27 (a) For the circuit of Fig. 4.34a, when switch Sw is open, find l c and V . (b) With Sw closed, find I c and V . Solution: (a) With the switch Sw open, lc = (1 + hFE )Icbo = ( 101 ) x 10 x 10 " 6 = X ma ' V = 20 v - l c (1000) = 19 v. (b) With switch Sw closed, refer to Fig. 4.34b. At first, disregard I cbo . This figure shows a simplified circuit for calculation. The equations for this circuit are 2.6 - 0.6 = 10,300 l B + 300 I c , I c = h FE I B . Substituting for l c , 2 v = 10,300 I B + 30,000 l B = 40,300 l B , I B = 50 pa, I c = 5 ma. Now include an additional l c component due to l CBO . Recall that Mc c lKfi P-VVAr-OVcc =20v ov = 100 / CBO = 10x10" 2.6 v (a) 0.6 v M CBO lKfi A/s/\, o 20v 2.6v Approximately A/ c = S A/ C bo- The stability factor must be calculated: R F + R B 10,300 R„+ Rb 300 + 10,000 ■S25. (b) Fig. 4.34 Transistor bias circuit for Prob. 4.27. Thus, Therefore, and + 1 """ ' ioo I c (due to lego) & 25 x (10 x 10" 6 ) = 0.25 ma. l c (total) = 5 ma + 0.25 ma = 5.25 ma F„ = 20 -(5.25) = 14.75v. 90Kft PROBLEM 4.28 For the circuit of Fig. 4.35, l CBO = 10 /xa, h FE = 100. Estimate /c ^ 7 . Solution: Initially, neglect the leakage component. Replace the resistance di- vider bias circuit by its Thevenin's equivalent source: 45 kO 45 V.„ = 30 eq 135 10 V, R E =5KQ R eq = 90 K 1 1 45 KO = 30 KQ = R B . This bias circuit feeds the input impedance R ln of the transistor. Using the approximate formula (see Table 5.1), K to is easily calculated: Fig. 4.35 Transistor bias circuit for Prob. 4.28. 94 Transistor Circuit Analysis 0V = 200 (a) R e<i =R B ~ R t +R 2 Km = Re (1 + /8dc ) = 5000 (1 + 100) £ 500,000 fi. At the base, the voltage is Since V BE =0.6 v, and in 500,000 „ „ r 10 v x = 9.45 v. 530,000 V E = 9.45 - 0.6 = 8.85 v / 885 177 ~, ' e = = 1 . 77 ma = / r . 5000 C Now calculate the current component due to l CB0 by setting R^ = R B and using the stability factor S derived in (4.38): S = R B +R, 30 KQ + 5 KQ 1 + /3 R ° +Rw ^™ + sm i\, ~ DC 101 The current component due to leakage is 7 x 10 = 70 fia. Thus, the total col, lector current is O V cc = and 20v 1.77 + 0.07 = 1.84 ma F o = 30-(1.84)(5) = 20.8v. PROBLEM 4.29 Referring to Fig. 4.36, determine the values of the resistors such that 7 C = 5 ma, V CE = 8 v, V E = 6 v, and S = 10. Solution: Use the previously discussed approximation techniques: Ie-Ic =5 ma.., V E = 6 v, 6 0.005 R in = 240 Kfl 1,200 Q, V BE = 0.6v -0-±l|— Using (4.38) for S = 10, S = Rr +r, R* + -Vcc Rl 1+A, 1200 + R B 1200 + R ^- 201 = 10. R,+J? 2 =R E (\+h FE )^ Solving, R B = 11,400 Q. This must equal the equivalent source resistance of R l and R 2 in parallel: (b) Fig. 4.36 Analysis of the transistor bias circuit of Prob. 4.29. R eq = R B = -^- = 11,400. R, + R 2 Refer to the Thevenin equivalent circuit, Fig. 4.36b. Note that V A = 6.6 v to account for the transistor base-emitter drop. Solving for V e , 6 240,000 11,400 + 6.6^0.9 v. Bias Circuits and Stability 95 Equate this to the Thevenin expression for 7 eq : R, x 20 V = 2 "eq = 6.9 v. R, + R 2 Combine with the previously developed expression for R eq : 6.9 R, = 20 R l R 2 R, + R 2 20 (11,400) =228,800; hence, 228^800 = 33j000 fl OV cc =-20v /3 DC = ioo Substituting and solving for R 2 , 6.9 R 2 ^ 17,40011, (a) 0.2 v For V CE = 8 v, V = 6 + 8 = 14 v. The drop across R L must equal 6 v at 5 ma, for R L = 1200 0. PROBLEM 4.30 For the emitter-follower of Fig. 4.37a, what value resistors are required for a quiescent operating (Q) point of I c =1 ma, V CE = 10 v, and S = 5? Solution: If V CE = 10 v, then V E = R B I E = 10 v. Since l E = 1 ma, R E = 10,000 H. Determine / B from (4.2): /„ = — t/c - Q3 D c + 1) 'cuol =7^-7^ (0005) - ° 005 ma - j8 D < 100 100 (b) Fig. 4.37 Calculation of quiescent operating point by estimation. Refer to Fig. 4.37b. It is necessary to determine R B = R eq , which can be established from the approximate expression forS: S = R P +R r R* + imQ £ dc + 1 Substitute R E = 10,000, DC = 100, S = 5, and solve for R B :R B = 42,000 0. Referring again to Fig. 4.37b, V ea = 10.2 + R B l B = 10.2 + (42,000) (5 x 10~ 6 ) = 10.4 v. This leads to the following relationships: -^- = ^1 = 0.52, -*^ R, + R 2 20 R, + R : Solving, R,, = 81,000 fl and R 2 = 87,500 0. lMfi = R B = 42,000 Q. OV rf: =20v OV Pdc= w0 / C BO = 5xl0 ~ PROBLEM 4.31 In the circuit of Fig. 4.38a, estimate the power dissipation in the transistor. Solution: Referring to Fig. 4.38b, the equivalent bias network, it is seen that 9.4 v 500 Kfl 10v /* = 18.8 /xa, 0.5 x 10 6 n Ic = finch + (/3 D c + 1) 1 cbo, l c = 1.88 + 0.50= 2.38 ma, y o = v cc ~ r l lc = 20 - (5000) (0.00238) = 8.1 v, 7 CE =8.1v. [4.2] (b) Fig. 4.38 Estimating power dissipation. 96 Transistor Circuit Analysis Therefore , Power dissipation = l c V CE = 0.00238 x 8.1 = 19.2 mw. a = 0.99 10Kfl-££ -AW PROBLEM 4.32 In Fig. 4.39, estimate I c . Solution: Since 10-0.6 10,000 = 0.94 ma, the collector current is l c = 0.99 I E + l CBO = 0.93 + 0.1 = 1.03 ma. 4.1 1 Supplementary Problems Fig. 4.39 Estimating collector cur- rent in a common-base circuit. PROBLEM 4.33 If R B = 100 in Fig. 4.10, does the current increase with temperature? PROBLEM 4.34 Determine whether the stability factor should be large or small for best stability. PROBLEM 4.35 Define the factors S, M, and N. PROBLEM 4.36 In the circuit of Fig. 4.17, how can a reduction of gain at audio frequencies due to the resistor R E be avoided? PROBLEM 4.37 (a) Determine the effect on S of a large resistance in the base lead in the circuit of Fig. 4.17. (b) What is the effect of a large resistance on stability? PROBLEM 4.38 Does temperature effect V BE ? PROBLEM 4.39 Determine if the static characteristics of a transistor vary with temperature. PROBLEM 4.40 If R, = 1800 11, R 2 = 680 O, R L =56Q, and R E = 68 fl in Fig. 4.17, find (a) S, (b) I c at 25°C and 70°C when I CBO = 1 ^a at 25°C. PROBLEM 4.41 Describe thermal run-away and its mechanism. PROBLEM 4.42 In the circuit of Fig. 4.38a, estimate the power dissipation in the transistor when a 5-Kfl resistor is replaced with a 10-Kfi resistor. SINGLE-STAGE AMPLIFIERS 5 CHAPTER 5.1 Introduction The performance of the single-stage audio amplifier will be calculated for small-signal conditions. It is assumed that bias voltages are set to establish a suitable operating point, and that the parameters corresponding to the operating point are known or available. The equivalent circuit techniques developed in Chap. 3 are directly applicable. Complete calculation procedures are presented for each of the basic transistor configurations, and a tabular sum- mary of formulae is provided in Table 5.1 on p. 113. The problem under investigation is defined in Fig. 5.1. A single-stage ampli- fier is energized by a small-signal a-c generator v t with an internal resistance R g . The load resistance is R L . The items to be calculated are input and output impedances, and voltage, current, and power gains, the calculation of which leads to methods of achieving desired circuit performance features. 5.2 Common-Emitter Circuit Figure 5.2 illustrates the common-emitter circuit adapted to the analysis of small a-c signals. Bias resistors are not included, since their effect on a-c performance is taken into account by combining them with the equiva- lent generator and load resistors, using conventional network theory. Also block- ing capacitors are ignored, since it is assumed that their impedances are negligible at the a-c frequencies in question. Their effect on amplifier bandwidth will be considered later. The a-c amplifier is most conveniently analyzed using the hybrid equivalent circuit discussed in Chap. 3. Note that for the common-emitter connection, i,- -. i b , PROBLEM 5.1 The A-parameter equivalent circuit of the common-emitter ampli- fier of Fig. 5.2 is shown in Fig. 5.3. Derive the following: Z„ Input impedance,* Z OI Output impedance,* A lt Current gain , A v , Voltage gain , A Power gain. Solution: Set up the basic equations for the circuit of Fig. 5.3: * Since reactances are neglected at low- and mid-frequencies, we can also speak of input and output resistances. Fig. 5.1 Single-stage amplifier con- nected to input generator and output load resistance. Vcc v,f> O v o Fig. 5.2 Simplified common-emitter circuit for small-signal a-c analysis. 97 98 Transistor Circuit Analysis v e = i t R e + i, h ie *■ v„ h re , h u i t = - v„ h c Solve (5.2) for output voltage, v ■ R, Substitute this expression in (5.1) and simplify: v* = ', («,^ l .- f t !"*; l Since from (5.4), v t = i, R g + v„ \ i + fto. Rl) This expression gives Z, directly: (b) Fig. 5.3 The h-parameter equivalent circuit for the common-emitter connection. From Ohm's law, (5.1) (5.2) (5.3) (5.4) (5.5) (5.6) The current gain is easily determined by substituting in (5.3) and simplifying: (a) r\y)v (b) Fig. 5.4 Determining the small- signal output impedance of the common-emitter transistor circuit. h,e l + A oe RL (5.7) Voltage gain is determined by combining previous results with the following fundamental relationship: -ht. 1 ± R Lh oe R L h te ft, 1 ""i^.,Rl; -h{ 9 Rl h te (l + h . Rl)-Rl h fe h re ' (5.8) It remains only to calculate output impedance, Z . This is done by short-cir- cuiting v g , and applying a voltage v across the output terminals. The current flow in the output circuit equals the applied voltage times the output resistance. Refer to Fig. 5.4. The current i is determined from the equation while i, is determined from *o = ft/« *i + v a h oe , v oh r »--i, (fc, e -, R e ). Single-Stage Amplifiers 99 Combine these two equations, and solve for Z„ - v /i : Z = ^ = 1 . (5.9) "oe h le + &i Power gain A p is simply the product of voltage gain A v and current gain A t . PROBLEM 5.2 Determine Z it Z ot A Y , and A t for a common-emitter amplifier using the 2N929 transistor, biased so that / c = 4 ma and V C b = 12 v. Assume R L = 5000 fi and R g = 500 Q. Use the hybrid parameters for the above circuit derived in Chap. 3: fi ie = 2200 0, /. re = 2xl0-\ ft fe = 290, h oe = 30 x lO -6 mhos. Solution: Find Z i by substituting in (5.6): Rl fife fire Z i = fife - 1+fioeKz, = 2200 - (5000) (290) (2 x 10~ 4 ) 1 + (30 x 10- 6 ) (5000) = 1950 O. This impedance is not substantially different from h ie . Similarly, calculate Z by substituting in (5.9): U hfe fire ^ie + R g 1 30 x 10- 6 (290) (2 x 2200 + io- 4 ) 500 = 118,000 fi. Voltage gain is given by 4 -fife ^L " v h le a+h oe R L )-RL fife fire Substitute the given values: A v 290 x 5000 2200 (1 + 30 x 10- 6 x 5000) - (5000) (290) 2 x 10- 4 -290 x 5000 2200(1.15) -290 -647. Current gain from (5.7) is A h <° 290 252. [5.8] l + h oe R L 1 + 30 x 10- 6 x 5000 PROBLEM 5.3 Describe qualitatively the effect a varying R L has on the input impedance of a common-emitter circuit. 100 Transistor Circuit Analysis Solution: Refer to (5.6): 7 t, Rl hfe Ke 1 = " " iTA^Rl • t5.6] For R L very small (output short-circuited), Z,^h ie (5.10) For R L very large, Zi=hie-~^. (5.11) The input impedance decreases between the limits of (5.10) and (5.11) as R L in- creases. PROBLEM 5.4 For the circuit conditions of Prob. 5.2, determine Z, as R L varies from to oo. Solution: Substitute numerical values in (5.6) to determine Z; vs. R L : 580 x 10- 4 R L Z, = 2200 1 + (30 x 10- 6 ) R L Values have already been found in Probs. 5.2-3 for R L = and 5000 Q. Now R L 100,000 and 1 Mfi. Then for R L = 100,000, nn 580 x 10- x 10- 1 + 3 = 2200 - 1450 = 750 fl. For R L = 1,000,000, Z t = 2200 580 x 10-" x 10 6 1 + 30 = 2200 - 1870 = 330 12. At R L = oo, substitute in (5.11): 7 99m 290 x 2 x 10-" ^ £r n Z t = 2200 _ — — __ = 265 fi . 30 x 10 -6 The values of input impedance vs. R L determined thus far are tabulated below: r l ,q Zi.Q 2200 5000 1950 100,000 750 1,000,000 330 265 Figure 5.5 shows Z t vs. R L plotted on semi- logarithmic coordinates, since this method provides the most convenient presentation over the complete wide range. For high £ transistors, R L is rarely above 10,000 Q. Thus, for all practical pur- poses, Z t will not vary significantly with R L . PROBLEM 5.5 For the circuit conditions of Prob. 5.2, determine Z for R 6 = 0, 500 fi, 10,000 fi, and «.. Single-Stage Amplifiers 101 2500 2000 1500 1000 500 1 Asymptote ^J*at 2200 fl ^v. 950 fl ,750 fl Asym ptote^i ,330(1 at 261 )fl 1 300,000 250,000 200,000 . 150,000 N — 100,000 50,000 10 2 10 3 10 4 10 5 10° R L . ft — 10 7 10° 10' — *^- 1 1 Asymptote "~ atZ = 275, 000 fl \ 118,000 fl ,39,600 n u c Jt z = 33,300 fl 1 Fig. 5.5 Variation of input impedance with load re- sistance for a common -emitter circuit. 10° 10 1 10 2 10 3 10 4 R g ,fl-^ 10 5 10° 10' Fig. 5.6 Variation of output impedance with generator resistance for a common-emitter circuit. Solution: Use (5.9) which was previously derived for output impedance: 1 1 h oe - hfe hre 30xl0- 6 - 290 x 2 x 10~" 2200 + R„ It is easy to determine Z for the extreme values of R g = and oo. At R g = 0, 1 30 x 10- 6 - — x 10- 22 = 275,000 fl. At R g = oo, Z„ = 30 x 10- 6 = 33,300 fl. Now substitute R e = 500 fl: 30 x 10- 6 - 580 x IP" 4 2700 = 118,000 fl. For R. = 10,000 fl, 30 x 10- 6 - 580 x 10" = 39,600 fl. 12,200 Tabulate the values determined thus far: Rg , fl Z , fl 275,000 500 118,000 10,000 39,600 33,300 Output impedance Z vs. R g is sketched on Fig. 5.6. 102 Transistor Circuit Analysis PROBLEM 5.6 For the circuit conditions of Prob. 5.2, determine A t and A v for R L = 0, 100 fi, 1000 O, 10 4 O, 10 s fl, 10 6 Q, and «, . Solution: Substitute the hybrid parameters of Prob. 5.2 in (5.7) and (5.8): A i = 1 +K*Rl 290 1 + 30 x 10- 6 R L ' -h te Rl fc/e (l + h oe R L )-R L h i6 h te -290 R L 2200 (1 + 30 x 10- 6 R L ) - 290 x 2 x 10~ 4 R*. -290 R L 2200+ 0.008 fl L -0.132 R L 1 + 3.64 x 10- 6 R L ' The values of gain, as determined by direct numerical substitution of values of R L , are listed below. Although not indicated, A v is negative in all cases. R L ,n A, 290 100 289 13.2 1000 282 131 10,000 223 1272 100,000 72.5 9670 1,000,000 9.36 28,400 oo 36,200 For the normally used values of R L (under around 10,000 0), current gain is relatively constant, while voltage gain is about proportional to R L . The change in voltage gain is largely due to the changing impedance of R L for a constant input current. The circuit is best described as a current amplifier, rather than a tallage amplifier. If a resistance R s is added to the emitter circuit (see Fig. 5. 7a), input im- pedance is increased substantially, but the resistor is not by-passed. This cir- cuit is most conveniently studied using the tee-equivalent circuit of Fig. 5.7b. 1 + /3 " 1 + /S A^V — • — — VW- 'L Vn (a) (b) Fig. 5.7 (a) Common-emitter circuit with feedback resistor in the emitter circuit, (b) Tee-equivalent circuit of (a). Fig. 5.8 Common-emitter circuit with total re- sistance r E in the emitter circuit. The usual parallel current source in the collector circuit is replaced by a series voltage source to sim- plify calculations. [See Fig. 5.7(b).] Single-Stage Amplifiers 103 PROBLEM 5.7 A resistor R E is inserted in the emitter circuit of the common- emitter amplifier analyzed in the previous problems. Calculate, for this modified common-emitter circuit, the following performance parameters: Z,, Input impedance , Z ot Output impedance, A t , Current gain, A v , Voltage gain. Solution: The tee-equivalent circuit is shown in Figs. 5.7b and 5.8. In Fig. 5.8, set r% = «" a + Rb, and replace the parallel current source in the collector circuit of Fig. 5.7b with a more convenient series voltage source. The use of the tee-equiva- lent circuit provides convenient formulae in terms of tee-parameters, which are then available for other applications. The basic circuit equations applicable to Fig. 5.8 are v e = (Rg + r b + r E ) i, + t% i ol 01 r c i, = t% »i + [Rl + . ° + t% ) Rearranging these equations and letting 1/(1 + /3) = 1 - <X , v e = (Rg + r b + r|) »i + te »o' = (r E - a r c ) i, + [R L + r e (1 - <X) + r* E ] i . Solving these equations for i, and i , using determinants, (5.12) (5.13) '< = A 1 * r E [R, +r q (l-a) + r*J v. [R L + r c (1 - a) + r B ] : — _» .. ___ , t ; A {R a + r b +r B ) r| - a r c - v a (fl - CX r e ) where A = determinant of system R g + r b + T E * tE t e -<Xt c R l + r c (1 - a) + r E - (Rg - r b - r* E ) \R l • r. : (1 - a) . rll - t% (r* - -»r c ). (5.14) Current gain is immediately determined from the ratio of i to i t : A -° A, = — : - (r| - «r e ) a r c - rs (5.15) », " Rl + f c (! + a) + rl R L + r c (1 - a) + r| This last expression can be simplified, recalling that r d = r c /(l + /S) = r c (l-ot) is much greater than r E . Thus, A,= 1-a Rl + i i + *l r c C 1 - a) r d (5.16) To find the input impedance Z,, for the moment substitute v t for v e and let /?, = in (5.12). Then solve for /',, using (5.14) for the system determinant A: v, [R L + r c (1 - a) + r* E ] 104 Transistor Circuit Analysis V( (r b + r E ) [Rl + r c (1 - a) +■ t E \ - r E (r E -Of c ) = r b + r E - Rl + r c (1 - a). r| (r| - a r c ) — r - + f E = * Tb + TE = * *b + IE = fb + *"E . Rl + r e (1 - a) r r| ! _ rl - a r c Rl + r c (1 - a) + rl Rl + r c - a r c t- te - 'e + a r c R L + r e (1 - a) + r| 7?l + r e «l +r c (l-a)+rS. , . Rl 1 + ^ L + f E (5.17a) (5.17b) Since R L + t * E « r c (1 - a), and 1/(1 - a) = 1 + /3 Z,=r b + r| (1 + Pi. It is now easy to find voltage gain, A v : (5.18) A. = . -»o ^L . Rl_ ■ v, i t Z, ' Z, Substitute (5.15) and (5.17) for A { and Z„ respectively: A r OC r c - te r l + r c (1 - a) + te -Rl * + TE Rl + r c l b Rl + r c (l-a)+r|. -(ar c -r|)ft L r b [Rl tr c (l-a)+ r|l (- r| (/?/. + r,) re H) r. -* r E L (1 - a) (W (5.19a) In terms of the appropriate equivalent tee-parameters, Rl TE r /s * 1 te A J3+1 r„Q3+l}J r b f 1 rl [P - 1 + 1"E I" R t + 1 + «£, r d (j8 + 1) '- Q3 + 1) (5.19b) Using the usual approximations in (5.19), K = -*±. (5.19c) The error in this approximation is estimated in Prob. 5.8. The last formula to be derived is an expression for output impedance. Set up the original equations (5.12) and (5.13) with v g = 0, and R L replaced by a voltage v B applied to the output terminals: = (R* + r b f r|) i, + T % i , (5.20) Single-Stage Amplifiers 105 v = (r| - a r c ) i , + [r c (1 - a) + f|] i c Solve (5.20) for i,i . !j *> — i «— j- «$ + ft + t e Substitute (5.22) in (5.21), and solve for Z a • v /i : (5.21) (5.22) Fot r|/r d « 1, '"o ^a + 'b + «"b Aj + fb + TB **** 1-+4*. # : 1 + R < t r " *• - «"*: P l + R£ + *b r* J (5.23) PROBLEM 5.8 Check the approximate formula (5.19c) against the exact formula (5.19a) for the voltage gain of the common-emitter amplifier with resistance R E in the emitter circuit. Use the 2N929 transistor at the previously defined operat- ing point, with the following parameters: r e =6.60, r b = 260 Q, r d = 34.5 kfi, j8 = 290 for Rt . = 5000 fi, r| = ioo n, a = 0.9966 , r c = 10 7 . Solution: In the exact formula, -R L /r* E is multiplied by the following factor: 100 Bias current v cc 0.966 - 5000 10 7 . 5000 260 1 +^T+777 10 7 5100\ 0.986. + 100 \291 + 10 7 / The approximate formula, A v = -R^/r E , is inaccurate by only 1.4%. Thus, an unby-passed emitter resistor provides excellent stabilization of voltage gain. This is a special type of feedback which will be described in more detail in Chap. 8. PROBLEM 5.9 Calculate Z„ Z ot A v , and A, for the circuit of Fig. 5.9. Assume that bias is set so that l c = 4 ma and V C e = 12 v, corresponding to the operating point of the previous problem. Solution: For the operating point of this circuit, the 2N929 transistor has the following tee-parameters: Ht- R L = 5000fl O Output 2N929 R E = 93.30 Fig. 5.9 Single-stage transistor am- plifier with an emitter resistor for stabi lization. 106 Transistor Circuit Analysis r e = 6.6 fl , R E = 93.4 Q , r c = 10 meg, r b = 260 fi , a= 0.9966, j8 = 290. These parameters are substituted directly in the appropriate formulae to calculate the required performance characteristics: -Rl 1 r E + R L * rb \r~-^~ ]+rB -5000 260 ( J_ + ±^±) + 100 \29C *"" ' 5100 \ ,290 ' 10 7 ) -49. This is just 2% less than the approximate value of /? L /r| = 50. Calculate input impedance: 1 Rl + r E = 260 + 100 (291) 1 + 5x 10 " 4 = 25,660 0. 1 5100 + 3.45 x 10 4 Note that the significant component of input impedance is the r% (1 + /S) term, which in itself gives a fair approximation. Compare the input impedance with the much lower value of 1950 O in Prob. 5.2 for the same circuit, but with R E = 0. Now determine output impedance: /3 1 | R 6 + r b 34,500 /1 + _?*L\ = 1.2 MO. 1 + Z^ 100/ It remains only to calculate current gain: A P I , Rl + r E + i?°_.251. 5100 34,500 Single-Stage Amplifiers 107 PROBLEM 5.10 By comparing the results of calculations on the circuits of Fig. 5.9 (Prob. 5.8) and Fig. 5.2 (Prob. 5.2) differing only in the presence of R E in Fig. 5.9, comment on the effect of R E on the principal amplifier characteristics. Solution: The amplifier characteristics are summarized in the following tabulation: Fig. 5.2 Fig. 5.9 z, 1950 n 25,660 Q Zo 118,000 Q 1.2xl0 6 S2 A, 252 251 A v -647 -49 The addition of R E increases the input impedance to nearly te (jS + 1). re- duces voltage gain to about Rl/te, and increases output impedance substantially. Current gain is essentially unchanged. The performance of the amplifier is sta- bilized since voltage gain becomes nearly independent of /3. 5.3 Common-Base Circuit The performance parameters of the common-base circuit, namely input and output impedances, and current and voltage gains, are derived in a similar manner to the derivation of the common-emitter parameters of the pre- ceding section. Using the hybrid equivalent circuit, Figs. 5.1 and 5.3 apply ex- actly, except that the subscript b (common-base) is used instead of the subscript e (common-emitter). PROBLEM 5.11 Derive formulas for Z,, Z , A,, and A v for the common-base am- plifier circuit. Solution: Modifying the previously derived common-emitter equations (5.6), (5.9;. (5.7), and (5.8) where the subscript b is used instead of e. I O h lb (1 + h b Rl) - Rl t*tb hrb PROBLEM 5. 12 A 2N929 transistor is operated in the common-base configuration at a bias point where V CB = 12 v and / c = 4 ma. For R 6 = 10 Q and R L = 5 KQ, calculate (a) Z„ (b) Z OI (c) A v , and (d) A,. The /i-parameters for this operating point have already been derived in Chap. 3: h ib = 7.57 SI, h fb = 0.27 x 10-\ h lb = -0.996, h ob = 0.103 x 10 -6 mhos. 10° 10= 10" 10' 10" 10 Zi.il- 108 Transistor Circuit Analysis / ^ 1 0*/ ~V^=9.7Mfi as R L = oo C-E as 267 n R L =<*> 3 1 iv *L = '</ = 33.3KQ — C-B = 267fl as Rl = oo 1 Solution: (a) To determine input impedance, substitute the above parameters in (5.24): z i = h ib - h ob + =- 7.57 + 0-996x0.27x10- = 7J Q 2 x 10- 4 + 0.001 x 10- 4 (b) Similarly, calculate Z by substituting in (5.25): 1 1 Z„ = htb A, 10" io 5 10 4 10 3 10 2 10 1 10 10 2 10 3 10 4 10 s r l , n-*> \a)Zj vs. load resistance z„,n— *■ c£* = r c = 9.7MQ = r d as Rg =0 1 S? S = r d = ^ 3.3 K$2 i = <*> — C-B «*^ 3 = 'd = 3.3 KQ = h/b as R t =0 as R« = » A ob - ""> "* ntw..^, (0.996) (0.27 X 10- 4 ) An, + K fi 7.57 + 10 = 614,000 0. (c) Substituting in (5.27), A v = 639 (voltage gain). (d) Substituting in (5.26), At = -0.996 (current gain). PROBLEM 5.13 For the 2N929 transistor of Prob. 5.12 with the same operating point and parameters, calculate and plot Z, as R L varies from to oo. Solution: Use (5.24): z i = A/6 - A/b h rb ,b + The general nature of the input impedance variation is self-evident from this equation. Letting R L = 0, 1/R L becomes infinite, so that the second term in (5.24) vanishes, and 10 10* 10' 10 1 1Q 10 2 10 3 10" 10 s (b) Z vs.R g a A v , Aj- io- 10 ' 4 V = 36,000 V as Rl = °°^^ C-E an i C-C d as 1 Ay =0 asR L = / C-B / as 1,=0 R L = •»■ # c Similarly, when /?£, and when 1/R^ = /, 6 , Z i = Aib- Zy = h, b - A/b A rb •^i = Ayb - A/b A ffc 2h nh 1 10 10 2 10 3 10 4 10 5 (c) /4i and A v vs. Rj, Fig. 5.10 Typical performance char- acteristics of single-stage audio amplifier. This last impedance value is the average of the values for R L = and R L = «.. The R L = and R L = <x conditions define asymptotes, which make it an easy mat- ter to sketch curves showing how Z t varies with R L . Substituting numerical values of the common-base A-parameters , Z = 7.57 + 0-996 (0.27 x 10~ 4 ) 1.03 x 10- 7 + JL Figure 5.10 shows a plot on logarithmic coordinates of Z t vs. R L . Note that R L has minor influence on Z it until it exceeds about 100 Kfi. Since such high values are usually not practical, Z f s h lb is a constant value in this common-base con- nection. For the tee-equivalent circuit, Z, = r e + r b (1 - a). PROBLEM 5.14 For the common-base connection and the operating point of Prob. 5.12, determine the variation of Z vs. R t . Single-stage Amplifiers 109 Solution: Use the common-base A-parameters of Prob. 5.12: Zo = - . [5.25] K b h ib + Rg As an aid to plotting, determine the limiting values where R e = and R g = 1 At R e = Z„ = htb h rt h "lb "rb "ob - *« AtR g = 00, ^ ob Substitute numerical values: Z„ = 0.103x 10- _/ -0.996 x 0.27x10- 7.57 + R e and Z is plotted vs. R 6 on Fig. 5.10. (Note that at R g = oo, Z = r c of the tee- equivalent circuit.) Although Z varies sharply with R e in the useful region where R e is of the same order as h lb , Z generally is not critical in circuit calculations. Thus the variation shown typically in Fig. 5.10 is not too troublesome. PROBLEM 5.15 For the common-base connection and the operating point of Prob. 5.12, determine the variations in A v and A, vs. R L . Solution: Use the same common-base h-parameters of Prob. 5.12: A l = ilJb A r hi 1 + h ob Rl -0.996 1 + 1.03 x 10- 7 R L ' -htb — — + h ib h ob - h tb h rb 0.996 — + 7.57 (1.03 x 10- 7 ) + 0.996 (0.27 x lO" 4 ) R L Current and voltage gains are plotted in Fig. 5.10. Note that current gain is almost independent of R L . Voltage gain, on the other hand, is very much a func- tion of load, as might be expected. 5.4 Common-Collector Circuit (Emitter-Follower) Proceeding as in the previous section, formulae for the common-collector single-stage amplifier configuration (emitter-follower) are de- rived simply by applying suitable subscripts to the hybrid parameters. 110 Transistor Circuit Analysis PROBLEM 5.16 Using the methods of Prob. 5.11, develop formulas for Z ol Z„ A,, and A r for the common-collector configuration. Solution: It is only necessary to change the subscripts of the A-parameters as indicated below: (5.28) (5.29) (5.30) z, = fl le Jtfur I z ■■i- K* »f. Kt ?\ + R e A A# e 1 + Kc Rl A r = - zh°*i . (5.31) "(e (1 + Aoe Rl.) - Rl ft/e A, c PROBLEM 5.17 Using the A-parameter formulae for the common-collector con- nection, determine Z, as a function of R L . Use the same operating point as in Prob. 5.12, and the corresponding numerical value of the A-parameters developed in Chap. 3: A )c = 2200 Q, A rc = 0.9999, A, c = -291, A oc = 30 x 10 -6 mhos. Solution: Substituting the given values in (5.28), Z, = 2200- ( - 291)(0 -" 99 > . 30 x 10- 6 + — Rl Note that the asymptotes for R L = and R L = oo described in Prob. 5.13 apply here, and are important plotting aids. At R L = 0, Z, = 2200 = A ic . At R L = oo, Z - 22 ° + 3bW= 9 - 7Mn - Substituting additional values for R L and calculating Z lt the required variation of Z { with R L is derived (see Fig. 5.10). For the practical range of R L , the expression for Z t may be simplified: Z t = h le - h tc R L . In terms of other familiar units, Z,sh„ + (fi+l)R L , since h ic = h le and h tc = - (1 + ft). Single-Stage Amplifiers 111 Input impedance obviously depends almost directly on R L in the useful range of operation. From the tee-equivalent circuit, Z t = h ic - h tc /h oc and Z, = r c as R L = «,. Therefore, r c constitutes a theoretical upper limit to input impedance. PROBLEM 5.18 Proceeding as in Prob. 5.17, and using the same operating point and equivalent h-parameters, determine Z as a function of R t . Solution: The applicable formula is *.«r A,« + R t Substituting numerical values and locating the R e = ~ asymptote, Z is conven- iently plotted in Fig. 5.10. A good practical approximation to Z is •r _ Rg + *ic . Z can never exceed r d of the tee-equivalent circuit and depends strongly on R a in the normal range of application. PROBLEM 5.19 Proceeding as in Prob. 5.17, and using the same operating point and equivalent A- parameters, determine A v as a function of Rl- Solution: The applicable formula is A r = -h, c RL ^ [5-31] h lc (1 + h oc Rl) - Rl h tc h rc Rewrite this expression: Since h re is the major term of the denominator, its more accurate value is required : (see Fig. 3.26c). Substituting, A 1 n v - h lc htcRL Using the /i-parameters for the common-collector circuit, and h re = 2 x 10~\ A 1 1 V " 1 - 2 x 10- + (3 ° * 10 " 6) (2200) + -g°°- 1.00002 + Z^I' 291 291 R L R l Observe that for R L » h le /h te = h ib , A Y si. As a matter of fact, the gain is very nearly unity over a wide and practical range of values of R L . It is therefore convenient to calculate the percent deviation from unity for values of gain ap- Percent deviation - (1 - A r ) x 100. 112 Transistor Circuit Analysis Percent deviation from unity is plotted in Pig. 5.11 as R^ varies from to ». Note- that when R L exceeds 100 /j Jtr , the deviation is less than 1%. This uniformity of gain of the common-collector or emitter-follower circuit is of great practical value. More will be said of this later. PROBLEM 5.20 Proceeding as in Prob. 5.17, and using the same operating point and equivalent /i-parameters, determine A, asa function of Rl,. Solution: The applicable formula is A, ifc l + fc oc /f L Substituting numerical values, i/c -291 1 + Aoc Rl 1 + 30 x 10- 6 R L Figure 5.12 shows A t vs. R L , as required. [5.30] 10.0 0.001 250 200 150 100 50 n 10° 10 1 10 2 10 3 10 4 10 s 10 6 R L ,Q.— +■ Fig. 5.12 Variation of Aj with R^ in common- collector amplifier. 10' Fig. 5. 1 1 Deviation from unity of voltage gain of common-collector amplifier with load resistorR^. As a convenience in carrying out analyses similar to the ones in this chapter, Table 5.1 summarizes the formulae for the performance factors of the single-stage audio amplifier. Single-Stage Amplifiers 113 Table 5.1 Single-Stage Amplifier Formulae A-parameters (Second subscript omitted) Common- base tee-equivalent circuit (see Fig. 2.31) Common-emitter tee-equivalent circuit (see Fig. 2.31) Note: rj = r. + R B ; R B = circuit resistance in emitter Common-collector tee-equivalent circuit (see Fig. 2.31) *,-■? V a*i.+(i-o)I» r . + '(> (l-«) + r t Rr. + f» = r. + r 6 (l-a), 1* « R *£ « 1 <•« r <f li r t + r;(l + /3) r,(l + /3) 1 + «£±£S ST 6+ r:(l + /3), R L + r; « r„ <■» + - (R;.+r.)(<3 + l) (R!.+r.)(/3+l) 2rr b + (R I .+r.)(/3 + l), r c »(R L +r.)(/3 + 1) K- h,h, h, + R t ar c + rt - 'b " 1 + r ' + R, r, r<j r« ^ R,+ r b rl - j + RgJ_Tb '.+ ^ 1 1+/3 1; R, + r t r„(l + /3) R t + r b up ' l + A R t <-i. + Rl 't + Ri. «1 >-t 1 + r « /3 <•: « r rf (1+0)- r. + Ri = 1 + /3, % + Rt « 'd A T = Z± Rl*i fti -(•-^): («*?)' «R L rt + R L <<c 1 r c + r 6 (l-a + ^)' " /3 + 1 r/(/3 + 1) r„(/3+l) r;(l + ) 8)V r„ / ~ Rl rl«r dl j3»l, Ri«fd 1 + H-H)*^ "('•-i^k l *Tf r t « 'c 5.5 High-Frequency Performance In this section, we will consider the behavior of the transistor at high frequencies, at which point transistor parameters become com- plex. In the border-line region where frequency effects just begin to appear, it is sufficient to use a complex forward current gain. For a useful representation sat- isfactory over a very wide frequency range, the hybrid-n- circuit is easily the most convenient. Consider the ft-parameter common-base equivalent circuit with complex forward current gain. Let h tbo be the low-frequency value of h tb . The following approxi- mate relationship based on intrinsic solid-state properties of the transistor holds at intermediate frequencies'. ft /bo h tb = (5.32) 1 + h lb in which t htb is the a (or h, b ) cut-off frequency where the absolute value of cur- rent gain declines to 0.707 of its low-frequency value. Now determine the complex forward current gain for the common-emitter con- figuration: h le = -h 1 + h tb (5.33) tb 114 Transistor Circuit Analysis BO O C Fig. 5.13 Single common -emi tter circuit. Substituting the complex expression for h lb in (5.33) and simplifying, '/bo A, 8 = l + /i /bo 1 + ■jt (5.34) { hfb (1 + h fbo ) If we let f htb (1 + h, bo ) = //,/e , the frequency where h te (or /S) is down to 0.707 of its low-frequency value (the /3 cut-off frequency), then h tB = Af.c (5.35) l+j '*■ /a Fig. 5.14 Hybrid-77 equivalent cir- cui t for common-emi tter connection. where h, eo is the low-frequency value of h t9 . The cut-off frequencies give a reasonably good idea where frequency effects start to become important. However, they are only a very approximate indication of a transistor's high-frequency capability. From (5.34), we see that the common- emitter configuration has a bandwidth smaller by a factor of 1 + h tbo than the common-base configuration. (Recall that h tbo is negative and nearly unity, so that 1 + h !bo is a very small number.) 5.6 Hybrid-7T Circuit The hybrid-77 circuit is superior to other high-frequency equivalent circuits, in that its parameters are relatively independent of frequency over a very wide range. Furthermore, we can measure all parameters directly at high frequencies. Figure 5.13 is the simple common-emitter circuit, represented for small signals by the hybrid-77 model of Fig. 5.14. The low-frequency conversion formulae from A-parameters are given by Table 3.1, with r b6 » arbitrarily chosen. Table 5.2 pre- sents a concise procedure for direct measurement of the hybrid-77 parameters. We now proceed to applications of the hybrid-?? model, calculating input im- pedance Z„ current gain A„ and voltage gain A v for given load conditions. Al- though the derived results are general, only a few simple cases will be investigated numerically, because of the extreme complexity of the computations. C b 'r. ; 1 h B' z s c ] JL zl-+- Z P z'l — I 1 — ( >— I \ )6 m V b ' (b) Fig. 5.15 (a) Hybrid-77 equivalent circuit set-up for calculating input impedance, (b) Substitution of complex impedances. Refer to Fig. 5.15a, and the somewhat simplified representation of Fig. 5.15b. For convenience, we replace the resistance-capacitance combinations with com- plex impedances. The base-spreading resistance r 6b / is initially neglected. The current source is replaced by a more familiar equivalent voltage source in Fig. 5.16a. The mesh equations are* ♦Note that in the remainder of this Chapter, rms values are used rather than instantaneous values. Single-Stage Amplifiers 115 v b , m = z p /, - Z p I 2 = Z p (/, - / 2 ), &* JV. Zl = - Zp h + &p + z . J z ^ f, - g m z p zi. a, - ij. z. 'l t z. 'i t - 1 A An v b ;2 •iH v b '. Z P |+ »— zL — 1 1 — t , i+fi»z L f* — - Z P zL z't—- 1+Sn.Zi. o — 1»— z. B r "' b' l+SmZ L , y w . _^y 6e VVe "zT Zp — 1 1 — zi (a) (b) Rearranging the second equation, - - Z p [1 + *„ Z£] / t + [Z p (1 + i„ Zl) + (Z£ + Z,)] /,. Dividing the above equation by 1 + g m Z£, the set of equations becomes V, - V*' 0— Z,/, «V. ^ + ^)' (5.36) (5.37) g m V b > e = 10- 1 x 10- 2 = 10- Th e currents may be summed at the node: a 1 ma. / 2 + / c •6 m V b (c) Vc A B r bb' o — VA — f c b '.; Figure 5.16b shows the equivalent circuit representation of these equations. The validity of the equivalent circuit may be demonstrated by writing its mesh equa- tions and comparing with those derived above. An obvious extension of Fig. 5.16b, in which r bb i is added to give an exact value of Z t without dependent sources, is shown in Fig. 5.16c. In Fig. 5.16d, the generalized complex impedances are separated into resistive and capacitive components. The real number A is introduced to simplify calculations. This cir- cuit can be used to compute Vb'* and /,, which are required to determine perform- ance characteristics of the amplifier stage* To determine l c and V c » , use the out- put circuit of Fig. 5.17b. which is derived from Fig. 5.15a. PROBLEM 5.21 Referring to the output circuit of Fig. 5.17b, where 7 2 = ;10- 4 a, V b ; = 10- 2 v, r ce =10 5 O, Z L = 5 x 10 5 Q, g m = 0.1 mho, calculate / c and V ce . Solution: Establish the current source: He real number (d) 1 T em ~ , Fig. 5.16 (a) Replacing the current source of Fig. 5.15(b) by an equiv- alent voltage source, (b) Further simplification of input impedance circuit with only passive elements. (c) Extension of (b) to include r bb >. (d) Circuit of (c) in which complex impedances are replaced by resis- tance-capacitance equivalents. CO EO 6m*Ve( I (a) CO- 6 m Vf m * o e < t) t: EO- where the current through r ce has the direction shown in Fig. 5.17b. Since l c = -V ce /R L ' v ce is the only unknown in the above expression. Substituting numeri- cal values and solving, V ce =-4.76(1- 0.1;) v, I fee 4.8 v. (b) Fig. 5.17 (a) Output portion of hybrid-TT equivalent circuit. (b) Expansion of (a). 116 Transistor Circuit Analysis TABLE 5.2 Direct measurement of Hybrid-77 parameters. (Refer to the hybrid-ff equivalent circuit of Fig. 5.14.) 1. Refer to Fig. (a). Capacitors C, and C 2 are effective short-circuits to a-c test signals. Set Rj_ and Rg for the desired operating (Q) point. The value of Rg ~S> Zj, so that bias current to the base is essentially constant. With the short-circuit from the collector to the emitter (due to C 2 ), C b ' e and C b ' c are essentially in parallel with r b ' e . Apply a sufficiently high frequency (say 5 times f b ( e ) from base to emitter, so that B is effectively short-circuited to ground. Measure /,- and v be to determine r bb '. More accurately, we can measure the resistive component of Z,- with a suitable impedance bridge. (Note that since r b ' c ^> r b ' e , there is negligible error in neglecting t b > c at the frequency in question.) 2. Compute the remaining low -frequency hybrid-'"' parameters from the previously derived formulae: h ie - T bb' *TsC 3 Ve 1 = h Ie — r bb' r bb' h oe u 1 + ftfeo -ftfe - n re — "ie - rbb' r\j "oe • [3.61] [3.60] [3.63] [3. 56] h ie -rbb 1 The circuit of Fig. (b) is used for measuring C b ' c . The emitter impedance is sufficiently large so as to present an essentially open emitter to a-c. Resistance R L is high enough so as not to shunt the collector-base capacitance too heavily. Using a capacitance bridge between points B and C, measure C ob , the output capacity for the common-base configuration with emitter open. The measurement is to be made at a frequency such that the reactance of C b ' c is much greater than r bb < and much less than r b ' c . In performing in the capacitance measurement, note that the dissipation factor is small, confirming that the test frequency has been properly selected. Output capacity C ob , thus measured, approximately equals C b < c . The measurement outlined here establishes the value of C ob = C b ' c , plus lead capacities, and the so-called overlap- diode capacity, which is the capacity between electrodes outside the active transistor region. These must be sub- tracted from the measured capacitance. Unfortunately, an accurate determination of C b ' c is not easy. Conveniently, however, C ob is usually specified by the manufacturer for high-frequency transistors. To measure C b ' e , refer to Fig. (c). Capacitance C is a short-circuit at the test frequency. Resistance R is a small resistor, chosen to measure current i c by its drop v R = i c R. The value Rg » hj e , so that i b is essentially a con- stant base current, i b = Vg/Rg. Since R is a very small resistance value, v ce = 0, so that from the point of view of input impedance, C b ' c is effectively in parallel with C b > e . Also r b ' c 3> r b > e . For these conditions, the hybrid-^ circuit corresponding to Fig. (c) is ap- proximately as shown in Fig. (d). The following expressions apply to the circuit of this figure,* Vr RUmVb'e, = R6n R 6 Ve + h/i*>(C b > e + C b ' c )] 6m v b'e By measuring Vr at low frequency, say 1000 cps, and at the much higher frequency where V^has dropped to 0.707 of V R at low frequency, we determine t b i e where V-= * • Thus, since C b ' c 2n-'h/e(<V e + C b ' c ) is known from the preceding step, 1 C b ' e + C b ' c =• 2"fhWb'e <V 2^fht B r b 'e C b > Manufacturer specification sheets usually give l h f e . Sometimes the transistor manufacturer gives ir~ h teoihte< a gain bandwidth product. In this instance, the a cut-off frequency is { T = flfeo l hfe = (\ + n feo> { hte = l h fb- * Note the use of rms vectors rather than instantaneous values in the following equations. Single-Stage Amplifiers 117 Solving for l c = -V ce /R L , I c = 0.95 x 10-' (1 - 0.1;) a; |/ c | = 0.96 ma. To complete the representation of the hybrid-;? by separate passive input and output circuits, we now derive an equivalent output circuit containing no depend- ent sources. Figure 5.18a shows the output circuit with the input terminated in a generator impedance, Z a . For convenience, complex impedances are used to rep- resent capacitor-resistor combinations. Impedances Z' p and Z' are defined on the figure. Referring to the simplified configuration of Fig. 5.18b, the circuit equa- tions are Combining, Solving for \ v Since V ea = - /, (Z. + Zp, 2' Eel -4(2. + gp z* + z' m Adding r ce , we derive the simple form for the output circuit of Fig. 5.18c. (5.38) <W B < I — W^ — •■ VI &mV b >J | 4 (a) b' z s 'i t Ji_ C v — i I 4 AntV.M J ce 1 » i ,i _ Obb'+ z g) z p '" 'bb'+Zg + Zp l + 6mZp (b) T z a l+g m Zr -oc -OE (c) Fig. 5.18 (a) Hybrid-77 equivalent output circuit, (b) Simplified equivalent of (a), (c) Final simplified representation. PROBLEM 5.22 Develop the input impedance network representation of the com- mon-emitter amplifier circuit of Fig. 5.19. Solution: The model of Fig. 5.16d is applicable and will be used for the solution to this problem. Direct substitution is all that is required: 118 Transistor Circuit Analysis = 20mv R g =2KQ, /u^\ Transistor hybrid-TV parameters r bb ' = 100 fi g m = 0.138 mho r b ' c = 10 7 fi r b ' e = 2100fi r ce = 435Kfi C,'„ 5/t/tf C b i e = 250/t/tf Fig. 5.19 Common-emitter circuit with parameters corresponding to Prob. 5.22. A = 1 + g n Rl r c R L + T ce Substituting numerical values from Fig. 5.19, A = 685. Continuing, Z L 5000 685 = 7.3 n, 435,000 685 = 635 n, ***-=—= 14,600 Q, A 685 ACy c = 685x5 = 3425ft /if. These calculated results are shown on the hybrid-n- circuit of Fig. 5.20. Sev- eral interesting results are evident. The 7.3 fi effective resistance from collector to base is so low that it acts as an effective short-circuit. Point C is therefore substantially at ground potential. Thus, as far as input impedance is concerned, AC b f e is in parallel with C b ' e . This multiplication of capacitance, leading to a comparatively very large capacitance in shunt across the input, is known as the Miller effect. It is an example of the effective amplification of an impedance in an active network. Although r b > c is also "transformed", it is relatively so large that its effect is usually ignored. = 2Kfi = ioo £2 v 6 = 20mv V h i ^=14.6Kfl A AC h i i(r h'c = 3425 /i/tf C b ' e = 250///tf '^- e = 635fi ' A > A = 7.3 Jl Fig. 5.20 Hybrid-77 equivalent input circuit corresponding to the configuration of Fig. 5.19. PROBLEM 5.23 For the amplifier circuit of Fig. 5.19, determine the frequenc, where voltage gain falls to 0.707 of its low-frequency value. Ri ^kd Solution: The circuit of Fig. 5.20, developed in the preceding problem, still ap- plies. Simplify as shown in Fig. 5.21 and calculate V b ' e : R,= r b'e r b'c/ A 18350 R, r b'e + r b ' c /A Ri = R 6 + r bb ' = 2100fi c » = c h' e + A Cb'c = 3675^/j.f Fig. 5.21 Simplified form of the in- put ci rcui t of Fi g. 5.20. JaCJ R 1 + - 'b'e JCO C, ^1 R,+ ■V* = R, jco c t R 2 (1 + ; co C,/?,) + R t R l + jco C, (5.39) Single-Stage Amplifiers 119 Now calculate I 2 , assuming that the collector is essentially at ground potential, and that r b ' c » l/(;'a)C b ' c ) in the significant frequency range. Thus, directly from Fig. 5.20, V b ' e h = 1 (5.40) ," eu C b > c A Referring to Fig. 5.17b, with V b < e and I 2 known from (5.39) and (5.40), and recog- nizing that r c e » Z L , 4m Vb'e =h+h (5.41) Substituting numerical values, V b ' e and I 2 are calculated, from which l c is directly obtained: 1835 1835 + 2100 (1 + jco 6.75 x lO" 6 ) V t , I t =jco 3425 x 10- 12 V b > e . Substituting in (5.41) and simplifying, 253 (1-0.0248 j <o x 10~ 6 ) c " 3935 (1 + 0.362 j co x 10" 6 ) * ' The frequency of interest, where the output falls to 0.707 of its low-frequency value, coincides with the 0.707 point of l c . This frequency is where the phase shift is approximately 45°. For simplicity, and because it is justified by the general accuracy of this type of calculation, we neglect the imaginary term in the numerator of the above expression for /«. . The required frequency occurs where corresponding to 0.362 wxl0-*= 1, / = 440 K Hz. v C c Fig. 5.22 Common-emitter amplifier circuit applicable to Prob. 5.23. Z„ -j 2 x10 s A ,690 Z' L j 5000 (at co = 10 6 ) = - ; 2 x 10 s , = - 290 fl, ,690 = 7.2 Q. O VNAr- PROBLEM 5.24 Calculate the input impedance Z { for the amplifier circuit of Fig. 5.22 at co = 10 6 radian/sec. Neglect r ce and t b ' c . Solution: Use the model of Fig. 5.16. Now, A=l + g m ZL = l + g m Z L . Since Z L = j <y L, A=l+g m QcoL). Substituting numerical values, A = 1 + (0.138) (y 5 x 10- 3 x 10 6 ) = j 690, 1 =2iooQ! — = -290Q A — vw, — c b > e = 250/i/if (a) 100 fi O Wr -327 Q« A = 7.2J2 (b) Cfc'e - = 250^1 i'b'e = j4000fi at W = 10 6 Fig. 5.23 (a) Equivalent input im- _,, . , , . , T-i- r r>-> m pedance circuit corresponding to The equivalent circuit representing input impedance is shown in Fig. 5.23. The the amp |jfj ero f Fig 5.22. (b) Sim- series resistances may be added and combined with t b ' e for an equivalent shunt plified equivalent of (a). 120 Transistor Circuit Analysis resistance of - 327 ft. Input impedance is then readily calculated as Z, = ~ 3980-/32.5. The input impedance is negative, which occasionally occurs when amplified im- pedance is reflected back to the primary. For this condition, the amplifier will behave as an oscillator. 5.7 Supplementary Problems PROBLEM 5.25 If h„ = 3001), h oe = 10- 5 ft, h re = 10~ 4 ft, and h ie = 2000ft in the common-emitter circuit of Fig. 5.2, calculate (a) Z, and Z Q for R L = R g = 10,000 ft, (b) the current and voltage gain for R L = R e = 10,000 ft, and (c) the value of R L which yields the maximum power gain. PROBLEM 5.26 Determine the effect of a change in R L on the input impedance Z j for a common-emitter circuit. PROBLEM 5.27 Calculate the common-emitter tee-equivalent circuit for the transistor of Prob. 5.25. PROBLEM 5.28 Find the effect on the tee-equivalent resistor of Prob. 5.27 when a resistor R E is added in series to the emitter. PROBLEM 5.29 A resistor R = 100 ft is added in series to the common-emitter circuit of Prob. 5.2. Calculate (a) Z it Z o! and the current and voltage gains when R l = Rg = 10, 000 ft, and (b) the value of R that yields the maximum power gain. PROBLEM 5.30 Discuss the advantages and disadvantages of the tee-equivalent circuit. PROBLEM 5.31 Derive the formulae for Z it Z o! A,, and A v for the common- emitter circuit using (a) A-parameters and (b) tee-parameters. PROBLEM 5.32 Why is the common-collector circuit useful? What are its main characteristics? PROBLEM 5.33 Define /3 cut-off frequency. PROBLEM 5.34 Find the gain change from to 1 MHz when a transistor with a /3 of 300 and a /3 cut-off of 10 MHz is used (a) in a common-emitter configuration and (b) in a common-base configuration. PROBLEM 5.35 In the circuit of Fig. 5.22, use a transistor whose hybrid-77 parameters are given in Fig. 5.19. (a) Calculate the input impedance and volt- age gain without neglecting r ce and iv c . (b) Compare the results of part (a) to those of Prob. 5.24. MULTI-STAGE AMPLIFIERS 6 CHAPTER 6.1 Introduction The techniques developed in previous chapters, particu- larly Chap. 5, may be applied to the analysis of multi-stage amplifiers. There are two basic analytical approaches: 1. We can replace each transistor by its equivalent circuit or model, and analyze the resulting multi-mesh network. 2. Or we can apply the formulae developed in Chap. 5 to each stage in turn, and account for the interaction between stages by using suitable input and output resistances. The effects of coupling networks can also be included in this approach. The first of these methods is extremely tedious, except in the simplest of cases, and is generally useful only for elementary multi-stage circuits with two or three transistors. The second method, on the other hand, is practical and easy to apply to all configurations. Consider, for example, the block diagram of a multi-stage circuit in Fig. 6. 1. For each stage, the output impedance of the preceding stage is its input or driv- ing impedance, while the input impedance of the succeeding stage is its output or load impedance. Stage 1 Coupling network 1 Stage 2 Coupling network 2 Stage 3 Coupling network 3 Fig. 6.1 Block diagram representation of a multi-stage amplifier. By means of the formulae of Chap. 5, the input impedance Z t of a stage can be calculated if the load impedance is known. As load impedance is generally specified only at the output stage, it is customary to start here. The input im- pedance obtained then becomes the load impedance of the preceding stage, and so forth. Similarly, the output impedance Z of a stage depends on its driving im- pedance. As driving impedance is generally specified only at the input stage, calculations start here. In turn its output impedance becomes the driving im- pedance of the succeeding stage, and so forth. PROBLEM 6.1 For the two-stage amplifier of Fig. 6.2, calculate the small- signal a-c quantities R it R ol and A v . Use the parameters for the 2N930 transis- tor given in Tables 6.1-2 and Appendix A. Solution: Figure 6.2 shows an emitter-follower stage feeding an output emitter- follower stage. Our first step is to estimate the bias voltages. 121 122 Transistor Circuit Analysis -°v cc = 20v 2N930 'fli -»v„ ■R„ TT Rj, R = input, output resistance, respectively, of two-stage amplifier ^i'i> "o! = input, output resistance, respectively, of first stage Rt 2 , R 0l = input, output resistance, respectively, of second stage C= coupling capacitor, assumed infinite R L = 2500(1 R g = 2000 ft Table 6.1 Type Mo. 2N930. Electrical Parameters l c = 13/ia / c = 4 ma Unit bib 2100 17 ohm Kb* 0.045 x 10" 6 0.056 x 10' 6 mho K e * 200 370 Kb* 1.5 x 10" 4 2.3 x lO" 4 u ** "Ic 420,000 6300 ohm U ** 9x10^ 20.8 x 10" 6 mho h tc -201 -371 U ** h rc 1 1 * Published data ** Derived data using conversion formulae of Chap. 3 R,= 1.78Mft R 2 = 2.27 Mft Fig. 6.2 Two-stage direct-coupled emitter-follower amplifier. Table 6.2 Type No. 2N930. Electrical /c = 1 ma 'c = 2 ma h- = ■3 ma Parameters 25° C 100° C 25° C 100° C 25°C 100° C Unit h ie 320 460 340 480 360 500 Kb 30 36 20 24 16 20 ohm Kb 0.076 x 10' 6 0.086 x lO" 6 0.11 x 10" 6 0.13 x 10" 6 0.15 x 10"' 0.18 x 10" 6 mho Kb 1.8 x 10" 4 2.5 x 10' 4 2.0 x 10" 4 2.8 xlO" 4 2.2 x 10" 4 3.0 x lO" 4 r c = l/h ob 13.2 x 10 6 11.6 x IP 9.1 x 10 6 7.7 x 10 6 6.7 x 10 6 5.57 x 10 6 ohm h ic =KbQ- + Af.) 9630 16,650 6820 11,600 5780 10,000 ohm A oc = A ob (l + h le ) 24.4 x 10" 6 39.6 x lO" 6 37.4 x 10" 6 62.5 x lO" 6 54 x 10" 6 90.5 x 10" 6 mho Ke =/i ib /i ob (l +h {e )-h rb 5.5 x 10" 4 11.7 x lO' 4 5.5x10^ 12.2 x 10" 4 6.4 x 10" 4 15 x 10" 4 T b = K-e ~ T e (l+h le ) 2230 2750 1710 2450 1330 1500 ohm r e = h re /h oe 23 1 30 15 19 12 17 ohm To determine the d-c bias at the base of the first transistor, point A, apply Thevenin's theorem to the voltage divider consisting of /?, and R 2 . The equiva- lent source voltage and resistance are V - R i V cc R e q R t + R 2 2.27 x 20 1.78 + 2.27 1.78 x 2.27 11.2v, 1MQ. R t + R 2 1.78 + 2.27 If base current is neglected, the potential at point A is 11.2 v. Between point A and the output V , there are two base-emitter drops in series. These are approximately 0.6 v each for silicon transistors at room tern- Multi-Stage Amplifiers 123 perature. Thus the potential at R L becomes 11.2 - 2(0.6) = 10 v. For R L = 2500 fi, the d-c bias current is 10/2500 = 4 ma. From Appendix A, h FE at 4 ma is 300. The base current I Bi is , 4 x 1Q- 3 l B , = = 13.3 ua. 1 300 Base current l B is, of course, the emitter current I E of the first transistor. For this current, h FE is approximately 150, so that 13.3 x 10" 6 „ - „„ This base current (previously assumed negligible) somewhat reduces the po- tential at point A. Since the Thevenin equivalent source impedance at A was calculated to be 1 Mil, the additional drop due to I Bl is about 0.09 v. The po- tential at A is more accurately 11.2 - 0.09 = 11.11 v. Again referring to Appendix A for base emitter drop, and assuming collector and emitter currents to be essentially equal, V BEl ~0.5v at / Cl = 13 pa, V BE2 = 0.61 v at I Cl = 4 ma, so that V Ei = 11.11-0.5-0.61 = 10 v. The first estimate of V E = 10 v is valid, so the above calculations need not be repeated. What we have accomplished thus far is a necessary first step in all amplifier calculations, namely, the establishment of d-c operating (quiescent) levels. We now have to determine from the transistor curves the small-signal fi-parameters corresponding to these levels. For the present problem, these parameters may be obtained from Table 6. 1. Now we calculate input impedance R; 2 of the second stage, which is easy since the load impedance R L is given: Ri, = &ic- hlch " • [5-28] Substituting numerical values from Table 6.1 and Fig. 6.2: R, = 6300 (-&V0-) = sge 300 Q , 20.8 x 10~ 6 + 0.4 x 10~ 3 which becomes the load impedance of the first stage, thus making it possible to calculate its input impedance: R. = 420,000 ( ~ 2 ° 1)(1) = 20 MO . ~ .— 1 9 x 10 -6 + 886,300 This is the basic a-c input resistance of the first stage. It is shunted by the bias resistors, R l and R 2 , which constitute an equivalent 1 MH shunt. The net input resistance of the first stage is Ri= ^p Mil =950,000 11. Now calculate A v , the gain from point A to R L . We obtain this by calculating separately the first and second stage gains, A Vl and A Vj , and multiplying: 124 Transistor Circuit Analysis A v = A Vl X A v ^. Again, using the formulae from Chap. 5 (after some simplification), 1 1 1+5" 1+ 2100 = 0.9976, [5.31] R L 886,300 A V2 = L— = 0.993, 2500 A r = A Vl x A Vi = 0.9976 x 0.993 = 0.9906. This figure is modified by the loading effect of R, on the generator source im- pedance R g . The loss in gain is 2000/950,000 = 0.00210. Including this attenua- tion component, the overall gain becomes 0.9885. To calculate the output impedance, proceed from the input to the output: *»!= TIT- [5 - 29] Substituting numerical values applicable to the first stage, R Ql = = 2060 0. 9 x 10~ 6 + ?2I 0.42 xl0 6 + 2000 The output resistance of the first stage R 0i becomes the equivalent source im- pedance for the second stage: *o 2 = — = 24.8 Q. 21 x 10" 6 + _ 6.3 x 10 3 + 2.06x 10 3 This is a very useful amplifier circuit, with an input impedance of about 1MO, and a 25 output impedance. Voltage gain is approximately 1.2% less than unity, making this two-stage device well suited for good isolation of source and load. 6.2 Capacitor Coupling In previous chapters, coupling methods were essentially ignored. Coupling and by-pass capacitors were assumed to be infinite, thereby providing zero impedance to a-c signals. In actual practice, however, coupling and by-pass capacitors introduce limits to the low-frequency amplifier response. These limits are now considered. Refer to Fig. 6.3, which shows a two-stage common-emitter amplifier using both capacitors for interstage coupling (C, and C 2 ), and by-pass capacitors (C E and C Ej ). The coupling or blocking capacitors prevent undesired d-c coupling between the bias circuits of the separate stages. The by-pass capacitors allow separate adjustment of d-c and a-c circuit parameters. As an introduction to the design characteristics of coupling circuits, it will be assumed in Fig. 6.3 that all capacitors, except the interstage coupling capacitor C„ are infinite. PROBLEM 6.2 In Fig. 6.3, with C El , C Ej , and C, infinite, analyze the effect of C, on the low-frequency response of the amplifier. Multi-Stage Amplifiers 125 ■OVr VAO[f\j z„ c 2 01 v A l(-t" Fig. 6.4 Equivalent circuit for in- terstage coupling of amplifier stages of Fig. 6.3. Fig. 6.3 Two-stage amplifier with coupling capacitors. Solution: First calculate the output impedance Z 0l of the first stage, and then obtain the input impedance Z« a of the second stage. With this information, the effect of a finite C, is easily found using simple circuit theory. The values of Z„ and Zj are readily determined by the formulae of Chap. 5, following exactly the procedures set forth. In Fig. 6.3, Z is the output impedance seen at point A , and Z t is the input impedance at point B. The equivalent circuit based on the above impedances is shown in Fig. 6.4. The following relationships are evident: V B = z>> v AO z, 2 + z 0l + 1 jcoC 2 Ztja c 2 (Z, 1 + Z 0l )iaC 2 +l Since Z l% and Z are resistive, let Z, a = R l} and Z 0l = R 0l . The time constant is r 2 = (K,- a + K 0l )C 2 (6.1) Letting <u 2 T, 'B _ V AO R h + R 0l oj a 1 + ; — This expression points up the effect of frequency variation in a particularly con- venient manner. At a « o) 2 (low frequency), V a O «2 \ R I 2 + R oJ Recalling that w 2 = l.'[(R, 2 + R 0l )C 2 ], "ao 126 Transistor Circuit Analysis At the intermediate frequency where <u = <y, , Vao i + ; At higher frequencies, co » (o t , *>> 2 R, a + R 0l Z45° Vao Asymptotes &) 2 (corner frequency) Fig. 6.5 Gain characteristic of in- terstage coupling network plotted very conveniently on log-log coordinates. R lt + R 0l The higher frequency characteristic corresponds to the condition where the re- actance of C a is negligible. Figure 6.5 shows the frequency characteristic on conventional logarithmic coordinates. This universal curve is applicable to any coupling circuit of the configuration considered here. The corner frequency corresponds to <u a in the above problem. The above analytical techniques are equally applicable in considering the effects of C„ assuming C 2 infinite. In place of R 0i and R ti , we use R e and R t , respectively. The voltage V 6 becomes attenuated and shifted in phase at the base of Q l in Fig. 6.3. The effect of C, would be plotted as in Fig. 6.5. However, there is a secondary effect, since a finite C, (acting like part of R e ) must introduce a complex component into Z 0l , the complex output impedance of the first stage. This complex component must affect the behavior of the inter- stage coupling network, and, in effect, it propagates through the entire circuit. This inconvenient feature is unfortunately common in transistor circuits. An exact analysis of the low-frequency response of such circuits becomes extremely tedious. Thus, notwithstanding inaccuracies at frequencies in the vicinity of and lower than the corner frequency, it is most practical to consider the effects of each coupling capacitor separately, and add the separate attenuations and phase shifts as though each were independent. In the higher frequency regions (w » &>,) where the amplifier is normally used and phase angles are small, the approximation approach is very satisfactory. The exact alternative, which is ex- tremely tedious, must be used if greater accuracy at very low frequencies is essential. PROBLEM 6.3 Refer to Fig. 6.3 and assume the following circuit parameters: R tl in parallel with R l2 = 1 MO, h le = 2200 0, Rt = lKQ, A/ e =290, C Bi'C Bi = C^oc, ft,e = 2xl0 -4 , r l, = R Ll = 5 K(2, h oe = 30 x 10" 6 mhos. What is the value of C 2 if gain is to be 3 db down at co = 500 radians/sec? Solution: Calculate R, t and R 0l , using the formulae from Table 5.1: R,-h lm - h " h » =2200 29 °( 2 * 10 ~ 4 > 1 30 x 10 -6 + 200 x 10" 1950 fi . hoe + r L Now find R n S-t- h oe - b " b " 3Q xl0 -«_ (290)2x10 fi, e + R 8 2200 + 1000 - = 84,400 0. Multi-Stage Amplifiers 127 This is in parallel with R Ll = 5 KQ, so that , = 5x84.4 g47Kfl 1 5 + 84.4 Now examine the universal curve (Fig. 6.5), noting that the gain is down by 3 db at the corner frequency where co = co 2 . The expression for co 2 is 1 [6.1] C 2 (R, 2 + R 0l ) Substituting numerical values and solving for C 2 , 1 o> 2 ( R i 2 + RoJ 1 C 2 = -v. -VW, o 500(1950 + 4700) = 0.3/itf. This is a typical coupling capacitor value in a common-emitter circuit. If R Ej were not by-passed, R, a would be larger and C 2 smaller, but this also results in a loss of gain. PROBLEM 6.4 For the circuit of Fig. 6.6, find C 2 , such that the low-frequency 3 db attenuation point is located at co = 500 radians/sec. Use the transistor parameters given in Fig. 6.6. Solution- Refer to Table 5.1 for the formula for input impedance of a common- , . ... h ib = 7.57fl - base circuit: h fb =-0.996 D h hfbhrb h. = 0.103 X 10" 6 mho R, = Aib -• h ° r b b = 27 x 10 - 4 ob+ R~ t> Fig. 6.6 Common -base transistor circuit fed with a constant cur- Substituting numerical values, rent emjner bias 0.996x0.27 x 10" R ,„, U.WQXU.Z/X w = 7>70 Qi ,_ 0.103 xl0- 6 + 200 xlO" 6 1 Since co, = , 7.70 C 2 C 2 = = 260 iii. 2 500 x 7.70 The low input impedance of the common-base circuit leads to an inconveniently high value of input coupling capacitor. PROBLEM 6.5 Referring to Prob. 6.3, determine C 2 for a 3 db low-frequency attenuation at co = 500 radians /sec, omitting by-pass capacitor, C Ej . Use the tee-equivalent circuit with the following parameters: = 290, R Ll =R L2 = 5KO, r e = 6.67fi, Rj|Ria = Kail|K M = 1Mil > r c =9.7MSi, R e = lKQ, f b = 26011, R Ea = 1KQ. Solution: Again use Table 5.1 to determine R h : -*28 Transistor Circuit Analysis 1 + Rl *i a = r b + tf(l +J 8) rd(1+ ^. l + 5t±£? Substituting numerical values, R, 2 =-265KQ, which in parallel with the equivalent bias network resistance of 1 MQ leads to a net Rj = 209 KO. From Prob. 6.3, /?o\ = 4.7KQ. Therefore , 500 [6.1] C 2 (209,000 + 4700) C 2 =- 0.0094 ni. With R E2 unby-passed, C 2 is conveniently small. However the second-stage gain is considerably reduced. PROBLEM 6.6 Refer to the two-stage common-emitter amplifier of Fig. 6.7. Calculate C 2 so that low-frequency gain is 3 db down at co = 500 radians/sec." Also calculate the voltage gain from point A to V . Use approximate methods where applicable in order to simplify calculations. Solution: The starting point is the calculation of R 0l and R h . We may make some simple approximations without significant loss of accuracy. Generally, for R L - 5KO, Ri 2 ~r b + (r e + R E )(l + /S). Estimate the value of r e from the d-c emitter current, l Ej . The bias at point B is estimated by considering R t and R 2 as components of a voltage divider across V cc : Bias at B = — ^— V cc = -JP_ x 25 = 5 v. R t + R 2 80 + 20 (This neglects base current in Q 2 .) Assuming a 0.6 v base-to-emitter voltage drop in Q 2 <a value characteristic of silicon transistors at room temperature), Ve 2 = 5- 0.6= 4.4 v. Emitter current is readily found: From (1.11), / 4.4 Ie 2 = TTtz ~ 2.9 ma. 500 + 1000 °-° 26 = 26*9 0. Ir 2.9 This value may indeed be neglected when added in series with the unby- passed component of emitter resistance R E2g . Also, r 6 is a negligible portion of R^. Hence, R> 2 = R E2a (1 + ft = (500) (101) = 50,500 . For the output impedance of the first stage, it is usually safe to ignore the output impedance of the transistor entirely and let R = R L = 5KO • Multi-Stage Amplifiers 129 © V r 0V V cc =25v R B = 680KQ C 1 = C Bl = <*> R g = lKQ C 2 (to be determined) Re 1 = 1.5 Kfl j8 1 = /3 2 = 100 R E2a =500Q R, =20Kfi r'l^Rl^SKQ Fig. 6.7 Two-stage common-emitter amplifier with emitter bias resistors only partially by-passed. 5Kfl AAAr Hf 50.5KQ -vs* — i ^JVxo Fig. 6.8 Simplified Thevenin's equivalent of interstage coupling circuit. (See Fig. 6.7.) The interstage equivalent circuit takes the form shown in Fig. 6.8. Note that resistor R p represents the bias resistors /?! and R 2 in parallel: Rp= 20x80 K = 16Kfl P 100 This must be added in parallel with R l% : 16 x 50.5 „ 66.5 12 KQ. To find the 3 db point on the low-frequency gain characteristic, 1 C 2 = C,(12,000 + 5000) 1 500, 17,000 x 500 = 0.12 /if. Returning to Fig. 6.7, the next step is to calculate the voltage gain from point A to V . The gain from the base of Q 2 to V is given approximately as 5000 = 10. [3.52] R E 500 (Note that if the second stage emitter by-pass capacitor were not "infinite" at this frequency, the input impedance would be complex, and even approximate calculations would be difficult.) _ At the corner frequency co 2 , the network attenuation is v2/2 = 0.707, so that voltage gain from A to V is 0.707 x 10 = 7, at a leading phase angle of 45°. PROBLEM 6.7 For the two-stage common-emitter amplifier of Fig. 6.7, let C 2 = oo and C Ej = 5 fii. All other parameters are as in Prob. 6.6. Calculate 130 Transistor Circuit Analysis = 50Kfi Rs 2fc (i+/3) = 100Kll Fig. 6.9 Equivalent emitter ci rcuit impedance. (See Prob. 6.7.) voltage gain from point A to V as a function of frequency. Specifically, calcu- late gain for a = 300 rad/sec, &> = 0, and a> = °°. Plot on logarithmic coordinates so that the frequency response shows the characteristic low- and high-frequency asymptotes. Use approximation methods where possible to simplify calculations. Solution: As before, the problem must be set up in terms of the output im- pedance of the first stage, and the input impedance of the second stage. The output impedance is approximately equal to R Li = 5KI2, as in the previous prob- lem. The input impedance to the second stage is approximately equal to the emitter impedance multiplied by /3 + 1 (see Table 5.1). The equivalent input im- pedance is represented in Fig. 6.9, where the relatively small emitter resistance is neglected. Figure 6. 10 shows the circuit of the interstage coupling elements. Resist- ance R p is the equivalent parallel resistance of the two bias resistors. This circuit allows the calculation of base current I bl in Q 2 , from which collector current and gain may be calculated. Calculate the impedances of Fig. 6.10, and simplify. The emitter impedance of Q 2 is (l+£) This simplifies to + (! + £)*-. Fig. 6.10 Interstage coupling cir- cuit set up for calculation. (See Prob. 6.7.) (l + £) '- 2b l+jcoC E R E + a + P)R B 2a 2b Apply Thevenin's theorem to Fig. 6.10, replacing the network driving the above impedance with a simpler equivalent series network: «e q =/?pl|i? Ll R P + R L ' r., = v A The current l bi is easily calculated: Re Rp + R Ll I b , R, R etl + (1 + ®R E2a + (1 + 0) - — _£ 1+lcoC 2b Ve q (l + Ja>C E R E ,J (1 + /9)K E2b + [1 + jo>C E2 R E2b ][R eq+ (1 + &R B J This can be put into a convenient standard form for plotting: V A R p (l + j(oC E2 R E2b ) f (R p + R L t ) [R eq + (1 + 0) (R E 2a + R E J\ b 2 1 + 2? eq + (l +J 8)R E2a "I jcoC E2 R E2b lRe q + (l + P)(R E2a +R E2b ) Let (6.2) (6.3) CO a = Re q +(l+P)(.R E2a +R E2b ) [« eq +(l + 0)i? E2a ](C B2 /? E2b ) (6.4) Multi-Stage Amplifiers 131 COb = B = C E 1 R E 2b (ftp + Rl. 1 )[*.q+ (1 + ff)(R E2a + R E2b )] Kn Substitute these new parameters in (6.3); 1b. . . CO 1 + ; — 1 w t B 1+/-SL 6>a (6.5) (6.6) (6.7) This is the desired form for plotting the frequency response of a network. Ap- pendix C describes the plotting techniques based upon the use of asymptotes on log-log coordinates. Now continue by determining V Q as a function of l bl to arrive at an expres- sion for gain as a function of frequency. From Table 5.1, /3/ba 'c s = 1 + Rl ~&t /S/ b2 i?r (6.8) Substituting (6.7) in (6.8), j8«L 2 B n ■ CO \ COb (6.9) This is the required expression for gain as a function of frequency. The plot of V /V A on log-log paper appears as shown in Fig. 6.11. This is a generalized plot of amplifier low-frequency gain as affected by capacitor C Bl . Of course, at very low frequencies, the coupling capacitors of Fig. 6.7, presently assumed infinite, become significant. Now substitute numerical values from Fig. 6.7 in the expression for gain: R p = fljflj = 16 KQ (as determined in Prob. 6.5), Hence, K eq = K P II R Ll =^|=3.81KO, lo + o B , ( R P + R lJ [i?eq + (1 + fi)( R E2& + R E2b) ] . B = Hi 5 - [3810 + (101) (1500)] [6.6] 16 21 16 x 155,300 = 204,000, 10 6 cab = R E 2b C E 2 1000 x 5 = 200 rad/sec, Ke q +(l+|8)(R E2a +K E2b ) [R eq + (l + /3)K E2a ]C E2 i? E [6.4] Fig. 6.11 Gain vs. CO on log-log scales. The asymptotes are shown as dashed lines; the actual curves are shown as solid lines. (See Prob. 6.7.) Note that V A CO |6Rl 2 "b B , co ' l +; — 2b 132 Transistor Circuit Analysis Substituting previously determined values for the numerator and 1/(C E R E ), the expression for co a becomes co a = 155,300 x 200 The bracketed term in the denominator is 3810 + (101)(500) = 54,300. Therefore, 155,300 x 200 „ n . , Wa = ' „ nn = 572 rad/sec. 54,300 Substituting the above values in the expression for gain, , . co , . CO Ys. = & L > 2>6. = (100) (5000) 200 Va b l + j^L ™ 4 >°° i + ;_^L co a 572 [6.9] i + ; 2.45 200 572 This is the final numerical expression for gain variation in the region where Ce 2 is most sensitive to frequency change. The responses for co = 0, co = 300 rad/sec, and <u = °o will now be calculated: At co ~ 0, V -?- = 2.45. At co = 300 rad/sec, As co = ^=2.45 20°- V A 300 572 Ik V A = 2.45 x 1.6= 3.92. £ * 2.45 x??l =2 . 45>< 572 Ka J_ 200 572 Refer to Fig. 6.11 for the plot of gain vs. frequency. The preceding problems have shown how C„ an interstage coupling capaci- tor, and C Bi , an emitter resistor by-pass capacitor, individually affect frequency response. For reasons which have been previously explained, the separate ef- fects cannot be superimposed unless the significant frequencies associated with each capacitor are widely separated. A really accurate investigation of the com- bined effects of two capacitors, such as C, and C Bj , requires an equivalent circuit analysis of both stages together, taking into account all interactions. Multi-Stage Amplifiers 133 PROBLEM 6.8 In the circuit of Fig. 6.7, if C 2 and C Ej = 5 /if, C t = 50 /if, and all other values are unchanged, show the exact tee-equivalent circuit for determining Po Ab- solution: Figure 6.12 shows the common-emitter tee-equivalent circuit to be used for computing the gain, and the numerical values of the circuit parameters applicable to this problem. VW — <>— V^~"° — ° v ° A = A = 10 ° R R = 680KQ R ?-- = ikQ R F = = 1.5 KQ R E 2a = R E 2 b = = 500 fi R t.= = Rl 2 = 5 c.= = 50/if c 2 = = 5 /if c E2 = 5/zf r b = 30011 r e = 9fi r d = 100 Kfl 5KQ ■-»► Oi Fig. 6.12 Equivalent circuitof two-stage amplifier. (See Prob. 6.8.) Note that r e «K E , r b «Rj, and r d » R L so that the transistor parameters can be neglected; hence, R t + R 2 PROBLEM 6.9 Using the circuit (Fig. 6.12) of Prob. 6.8 and formulae from Table 5.1, calculate gain V /V e as a function of frequency. Make assumptions and suitable approximations where necessary to simplify calculations. Consider the frequency range from 1 cps to 10,000 cps. Solution: Calculate the input resistance of the first stage. by the second stage, Neglecting loading R ll £r b + (r a + R Bl )(l+P)- = 300+ (1509) (101) 1 + 1 + 0.0005 1 + 0.06 R L +R E + r ( 144 Kfl. (1 + /3) Note that the terms in the expression for R tl , which include R L , actually have less than a 6% influence on the result of the calculation. It is not neces- sary to know R L to a high degree of accuracy in order to determine R v This is fortunate since the effective R L is equivalent to R Ll in parallel with the effec- tive frequency-sensitive impedance to the right of R Ll in Fig. 6.12. An exact computation would be extremely tedious. 134 Transistor Circuit Analysis (a) R eq =5KQ eq- -VV/Vr 3.13 V g G (b) Fig. 6.13 (a) Equivalent input cir- cuit, (b) Thevenin's equivalent cir- cuit for the first amplifier stage. (See Prob. 6.9.) For R ix = 144 KQ, we can calculate the attenuation characteristics of the first stage input circuit. Refer to the equivalent input circuit in Fig. 6.13a. The gain of this circuit falls 3 db at that frequency where the capacitive reactance equals the equivalent series resistance: + R il R B co C t R^ + R B Now substitute numerical values and solve: -L = i + (|» = 121KQ . <y C\ 144 + 680 For C, = 50 iii, oi = 2 nl = 1/6 rad/sec and f = 0.027 Hz. At / = 1 Hz, the capacitive reactance is relatively small (3170 Q). Its attenuating effect on gain is therefore negligible in the frequency range of interest. Thevenin's theorem is readily applied, replacing the first stage by its open- circuit output voltage, V A , and its equivalent output impedance, R 0l : R 0l ~r d (l + Substituting numerical values, R 01 ~ 100,000/1 + 1 + R e + Tb r e + R El 100 \ 1 + 1300 j 1509/ = 5.5 MQ, This is much greater than R L t , and the effective output impedance i? eq is there- fore approximately equal to R Li = 5 Kfl; hence, voltage gain from (3.52) is A.-&S ** V D T e + R Ei 5000 1509 3.15. This figure for voltage gain must be multiplied by the attenuation of the input network to determine the over-all gain of the first stage: V D = V eq R t RiAR B + R e where R, t \\r b is the equivalent parallel resistance of R tl and R B . Therefore, *-*i£-"«*- Substituting, Va (open circuit) = 3.15x0.995 = 3.13. The first stage can be replaced by the equivalent network shown in Fig. 6.13b. Now calculate the input impedance of the second stage: z i 2 = T b + R E 2 B +R E 2b jco C J (1 + /3) Rl + Re 2s + «E 2b + R r 1 + Multi-Stage Amplifiers 135 The term in the last pair of brackets is 1 + 0.0005 5500 + 1 + The expression **2b 1 + j<oC Ei xR E2b 99,000 '2b 1 + joC B x R E 2b can never exceed R E2b = * KQ. Thus, the entire bracketed expression merely introduces an approximately 6.5 % reduction in the value of Z, 2 . In the interest of simplicity, this entire bracketed expression will therefore be neglected, so that Zi 2 ~r b + R* + R, 2b 1 iaC, Bi (1+0). The simplified equivalent circuit for the two-stage amplifier is shown in Fig. 6. 14. The voltage at point A is the output of the first stage when loaded by the second stage. It is straightforward, though somewhat tedious, to determine l b% over the frequency range. From / bj , the output voltage V ot and therefore the over-all voltage gain, are determined. As a good approximation, current gain of the second stage is (Table 5.1) Ai P r rf so that A, 100 1 + 0.05 = 95, Ys. = A,R L = 95 x 5000 = 475,000. The frequency dependency of the multistage amplifier is now confined to the variation of l bj , determined from Fig. 6.14. To aid in carrying out the calcu- lations, note the principal circuit time constants. As a first approximation, the loading of the first mesh by the high-impedance second mesh across R p will be neglected. The first mesh time constant T a equals (R eq + R P )C 2 ~ 0.105 sec. The time constant T b = C E2 R E2b = 5 x 10 -6 x 10 3 = 0.005 sec. 3.13 Rfi 2a (l+/8) = 50Kfi AAAr ioo kD Fig. 6.14 Simplified equivalent circuit of two-stage amplifier. (See Prob. 6.9.) 136 Transistor Circuit Analysis The wide difference in time constants permits an analytical simplification. At very low frequencies, where T a is in the range of interest (co = 10 rad/sec), the time constant associated with C El has negligible influence. Capacitor C El may well be considered open-circuited. At higher frequencies, where T b is in the range of interest (co = 200 rad/sec), the reactance of C 2 may be considered negligible. The circuit therefore can be treated as though it contained two separate and isolated time constants. The circuit of Fig. 6.14 is most easily solved by considering the various frequency ranges of interest separately. At very low frequencies, C E is es- sentially open-circuited, and 3 13 V* h~ ±~* = 3.1370, C 2 V e , JcoC 2 '*, = / R, ■■ 3.13 V g jco C 2 R P R P + (R E2a + RE 2b )(l + ® 3.13 V 6 jco x 16,000 x 5 x 10"' 16,000 + (101) (1500) 1.5 jcoV efia . At a somewhat higher frequency, where C 2 is negligible while C El is still re- garded as an open-circuit, I bi is easily calculated from the all-resistive network as/ b2 = 15.4 V gi ia. At higher frequencies, both capacitors act as short-circuits. Solving again f or/ 6 „ 3.8K12 -AAAr 50Kfi -AAAr- {r^J ' Ib > ) cos ^f^ Fig. 6.15 Simplified Thevenin's equivalent circuit, with re- actance of C 2 essentially zero. (See Prob. 6.9.) /* 3.13 V e 16 ~ z 5000 +16,000 || 50,000 50 = 58.1 V g fia. Consider now the intermediate frequency range, where C 2 is essentially a short-circuit, while C Ei is not. Apply Thevenin's theorem to eliminate the first mesh. Figure 6.15 shows the simplified form where the reactance of C 2 is taken as zero. This circuit is easy to solve for I bi . First determine the impedance of 0.05 /zf in parallel with 100KQ: 2 P = 100,000 x - jco 0.05 x 10~ 6 100 Kfi 100,000 + 1 jco 0.05 x 10" 100,000 1 + jco 0.005 Now solve for I b /„ = 2.38 V„ 53,800 + 100 ' 000 1 + jco 0.005 15.4 7^(1 + jco 0.005) 1 + jco 0.00175 tia. Multi-Stage Amplifiers 137 Recalling that V /I bl = 475,000, we now have sufficient data to sketch the amplifier attenuation characteristics. This is most easily done by a plot on logarithmic coordinates as in Fig. 6.16. Note that this problem was substantially simplified by the separation of time constants, T a and T b . This allowed us to consider the separate frequency re- gions independently. If these were not separated by about a factor of 10, more precise calculation methods based upon the formal solution of the network of Fig. 6.14 would be required. A more accurate solution is necessary only when a simplified sketch based on asymptotes and corner frequencies is insufficient. Usually, the simplified sketch is quite satisfactory. PROBLEM 6.10 If, in the two-stage amplifier of Prob. 6.9, C 2 is changed to 0.25 j^f, all other parameters remaining the same, find the attenuation character- istics of the amplifier over the region where C 2 and C Ei have significant re- actances. Solution: The time constant due to C 2 has been reduced by a factor of about 20, so that both capacitors influence amplifier gain over approximately the same frequency range. An accurate loop analysis is necessary to determine the effec- tive circuit time constants. Referring to Fig. 6.14, the mesh equations are 3.13 V t = 21,000 + 10 6 0.25 j co I, - 16,000 l„ 2 , = - 16,000 / t + 66,000 + 100,000 1 + jco 0.005 These equations may be solved for / bj by eliminating /, (or with determi- nants), leading to the following simplified solution: l b2 = 0.075 V 6 \ 200, \ 192/ \ 612/ na. Calculating output voltage, V e =A,R L l bt = 9Sx5QQOl ba = 475,000 /„,. Substituting the expression for / bj , and solving for voltage gain, = 0.0355 io> [l + ; 200 ( 1 + jJl)( 1 + jJ!l) \ 192/ \ 612/ This is the required expression for gain as a function of frequency. 10.25 200 572 0.746/« 1 + ; 200 i + ; 10.25/ \ l +i 572 Fig. 6.16 Asymptotic diagram and accurate curve of output voltage vs. frequency. (See Prob. 6.9.) Certain deductions may be drawn by observing and comparing the results of Probs. 6.9-10. The high frequency gain, where capacitive reactances are negli- gible, is unchanged by our approximations. The approximate method of con- 138 Transistor Circuit Analysis sidering corner frequencies individually, leads to somewhat erroneous values for these frequencies, but the inaccuracies are rarely important in practical amplifier 6.3 Transformer Coupling Transformer coupling between the stages of a transistor amplifier offers significant advantages such as: 1. Good bias stability. 2. Simple impedance matching for optimum power gain. 3. Isolation of stages. The disadvantages are: 1. Relative high cost. 2. Bulk and weight. 3. Limited frequency range. 4. Nonlinearity, due to nonlinear magnetic core. The most significant advantages have to do with impedance matching. The equivalent circuit of the transformer is a linear network which represents the transformer in a convenient manner for purposes of calculation, as illustrated in Fig. 6.17. The figure also defines all symbols for use in this section. The idealized output transformer at the output terminals of the equivalent circuit pro- vides the required amplifier impedance matching. Fig. 6.17 Trans former equivalent circuit. Note that R t = primary d-c resistance; R 2 = sec- ondary d-c resistance; n = primary /secondary turns ratio; L t = primary inductance; K = co- efficient of coupling of primary to secondary; KL 1 = mutual inductance, primary to second- ary; (1 — K)L l = leakage inductance, primary or secondary; R^ - equivalent core loss re- sistor; C w = equivalent distributed capacitance referred to secondary winding. Leakage in- ductance L t is measured with the secondary open-circuited; secondary inductance L 2 can be measured with the primary open-circuited: L t = n 2 L 2 ; R L is relatively independent of frequency except at frequencies in the low audio range; K is usually slightly less than unity. PROBLEM 6.11 For the circuit of Fig. 6.18a, determine the variation of gain with frequency. (The transformer is effectively coupling a driving source R a to an output load R t .) Solution: In complex problems of this type, it is most convenient to carry out separate calculations for the low- and high-frequency regions. Consider first the low-frequency region. Since K = 1 for a reasonably efficient transformer, uMl -!:)«£„ n a R 2 at very low frequencies (<u small); thus the series inductance terms may be neg- lected. Similarly, for very small o>, the R L term may be ignored in comparison with the reactance of KL X . Also, the capactive reactance C„ becomes very large, Multi-Staga Amplifiers 139 Li(l - K) R 1^(1 - K) n 2 R 2 Fig. 6.18 (a) Equivalent circui t showing an interstage transformer coupling the output of one stage to the input of the next, (b) Simplified equivalent circuit for low-frequency condition. and may therefore be neglected. The simplified equivalent circuit corresponding to low-frequency operation takes the form of Fig. 6.18b. Now using Thevenin's theorem, _ - xn 2 R, (- V 0l (JR i + R 0x )(}K<oL l ) (R l + R 0i ) + ;KftiL 1 + n*<R, + R h ) jKoiLytiR! («i + RoJQKcoLJ + n'OJj + R^fiR, + /?„,) + jK<oL l ] jKci)L 1 nR i n 2 (R t + R^XR, + R 0i ) + jKa>L l [n\R x + R, 2 ) + (R, + R 0l )] jKcjL.R^ n{R 1 + R t )(R 1 + R 0l ) 1 + jKcoL l R l + R 0l n 2 (R 2 + R i2 ) ]KaL t R. nd^ + Rj^i + R,,,) 1 + / * l [Cfix + R^nHRa + Ria) . (6.10) Equation (6.10) shows the output increasing uniformly with frequency, and then leveling off as the second term in the denominator becomes dominant. The comer frequency, where attenuation begins to level off with increasing frequency, oc- cuts where Above this comet frequency, the frequency-sensitive term in the denominator of (6.10) becomes dominant. In this intermediate or mid-frequency range, "i, nR, V 0i (R 1 + R 0l ) + n 2 (R 2 + R il ) The equivalent circuit for the mid-frequency range is given: in Fig. 6.19, which yields V, /V almost by inspection. Frequency response is flat in this region, which is generally the useful operating region of the transistor amplifier. 140 Transistor Circuit Analysis K„i + Ri / R °i$ /j? niR i fii. -n'R, n: 1 Fig. 6.19 Simplified circuit showing transformer coupling. Fig. 6.20 Interstage-coupling equivalent circuit for high-frequency condition. Loading on the transformer output terminals is assumed negligible. Resonant peak due to tuning of C w Fig. 6.21 Gain characteristic of a transformer coupling network. The high-frequency region of the frequency-response characteristic is dis- tinguished by the increasing importance of the series inductors, L x (1 - K). The shunt inductor KL l leads to a very high reactance at high frequency, and may be neglected. The equivalent circuit for high-frequency conditions is shown in Fig. For simplicity, capacitance C„ will be neglected for the present. This is usually a practical assumption in transistor circuits, as R, generally provides a relatively low-resistance shunt at the upper frequencies of typical matching trans- formers. From the equivalent circuit, the L/R time constant is determined by inspection: 2(1 -K)L, (6.11) R « 0l + *, + «' (a,, + jy" The corner frequency corresponding to this time constant is With the upper and lower corner frequencies determined, as well as the level of the intermediate frequency region, the gain characteristic may be plotted on logarithmic coordinates as in Fig. 6.21. Sketches of both asymptotes, as well as the general shape of the gain curve itself, are shown. Thus far C w has been ignored. However, it is possible for C w to series-reso- nate with the equivalent transformer leakage reactance somewhere in the high- frequency region. Should this be the case, the gain curve is modified to show a characteristic resonant peak (Fig. 6.21), whose amplitude is determined by the "Q" of the circuit. Above this resonant peak, the shunting capacitor acts to at- tenuate the output even more rapidly than before. Commercially, K, n, and L, are not generally given in transformer catalogs. However, it is often possible to estimate these parameters. For a matching trans- former, n = (6.12) where R, and R L are the driving and load impedances, respectively, to be matched. If the lower corner frequency a L is given, for K - 1, R, = R 2 = 0, and letting R Bl - R„ Ri t - Rl (see Fig. 6.18b), R. Rl n 2 Rl R„ u L = L, (R. + n I R L )L l 2R„L l 2 L t 2 £u L * (6.13) Multi-Stage Amplifiers 141 PROBLEM 6.12 A manufacturer's catalog lists the following data on a trans- former: R s = 10,000 Q, R L = 100 a co L = 500 rad/sec. Estimate h x and n. Solution: From (6.12) and (6.13), 2 a,. 2x500 Resistors J? t and /^ can be measured as the d-c resistances of the trans- former windings, or very roughly estimated as R A 2N929 4 c ma mr—] P = 290 h oe = 33 Kfi Fig. 6.22 Output transformer to match load in a common -emi tter ampli fier. 33 KQ -WAr e j8/b L > 3Kfl (a) PROBLEM 6.13 For the circuit of Fig. 6.22, estimate the required transformer primary inductance L 1 , such that the low-frequency response is 3 db down at 60 Hz. Solution: From Table 5.1, the transistor output impedance is easily estimated: R = h oe = 33 K a. A simplified but satisfactory equivalent circuit is given in Fig. 6.23a. Figure 6.23b shows the circuit rearranged to simplify determination of the corner fre- quency. The effective resistance for the computation of corner frequency is 33 K fi|| 3 K fl, or2.75Kfl. The lower corner frequency, co L = 2 n II = 2 n- x 60 = 377 rad/sec. The time constant of the inductive circuit must equal the reciprocal of co L : 2750 1 377' L t -2750. 7.3 h, 1 377 for I c = 4 ma d-c through the transformer primary. PROBLEM 6.14 Using reasonable approximations for the circuit of Fig. 6.24 at the specified operating conditions, determine (a) R t , for correct bias, (b) C lf for a corner frequency co = 10 rad/sec, (c) output transformer inductance L i for a corner frequency, co L = 200 rad/sec. Assume that the d-c resistance of the transformer secondary is 10% of the load resistance. Solution: For the conditions of Fig. 6.24, l E = I c = 10 ma. The emitter voltage is 1.2 K fl x 10 ma = 12 v d-c. To compensate for the base-emitter drop, the volt- age at the base V B = 12.6 v. L i '2.75 Q, Fig. 6.23 (a) Simplified equivalent output circuit, (b) Simplified equiv- alent circuit for estimating corner frequency. (See Prob. 6.13.) 1.2 Kfi Fig. 6.24 Amplifier circuit with transformer-coupled output. At I c = 10 ma: ^ = 380, r e = 1 1 0, r b = 700 SI, r d = 5250ft. 142 Transistor Circuit Analysis It is convenient to estimate the input impedance R, of the base circuit. From Table 5.1, R, = (1 + f>/e) Re = 381 x 1200 = 458 K fi. Since this high input impedance is in parallel with R,||/? 2 (where /?, = 22 Kft) at the base input, it may be neglected without introducing more than a few percent error. Thus, neglecting base current drawn from the voltage divider formed by i? t and R 2 , it is easy to calculate R 2 : 12.6 = R, R t + R 2 -x24. Since R, = 22 K 0, R 2 is calculated as 24 K fi. We can now determine the value of C, for a corner frequency of 10 rad/sec. The RC time constant must be 0.1, so that * JI *'=ifir 1L5Kfi ' R, = 900 fi -VW- n 2 R 2 = 900O -WV 1 ► 1.18MQ n'R L = 9000 n' Fig equi 6.25 Transformer-coupled valent circuit calculation. (See Prob. 6.14.) c,= 10 11,500 = 8.7 n f. Now examine the transformer in the collector circuit of the transistor. The transistor is effectively a high impedance current source, in comparison with the relatively low load impedance. A simplified but fairly accurate equivalent cir- cuit is shown in Fig. 6.25. The effective time constant is n*(R 2 + R L ) 9900' This must equal the inverse of the specified transformer low-frequency corner: 9900 1 200 20 PROBLEM 6.15 For the amplifier of Fig. 6.24, estimate voltage gain at o = 200 rad/sec. Solution: Neglecting the effect of transformer inductance, from Table 5.1, A = Rl v ~Rl' where R L is the effective a-c resistance in the collector circuit. From Fig. 6.25, this is 10,800 Q. Substituting, A v = ^m = 9.0. v 1200 This, however, is the voltage on the primary side of the transformer, reduced on the secondary by a factor of 30 by the transformer step-down ratio. An addi- tional attenuation of 9000/10,800 is introduced by the transformer winding re- sistance. Further, the gain is reduced by a factor of ^2/2 = 0.707 at the corner frequency, co L = 200 rad/sec. Therefore, the voltage gain at co L = 200 rad/sec is Multi-Stage Amplifiers 143 O)=200 9.0 x-Lx-^-x 0.707 30 10,800 0.176. In the design of signal amplifiers (as contrasted with power amplifiers dis- cussed in Chap. 7), it is not only necessary to verify that gain is adequate, but one must also verify that the required "swing" of the output voltage is restricted to the linear region. This problem is important in the output of multi-stage am- PROBLEM 6.16 For the circuit of Fig. 6.26, design the bias circuit to permit a distortion-free output voltage of 2 v rms, while keeping the stability factor S < 4. Solution: We must determine the range of I c . For Ic f_0, the collector is at a 12 v potential. Recall that 2 v rms corresponds to 2 x 2 \/2 = 5.66 v, peak-to-peak. Thus, the collector potential may be as low as 12 - 5.66 = 6.34 v for peak col- lector current. This value of peak current = 5.66/5000 = 1.13 ma. This reasoning indicates a value of collector bias current of about 0.6 ma, swinging from nearly zero to a value somewhat under 1.2 ma. A load line for this condition is shown in Fig. 6.27. For the 2N929 transistor, a minimum V C e of 1 v assures satisfactory opera- tion (see Fig. 2.5b). Thus, Ve = Vcc - Rl Ic, Substituting values, V B =U-[ 5000 x 0.0006 + - V CE min 5.66 -1 =6.17 v. Assume, as a convenient approximation that 7 B = 6v for the d-c operating level. Since I E =I C = 0.6 ma, + R Eb 0.0006 10 K a. Thus, R E „ = 500 ft and R Eb = 9,500 ft. The base voltage is 6 v + V BE = 6.6 v. Base current is also easily found: /s= 7T Ic 0.6 x 10- 300 -= 2/* a. From (4.16), For S = 4, S=l + R, Re ' R p = 30 K ft. If we assume that the 2 fi a of base current is negligible compared with the current drawn by the /?,, 2? 2 divider, then R, R 1 + R 2 However, Vcc = 6.6 v. RiRi R t + /? 2 R F =500 ft r e =-- lOMft /3 = 300 Fig. 6.26 Load line superimposed on idealized collector character- istics. (See Prob. 6.16.) 1.4 \ 1.2 V b = 2/b ° 1.0 t" \ f 0.8 ! \ o i \ E . 0.6 6 0.4 i !\ 0.2 ! ' \ n ! ! vs = o 2 4 6 8 10 12 V CE , volt — ►• Fig. 6.27 Load line superimposed on idealized collector character- istics. (See Prob. 6.16.) 144 Transistor Circuit Analysis Substituting, Solving, using known values, R *1 2-Vcc = 6.6 v. R, = 54.5 K fl, R, = 66.5 K fi. It may be verified, if desired, that divider current is much greater than base in- put current. PROBLEM 6.17 If, in Fig. 6.26, a large capacitance is connected from the out- put terminal to ground, what is the maximum undistorted rms capacitance current? Solution: Since bias current is 0.6 ma, this is the maximum instantaneous peak collector current that can flow without the collector current actually reaching zero. Therefore, 0J5 = 0.424 ma rms. PROBLEM 6.18 In Prob. 6.13 (Fig. 6.22), what is the maximum undistorted volt- age across the transformer primary? 10 8 6 o 4 35/ia 30 /xa 25fJa 2C X2v^ _ P IS/ia -10 \tar -^£* >9p0 — 5fia- i B =o 10 15 20 V CE , volt — 25 30 35 Fig. 6.28 Collector characteristics with superimposed load line for the circuit of Fig. 6.22. Solution: The transformer primary voltage measured at the collector can theo- retically vary from to 24 v, since the voltage may be either plus or minus (see Fig. 6.28). However, since the transistor drop V CE must not be less than 1 v and the collector current not be allowed to go to zero, the distortion-free primary voltage can typically vary from 1 v to 23 v. Actually, referring to Fig. 6.28, V mln = 0.8 v, so that the full sinusoidal voltage swing is 2 x 11.2 = 22.4 v peak-to-peak, or 7.85 v rms. 6.4 Direct Coupling Direct coupling of amplifier stages leads to the following advantages: 1. It avoids large coupling capacitors, and does not limit the low-frequency response or allow low-frequency phase-shift. 2. Quiescent output voltages provide input bias to subsequent stages, avoid- ing bias networks. This allows higher base-circuit input impedances. Multi-Stage Amplifiers 145 3. Feedback around several stages can lead to bias stability factors, S, less than unity. In this section, we will examine some two- and three-transistor direct-coupled amplifiers, which may then be cascaded, as desired, using a-c coupling methods. PROBLEM 6.19 Making reasonable assumptions, analyze the circuit of Fig. 6.29 for the variation of / c , due to changes in leakage current Icbo with tem- perature. Solution: This is a two-stage direct-coupled amplifier with d-c feedback from the emitter circuit of the second stage to the input of the first stage. Now first examine the nature of the feedback itself. If Icbo, increases, the emitter current of Q 1 increases correspondingly, thus increasing the base voltage of Q x . The base potential of Q t decreases, acting to reduce l B . In a similar manner, the circuit automatically tends to compensate for changes in Icbo.- Note that the by-pass capacitor C E prevents a-c feedback which would reduce amplifier gain. OVr cc M ICBO^ + Pi> + /V B2 Rb, + r b 2 Rf, - Rr + R Bib Fig. 6.29 Two-stage direct-coupled amplifier. To analyze the circuit of Fig. 6.29, use the techniques of Chaps. 3-4 to set up the d-c equivalent circuit of Pig. 6.30a. The circuit is simplified, in that the usual resistances (r^) across the two current generators are omitted without in- troducing appreciable error. Further simplification is achieved by using Thevenin's theorem (see Fig. 6.30b). The circuit equations are l B t Ke 2 + V B e 2 = V a -(Rl, + r ba )l Bl , v v ' V A = Vcc-RlJc 1 , I Cl -/cbo 2 (! + &)_ /e /„ » Similarly, f H = & i + fe- /c *<v <6.14) (6.15) (6.16) (6.17) (b) Fig. 6.30 (a) Simplified d-c equiv- alent circuit of the two-stage am- plifier of Fig. 6.29. (b) Simplified Thevenin's equivalent circuit for the input to Q 2 . 146 Transistor Circuit Analysis , * Re, Ie. - Re. Ie, - Vbb, l "i = „ ' L - (6.18) ^ V J Equation (6.16) may be substituted in (6.14) to eliminate / Bj , and (6.18) may be substituted in (6.17) to eliminate / B| . There remain two equations in unknowns f e, and ls 2 , with Icbo 1 and /cso, as independent variables: / Bj [ RB i + i^o-) + V BE i = V C c -RlJe, ±R*Icbo s , (6.19) « *£, /«, - Fbe, = / El (i^" + *■.) ~ 'c»o, *.* Solve (6.20) for / Ei and substitute in (6.19) to develop a formula for l E + Vbe, (6.20) '«. * E , + r ^_ + . v,« u + ft if/ Fbe, -ry Equation (6.21) contains the desired information relating changes in l E to changes in leakage currents. Use derivatives to get a convenient expression for dI Ej in terms of dIcBo t and <ScBo t - This expression is "*'-. \ * d!cBo x + Rl <Hcbo 2 , R; 1 + /3,, "'"■I mT £^ <" 2 > "-TW «fp_.*|Y PROBLEM 6.20 The component values for the circuit of Fig. 6.29 are R„ = 1000 n, Re 1& = R s 2a = 100 Q *«lb = 1900 0, ^cc = 12 v, / c 1 = 1 ma, Ic The required distortion-free output voltage is 2 v rms, minimum. Determine the remaining circuit parameters to meet the specified requirements. Solution: The emitter potential of Q, is V Ei = 0.001 x 2000 = 2 v, so that V B< = V E , + 0.6 = 2.6 v. Multi-Stage Amplifiers 147 For the 2N930 transistor, h FEl = 280 at 1 ma; therefore, I - 001 o CQ l B = — =3.58ua. Bl 280 p Proceed now to the second stage. A 2 v rms output corresponds to a 5.67 peak-to-peak voltage swing. Allowing a minimum of about 1 v for Vce 2 • Ve 2 = 12- 5.67-1 = 5.3 v, 5.3 _ 5.3 R Ei = = ic In Therefore, Now determine R L : jr e = 5 - 3 1060 fi. 2 5 x 10- 3 ^l 2 fc 2 = 2.84 v (|-of peak-to-peak value). Solving, r l = -2^1=567 0. L * 0.005 To determine a, we must at the same time choose R l for a required stability; i?i is the principal component of R*, the d-c resistance in the base circuit. In Chap. 4, it was pointed out that this resistance must be low for reasonable sta- bility in the first stage. On the other hand, R* acts as a load on the input. A compromise is necessary. Assume, somewhat arbitrarily, that R* = 30 K O. Then, For R* = 30 K Q, a = 0.492. Since V A = Vb 2 = ' e 2 Re 2 + 0.6 = 5.9 v, V C c- lc x Ri,, = 12 - 0.001 R Ll = 5.9, so that R Ll = 6100 Q. PROBLEM 6.21 For the parameters of the preceding problem, determine the sta- bility factors d lg t /d /cbo,> d /e, '3 'cbo,- Solution: From (6.22), the following expressions are derived by simplification: (6.23) *?*. ~<t+&) d $.24) Making the approximation that r b «Rl, and substituting previously derived nu- merical values, — z.i, ° 'cso. 148 Transistor Circuit Analysis <?/ dl 2L— .32.5. CBO. PROBLEM 6.22 Referring to the circuit of Fig. 6.29, develop an expression for the stability of l Bf with respect to changes in Vbb^ and Vbe ,• Solution: By simplifying and differentiating (6.21), the required expression is -dV BBi + \ K CB~ m *i,a+A) *£, 1 + Rt, a *s: L i + /3i, R? d + ft)*./ (6.25) PROBLEM 6.23 For the circuit of Fig. 6.29, using (6.25), determine A I Bi due to changing V BE for a 100 °C change in temperature. Solution: As pointed out in Chap. 1, A V BB - -2.2 mv per °C increase in tem- perature. Substituting numerical values in (6.25), A I E s (-85 + 246) x 10- 6 = 161 ft a. The calculation of multi-stage amplifier performance is so cumbersome that maximum use must be made of reasonable approximations for achieving practical results. Little is lost by |hese approximations, however, since the values of the transistor parameters themselves vary widely with temperature and among tran- sistors. In the following problems, idealized transistors are used, where r B = 0, r c = oo , and r B = 0. PROBLEM 6.24 In the circuit of Fig. 6.31 and its equivalent circuit (Fig. 6.32), determine the stability of the operating point of @j with variations in 1 C bo- Solution: Refer to the equivalent circuit Fig. 6.33 of Fig. 6.31. Let A/ c be the change in collector current due to the change in 1 C bo- It is necessary to determine A / c . = A In. + A l c . .+ A Ir (6.26) R ■fc- 'CC Fig. 6.31 Three-stage, direct-coupled common-emitter transistor amplifier. Rxi I c l \iR Ll R 2 OV, CC ov„ Fig. 6.32 Simplified equivalent circuit of three-stage amplifier for bias calculations. (See Fig. 6.31.) Multi-Stage Amplifiers 149 56.6K12 10Kfi> R Ei-> R o _^ 3K " 1 L-3I Fig. 6.33 Three-stage, direct- coup led common-emitter transistor amplifier of Fig. 6.31, with calculated resistance values and bias voltages. where the separate components are due to changes in I cbo in stages 1, 2, and 3, respectively. Recall that, by definition, A / Cl _, =• S, A Icbo x > The voltage at the collector of Q t is reduced by S! Rl 1 A /cso,» which corresponds to a reduction in the base voltage of Q 3 . Since the d-c input impedance of 0, is approximately equal to Sj Rl x A /cbo, A/- — Since A / Cji = 0, A l Bl , Mb - — X ifiz!_^i_ Rl, */?«,(! + ft)' Therefore, combining terms, A/ Cs _ x = / 8,A/ Sj = This is the change in the collector current of Q s , due to A I cbo • Similarly, the contribution from A 1 C bo is «. A ft A'cso, Re. l + -J(l + fc) 1 + — lou/3,) (6.27) (6.28) (6.29) * 'c,_, = - j8, S a A /cbo, A/ c,_, =S } A/cbo 3 . (6.30) (6.31) The total variation in l C) is determined by substituting in the expressions for the components of A/ c in (6.26); S may be calculated using expressions de- veloped in Chap. 4. Since A/ c is negative, some compensation exists, but this is not significant in practical cases. Note that the principal drift com- ponents are introduced by leakage changes in the first stage. PROBLEM 6.25 For the circuit of Fig. 6.31 and its equivalent circuit (Fig. 6.32), determine the stability of I c due to variations in V BB . Assume l CBO = 0.01 jua. Solution: Proceed from stage to stage, as in Prob. 6.24: 1^0 Transistor Circuit Analysis A 'c,--0. AV EB l R Bi + R El (1 + ft) is the current change in Q, due to A V EB . As before, A'c^-A/c, l Rl i& Rl, + Re, (1 • ft) At the third stage, A/ c .Ale ^ • ^R^-R^d + ft) Combine terms: -ft ft ft J? tl Rl 2 A V EBi A/ t [»■» + *«!(! + ft)] [Rt, + Re, (1 + AM [Rl 2 + Re 3 (1 + ft)] Similarly, we may derive expressions for A l c and l c _ , which may be summed to give the total change, A I c : Air = -A^ft^Rt 2 A7 fl£l ' 3 IRb 1 + Rb 1 (1 + ft)] [R Ll + Re 2 (1 + ft)] [R Li + R Ei (1 + ft)] + ftftRi,, AK BEa ft A V BE3 f«L, + Re, (1 + ft)] [R Ll + Re, (1 + ft)] " R Lj + R Ej (1 + ft)" (6 " 32) This expression shows some compensation. It is possible to take advantage of this by adjustment of parameters. PROBLEM 6.26 Design an amplifier for maximum output voltage, using the cir- cuit of Fig. 6.31. Let l Cl = 1 ma, R Ei = R E2 = R E3 = 3 K «, 'c 2 = 2 ma, R 2 = 10 K Q, lc 3 = 3 ma, C Ei = C El = Ce 3 = 3000 p f , V cc = 24 v. Use approximate design procedures. Considering the sensitivity of I C bo and V BE to temperature, calculate A l Cl resulting from changes in these two param- eters as temperature increases from 25 °C to 100 °C. Solution: Assume base currents are negligible compared to collector currents. Then, V El = I Cl Re, = 0.001 x 3000 = 3 v, V Bl = V El + V BEi = 3.0 + 0.6 = 3.6 v. Calculating R, from the voltage division ratio of the bias divider, R 1 = 57 Kfi. Then, Pe 2 = l Cj Re 2 = 0.002 x 3000 = 6 v, Vb 2 = V El + V BE2 = 6.0 + 0.6 = 6.6 We may now determine R L : Vcc-1 Ci Rl^V Bi . Substituting and solving, R L = 17.4 K 0. Continuing the above with bias calculations, Multi-Stage Amplifiers 151 V E% = I Ci R E3 = 0.003 x 3000 = 9 v, Vb 3 = V E3 + V BEl s 9.0 + 0.6 = 9.6 v, 24 - 0.002 R Li = 9.6, R Li = 7.2 K 0. The choice of R L is based on the requirement for developing maximum out- put. We require the maximum swing in collector voltage, V Ci : where whole V C e ■ * s the minimum V C e for 03 for correct transistor operation, and is assumed to be 1 v. Therefore, V c . = 9 + 1 = 10 v, 24 v. The collector volt- and when the transistor is nearly at cut-off, V c , age may swing from 10 v to 24 v, with a quiescent level of 17 v. The peak-to- peak swing is 14 v, and the peak swing = 14/2 = 7 v: Rr.Ir Rl, = 3 0.003 = 2330 0. The final circuit with all parameters included is shown in Fig. 6.33. Now calculate the temperature sensitivity of I c , as Icbo and Vbb vary. The assumed Icbo = 0-01 ^a. From Fig. 4.2, Icbo increases by a factor of 17 atlOO°C. Thus, A/ CBO =/ C bo = A/cbo 2 = A / C bo 3 = 0.16 ji a. Substitute in (6.29), (6.30), and (6.31), with known A/ ceo : A i c = _ 1 + P<l + &) /8, S 3 A Icbo 1 + ^fi (1 + ft) + S 3 A Icbo- (6.33) Calculating stability factors, S, = 1 + R*. = 3.8, 1 + ^=6.8, Re, S, s 1 + Substituting in (6.33) and solving, Al r 71.1 A Icbo = 11.4 pa at 100°C. This is negligible since I Cl = 3 ma, and A Ic 3 corresponds to only a slight bias shift. The result confirms the validity of our approximation methods; high cal- culation accuracy is not justified. Now determine the component of change, A l c \> ^ ue to AV B e over the speci- fied temperature range: A V BE = (75°C) (- 2.2 mv/°C) = - 165 mv. Substituting in (6.32), 152 Transistor Circuit Analysis A I' c ' 3 = 0.69 ma. While this is much more significant than the A l' c resulting from A/ C so. it is still not a serious shift. The stability of the multi-stage direct-coupled amplifier is very satisfactory. We may make an estimate of the effect on / c of a change in /3: A/c. f A(H ■ sS [4.15] Assume, as the basis of estimating, that j8, varies from 325 at 25 °C to 460 at 100 °C, with an average value of 390. The value of A a /a corresponding to this $ x change is 0.9 x 10 -3 . Therefore, for the first amplifier stage, A/ f S t (0.9 x lO" 3 ), and since S t = 3.8, A/ f ^■= 0.34 x 10- Due to the change in j8,, / Cj changes by 0.34%. This is also negligible. The effects of changes in /3 2 and /3 3 may similarly be estimated as negligible. Only the temperature effects of changes in V BE are significant. These can be compensated for by the methods of Chap. 4. PROBLEM 6.27 For the amplifier circuit of Fig. 6.33, estimate the input im- pedance and voltage gain at 1000 Hz, and the maximum undistorted peak-to-peak output voltage at 25°C and 100°C. Specify C, so that voltage gain is 3 db down at 60 Hz. Note that the circuit parameters are those derived in Prob. 6.26. Solution: This problem provides an example of a step-by-step series of calcula- tions of input and output impedances, starting with the last stage. At 1000 Hz (gj = 6280 rad/sec), all emitter by-pass capacitors are essentially short-circuits. The transistor parameters used here are summarized in Table 6.2. Practical ap- proximations are made throughout. From Table 5.1: R i = r b + (r e + Re a) 1 + Rl + Rb b ,. 1+ (1 + /3) Table 6.3 25°C 100°C Unit R„ 6310 9830 ohm *L. 4640 6290 ohm *.. 8800 13,850 ohm R-, 5280 8550 ohm *I> 3040 3910 ohm with R L « r c , and r e « r c /(l + /3). The portion of emitter resistor not by-passed by C E , Re b , is zero. Now substitute numerical values from Table 6.2 in the above expression to find the input impedance of the third stage: At 25°C, Ki 3 = 5280Q, At 100°C, R l} = 8550 Q. These input impedances, paralleled with R Li , establish the effective load, R* 2 , on the second stage: At 25°C, i?£ 2 =3040fl, At 100°C, Rl 2 =3910fl. Proceeding in this manner, the required additional items may be calculated, and are summarized in Table 6.3. These include the required amplifier input impedance. Multi-Stage Amplifiers 153 Now calculate the currents and voltages in successive stages as required to determine over-all gain. If the current gain of the input stage is A ti , then R Ll /, '», A h **.+*', Substituting the expression for current gain from Table 5.1, and replacing I bi by v,/R h , V, ft R Ll or R >> h x p.rl, Rt l + -JH(l + ;8i ) R Ll + R h R <> 1 ' ik ™ l + ^d + A) . f «i J KL. + Kij (6.34) Similarly, h, ft j Rr. 2 1 + —1(1 + ft) . f <= 2 Ri. 2 + ^y 3 (6.35) Note that ftRi., 1 + ^1(1 + ft) (6.36) Substituting numerical values for the circuit parameters (see Table 6.1), the required performance data summarized in Table 6.4 are obtained. The voltage gains in Table 6.4 are part of the required solution to the problem. Note that although j8 increases by about 40% with increasing temperature, suggesting an over-all increase in gain of about (1.4) 3 = 2.75 times, the actual gain increase is only about 15%. The separate effects of changing ft as shown in the gain equations, partially cancel. Determine C, such that gain is 3 db down at 60 Hz. At this frequency, the capacitive reactance must equal the minimum (25 °C) input impedance: X Cl = R h = 8800 n, C, = 0.302 n f. Determine the maximum undistorted peak-to-peak output voltage from the bias point. At 25 °C, the quiescent conditions are /c 3 = 3 ma, and since R Li = 2330 and R Ej = 3000 Q, V E = 0.003 x 3000 = 9 v, V c = 24 - 7 = 17 v, V CE = 24 -(9 +7) = 8 v. The resistance i? Ej is by-passed, so that the voltage V Ei is a fixed 9 v re- gardless of the a-c signal component. A load line for this amplifier stage is drawn for an effective d-c voltage of 24 - 9 = 15 v, and a resistance of 2330 Q, Table 6.4 25°C 100°C 17.8 x 10"' 2.83 2.1 x 10 s 17.5 x 10" ! 2.54 2.4 x 10 8 154 Transistor Circuit Analysis (see Fig. 6.34). If a minimum V C e for satisfactory transistor operation is set at 1 v, it may be seen that the peak-to-peak a-c voltage swing is 14 v. Re-examine the peak-to-peak output range for 100 °C operation. As previously calculated in Prob. 6.26, I c% = 3.69 ma at 100 °C. The emitter voltage is 11.1 v, so that the effective voltage for the load line is 12.9 v. The modified load line is indicated in Fig. 6.34, showing a possible peak-to-peak output excursion of 11.9 v. This reduction in output voltage range is the principal reason for main- taining bias point stability. If l c increases to 6 ma, the amplifier actually be- comes inoperative. PROBLEM 6.28 For the amplifier of Fig. 6.33, connect a 1000 output load to the collector of Q 3 through a blocking capacitor C 2 having negligible a-c react- ance. What is the maximum undistorted output swing? Solution: Since the 1000 U, load does not effect the d-c operating point (because of the blocking capacitor), this point remains the same in Fig. 6.34. The a-c load line, however, corresponds to an effective resistance, Rf 3 , equal to 2330 Q in parallel with 1000 fi, or 700 Q. The maximum undistorted output swing is limited by the / c = axis, and is 8. 56 v peak-to-peak. 10 E o2 35 /ia 30fla 25 fia. 10 /ia 15^a ■f— »o L00°O- L*-— ^ v5v (25 ° c) lOj^a — 5fJa— ^ ^. \s 10 15 20 V CE , volt — 25 30 35 Fig. 6.34 Load lines superimposed on 2N929 common- emitter characteristics; Vg = 9 v. 1000 Q Fig. 6.35 Output transistor circuit driving a 1000 Qa-c load. PROBLEM 6.29 With the amplifier of Fig. 6.33 and the load of Fig. 6.35, find the voltage gain at 25 °C. Solution: From (6.36), y,~v, Pt 7~RT t ' 1 + __L (1 + ) [6.36] We replace Rl 3 by R* 3 = 700 Q. Substituting numerical values from Prob. 6.27, V 700 -2. = 2.83 x 360 = 690,000, V, , 700x361 1 + 6.7 x 10 6 as compared with over a million before the 1 K O load was added. Multi-Stage Amplifiers 155 PROBLEM 6.30 Repeat Prob. 6.27, but with 100 fi (see Fig. 6.36) of each emit- ter resistor unby-passed. Calculate only V /V l . Solution: Proceed as in Prob. 6.27, using the formulae summarized below: 1 Ri = t„ + {f. + R Ea ) (1 + /3) l + *£.(l + j 8) for R Ea « r c /(l + j8). Then, ' bl ft R Ll V, *,, sk " l + ^Ni + ft) [Rl^*!,] 7, V t Pl 1 + ^11(1 + ft) Rl 3 [Rt, + R,J [6.34] [6.35] [6.36] By substituting numerical values, the partial results and the computed voltage gain are determined and listed in Table 6.5. Observe the much reduced gain by comparing the data with that of Table 6.4. Although the variation in gain with temperature is somewhat improved by the unby-passed emitter resistances, the principal benefit is the reduced sensitivity of circuit gain to variations in the j8's of individual transistors. =3000 fit R,. = 100Q R 2b = 2900 Fig. 6.36 Modified portion of the circuit of Fig. 6.33, with portion of emitter resistor unby-passed. Table 6.5 25°C 100° C Unit R« 37,200 50,000 ohm R L2 6040 6300 ohm R,2 33,710 43,500 ohm RL 11,500 12,400 ohm Rn 33,000 43,000 ohm I b */Vi 2.62 x lO" 3 2.06 x lO" 3 amp/volt i b yv, 111 x lO" 3 88 x lO" 3 amp/volt Vo/V, 8.3 x 10 4 8.4 x 10 4 6.5 Complementary Transistors The availability of complementary pairs of transistors (n-p-n and p-n-p types with otherwise similar characteristics) has made it easier to design direct-coupled amplifier stages. They overcome the difficult biasing problems of direct-coupled stages using only one transistor type. The following example shows how easy it is to devise direct-coupled stages using complemen- tary pairs. 156 Transistor Circuit Analysis PROBLEM 6.31 Referring to Fig. 6.37, determine the load resistors, R L and Rl 2 , for the specified operating conditions; select Rl 3 for maximum undistorted output. Solution: For the specified conditions, V El = /e, R El = I Cl Re, = 10-' x 3000 = 3 v. Allowing 0.6 v for the base-emitter drop of Q lt V A = 3.0 + 0.6 = 3.6 v. Since high fl transistors are employed, we may neglect the base current of Qi in determining R, for the required V A : V A = Vcc ^^ R, + /? 2 The validity of this approximation may be verified later, if desired. Substituting numerical Values for V A , R 2 , and V cc , R t is determined to be 15 Kfl. Continuing to the second stage, V Bi = 9- (0.001) (3000) = 6 v. The base potential, allowing the usual 0.6 v drop from the emitter, is 5.4 v. This, of course, must be the collector potential of Q t . Thus, F Cl = 5.4=9-(0.001)i? Li . Solving, R Li = 3.6 K Q. Continue in a similar fashion to the third stage: V E} = (0.001) 3000 = 3 v, Vc 2 = 3 + 0.6 = 3.6 v, 3.6 = (0.001) R Li , R Li = 3,600 fl. This completes the determination of all parameters except R L} , which is found from the requirement for maximum undistorted output. Allowing a minimum V CE of 1 v f°r Qit V C3 varies from 4 v to a cut-off 9 v, for a peak-to-peak swing of 5 v. Set Vc 3 to the average of 6.5 v. Thus, a 1 ma swing of collector current corresponds to 2.5 v, so that K Lj = 2.5 K fl. For greater gain, R Lj can be increased until the limiting condition is reached where V Cl = 4 v maximum for Rl 3 = 5 K fl. For this condition, the gain is doubled and the peak-to-peak undistorted output is zero. Previous multi-stage amplifier calculations relating to performance and tem- perature sensitivity apply to amplifiers using complementary transistors, except that there are no d-c polarity inversions between stages. Therefore effects of changes in V BE , I CB o, and /8, always add. A-c calculations are identical to those illustrated earlier in this chapter. PROBLEM 6.32 For the circuit of Fig. 6.38, determine the unspecified parameters for the given operating conditions, as well as a-c gain and maximum power ab- sorbed by R L . Solution: Assume, as a crude estimate (which may be re-examined after an ap- proximate analysis is completed) that about 5% of the collector current of Q 2 is diverted by feedback to the first stage. Thus, if current through R L is specified at 95 ma, and current through R 3 at about 5 ma, then l c = 100 ma. Therefore, ^c 2 = 31x(-0.095)= J -3 v. Multi-Stage Amplifiers 157 ■ high /8, /3>100 Q x , Q 3 : n-p-n silicon transistor Q 2 : p-n-p silicon transistor R Ei =RE 2 =RE 3 =3K-ti, R 2 = 10K(2 C E > C E > C E , C, : essentially infinite l C x = Ic 2 = J C 3 = 1 ma Fig. 6.37 Direct-coupled amplifier using complementary transi stors. g 2 : 2N2907 I Cl = 100 ma, r e2 = 0.26 fl, r bj = 74 Q /S 2 = h fe2 = 100, h FE2 = 110at/c 2 = 100ma Ce , Ce : essentially infinite ' cc = 10v, R L =31 fl Dynamically, K c , may vary from zero, near transistor cut-off, to approximately -6 v maximum. Allowing for the usual 1 v minimum V CE , the drop across R El is 10-(6+l) = 3v. For I E = 100 ma, R E = 30 0. Continuing, since Ve 2 = - 7 v, allowing 0.6 v for VbEj, Vc, =-6.4 v. At lc 2 = 100 ma, the 2N2907 transistor has a d-c current gain of h FE = 110, so that a bias current of less than 1 ma is required. Thus, l Cl = 10 ma should be ample drive. This gives Now evaluate the parameters of the feedback circuit containing R 3 . Feedback resistor R 3 tends to stabilize collector current by resetting the bias of Q, so as to automatically oppose changes. For correct feedback, V c% must be greater than V E . Assume, consistent with earlier calculations, that V El = 2 v, and current through R 3 is 5 ma. Then, 3-2 0,: 2N930 r ei = ll £2, r Cl =2 XlO 6 A = fife, =380; h FEl =380atI Cl = l-0™* Fig. 6.38 Two-stage di rect-coupled ampli fier using complementary tran- sistors, and d-c feedback for bias stabilization. R, = 0.005 With current through R Ei equal to Ic l + 5 ma 2 v = 200 fl. Rr. = '» 0.015 133 Q. Now calculate the base bias resistors of Q x . Since V El = 2 v, Va = 2 + 0.6 = 2.6 v. Assume (as a starting point) 1 ma through R 2 . Thus, R 2 = -^-=2600Q. 0.001 The d-c input impedance to the first transistor may be estimated: ^58 Transistor Circuit Analysis R, 'lld-c = (1 + h FE ) R Ei = 381 x 133 = 51 K ft. We may safely ignore the base drain of Q t in calculating Rj r77r- 2 Vcc = v *- Substituting and solving, R t = 7400 0. If the shunting effect of the input imped- ance of Q l is considered, R, becomes 7050 ft. (Actually, final bias adjustments are best done experimentally.) For Q lt a 2N930 transistor, r e = 11 ft, h fe = 380, and r c = 2 x 10 6 . Thus, con- tinuing, R i t s (1 + hte) r e = 381 x 11 = 4200 ft, / =A = _i±_ bl R, 4200' A,= i - j8 380 l+^-(l+j8) 1+' ■-358, [Table 5.1] j _ 358 V, f o 2 x 10 6 V, 4200 ' i.-L RL < Rl, + R l2 R i t = r bl + r„ 2 (1 + h, H ). [Table 5.1] Substituting numerical values for the 2N2907 transistor, R ti = (0.26) (101) + 74 s 100 ft. Continuing to substitute numerical values, ^• 2 =100, I C2 = A i2 l b2 = 6.7 V h For a-c signals, R 3 is in parallel with R L so that effective load resistance is 31||200, or 26.8 ft. Therefore V = 6.7 V, x 26.8 = 179 V t ; hence, the required voltage gain is v t- ™- The instantaneous voltage across R Li can go from 9 to 6 v (leaving a 1 v minimum for V CE ). Thus, maximum undistorted output is 6 v peak-to-peak, or 2.12 v rms. Maximum undistorted power output P is p _Vl (2.12)* f = -— = — — = li5 mw. Rl 2 31 Under quiescent (zero a-c) conditions, the power in the collector junction of Q 2 is 0.1 a x (7 - 3) v = 400 mw. From the data sheet of the 2N2907, the tran- sistor power rating at T °C is Power rating = 1.8 - 10.3 x 10"' (T - 25°C). At T = 100 °C, the rating is approximately a watt, so we are well within the rating for the calculated 135 mw dissipation. As shown in the next chapter, the dissipa- tion in this type amplifier decreases with increased a-c input signals. Note in Fig. 6.38 that if only part of R Ei is by-passed by a capacitor, a-c Multi-Stage Amplifiers 159 feedback is introduced which simultaneously lowers and stabilizes the gain, and increases input impedance. (Feedback is considered in greater detail in Chap. 8.) 6.6 Supplementary Problems PROBLEM 6.33 In the circuit of Fig. 6.3, let K n = R l2 = 1 Mfl, R s =100 0, Rl, = Rl 2 = 6.8 Kfl, h ie = 2200 fl, h ie = 100 O, h re ~ 0, and h oe ~ 0. Calculate C 2 for a break frequency of 25 Hz. Assume that the influence of all other capaci- tors is negligible. PROBLEM 6.34 For the circuit of Prob. 6.33, calculate the frequency response if C E =5 fiF and C 2 is equal to the value obtained in Prob. 6.33. Assume C El = Cj = °o. PROBLEM 6.35 In the circuit of Fig. 6.24, let R, = R 2 = 10 Kfl, R E = 1.5 Kfl, C-, = 10 fiF, and n = 3Q. The load resistance R L = 10 fl with all other transistor characteristics the same as in Fig. 6.24. Calculate (a) the frequency response of the amplifier if it is driven by a zero-resistance generator, (b) the power gain at 1000 Hz, and (c) the maximum power output using the characteristics of the 2N930 transistor in Fig. 6.28. PROBLEM 6.36 In the circuit of Fig. 6.38, let R L2 =18fl, R Ei = 33 fl, C Ez = 100 ixF, R Ei = 120 fl, C Ei = 100 /xF, R Ll = 270 fl, R 3 = 180 fl, R 2 = 270 fl, and R Y = 7500 fl. Then the transistor characteristics of the Q 2 : 2N2907 and the Q x : 2N930 transistors are as follows: Q 2 : 2N2907 Q,: 2N930 hi. = h FE = 100 h le = h FE = 400 r e = 0.25 fl r e = 10fl r b =100fl r b = 0fl Determine (a) the frequency response when feeding R L ^ if C, = <x> and R g = 0, (b) the maximum undistorted power absorbed by R Li , (c) the d-c gain in R Li with zero input, and (d) the power dissipated in Q 2 with zero input. 7 CHAPTER POWER AMPLIFIERS 7.1 Introduction Up until this chapter we have focused our attention on the transistor small-signal amplifier, with emphasis on voltage and current gains. But in an amplifier system that delivers appreciable output power, voltage and current gains are only important considerations in the design of the preamplifier stages, not in the power stage. The prime objective in the design of the power stage is the achievement of a required power output with a specified efficiency and per- missible distortion. The power transistor must usually be operated over its full range of output characteristics, which includes regions of nonlinearity. Because of this latter factor, power stage design by means of equivalent circuits is less useful. Graphi- cal methods are much more suitable, as will be illustrated. A more detailed knowledge of transistor characteristics is needed for the analy- sis and design of power stages than for small-signal stages. Refer to Fig. 7.1 which shows the transistor output characteristics for the common-emitter connec- tion over their full range of interest. In particular, let us examine the character- istic curves in the high-voltage region. Saturation^ region Saturation . resistance line Cut-off ~f BVcEO region 'CBO V C E- BV, CES Fig. 7.1 Permissible operating limits of power transistors, showing non- linear regions. 160 Power Amplifiers 161 Consider the case where the emitter is open, and a normal reverse bias is ap- plied to the collector-base junction. As described in Chap. 1, minority carriers are accelerated across the depletion layer and impinge on other atoms in the crys- tal. If the applied voltage is sufficiently high, these impacts lead to the genera- tion of additional electron-hole pairs, thus increasing the current flow. The newly created carriers, themselves accelerated, in turn may generate even more electron- hole pairs. A multiplication or avalanche effect thus occurs, which may be de- scribed analytically by the following expression: where a is the current gain of the transistor in the common-base circuit, M is a dimensionless multiplying factor, and a* is the current gain at higher voltages in the vicinity of breakdown. The multiplying factor is described by the expression -GT where V B is the breakdown voltage of the collector-base junction with the emitter open-circuited (BV C bo). and V is the applied junction voltage. In the breakdown region, the collector-base current is multiplied by M. An expression for the effective forward current gain for the common-emitter connection is easily derived: a* -an- h * F l Solving for h FB , h * = _f^_. (7.3) '""i-ctir; For zero base current, 'c=Icbo a+h* FB )M M = 1 CBO 1- cxM Note that n and V B are constants for a particular transistor. Typically, n = 3 for a germanium p-n-p transistor, and n = 5 for a germanium n-p-n transistor. From (7.4) when l c = », + a= l. m Substituting typical numerical values, e.g., a = 0.97, n = 3, and solving, V/V B = 0.31. The voltage V, for which the collector current approaches infinity, is re- ferred to as BVceo (collector-emitter breakdown voltage with base open). Thus, for this particular case, BV CEO = 0.31 BV CB o- In the interest of low distortion and as a safeguard against damaging transients, the collector voltage is best limited to a maximum of BV ceo . Collector current, on the other hand, has no such well-defined limit. Practi- cally, the maximum current is limited by the fall-off in h FE , as shown in Fig. 7.2. 162 Transistor Circuit Analysis 500 Fig. 7.2 Typical current gain characteristics of a medium power germanium transistor. Results are normalized for convenient representation. High current occurring simultaneously with high voltage is limited by junction temperature. Junction temperature depends directly on the product of power dissipation and the thermal resistance between the collector-base junction and its environment. The thermal resistance from the junction to the transistor case is normally speci- fied by the transistor manufacturer. The unit for thermal resistance is °C/watt. Additional components of thermal resistance are the resistance from transistor case to mounting surface, and the resistance from mounting surface to the en- o W\r •VW l i junction temperature ■AAA* — o 1000 T c = case temperature T g = heat sink temperature T a = ambient temperature 9j c = thermal resistance, junction to case ^c» = thermal resistance, case to heat sink "sa = thermal resistance, heat sink to ambient temperature All temperatures, °C All thermal resistances, °C/watt Total thermal resistance 100 X Units mounted in the cen-i ter of square sheets of 1/8-inch thick bright alu-' minum. Heat sinks were held vertically in still air.' (Heat sink area is twice, the area of one side.) perimental average! e is = e, e + (a) 1.0 d sa , therma 10 10 1 resistance, °C/watt *» (b) Kit No. Insulating Washer Typical Mounting Thermal Resistance (#cs) °C/w (includes contact resistance) Dry With DC4* MK-10 MK-15 MK-20 No insulator Teflon Mica Anodized Aluminum 0.20 1.45 0.80 0.40 0.10 0.80 0.40 0.35 m .„j.j . i •!*.., t "™ *"-' ° r e 1 UI vaient is mgniy recom- mended especially for high power applications. The grease should be applied in a thin layer on both be added W " her - Whe " transiat °" ■» replaced in the sockets a new layer of the grease should (C) Fig. 7.3 Basic components of thermal resistance, (a) The resultant thermal resistance is the sum of the separate series components, (b) Thermal resistance to the environment of an aluminum plate, (c) Junction to case resistances of typical Motorola components. Power Amplifiers 163 vironment. Figure 7.3 shows the series addition of thermal resistances, as well as typical numerical values. Based on these considerations, where P c = maximum permissible collector junction dissipation (watt), T imttK = maximum permissible junction temperature (°C), T a = ambient temperature (°C), and 8, a = total thermal resistance from junction to environment (°C/watt). Saturation is another limiting transistor characteristic (Fig. 7.1). In the satu- ration region, V C e falls below the value required for proper transistor action. Since this region is bounded by an approximately straight line whose inverse slope has the dimensions of ohms, we may define a saturation resistance, R s • Figure 7.1 also shows a cut-off region. When I c <Icbo, Ie reverses sign, the emitter-base diode blocks, and transistor action ceases. At I c = 'cboi ^b = - 1 CB0 and I E = 0. The transistor is said to be cut-off. The transistor can be driven to this condition if the driving source supplies a negative base current. Thermal runaway, covered in Chap. 5, sets a further limit on transistor opera- tion. For safest results, the thermal runaway condition should not exist anywhere on the load line. Thermal runaway cannot exist if V CB < - V cc . Also relative to some of the above items is distortion, which is often a limit- ing factor in transistor power amplifier design. While theoretically it is possible to operate between points P, and P 2 on the load line of Fig. 7.1, distortion is considerably less for operation between points P, and P 4 . PROBLEM 7.1 Collector power dissipation in a transistor is given as the prod- uct of collector current and collector-base voltage, or P c = V CB I C * The locus of constant power dissipation may be plotted on the common-emitter collector characteristics as a hyperbola: , Pc -v Pc "cB V CE K '' DJ Show that maximum power output is equal to 1/2 P c , and that the load line is tangent to the hyperbola corresponding to the maximum permissible transistor dissipation. Neglect leakage current and saturation voltage. Solution: Refer to Fig. 7.4. The equation for the load line is 'c='< Vce ; m « R _ *$(>.':,''. , , v.; ^.•;.>,.;.;"i;s-^ , :i, ; :' where R L = V mttx /I c ■ Since the constant power hyperbola is defined by (7.5), we may solve for V CE : JjL=I c -I".. (7.6) Equation (7.6) is a quadratic. Solving for V CE , 4— V C ,: /<. " Pc - U. or + '<*■„ ± ]/ /J c m „ 17 ' 4P C •Dissipation at the emitter-base junction is usually negligible compared with dissipation at the collector-base junction. 164 Transistor Circuit Analysis 2N1537 Fig. 7 20 36 40 60F max 80 100 120 140 Vce / volt ». 4 Collector characteristics of the common-emitter circuit with superimposed load line and constant power hyperbola. Since the load line and hyperbola intersect only at the point of tangency, the radi- cal term goes to zero, and = 4 r7' or v A ''max *• The load power for sinusoidal signals is given as "-^M-fcW'-', Changing parameters, Comparing transistor dissipation and load power, P^IPa. (7.7) (7.8) (7.9) This relationship is true regardless of R L , so long as the load line is tangent to the P c curve. An elementary design procedure to follow is to draw a load line on the family of collector characteristics, tangent to the hyperbola of permissible power dissi- pation. This will insure that the maximum rated collector dissipation for the transistor is never exceeded. PROBLEM 7.2 Design the power amplifier circuit of Fig. 7.5 for maximum power Power Amplifiers 165 output using the transistor characteristics of Fig. 7.6. The additional parameters •* j max = 80 C, r a = 25°C, <9 ic = 0.6°C/w, 6> cs = 0.2°C/w, Y See Fig. 7.3 sa = O.5°C/w.. Use approximate methods of analysis. Neglect distortion and leakage currents. 5 ic -^r^\ O Vcc Q lt 2N1537A Fig. 7.5 Simplified power amplifier circuit. 20 40 60 80 100 120 140 V CE , volt — +■ Fig. 7.6 Superimposed load line and Pq = 42 w hyper- bola. The current at Q = 1 .05 a. Solution: First determine the permissible power dissipation. Total thermal re- sistance 6 ja is e ja = 0.6 + 0.2 + 0.5 = 1.3°C/w. Hence, — ' niax ■* a 6 i» 80-25 1.3 = 42 w. The hyperbola corresponding to this power is plotted on Fig. 7.6. A load line (R L = 38 fl) is sketched from V max = 80 v tangent to the hyperbola, and the op- erating point Q is noted. This load line avoids both the high-voltage and high- current regions of the characteristics, where excessive curvature occurs. Maximum load power, from (7.9), is 1/2 P c = 42/2 = 21 w. Total quiescent power, from the bad line, is P TOT = Vcc x lco = 80 x 1.05 = 84 w. This high input power is required since 42 w are dissipated in R L . To avoid this dissipation component, power amplifier stages almost always use transformer coupled outputs. PROBLEM 7.3 In the circuit of Fig. 7.7, the transistor is operated at the same quiescent point as in Prob. 7.2. What is the total quiescent power? Solution: Assuming an ideal transformer, the d-c load line sees only the tran- sistor drop, and extends from V C c = 40 vertically to Q. The a-c load line passes through Q as before. Power input is half of the previous value, or 42 w, corre- sponding to the maximum transistor dissipation. cc Fig. 7.7 Power amplifier ci rcuit using transformer-coupled output. 166 Transistor Circuit Analysis 7.2 Distortion 240 200 80 2N1537A Vc ;e = r 2v 2 3 fc , amp Fig. 7.8 The fall-off in h FE at high Iq is a distortion-producing factor in power amplifi ers. Although one avoids the region of collector current multil plication, where I c increases rapidly with increasing collector voltage, there an nevertheless two additional factors introducing distortion: 1. The rapid fall- off of h FE at high I c (see Fig. 7.8). 2. The highly nonlinear I B vs . V EB characteristic (see Fig. 7. 9a). From inspection of the above two figures, it becomes apparent that the two com! ponents of distortion-generating nonlinearity tend to cancel one another. Th« resultant l c vs.V EB characteristic is more linear than either component. Figur 7.10 shows this characteristic for the 2N1537A germanium power transistor. Th«l ratio g FB = lc/V B E, the transconductance, is the parameter used to describe! this type of curve. Operation at a bias voltage of V BB = 1.1 v leads to excellent linearity for the approximate range, 0.3 < V EB < 2.0 v, thus permitting a 1.7 v" peak-to-peak input voltage swing. In the above instance, the transistor base is driven from a low-impedance voltage source. For other transistors, optimum linearity may be obtained at some particular value of driving resistance. As an extreme condition, observe the "re- verse" nature of the resulting distortion in Fig. 7.9b when driving from a sinusoidal current source, where V BE (and / c ) take on a decidedly nonsinusoidal character 0.3 0.2 0.1 0.3 2N1537A V C E 1 = 2v \ 1 1 1 1 1 i v EBi volt Fig. 7.9 (a) Base current distortion with low impedance source (voltage drive) for the 2N1537A transistor, (b) Distortion in output current of the 2N1537A transistor with sinusoidal case current drive and high impedance source. Fig. 7.10 A plotof collector current/ G vs. emitter-base voltage V EB . PROBLEM 7.4 Given a set of transistor common-emitter collector and input characteristics, devise a simple graphical construction for determining output voltage distortion in the circuit of Fig. 7.11. Solution: The required construction is shown in Fig. 7.12. Note how selection of the driving or source resistance R g and load resistance R L can lead to minimi- zation of distortion through reduction and cancellation of distortion components. As previously noted, the load resistance R L is wasteful in dissipating d-c quiescent power, so that transformer coupling to the load is normally used. The transformer primary d-c resistance leads to a much smaller quiescent power loss. Figure 7.13 shows a transformer-coupled amplifier. Note that R E is added for bias Power Amplifiers 167 Transfer Output M r o V BB -ZZ Fig. 7.11 Biased power stage with quiescent current going through the load. (See Prob. 7.4.) Fig. 7.12 Graphical analysis of typical operating conditions using current gain curves. stability. This is usually not by-passed in power amplifiers due to the awkwardly large size of the required capacitor. In addition, an unby-passed R B will provide improved a-c performance. Though gain is reduced, distortion and gain stability are improved. The circuit of Fig. 7.13 may be investigated in nearly the same manner as that of Fig. 7.11. The significant difference is that the input to the base now in- cludes the R B drop as well as Vbe- PROBLEM 7.5 Devise a simple graphical construction relating output voltage to applied base voltage for the circuit of Fig. 7.13. Use the same transistor char- acteristics and general approach as in Prob. 7.4. Solution: Refer to Fig. 7.14. Note the d-c and a-c load lines superimposed on the collector characteristics. The l c vs. l B curve is easily plotted as before. For each l c , there is a drop IeRe = 1 cRe in the emitter resistor. This drop must be Fig. 7.13 Typical bias amplifier with transformer- coupled output. Vb6 VlT *V<'Y TV" slope = --=r- *^ ( lt — Input V B e. Vb Fig. 7.14. Solution to Prob. 7.5. 168 Transistor Circuit Analysis added to V BB at corresponding points to come up with the resultant V BB vs. / s characteristic. Figure 7.14 also shows how input and output voltages for the am- plifier of Fig. 7.13 are compared in order to determine gain, power output, and distortion. In the accurate analysis of power transistor circuits, it is best to use transis- tor characteristics corresponding to the actual operating temperatures. These are often unavailable, however, in the manufacturers' literature. Figure 7.15 shows how the significant transistor characteristics vary typically with temperature, for estimating the order of magnitude of inaccuracies which might be expected from using room temperature characteristics. o. E 25 20 15 10 _J >+25°c ^y^ + 50°C// ' ,X '+75 C !, J f '/ • if L_ f V CE =2v 0.4 0.8 1.2 1.6 2.0 2.4 Ig , amp » 2N1167 Collector current vs. base current (a) 1.5 1.0 A 0.8 i 0.6 o > « 0.4 0.2 + 50°C •+2 5^ r +1 0°C V C E =2v ft. 0.5 1.0 1.5 2.0 2.5 I B , amp »- 2N1167 Base-emitter voltage vs. base current (d) 80 60 40 20 V C E=2v <J 130 110 a 6? 25 20 15 10 01 *. tl fl / i j + 75° Cy >/ 1 ^// >/' 25°C /. V s m -"■ V CE =2v 5 10 15 20 Iq i am P — *- 2N1167 Current gain vs. collector current (b) 25 0.2 0.4 0.6 0.8 1.0 1.2 Vbe , volt — ► 2N1167 Collector current vs. base-emitter voltage (c) + 100 ApE vs. temperature (e) Fig. 7.15 Characteristics of power tran- sistors, showing some of the effects of temperature, (a) Collector current vs. base current, (b) current gain vs. collec- tor current, (c) collector current vs. base- emitter voltage, (d) base-emitter voltage vs. base current, (e) h FE vs. tempera- lure, and (f ) 6j?e vs. temperature. + 60 +100 T c , °C— +■ &FE vs. temperature (f) PROBLEM 7.6 Design a common-emitter power output stage to deliver 17 w of maximum power to a 10 fl loudspeaker coil. Ambient temperature is 25 °C; maxi- mum junction temperature is limited to 80 °C. Use a 2N1537A germanium transis- tor mounted on a cooling plate which results in 6 ja = 1.3°C/w of dissipated power. Check bias stability and the possibility of thermal runaway. Assume I C bo = 5 ma Power Amplifiers 169 at 80 °C. Aim for a stability factor S of about 5. Assume that the output coupling transformer d-c resistance is 10% of the load resistance referred to the primary. R^ = lon Fig. 7.16 Transformer-coupled power amplifier. (See Prob.7.6.) Solution: Use the configuration of Fig. 7.16. Note the unby-passed emitter re- sistor R E used for bias stabilization. Now following the procedures of Prob. 7.2, we establish a Q-point at V C b - 40 v and lc - 1.05 a. To minimize shifts in the operating point with temperature, restrict A/ c due to increased Icbo to less than 25 ma, corresponding to a maximum S of 5. Letting the reflected load = 38 fl, the transformer resistance is R c = 0.1 x 38 = 3.8 Q. Resistance Re may be chosen on the basis of its quiescent power loss. To minimize power dissipation in R E , let R E = 4 Q. Then, S = 4 + R B 4 + R B 1 + A 5, [4.38] where Rn = FE R l R 2 R, + R 2 Since h FE = 100 at the Q-point (see Fig. 7.8), we may substitute and solve for R B = 16.8 fl maximum for S = 5. From the l c vs. V BE curve (Fig. 7.10), at I c = 1.05 a at 25°C, V BE = 0.5 v. The voltage at point A is therefore V A = 1.05 R E + V BE = 4.2 + 0.5 = 4.7 v. However, the diode drop cancels V BE , so that a drop of only 4.2 v across R 2 is required. Since V C c = Vce + R c lc + Re / c = 40 + (3.8 x 1.05) + 4.2 = 48.2 v, we may solve for the resistances R t and R 2 : R,= 193 0, R 2 = 18.4 n. The current in the bias network is approximately 48.2 - 4.2 193 0.23 170 Transistor Circuit Analysis Since this is quite large, we might wish to redesign the bias network, con- tenting ourselves with an increased value of S, or else use a separate bias sup- ply as shown in Fig. 4.22. From Prob. 7.2, R L = 38 fl has been determined as optimum, and since R c = 3.8 fi, the load impedance observed at the transformer primary is 38 - 3.8 = 34.2 fi. The transformation ratio, primary to secondary, is therefore V34.2/10 = 1.85. The rms collector current is A/ c . . 2V2 where A/ Cp p is the peak-to-peak collector current. Thus, maximum load power is 2.05 2x/2 34.2 = 18 w. To check for the possibility of thermal runaway, use (4.49): 1 SO, > 0.07 Icbo [Vcc - 2 / c (Re + Rc)] . Substituting numerical values, 1 5x 1.3 = 0.15 > 0.07 x 5 x 10- 5 [48.2 - 2(1.05) (4 + 3.8)] = 0.014. The safety factor is greater than 10. This should provide sufficient margin, even allowing for the approximate nature of the calculation. PROBLEM 7.7 In the amplifier of Prob. 7.6, determine the maximum current out- put without clipping due to saturation. Also determine the distortion for this output. Solution: Figure 7.17 gives the transistor collector characteristics with d-c and a-c load lines and the quiescent operating point Q (which was determined in Prob. 7.2). Figure 7.17 also shows / c vs. V EB (V EB = - V BE = a positive number). E o O Rc+Re I B = ~ 5 ma Fig. 7.17 Excursion of collector current with sinusoidal base-emitter voltage drive. Power Amplifiers 171 From the saturation point P w we determine V BB at saturation. If we apply a symmetrical sinusoidal base-emitter voltage about Q from a zero impedance source, point P 2 is established. The output I c varies from 0.1a to 2.05a about I c = 1-05 a at point Q. This is the range of the a-c output current swing. Now determine distortion. This is accomplished by the methods of Appendix D. Use the following simplified formula, applicable to waves exhibiting primarily a moderate degree of second harmonic distortion: D 2 = ^xl00 = /l + /2 ~ 2 '° xlOO. B x 2(1,-1,) In the present instance, /j = 0.1 a, I 2 = 2.05 a, and l Q = 1.05 a. Substituting, D 2 = 1.3%, second harmonic distortion. PROBLEM 7.8 For the amplifier of Prob. 7.6, determine the driving current and voltage. Solution: Refer to Fig. 7.17, and interpolate to establish base currents: / c = 0.1 a, / s = — 3 ma, I c = 1.05 a, I B = 10 ma, l c = 2.05 a, l B = 29 ma. Peak-to-peak base current is 29 -(-3) = 32 ma. (From the above tabulation, note that base current, is not sinusoidal, although input and output voltages are nearly so.) Peak-to-peak base-emitter voltage, from Fig. 7.17, is 0.7 v. Thus, input power may be approximately calculated: i ♦ °- 7 °- 032 oa Input power = = x -=2.8mw, * 2 V2 2 V2 18 Power gain = = 6400. 0.0028 This power calculation neglects distortion. However, it is certainly suffi- ciently accurate for the purposes of designing a driver stage to precede the power stage. 7.3 Power Amplifier Design Equations Manufacturers' data on power transistors are rarely suf- ficient for meticulous design calculations. The principal deficiencies relate to differences among transistors of the same type, and the effects of temperature. As a result, practical design procedures are based on idealized transistor char- acteristics. The design may then be refined by more detailed graphical methods and/or laboratory adjustments. In this section, we examine the idealized transistor characteristics. Derived design formulae are tabulated, and a complete list is developed. The formulae are summarized in Table 7.1. Refer to the idealized characteristics of Fig. 7.18 and note the following: 1. Total equivalent saturation resistance R T = Rs + Re + Re- 2. Maximum usable voltage BV mmx = 2 Kcc - Vcs- t . D (Vcc-V C s\ I D (Vcc-Vcsf 3. A-c power output P = I ^ j I Q , or P a = ^- f 4. V = gy m .x + 2Rr^ 172 Transistor Circuit Analysis N. /Rr = R S +R B + Rc a/ 1 Slope — — 'Q '1 ^ / ! ' c = - Ic sin Oit 2V CC -Vcs = BV m „ i?S = Transistor saturation resistan ce Re = Equivalent d-c emitter ci rcui t resistance R c = d-c resistance of trans- former primary winding Fig. 7.18 Load line construction for Class A amplifier operation. Fig. 7.19 Single transistor power amplifier stage with transformer- coupled load. (See Prob. 7.9.) TABLE 7.1 Class A amplifier design formulae.* Item Formula Formula 0?r*0) (Rr= 0) p a^max'o RtIq 4 2 p c 2 (7.10a) P CE max VccIq = p c p c (7.10b) V C c BV m „ + 2R T I Q 2 2 (7.10c) Vm»x 5 R ' L 0.5 (7.10d) °l + 2 Rt RL BV max — t\ t 2I Q S^max 4 p c (7.10e) p CE max Vopt) BV m . x 4R T 0.5 , 2P C " * max (7.10f) (7.10g) v/l + 16 V r -l V BV f " r max P cc max V CC Iq = Pc ^cc'o = P C (7.10h) 'max 21 Q 2/g (7.101) Transformer coupling to load assumed. ce Ey integrating the expression for instantaneous transistor power over a com- plete sinusoidal cycle of collector current, the following formula is obtained: (transistor dissipation) - V cc !q-& j^°9" Vc *\ . (7.11) This expresses the almost obvious relationship that transistor dissipation is the difference between the (approximately) constant input power ana the output power. Dissipation is minimized when power output is at a maximum. ' Within limits imposed by BV mttX and l 0a „ , the transistor dissipation is de- termined by the hyperbola of constant power dissipation. The approximate design formulae for setting the operating point and estimating efficiency are listed in Table 7.1. PROBLEM 7.9 Using a 2N930 transistor in the circuit of Fig. 7.19 having a thermal resistance of 500°C/w and a permissible junction temperature of 175 °C, calculate/ Q opt , V C c, P Q , rj (the collector circuit efficiency), and/?!.'. UseBV max = 40 v, R s = 50 Q, and assume an ambient temperature of 70 °C. Solution: Start by determining the permissible transistor dissipation: 175 °C - 70°C = 500°C/w x P c . Solving, the maximum permissible dissipation P c is 0.21 w. Also, R T = R c + R E + R s = 150 + 50 + 50 = 250 Q. To determine the required performance characteristics, merely substitute in the formula of Table 7.1: '<? opt = %f(/ i t 16 Pc Rt 1 [7.10g] Power Amplifiers 173 Substituting numerical values, Iq opt = 9 -4 ma. Continuing, we obtain by direct substitution the following quantities: = BV max + 2R T l Q ^ 2 P„ = Rl'=Rl = BV m »*I Q Rrh BV n 21, 0.084 w, - R T = 2130 fi, Rl 2(R h '+2R T ) = 0.405, or 40.5%. [7.10c] [7.10a] [7.10e] [7.10d] 7.4 Common-Base Connection The common-base collector characteristics of transistors with high h FB are quite linear as shown in Fig. 7.20. In contrast with A^Ei firs i« almost exactly unity we* 3h* lull operating range of foe transistor. For the common-base csnaeeMoa, leakage ewrent is negligible. The maximum voltage (BYcboI is gteaie* then for the common-emitter or common-collector configura- tione< Because t e is very nearly equal to le, the only distortion introduced is by the nonlinearHy of the emitter input chcuit. If the input is a high impedance cur- rent source, such as is usually the case when the input is not transformer-coupled, distortion is very low. If a transformer-coupled input is used to increase power gain, distortion is sharply increased. To determine this distortion, the V B g vs. Ib characteristic, not usually available, is required. However, the Vbe vs. lc characteristic may be used without significant error. ICBO 1 E = -v. Fig. 7.20 Common-base characteristics of the 2N1537A transistor at 7) = 80°C and BV CBO = 100 v. Fig. 7.21 Common-base trans former- coupled output stage, BV CBO = 100 v. PROBLEM 7.10 Design a transformer-coupled common-base power output stage under approximately the conditions of Prob. 7.6. Assume the transformer d-c re- sistance is 10% of the reflected load resistance and determine the optimum output transformer ratio. Calculate maximum load power, distortion, input current, volt- age, and power, and approximate input impedance, with current drive and voltage drive. Determine V C e and compare with the common-emitter amplifier of Prob. 7.6. Solution: The circuit configuration is shown in Fig. 7.21. From Prob. 7.6 and the manufacturer's data in Appendix A, extract the following parameters: Tj = 80 °C, allowable junction temperature, T a = 25 °C, ambient temperature, d ja = 1.3°C/w, thermal resistance, 174 Transistor Circuit Analysis l CBO =5 ma at 80°C, BV CBO = 100 v. Using these parameters, calculate permissible junction dissipation: p T j -T B 80-25 Now refer to Fig. 7.20 where the 5 ma l CB0 is shown on an exaggerated scale, since otherwise it would not be discernible. For the common-base connection, voltage breakdown occurs at BV CBO , A load line may therefore originate at lOOv on the horizontal axis. For a permissible transistor dissipation of 42 w, the load line must be tangent to the P c = 42 w hyperbola at 0(50 v, 0.84 a). This is the operating point. The load line, therefore, corresponds to R,-^- 59.5 0. We have assumed that 10% of the apparent load resistance occurs in the out- put transformer winding, a reasonable assumption related to transformer efficiency. Thus, R c , the transformer resistance, is 5.95 0, and the effective load resistance must be 59.5 - 5.95 s 53.5 O. Since the load is actually a 10 resistor, the turns ratio may be determined: V 10 5 ^-23 10 Calculate power output P to the load. The rms current is l = °-4^ =06 V2 (neglecting distortion). Therefore, P = P^-l x 53.5 = 19.2 w. V2 Saturation actually reduces the power output slightly. Now establish V cc : V C c = V q + R c Iq = 50+ 5.95 (0.84) = 55 v. For sinusoidal input current, output distortion is negligible. This is a direct consequence of the current gain characteristic of the common-base circuit, where i hpE , , . . , 'c=- l E , h FE »l. l + n FE However, there is no current amplification. The driving voltage, on the other hand, is decidely nonsinusoidal. The V EB vs. I c curve, Fig. 7.22, is used instead of the usually unavailable V EB vs. l E curve, with little loss of accuracy. Note that the curves used should correspond to the actual junction temperature. From the operating points superimposed on Fig. 7.22, the evidence of voltage distortion is apparent. The lower voltage extreme is 0. 45 v below the quiescent value, while the upper extreme is 0.3 v above the quiescent value. Using our ap- proximate formula for second-harmonic distortion from Appendix D, D, (%) = E i + E 2~2E Q > 2(E 1 -E l ) Substituting numerical values, Power Amplifiers 175 The negative sign has no meaning; the second harmonic distortion is 10%. The fundamental component is similarly evaluated: Fundamental E 1 + E 2 then substituting, the fundamental voltage is + 0.75/2 = 0.375 v peak, or 0.265 v rms. Since / ; = l = 0.6 a rms, input power is P, = 0.265 x 0.6 = 0.158 w. Power gain is 19.2/0.158 = 120. This is much less than the power gain of the comparable common-emitter circuit, explaining why the common-base circuit is not often used. Input impedance is readily estimated: 0.265 v *, = ■ -= 0.44 fl. 0.6 a This is much lower than for the common-emitter connection. The only significant advantage over the common-emitter connection, apart from the somewhat increased output power rating, is the reduced distortion. This, however, can also be achieved by negative feedback from the power stage output to earlier amplifier stages (see Chap. 8). The high input current requirement of the common-base amplifier is a serious disadvantage. Now consider the same common-base amplifier circuit, Fig. 7.21, driven from a sinusoidal voltage source. This could be, for instance, a transformer input with a sufficient voltage step-down to reduce generator resistance Rg to a negligible value. o 5 2N1537A V CE =2v 7>80°C 3 2 1.70 X 0.85 y i S i / 1 0.4S ( 0.75 1.0 2.0 3.0 'EB volt . Fig. 7.22 Emitter-base voltage vs. collector current for T= 80°C. Fig. 7.23 Influence of generator source re- sistance on a 2N 1537A transi stor stage characteristic. Figure 7.23 shows how the now distorted input current is obtained by graphi- cal construction. From the figure, we see that a maximum V t = 0.3 v peak (corre- sponding to 0.212 v rms), which avoids saturation clipping. Second harmonic current distortion is (from Appendix D) D 2 (%) '» + '.-2/< 100. 2 (/,-/,) Substituting numerical values from Fig. 7.23, D z = 10%. 176 Transistor Circuit Analysis Neglecting harmonic distortion, input power is approximately 0.212 x 0.545 = 0.115 w. The fundamental power output is P = (0.545) 2 53.5 = 15.8 w. Thus, power gain is 15.8/0.115 = 138. We may see the effect of introducing some moderate value of source resistance R g . Figure 7.23 shows an input load line corresponding to an R e of 0.93 fl. In- put voltage must be increased to 1.08 v peak, or 0.765 v rms, leading to an in- creased power input, P l = 0.545 x 0.765 = 0.418 w. Distortion is essentially un- affected by introducing R g . Much greater values, approaching a current drive input, would be required. The input resistance for voltage drive is K,-**" 0.4 0, 0.545 nearly the same as for current drive. 7.5 Common-Collector Power Amplifier Stage The common-collector or emitter-follower circuit has a characteristic voltage gain of somewhat less than unity. The techniques for the design of common-emitter amplifiers may be directly transferred to the design of the common-collector amplifiers. The load, in effect, becomes R B , with R L = 0, and the transformer d-c resistance now serves as a d-c stabilizing emitter resis- tance. Input impedance approximately equal to (1 + h FE )R E is high. The common- collector connection is seldom used for power amplifiers. Vcc Fig. 7.24 stage Common -co I driving an 8 lector power load. PROBLEM 7.11 Using the 2N1537A transistor with the same characteristics as in Probs. 7.6 and 7.10, design a common-collector power stage to drive an 8 fl loud-speaker coil without an output transformer. Calculate maximum power out- put, distortion, input power, and power gain. Determine V C c and Fee. If the transformer has a secondary resistance of lfl, i.e., R B = 10, calculate the sta- bility factor S. Source impedance R e = 0. Solution: Refer to Fig. 7.24 which shows the configuration of the power-stage circuit. Using the thermal characteristics of Prob. 7.6, permissible transistor dissipation is 42 w. Now draw a load line on the collector characteristics of Fig. 7.25. Since collector and emitter currents are nearly equal, the load line is drawn to correspond to an 8 load. Therefore the figure will show the load line and the operating point Q corresponding to a 42-watt dissipation. Limiting the total output signal swing to avoid clipping due to saturation, voltage ranges from 3.5 v to 36.3 v, and the current swing is 4.2 a, p-p. Collector supply voltage V cc is thus 36.6 v. Neglecting distortion, P„ = 33 4.2 17.3 w. 2 V2 2 V2 Figure 7.26 shows the l E (= l c ) vs. V E b characteristic, the relationship From this curve and V, = lc Re + V EB (Vj = base voltage), the required swing of V t is determined as ± 17.6 v around a bias point of V BE = 17.6 v. Because of this symmetry, there is no distortion to affect the accuracy of our calculations. Power Amplifiers 177 2N1537A 2N1537A 4.2 4 E o J" V CE = 2 V Tj = 80°C 1.0 2.0 3.0 140 I , volt Fig. 7.25 Collector characteristics with superimposed load line for Prob. 7.11. Input (base) current is readily determined from Fig. 7.25: / BffiM =0.095a, /bq = 0.026 a, Z Bmin = - 0.005 a. It can be seen that l B is quite nonlinear, so that a low source impedance is nec- essary to insure low output distortion. Using the methods already applied several times in the preceding problems, it may be determined that a 40 fi source imped- ance leads to about 5% distortion. The 1 A transformer resistance leads to negli- gible distortion. Input current is very roughly Fig. 7.26 Emitter-base voltage vs. collector current. (See Prob. 7.11.) 0.095 = 0.0475 ma peak, or 33.5 ma rms. This neglects distortion. Thus, R,= 17.6 If Continuing, 33.5 x 10- 3 ^x33.5x 10" V2 373 0. 0.418 Power gain is 41.5, by far the lowest figure of each of the transistor configurations. To calculate S, use (4.38): Ssl + R f 1+-= 1.125. 8 This low stability factor is characteristic of transformer-coupled input circuits. To calculate collector efficiency 77, note that quiescent battery power is 36.6 x 2.1 = 77 w. Efficiency, the ratio of power output to input d-c power, is 17.3/77 = 22.5%. This poor efficiency is primarily due to the d-c dissipation in the 8 Q emitter load resistance. 178 Transistor Circuit Analysis 7.6 Push-Pull Amplifiers Fig. 7.27 Push-Pull transistor am- plifier with transformer-coupled in- put and output. Fig. 7.28 Classes of amplifier oper- ation determined by portion of cycle in whi ch the transistor is cut off. For relatively high distortion-free output power, the push- pull amplifier is used (Fig. 7.27). As a consequence of its circuit symmetry, there exists a characteristic symmetry of the output waveform, such that even harmonic distortion components are cancelled. Additional advantages, not im- mediately obvious, are the d-c cancellation in the output transformer, and greatly increased efficiency as compared with single-ended devices. Depending on the bias voltage of the circuit of Fig. 7.27, we get different classes of operation. These classes may be defined most easily by reference to Fig. 7.28. Classification is based primarily on the degree of clipping of collector current during operation. The principal advantage of clipping is improved efficiency. Class AB opera- tion is more efficient than Class A, and Classes B and C are even more efficient. However, the latter two modes lead to relatively high distortion, and are not em- ployed except for tuned loads. Because of the numerous advantages of the common-emitter circuit as a power stage, we will confine our discussion to this configuration. However, the method- ology developed below is equally adaptable to the other basic configurations. 7.6a Class A Push-Pull Amplifier Because of simplicity of design, we will initially con- sider Class A operation, although more efficient Class AB operation is usually preferred in practice. Actually, Class A push-pull operation is about the same as Class A operation of two separate but symmetrical transistor circuits, whose out- puts are combined by the action of the output transformer. PROBLEM 7.12 Design a push-pull Class A amplifier by combining two ampli- fiers of the type designed in Prob. 7.6. Draw load lines, determine its power out- put to a 10 load, and determine its second harmonic distortion when operated to the point where clipping barely occurs (maximum output). Also determine its ef- ficiency, and calculate the output transformer turns ratio. Solution: The push-pull circuit is shown in Fig. 7.29. Figure 7.30 shows the composite transistor collector characteristics, obtained by superimposing the two sets of characteristics to show push-pull behavior. Resistances R A and R B are used to adjust bias, as well as to optimize diode temperature compensation. The diode provides a compensating voltage drop for the variation in V BE with tem- perature. From Probs. 7.6-7, the following quiescent operating conditions for the sep- arate transistor circuits are established: I Ct = / c , = 1-045 a, ' CE 40 The voltage drive is 0.7 v p-p, and bias is adjusted for the required Q-point. The design is similar to the design of single-ended Class A amplifiers, except for the effect of the output transformer, which couples the separate circuits. The basic transformer equation is /„ = nCr Cl -/ C2 ). Since the separate outputs are 180° out of phase, I c is maximum when I c is minimum, and vice-versa. Power Amplifiers 179 r[ =2 nR L =38 Q Fig. 7.29 Push-pull power amplifier showing tempera- ture-compensated bias circuit. Substituting extreme values, I = n(2.05 - 0.1) = 1.95 n = I 2 , positive maximum output, l = n (0) = = Iq, quiescent output , l = n (0.1 - 2.05) = - 1.95 n = /,, negative maximum output. It may be immediately noted that current swing is double the value for the single- ended circuit. Because of the symmetry of the peak output currents, the second harmonic distortion is zero (assuming identical transistors and ideal transformers). This freedom from even harmonic distortion is characteristic of symmetrical push- pull stages. The voltage swing across the load resistance R L is (38 + 38)/n v p-p. Calcu- late power output: / B =-5m 1.95 n 38 V2 nV2 -=— x = 37w, assuming an ideal output transformer. From Fig. 7.30, since R£ = 2n 2 R L and R L = 10, K£ = 38. We may solve for the turns ratio (1/2 of primary to secondary), n = JT3) = 1.38. With regard to efficiency, note that the output voltage swing is unchanged, and the output current swing is doubled, in comparison with a single-ended stage. Also, the supply voltage V cc is unchanged, and I Q is doubled. Thus, efficiency is the same as for a single-ended stage. We have assumed in Prob. 7.12 that the transistor load lines are perfectly straight. This is not exactly true for push-pull amplifiers, although the deviation is usually small. The curvature is most pronounced where current gain varies substantially with l c . Figure 7.31 shows a characteristic exhibiting relatively high curvature. 70 ma — 95 ma — 120 ma - Fig. 7.30 Superimposed collector characteristics for evaluation of push-pull operation. Load lines are shown for a Class A circuit. 0.8 | A S\ A ^ %v Q ,„ B • J^« ^4 0.2 -v^ — ~^M, \\ — """n n v 10 20 30 40 50 60 Vce - Fig. 7.31 Curved load line in push- pull transistor circuit. PROBLEM 7. 13 Draw the equivalent circuit for a push-pull Class A amplifier. Reduce it to a simple format for easy analysis. Solution: The equivalent circuit is shown in Fig. 7.32. Because of symmetry, the circuit may be reduced to the single-transistor circuit of Fig. 7.33, where the circuit parameters are modified appropriately by factors of two. The equations for the single-transistor circuit of Fig. 7.33 are identical to the equations of the com- 180 Transistor Circuit Analysis & b *b O VVAr -WAr ±5\'° 2n:l ^P mbbJLm* mmm l mm r b — VW " VW r d Til, L0J (3r b Fig. 7.32 Equivalent circuit of a push-pull Class A amplifier. 2' 6 |8 O ^ VAr- -vs/v — * r d n:l Fig. 7.33 Simplified single-transistor equivalent circuit of push-pull Class A amplifier. plete circuit of Fig. 7.32. The reflected load impedance is n*R L , half of the load for each transistor. 7.6b Class B Push-Pull Amplifier Where high power efficiency is required, Class B opera- tion is recommended. For Class B operation, bias is adjusted for approximately zero quiescent current. Transistor conduction is essentially zero for half of each cycle. However, in the push-pull connection (Fig. 7.34) with each transistor con- ducting half the time but coupled through a common transformer to the load, the complete output wave is generated. -4 .^ y.i, x+ ©L 2n:l Fig. 7.34 Push-pull Class B amplifier. Load = n 2 R L =R' L Vcc 2V r Fig. 7.35 Load line for a Class B amplifier, cut-off for half a cycle. Figure 7.35 shows the typical load-line for the individual transistors. The load lines are combined in the composite characteristic of Fig. 7.36. Note from Fig. 7.35 that current is zero when voltage is at its maximum, thus minimizing dissipation. Each transistor effectively amplifies half the input wave. Figure Power Amplifiers 181 45 ma Fig. 7.36 Superimposed collector character- istics for evaluation of push-pull operation. ,t t v ec 2 J £l_z\. ^j - -^ Double amplitude = 2V CC Double amplitude = 2V CC g. 7.37 Waveforms in a push-pull Class B amplifier. (a) Time (b) Fig. 7.38 (a) Superimposed 2N1537A transistor characteristics showing distortion in push-pull output. This distortion is due to an equivalent offset in V B £. (b) Typical crossover distortion in a Class B push-pull amplifier. 7.37 shows current and voltage waveforms for the Class B connection. Since only one transistor is conducting at a time, the load line corresponds to the resistance Figure 7.38 shows the composite I c vs. Vbe characteristic. A sinusoidal in- put wave leads to a distorted output current. This is called cross-over distortion. To avoid cross-over distortion, it is advisable to operate slightly in Class AB with an added bias, above the cut-off value. This provides for conduction, during slightly more than half of each cycle. The bias differential is about 0.15 v for germanium and 0.6 v for silicon. Preferably, the bias is provided by a diode to compensate for temperature. The improved composite characteristic is shown in Fig. 7.39. The improvement in linearity is obvious. 182 Transistor Circuit Analysis TABLE 7.2 Class B push-pull amplifier, design formulae. (a) (b) Fig. 7.39 (a) Superimposed 2N1537A transistor characteristics showing how crossover distortion is elimi- nated by modifying the bias of push- pull transistors. Compare with Fig. 7.38(a). (b) Class A-B push-pull am- plifier biased to eliminate crossover distortion. Item Formula (R T hO) Formula 0?r=0) Equation "max CE max *?max RL Po "max P CE max V cc Per 00 max VccRL Vcc 4RL Vcc n 2 RL 0.785 Vcc n 2 P c n 2 — = 2.466 4 BV v cc = ^f* Vcc nRL (7.12a) (7.12b) (7.12c) (7.12d) (7.12e) (7.12f) (7.12g) 4(R T + Rlf Vcc n 2 Ri rr RL 4 R T + R[ V% c n 2 P C * 1 RL V 4\R T + RL) 2 V cc -BV max V cc + n 2 P c R T V cc = BV max P c R T n 2 Vcc tt{R t + RL) For l c = kl c , k<l ^ '-'max ' Pec V Po max v cc kV% c nRL (per transistor) kn T k 2 V 2 cc 4RL (per transistor) V CC RL (7.12h) (7.12i) (7.12J) (7.12k) R T + R L Design formulae for Class B operation are readily derived by reference to the circuit diagram of Fig. 7.40 and the idealized collector characteristics of Fig. 7.41. All significant symbols are shown on the figures. Although the derivations are not presented here, the design formulae are given in Table 7.2. Comparing Class A and B operations for the same permissible transistor dis- sipation, power output is about five times greater for Class B operation than for Class A. The theoretical efficiency at maximum output is 78.5% for Class B as compared with 50% for Class A. Furthermore, d-c power for Class B approaches zero as the input signal is reduced, so that in normal use, the average d-c power Power Amplifiers 183 I?T ~ Rs "*" ^C "*" ^£ Ic sin cot Fig. 7.40 Class B push-pull circuit rearranged for convenient ca I cu la ti on s . is far less than the maximum. This is often a most important consideration. These reasons favor the use of Class B operation for most high power applications. PROBLEM 7.14 For the Class B circuit of Fig. 7.42, determine V cc , P ot n, rj, Ri, R lf and R 2 . Use a maximum collector-emitter voltage, BV mBX = 45 v. Thermal resistance 6 ja = 500°C/w, and T, = 175°C, max. Assume that the maximum am- bient temperature is 70 °C. Vcs V CC BV max 2V C c-V C S=BV max Fig. 7.41 Load 1 ine characteristics appl cabl e to the push-pull Class B circuit. AAAr R c =30O Fig. 7.42 Push-pull Class B transistor amplifier. (See Prob. 7.14.) Solution: Estimate allowable collector dissipation P c : Pc = 6 ja (175 - 70) = ^(105) = 0.210 w. Saturation resistance R s = 50 (see Prob. 7.9). No emitter resistor is used, since for Class B biasing, minimum power loss is more important than stability. Therefore, R T = R c + Rs = 50 + 30 = 80 12. To find Vcc, refer to Table 7.2: 2V CC - BV max V cc + rr 1 P c Rt V C c = BV max P c Rt n\ [7.12f] Substituting numerical values and solving, V cc = 25.14 v. For our purposes, it is sufficiently accurate to let V cc = 25 v. 184 Transistor Circuit Analysis 0.1 0.2 0.3 0.4 0.5 'be . >lt Fig. 7.43 Approximate method to determine bias offset to eliminate crossover distortion. Again, referring to Table 7.2, i? I ,'=-l£^=302n, For both transistors, P a Continuing, n 2 P Vcc Rl 4(R T+ R' L y = 0.646 w. 0.323 w per transistor. [7.12d] [7.12a] Rl 4 R T + R L ' = 0.62 or 62%. [7.12c] The transformer turns ratio is obtained from the relationship Rl^ti'Rl, so that n = 0.376. To minimize cross-over distortion, particularly important at small-signal am- plitudes, a bias of V BE = 0.25 v at T ; - = 175 °C is required (see Fig. 7.43). This curve is actually derived from the 125 °C curve, since the curve for 175 °C was not available. The required characteristic is derived from the 125 °C h FE vs. / c characteristic. (Approximations in our method are acceptable, since h FB nor- mally varies more among individual transistors than with temperature. Further- more, refinements in adjustment are best made experimentally.) The sloping straight line approximation moves 2 mv/°C to the left for increasing T, . We may use a silicon diode for R 2 in Fig. 7.42. The diode should carry about ten times the sum of the base bias currents for the two transistors, or about 5 ma. Since V cc = 25 v, Ri 'cc 5x 10- ; , 2 L . = 5000 Q. 5 x 10- 3 If we do not use the temperature compensating diode, R, 0.25 0.005 = 50 0. PROBLEM 7.15 In the circuit of the preceding problem, what is the input power for maximum output? What is the power gain? TABLE 7.3 Measured parameters on a 2N930 transistor. /c (ma) ?,= 175°C T t = 25°C T, = 175°C h FE /fi (MB) V BE (v) v BE (y) 1 460 2.2 0.515 0.215 2 490 4.1 0.54 0.24 5 500 10 0.56 0.26 10 490 20 0.58 0.28 20 425 47 0.61 0.31 30 380 79 0.635 0.335 40 335 120 0.655 0.355 50 310 161 0.67 0.370 60 290 206 0.68 0.38 70 250 240 0.695 0.395 Power Amplifiers 185 Solution: Refer again to Table 7.2: f Cmai = Vcc =— = 0.0655 a. [7.12k] R T + R L ' 382 From Fig. 7.43, this yields V BE = 0.395 v, and from Table 7.3, l B = 240 ft a. Neg- lecting distortion, we need V BE = 0.15 v peak at I B = 240 /xa peak, for a collector current of 65.5 ma. Thus, p _ 0.15x_240x_10- 6 1Wfl . V2 x V2 Power gain = - — = 36,000. 18 x 10 7.7 Supplementary Problems PROBLEM 7.16 Define the terms I C bo, BV ceo , and BV CES - PROBLEM 7.17 Explain the avalanche effect. PROBLEM 7.18 What is the effect of temperature on h fe and / co ? PROBLEM 7.19 (a) Define thermal resistance, (b) Describe .mathematically the effect of change in temperature on the power that is transferred to a body. PROBLEM 7.20 Derive the optimum bias point of a common-emitter amplifier for maximum power output. PROBLEM 7.21 What is the maximum efficiency of a class A amplifier? PROBLEM 7.22 Design a power stage for maximum power output using the cir- cuit of Fig. 7.5 and the transistor characteristics of Fig. 7.6. Ignore distortion and assume negligible leakage at the temperatures of interest. Maximum junc- tion temperature is 150°C; maximum ambient temperature is 50°C. The total thermal resistance from junction to ambient is 2°C/w. PROBLEM 7.23 Repeat Prob. 7.22 for the circuit of Fig. 7.7. PROBLEM 7.24 Consider a 2N1532 transistor in the circuit of Fig. 7.19. Let R c = 12, R E = 100 fl, R s = 2 £2, and BV CEO = 50 v. The junction tempera- ture is 90°C, the ambient temperature is 50°C, and the thermal resistance 6 JC is 0.6°C/w. Evaluate Iq opt , V cc , P , r], and R L . Compare the results with those of Prob. 7.9. (Cf., App. A for the characteristics of a 2N1532 transistor.) PROBLEM 7.25 Repeat Prob. 7.24 using the push-pull circuit of Fig. 2.42. 8 CHAPTER FEEDBACK 8.1 Basic Concepts of Feedback s—d ^ e . A V > O I P + J Fig. 8.1 Block diagram of a feed- back amplifier. described by its equivalent circuit parameters. The parameters can then be use*. to calculate various properties of the amplifier, such as gain, and input and out^ put impedances. For a given amplifier, the properties are fixed. In certain as* plications, however, we may have to alter such properties as gain, or input and output impedances, or frequency response. Obviously we can achieve this by redesigning the amplifier itself. However, there is a much more efficient method' that can be used. When a signal proportional to either die output voltage or current (or soma' combination thereof) of an amplifier is fed back to its input, we find that thai amplifier thus formed has properties quite different from those of the original amplifier. This process of adding a portion of the output of an amplifier to its ift£ put in order to alter its performance is called feedback. In its simplest form, the effects of feedback can be seen with the kelp of tbj block diagram of Fig. 8.1, in which block A denotes the amplifier, and block ft ;■• the feedback network. The relations for the system are * The gain of the new amplifier, i.e., the feedback amplifier, is defined by A tb = — (8.2) Solving (8.1) for e in terms of e L , .•*..•»: The gain therefore becomes I'-fiA' (8.3) iU» 1-0/T (8.4) The term /3A is called the feedback factor. The amount of feedback, expressed decibels, is usually given as lit db of feedback - 20 log |1-0A|; where the logarithms are to the base 10. = -201og|l_/34|, (8.5) 186 Feedback 187 The feedback is termed negative or degenerative when it reduces the magni- tude of the gain, i.e., when |1 - &A\ >1. When @A is real, it becomes pA <0. The feedback is called positive or regenerative when it increases the magni- tude of the gain, i.e., when |1 - /SA| <1. However, it is also seen that, when j8/4 = +1, the closed-loop gain becomes infinite. (Closed-loop refers to the amplifier with the feedback in the circuit, and open-loop refers to the amplifier with the feedback removed.) This means, in a practical sense, that the feedback amplifier breaks into spontaneous oscillation. Thus an amplifier employing feedback must be closely examined for the presence of any instability. It will be seen that negative feedback has the following properties: 1. It stabilizes the gain of the amplifier against component and bias supply variations. 2. It extends the frequency response. 3. It reduces the level of noise generated within the amplifier A. 4. It reduces the harmonic distortion in the output signal. Suppose that some arbitrary parameter in the amplifier A changes, thereby causing a change in the gain. The fractional change in the gain can be expressed as dA/A. In turn, this will cause a change in the gain of the feedback amplifier which can be written as tM ft A4«,. Differentiating (8.4) with respect to A, dA tb dA (l-j84)» 1-fiA Thus the fractional changes are related by (l -pAJA Mtb 1 dA (g g) A lb 1-PA A ' It is therefore seen that negative feedback reduces the effects of the parameter variation; i.e., negative feedback stabilizes the gain, while positive feedback has the opposite effect. In addition, it will be seen that feedback can also be used to alter impedance levels and shape frequency response. PROBLEM 8.1 Show that negative feedback extends the frequency response of the amplifier A when the gain is AUa ) = _ A o_, (8.7) ..,'•. ^l:;;;t^B';:' , -r :,;; ;V : --\" : - .'...- where A B is the middle frequency gain, and <u„ is the high-frequency cut-off point. Solution: This can be shown by substituting (8.7) into (8.4) and observing the frequency response of the feedback amplifier: l + 22»" An {)<»)- 1 - j8 A a- i + B. A„ l-fcl. + £ 1-/8*. 1+ . (8.8) (1 - PA O )0 S 188 Transistor Circuit Analysis Without tb With tb «o <4> fb Fig. 8.2 Frequency response of an amplifier with and without feedback. Define Ait,. =■ 1-/34, Then, Atb(jco) = A tt>„ /ft) (8.9) (8.10) For negative feedback, i.e., £}A 9 <0, we see that <o tb >w ; thus the high-fre- quency cut-off is higher for the feedback amplifier than for the open-loop ampli- fier. Figure 8.2 shows the comparison of the frequency responses with and with- out negative feedback on logarithmic coordinates. The fact that the two curves merge for very high frequencies can be seen by assuming that a> is very large in (8.7) and (8.10), Then, u ib a <u 4(/o>)^ii . jfft) + fi , 1 ' Fig. 8.3 The effect of a distur- bance, such as noise, on a feedback amplifier. Thus the two expressions become identical at very high frequencies. A similar analysis can be performed to show that negative feedback decreases the low-frequency cut-off point. Now consider the effect of feedback on a disturbance (e.g., noise) generated within the amplifier A. Examine the block diagram of Fig. 8.3. This amplifier is similar to Fig. 8. 1 except that the amplifier A is broken into two parts with gainfe A t and A 2 , such that A = A i Aj . The voltage e d represents a disturbance that can be thought of as occurring within the amplifier A in Fig. 8.1. PROBLEM 8.2 Find e„ in terms of e, and e d for both the open-loop and closed- loop cases. Solution: First consider the open-loop case which is equivalent to setting 6 = 0. We see immediately that e„ = A x A t e t + A t e d . For the closed-loop case, e, = e + |8e , li^Siflii|llil9i^Btti^.,,, .,,,, Solving these equations for e„ in terms of ej and e d , e + j9e , "V - At e + e', L e„ = e, + l-pA l A 1 1-/3/4.A eji (8.11) (8.12) (8.13) Now define e os as that part of the output caused by e /F and e od as that part of e 6 due to the disturbance e d . Then for the open-loop case, Feedback 189 (8.14) And for the closed-loop case, e' = "i "a A* Oi e d (8.15) ° d 1-/8A,/*, where the primes have been used to distinguish between the two cases. A signal to noise ratio can be defined as the magnitude of the ratio of the output signal to the output disturbance. (Noise is defined as any unwanted or undesirable signal.) Thus for the open-loop case, P = e o<* M,4.e,j l^e,) l^e„| ie*r And for the closed-loop case, p'- B od |-4»eil (8.16) (8.17) Note that the two ratios are the same, and outwardly at least no advantage is gained by employing feedback. However, looking at (8.14) and (8.15), II -P^ it, J and for negative feedback, this ratio is less than unity. Now suppose that the input to the feedback amplifier is altered by a factor 1 - ]8A t i4, ; i.e., the new input is eJ-d-jBiMOe, • Then the signal output of the feedback amplifier becomes ^lA—e' l =A l A 2 e l 1-/S/M, (8.18) In other words, the output signal level is raised until it is the same as for the open-loop case. Then the signal to noise ratio for the closed-loop amplifier becomes -\l-PA t A t \ Jil*l-|l- p A t A,| p. I«d| P' = ei. e ad (8.19) Thus, when negative feedback is employed, the signal to noise ratio is increased by a factor |1 - j8A,i4,| > 1 when the input signal levels are arranged so that the output signal levels are the same for the open- and closed-loop cases. Moreover, the disturbance e d can be thought of as representing the nonlinearity in the am- plifier A; i.e., e d may represent the origin of the second and higher harmonics that appear in the output signal. Then it can be seen that, when negative feed- back is employed, harmonic distortion can be effectively reduced by a factor !i-/8<M a |. 190 Transistor Circuit Analysis 8.2 Types of Feedback The principal types of negative feedback are voltage and current feedback. Voltage feedback means that part of the amplifier output signal voltage is fed back to the input Similarly, current feedback means that voltage proportional to output signal current is fed back to the input. Negative voltage feedback tends to make output voltage independent of load; i.e., it reduces output impedance. Current feedback tends to make output current independent of load by increasing output impedance. The higher the gain through the amplifier and around the feedback loop, the closer the output characteristics approach the idealized zero and infinite impedance conditions for voltage and current feedback, respectively. In many instances, voltage and current feedback occur simultaneously, and a clear-cut distinction may not be possible. However, this consideration does not influence the accuracy of our calculations. PROBLEM 8.3 Referring to Fig. 8.4a, state what type of feedback is used. Calculate input and output impedances, and voltage gain. Vcc = 30 v (a) O •■ r\> I K B v (c) Fig. 8.4 (a) Feedback amplifier, (b) Equivalent ci rcuit of (a), (c) Transistor equivalent circuit. Solution: The derived feedback is proportional to the output voltage. This is voltage feedback. The feedback voltage acts to inject current into the transistor base circuit, in parallel with and subtracting from the applied current, /,. Figure 8.4b shows a simple equivalent circuit, in terms of input and output impedances, of the common-emitter transistor stage. The hybrid circuit for the transistor itself, including the bias resistors, R B and R s , is illustrated in Fig. 8.4c. To calculate voltage gain from Fig. 8.4b, it is necessary to determine A v , the voltage gain in the absence of feedback. An effective load resistance R£ Feedback 191 must include the separate parallel paths of R L and R t . Resistance Ri must be known in order to calculate the open-loop gain A v . Fortunately, R t is so high that we usually ignore the contributions of R t in determining R' L . Now calculate A vL , the voltage gain with R L connected: A vL = Ys. = , z5?Jlh _ . [5.8] V, Calculating R' L as described above, '!♦*(*- -^J or, substituting numerical values, R' L = 3220 Q. Substituting in (5.8), the expres- sion for voltage gain in the absence of feedback is A vL = -420. Define a voltage feedback factor, p\ , such that ft = — ?i—, (8.20) r; + r, where Rj is the parallel impedance of Rt and R B , the bias resistor. From (5.6), Ri = h le fife Ke ' h 1 /l oe+- F r Substituting numerical values in (5.6), R, = 2030 Q. This is essentially equal to Ri , since the contribution of R B is negligible. Therefore, 2030 Pi ~ 100,000 + 2030 = 0.0199. We may use this feedback factor to develop a convenient formula for input impedance including the effects of feedback, R if : R Vi K ii = -J-> V n lf = A vL R', Ri + Rt These equations may be combined to yield an expression for R lt : R lf = Ii= ^i , (8.21a) " Ri + R< or Rlt = ! Jl*L . . (8.21b) . ^ IbAyL Rj b R' i + R, ^92 Transistor Circuit Analysis Cancelling l b and substituting the expression for /S, in (8.21b), Rlt = 1 R A ' a ■ ( 8 - 22 > Recalling that A vL is negative, note that parallel feedback decreases the in- put impedance of an amplifier by a factor of 1 - A vL 8 lr or, more generally, 1 - A v 8, where A v is voltage gain, and 8 is the fraction of fed-back voltage. Substituting numerical values, 1 - A vL ft = 9.35, and R i{ = 2030/9.35 = 217 Q, a very large reduction due to feedback. Consider the effect of feedback on voltage gain: t A -^- Substituting numerical values, Zi =_420 ilZ_=_75. V g 1217 Without feedback, R t ' = 2030 Q is used instead of R if in the equation for gain. Numerically, £«^*L_ --420*30.-282. v e r; + R g 3030 Feedback has led to a substantial loss in gain. However, gain variation and noise have been reduced by the same factor, achieving important advantages, as previously explained. The output impedance remains to be calculated. To do this, "cut" the output circuit as shown in Fig. 8.4b, and determine the impedance seen looking back into the amplifier from R L . Define a voltage gain, A vo , analogous to A vL , but with R L removed. A new parameter, /?/', is the apparent input impedance seen looking back toward the input through R t : R,'+R a Therefore, V Jf" V, = p ' = V B The basic transistor output impedance in the absence of feedback, R (Fig. 8.4c), is determined as R ° = l -nr- • [5 - 9] "oe hie + Rg Referring again to Fig. 8.4b, R is parallelled by R s and R t + R 1 ,'^ R f to give an effective output impedance, /?„, still omitting the effect of feedback. Similarly, we may determine the open-circuit voltage gain, A vo , with R L removed. Gain A vo is computed using the previously developed gain formula with a load resistance d" _ RfRs R t + R s the resistance parallelled with R in calculating £„. The following formulae are applicable: Feedback 193 "vo - v, A*. -h fa R'l 1 + R, Ke ~ hfe Ke He [5.8] R a = [5.9] Ke ~ hfeKe h i6 + R g Now include the effect of feedback to calculate R of , the output impedance in the presence of feedback. The feedback voltage, with V applied to the output and V e shorted, is v, (8.24) where R,"is the parallel impedance of /? f "and R e . The ratio, Ri'/{R" + R { ), is an- other feedback factor, /3 2 , specifying the proportion of output signal fed back to the input. This fed-back voltage is, of course, amplified and applied to the out- put as V o P 2 A vo . Output impedance, R ol , is easily calculated from output cur- rent, l L : V o — " o Pi "vo Solving for R ol = V /I L , It." Rot R' R'o 1 — A YO p 2 (8.25) This formula shows how output impedance is reduced by the factor, 1 -A vo fi 2 , due to voltage feedback. Substituting numerical values in the above formulae, R'i = 4750 Q, R/' = 705Q, A vo =-618, R = 84 Kfl (transistor alone), K o ' = 4500Q, l-A vo p 2 = 1.425, R o/ = 3160 11. Note that the decrease in input impedance is not the same as the decrease in output impedance. PROBLEM 8.4 Repeat Prob. 8.3, using the circuit of Fig. 8.5a. Solution: Figure 8.5a shows a common-emitter feedback amplifier. Feedback is taken across a resistor in series with the load. The fed-back voltage is propor- tional to load current; this is current feedback. Figure 8.5b shows the equivalent circuit of this feedback amplifier. The fed-back voltage is summed at the input in series with the applied input signal. The amplifier equivalent circuit without a load termination, but including bias resistors, is the same as given in Fig. 8.4c. Now start by calculating voltage gain, and then the input impedance. Use roughly the same procedures as in Prob. 8.3. However, we can use simplifying approximations more freely. Highly accurate calculations are rarely justified. 194 Transistor Circuit Analysis K cc =30v R L = 9.5Kfi (a) (b) Fig. 8.5 (a) Amplifier with feedback voltage proportional 1o load current, (b) Equivalent circuit of (a). If we neglect a small amount of forward feed from input to output through R t , and ignore the loading effect of the 100 KO feedback resistor R f , then the gain,' A rL =-420, is the same as in Prob. 8.3. The feedback factor is easy to calculate': ft-- Rt 1 R L + R t 20 Again, by referring to Prob. 8.3, R t ' =2030 0. We must now calculate R„ = Vt/I,. From Fig. 8.5b, V, = V be -V t , I, Combining and solving, V, = V be -A vL Vt.fr = V b . (1 -A vL fr), ^■a. ^o-A. lft ) . _ 'i v b. 'b. r: (8.26) The series feedback at the input increases the input impedance by a factor of a~A rL fr). Substituting numerical values, i*,r, A = (-420)(l/20) = -21, R u = 2030 (1 + 21) = 44.7 Kfi. Now determine the closed-loop voltage gain: V ° _ VbeA vL V, A vL But so that v t y t o.-A VL fr)v e ■ * = v * ^~r . Ru + R e Feedback 195 1 V L Rl V g (1 - A vL ft) R if + R g For substantial feedback, where A vL /3 t » 1, and Rn » R g , (8.27a) V ^ 1 ^ R if ^_ _1_ (8.27b) ^ ft *?„ + /?«, ft Thus the closed-loop voltage gain becomes, to a large extent, only a function of the feedback factor ft. Substituting numerical values, solve for A vl > the volt- age gain in the presence of feedback: «i™ (-420) 7 45,700 A vt = 77- ^ .l-(-21) = -19.1. The approximate value, A vl = -1/ft = -20, checks very well. Without feedback, the gain is «/ 1 vL= ^(-420) = -282. r; + R g v " 3030 The reduction in gain is accompanied by greatly improved stability. Very large percentage changes in the transistor characteristics are accompanied by very much smaller percentage changes in amplifier gain with feedback. Where very high gain is required, it is much better to cascade stages, while employing generous amounts of feedback. Now we calculate the output impedance. Proceeding as before, we open the output circuit by disconnecting R L , and substituting an externally applied volt- age) V . Source voltage Vg is short-circuited. For this condition, the voltage gain, as before, is A vo = -618 (neglecting the small influence of R { ). Hence, V 1 = R t I L = — R +—V = P 2 V , Ri + Rl V -R t A vo I L +R^I L +R t I L R °' = rr r L ' R ol = Ro + R,d-A vo ). (8.28) Current feedback has increased the output impedance by R/(l - A vo ). Substituting numerical values, R o( = 4500 + (500) (421) = 25,500 Q . The preceding material points oat the important benefits that ate derived by the use of feedback. Amplifiers with gain stabilities to one part in 10,000 are not uncommon. Feedback allows the use of wide-tolerance transistors in precision However, feedback cannot be applied indiscriminately. Note that from (8.4), »tt l-A/3 This shows the characteristic way in which gain is affected by feedback. When A j8 = 1, A^ becomes infinite. This means that the amplifier develops an output with no applied input. If A )3 is a function of frequency (as it almost always is), 196 Transistor Circuit Analysis then even a theoretically correct low-frequency feedback factor can become unity at some high frequency, in which case, self-sustained oscillations occur at that frequency. This, of course, is the basis of operation of oscillators. But in high- gain amplifiers where it is desired to incorporate feedback, the possibility of oscillations is very real, and it is important to carry out a design with this reali- zation in mind. 8.3 Stability rt_ A e > C e ib < j> P e = ej — e^j, , in the closed-loop circuit Fig. 8.6 Open-loop configuration of feedback amplifier circuit. Unstable Marginal stabi lity P = Nyquist point at unit distance from origin and 180°. Fig. 8.7 Polar coordinate Nyquist plot of open-loop gain. Increased gain leads to oscillation at fre- quency ojj. If gain is set to the un- stable value, resulting oscillation leads to a gain loss so that the curve passes through P. ' Let us consider the subject of stability in greater detail. Refer to the open-loop circuit of Fig. 8.6. It is apparent that if the voltage gain around the loop is such that e /6 is equal to e, with a 180° phase shift, then self- sustained oscillations will occur on closing the loop. This may be deduced from the basic gain equation Gain = ■ l-Afi [8.4] where gain becomes infinite when A (3 = I, so that an output (the sustained oscil- lation) occurs in the absence of any input. The frequency of oscillation is that frequency where open-loop phase shift is 180°. An oscillator differs from a feed- back amplifier only in that the feedback characteristics are chosen to assure a stable oscillation at a desired frequency, rather than a stable gain. It is important to note that oscillations also occur when e tb exceeds e, with the required 180° phase shift. In this instance, oscillations build up in amplitude, until they are limited by amplifier saturation. For the steady saturated condition, loop gain is reduced to unity, since obviously the open-loop output is identical with the amplifier input. Nyquist devised a fundamental method for analyzing the conditions for insta- bility. The method consists of plotting complex open-loop gain on polar co- ordinates as a function of frequency varying from zero to infinity. The point whose polar coordinates are (1, 180°) is called the Nyquist point. If the gain vs. fre- quency curve passes through this point, then the conditions for oscillation de- scribed above, unity gain, and 180° phase shift exist, and self-sustained oscilla- tions occur. Somewhat more generally, if the open-loop gain characteristic encircles* the Nyquist point, the closed-loop will oscillate. As the loop satu- rates, the Nyquist characteristic contracts until it lies on the minus one point, as in Fig. 8.7. The frequency of oscillation corresponds to the frequency o^ on the gain characteristic at the Nyquist point. PROBLEM 8.5 Refer to Fig. 8.8 which shows several Nyquist complex gain plots, and indicate what plots correspond to stable closed- loop conditions. Solution: The required solution is noted directly on the figure. Normally, in the mid-frequency range of amplifier operation, gain is relatively uniform with frequency, with phase shift approximately zero. It is only in the low- and high-frequency regions of operation that large phase-shifts occur. Thus, in the design of feedback amplifiers, much attention must be given the low- and high-frequency gain and phase-shift characteristics to avoid instability, even though normal operation may be required only in the mid-frequency region. There is a practical point to be observed in relating open-loop and closed- loop characteristics. When breaking the loop, care is necessary to ensure that terminating impedances are unchanged. The load on the open feedback circuit * The intuitive concept of encirclement becomes somewhat ambiguous for complex circuits, and thus Nyquist's stability ctlteilon must be extended. However the topic, amply covered elsewhere, is outside the scope of this book. Feedback 197 must correspond to the effective impedance seen by the feedback circuit when the loop is closed. Similarly, the effective impedance in the amplifier input cir- cuit must be unchanged for the open-loop analysis. PROBLEM 8.6 Consider the simple multi-stage amplifier of Fig. 8.9 which shows the individual amplifier stage and the three cascaded stages, with identical interstage coupling networks. The fraction of output voltage fed back to the input is designated as /8. For simplicity, it is assumed that /8 does not affect either the load seen by the amplifier output, or the driving impedance seen by the input. For /S = 1/100, determine whether the amplifier is stable. Use Nyquist's criterion, with suitable approximations to simplify calculations. Solution: The key to a reasonable and simple solution is to analyze circuit behavior separately at low and high frequencies. The wide separation between low- and high-frequency regions makes this procedure possible. Gain falls off at the extreme frequencies. At very low frequency, gain is attenuated by the series capacitor, C,; at very high frequency, R in combination with the shunting capacitor, C 2 , serves to attenuate the gain. Consider the interstage network at low frequency, where C 2 <3C C t may be neglected in comparison with R,, and the reactance of C l . The interstage network, at low frequency, therefore introduces an attenuation (for three stages) of Ki = Ri + jcoCu 1 + jcoR i C 1 (8.29) Substituting numerical values, K l = ja> 1 + jco Similarly, the attenuation at high frequency may be approximated. The net- work components of importance in this region are R and C 2 . In the high-frequency region, C i is essentially a short-circuit, while R, is much higher than the re- actance of C,. Therefore, K h JOii 1 + joC 2 R, (8.30) jcoC 2 ■'9 (d) 8.8 Typical Nyquist frequency plots on polar coordinates. r, = iookQ, R o =100£2 (a) o * Stage 1 R i< \ r> <-> r 0e i /3 Stage 2 Stage 3 ^ C, = 10 /if, C 2 = 0.1^f (b) Fig. 8.9 (a) Simple multi-stage amplifier, and (b) its three cascade stages with identical interstage couplin g networks. 198 Transistor Circuit Analysis Fig. 8.10 Nyquist plot of transfer function of Prob. 8.2. Note that gain is 1.25 at 180 phase for high- and low-frequency regions. Substituting numerical values for R C 2 , K h = 1 + ;<u 10" Re Compare (8.29) with (8.30). For (8.29), as frequency increases to the rela- tively flat mid-frequency region, K/ = 1. Similarly, for (8.30), as frequency de- creases to the mid- frequency region, K h = 1. These limiting high- and low-fre- quency conditions must, of course, lead to the same mid-frequency gain if our approximations are valid. Open-loop gain, used in plotting the Nyquist characteristic, corresponds to the attenuations calculated by (8.29) and (8.30), multiplied by the feedback factor )8, and the gains of the amplifier stages, (10) 3 . Figure 8.10 shows the Nyquist characteristic plotted over the entire frequency range. Note that the Nyquist point is encircled not once, but twice, in both the low- and high-frequency regions. The loop is clearly unstable. Note particularly that (for this special case), the high- and low-frequency regions of the curve exhibit the same gain at 180° phase shift. Reducing the gain by 80% (by reducing /8 from 1/100 to 1/125) leads to a marginally stable system. Of course, this marginal stability is not practical; at least 6 db of gain margin are normally recommended for the 180° phase-shift point. The simple example illustrated above shows how neatly the Nyquist charac- teristic describes the amplifier frequency characteristic in the presence of sub- stantial amounts of feedback. The amplifier designer must shape the Nyquist curve to avoid the - 1 point by an adequate margin, while maintaining the desired degree of feedback required for gain stabilization in the mid-frequency range. Without the visualization provided by the Nyquist plot, stability analysis is relatively tedious. PROBLEM 8.7 For the feedback amplifier of the preceding problem, determine the frequency and gain at which phase-shift equals 180° at both ends of the frequency band. Solution: The loop gains for j8 = 1/100 are given by the following equations: Ki (loop gain) =10' ,co K h (loop gain) = 10 1 + /ft) 1 ^l + /ft)10- 5 The frequency of Ki has a 180° phase shift when 270° - 3 arctan w/ = 180°, or arctan coi = 30°; that is, when ft>; = 0.577. Gain at this point may be calculated by substituting to/ = 0.577 in (8.29) 0.577 3 [8.29] [8.30] K l \ = 10 1.25. [1 + 0.577 2 ] 372 Similarly, K h has a 180° phase shift when 3 arctan 10 -5 co h = 180°, or tan 60° = 10" 5 <o h = y/3. Solving, co h = V3 x 10 ! . Feedback 199 To calculate gain, substitute <y h = V3 x 10 5 in (8.30) 1 10 IK* I = 10 [1 + 3] 3/2 8 = 1.25. The results calculated above check with those shown on the Nyquist curve of Fig. 8.10. In all but the simplest cases, calculations of the type performed in this problem prove tedious and impractical. Nor do they provide the visual picture of circuit behavior afforded by curves such as the plot of Fig. 8.10. Nevertheless, even the Nyquist characteristic is difficult to use. The polar coordinate plots are tedious to make. The effects of circuit changes to improve specific aspects of performance require equally tedious evaluation. The Nyquist approach is not a convenient tool for synthesis. 8.4 The Bode Diagram The Bode method of plotting frequency response is based on the use of logarithmic coordinates for gain (or attenuation) and frequency. This type of coordinate system leads to easily derived asymptotes which provide a surprisingly accurate representation of the gain function. Appendix C describes how to plot Bode diagrams. PROBLEM 8.8 Solve Prob. 8.6 using the Bode plotting methods of Appendix C. Solution: The asymptotic solution is shown in Fig. 8.11. The phase shift needs but to be sketched in, taking accurate points only in the vicinity of 180° phase shift. A convenient stability check, well-suited to the asymptotic method of plotting gain, is to calculate phase shift at those frequencies where asymptotic gain falls to zero db, as shown in Fig. 8.11. If phase shift exceeds 180°, insta- bility occurs. As shown on Fig. 8.11, phase shift does indeed exceed 180° when gain has fallen to unity. Fig. 8.11 Bode plot of open -loop gain of amplifier ci rcui t of Fig. 8.9. PROBLEM 8.9 For the feedback amplifier of Prob. 8.8, determine the feedback factor j8 for the condition of marginal stability. Also find j8 for a 135° phase shift at the asymptotic zero db point (45° phase margin). Solution: From Fig. 8.11, j8 = 1/125 and 1/355 for 0° phase margin and 45° phase margin, respectively. These results are necessarily compatible with those obtained from the Nyquist plot. 200 Transistor Circuit Analysis 8.5 Operational Amplifiers Operational amplifiers are feedback amplifiers with very high loop gain. The gain of an operational amplifier depends almost entirely on the feedback factor j8 : Gain = . 1-/4)8 where A is the amplifier gain in the absence of feedback. When A/8 » 1, As a result of this feature, operational amplifiers are used to achieve precise gain characteristics by means of suitable feedback networks. The correct opera- tion of the operational amplifier presumes that factors relating to stability have been taken care of in the design of the open- loop circuitry. Figure 8.12 shows a typical operational amplifier designed to accept three separate input voltages, taken with respect to ground. The impedances of the amplifier can be pure resistances or complex impedances. • Oe Fig. 8.12 Operational amplifier, including input summing circuit. PROBLEM 8.10 For the circuit of Fig. 8.12, find the expressions for gain eo/e/j, e /e, 2 , and e /e, 3 . Solution: Let the voltage at the amplifier input be e B . Then, by summing currents, e i, - e a e i2 - £j %2 Z$ ^F ^i Rearranging, e ', e t 2 e h ( 1 1 1 1 1 \ — - + — ? + — = e, — + — + — + — + — "l *"1 "I \"l ^2 ^3 "I "Ft ** F However, e = —Ae a (the negative sign means negative feedback). Therefore, " 1+ Z f /J_ + J_ + J_ 1_ 1 4 \Z t + Z 2 + Z, + Z, + Z z, Zi z 2 Is. z, (8.31) This is the fundamental equation for the required gains of the separate inputs. For the operational amplifier, gain A is very large (assume «), so that Zj Z a Z 3 z F (8.32) Feedback 201 Equation (8.32) is a very accurate expression for the gain of operational amplifier circuits. The error in gain due to assuming A = «> is 100/, Z F Z F Z F \ ^1 **\ ' '*?■ PROBLEM 8.11 For the feedback amplifier of Fig. 8.12, A = 100, Z t = 100 KQ, Z 2 = Z 3 = oo, Z, = 100 KQ, Z F = 1 MQ. Find the exact gain, and compare with the gain calculated by the approximate formula (8.32). Solution: The gain is obtained from e 'i __ e p 10 5 10 6 1 + 100 — I— + —+ —\ 100 \10 5 + 10 5 + 10 6 / Simplifying, e fl = - || (1.21) = -0.121 e OJ — = gain =-8.25. % The gain is —10 by the approximate formula. The approximation is poor because of the low gain, A = 100. PROBLEM 8.12 Repeat the preceding problem for A = 10 s . Solution: Substituting, as before, e 'i ^ e p 10 5 10 10 5 \10 5 10 5 10" Simplifying, e, = - ?2. (1.00021) , -52. = ~ 10 1 10 e,, 1.00021 The error introduced by using the approximate formula is 0.02%. Note the applicability of the approximate gain formula to high-gain opera- tional amplifiers. Gain is almost entirely dependent on the summing and feedback resistors. Stable resistors mean correspondingly stable gain. PROBLEM 8.13 Refer to Fig. 8.12, with Z, = 100 KQ, Z 2 = 1 Mfl, Z 3 = 10 KQ, Z F = 10 6 . Let A be essentially infinite so that we may use (8.32). Find e in terms of e^, ei 2 , and ej 3 . Solution: The following is the required expression: 10 s 10' 10 4 10 6 ' Simplifying, -e = 10e il + e f2 + 100 e, 3 . The voltage e„ is the sum of the inputs, each input with an appropriate scale factor. The arrangement of input resistors is called a summing network. PROBLEM 8.14 In the circuit of Fig. 8.10, Z F = l/Qa>C F ), where C F is a feed- back capacitor, Z x = \ MQ, Z 2 = Z 3 = oo. Assume A is essentially infinite. Find the gain. 202 Transistor Circuit Analysis Solution: Substituting in (8.32), -fL = 10 6 jcoCf jcoC F e , o)C, 10- The gain varies inversely with the frequency of the input. This is a characteristic of an integrator. Output represents the integral of input. This circuit is widely used in analog computers to carry out the critical integration function. 8.6 Supplementary Problems PROBLEM 8. 15 What type of feedback produces low input impedance and low output impedance? PROBLEM 8.16 What type of feedback produces high input impedance? PROBLEM 8.17 What type of feedback is commonly used in operational ampli- fiers? (Cf., Fig. 8.12.) PROBLEM 8.18 Discuss the advantages of negative feedback. 100(1000 + jco) PROBLEM 8.19 If the voltage gain of an amplifier is A v = — ; rTr , find 1 + 0.1;<y (a) the d-c gain, and (b) the gain at 1 MHz. PROBLEM 8.20 Using the block diagram of Fig. 8.3, let A, 1000 jo + 1 ' 0.15 +1 e e A 2 = n »,,. — , . and j8 = 0.1. Calculate (a) — (<u), and (b) — at 1 Hz. 0. 015 + 1 e d e . Fig. 8.13 The circuit of Prob. 8.24. PROBLEM 8.21 Using the block diagram of Fig. 8.3, let A l 0.15 + 1 _500_ 0.015 + 1 G G , and /3 = 0.1. Calculate (a) — at 1 Hz, and (b) — (&>). e, e. PROBLEM 8.22 Using the system of Prob. 8.20, determine the (a) forward gain at 1 Hz, (b) zero-frequency forward gain, (c) feedback gain at 1 Hz, (d) zero-frequency open-loop gain, (e) closed-loop gain at 1 Hz, (f ) zero- frequency closed-loop gain, and (g) frequency at which the closed-loop has dropped 3 db below its initial value at zero-frequency. PROBLEM 8.23 For the circuit of Fig. 8.9b, determine (a) the closed-loop gain at &> = 200, (b) the change in the closed-loop gain at co = 200 if the stage gain increases from 10 to 20, and (c) what happens if the stage gain is reduced to 7. PROBLEM 8.24 For the transistor Q^ in the circuit of Fig. 8.13 the parameters are /3 = 100, r„ = 10 ft, r b = Oft, and r c = ■», Determine (a) the nature of the feedback used, (b) A v = e /e it and (c) Z in and Z . TRANSISTOR CHARACTERISTICS A APPENDIX A.1 Types 2N929, 2N930 n-p-n Planar Silicon Transistors* FOR EXTREMELY LOW-LEVEL, LOW-NOISE, HIGH-GAIN, SMALL-SIGNAL AMPLIFIER APPLICATIONS • Guaranteed h F E at 10 pa , T A =-55°C and 25°C • Guaranteed Low-Noise Characteristics at 10 fia • Usable at Collector Currents as Low as 1 \ia • Very High Reliability • 2N929 and 2N930 Also Are Available to MIL-S-19500/253 (Sig C) ^3 - COLLECTOR ALL DIMENSIONS ARE IN INCHES UNLESS OTHERWISE SPECIFIED THE COLLECTOR L£ IN ELECTRICAL CONTACT WITH THE CASE ALL JEDEC T0-18 DIMENSIONS AND NOTES ARE APPLICABLE Fig. A.l JEDEC registered mechanical data. TABLE A.l JEDEC registered absolute maximum ratings at 25 C free-air temperature (unless otherwise noted). Collector-Base Voltage 45 v Collector-Emitter Voltage (See Note 1) 45 v Emitter-Base Voltage 5 v Collector Current 30 ma Total Device Dissipation at (or below) 25°C Free-Air Temperature (See Note 2) 300 mw Total Device Dissipation at (or below) 25°C Case Temperature (See Note 3) 600 mw Operating Collector Junction Temperature 175°C Storage Temperature Range -65°C to +300°C This value applies when the base-emitter diode is open circuited. Derate linearly to 175 C free-air temperature at the rate of 2.0 mw/C . Derate linearly to 175 C case temperature at the rate of 4.0 mw/C°. * TEXAS INSTRUMENTS Incorporated. 203 Transistor Circuit Analysis TABLE A. 2 JEDEC registered electrical characteristics at 25 C free-air temperature (unless otherwise noted). PARAMETER TEST CONDITIONS 2N929 2N930 UNIT MIN MAX MIN MAX BVceo Collector-Emitter Breakdown Voltage lc = 10 ma, l B = 0, (See Note 1) 45 45 V BVebo Emitter-Base Breakdown Voltage l E = 10 no l c = 5 5 V Icbo Collector Cutoff Current Vcb = 45 v, l E = 10 10 na Ices Collector Cutoff Current (See Note 2) V CE = 45v, V K = 10 10 na Vce = 45 v, V M = 0, J A = 170°C 10 10 /na Iceo Collector Cutoff Current Vce = 5 v, 1, = 2 2 na Iebo Emitter Cutoff Current Veb = 5 v, lc = 10 10 na h F E Static Forward Current Transfer Ratio V CE = 5 v, l c = 10 /to 40 120 100 300 Vce = 5 v, lc = 10 fia, T A •= — 55°C 10 20 V C e = 5 v, l c = 500 jiia 60 150 Vce = 5 V, l c = 10 ma, (See Note 1) 350 600 Vbe Base-Emitter Voltage Ib = 0.5 ma, l c = 10 ma, (See Note 1) 0.6 1.0 0.6 1.0 V VcE|i.t] Collector-Emitter Saturation Voltage l B = 0.5 ma, l c = 10 ma, (See Note 1) 1.0 1.0 V . Small-Signal Common-Base ib Input Impedance V C b — 5 v, l E = — 1 ma, f = 1 kc 25 32 25 32 ohm Small-Signal Common-Base rb Reverse Voltage Transfer Ratio V C b = 5 v, l E = — 1 ma, f = 1 kc 6.0 x io- 4 6.0 x io- 4 Small-Signal Common-Base n ° b Output Admittance Vcb = 5 v, l E = -l ma, f = 1 kc 1.0 1.0 jLtmho Small-Signal Common-Emitter f * Forward Current Transfer Ratio V C e = 5 v, l c = 1 ma, f = 1 kc 60 350 150 600 1 1 Small-Signal Common-Emitter 1 '*' Forward Current Transfer Ratio Vce = 5 v, l c = 500 (jua, f = 30 mc 1.0 1.0 - Common-Base Open-Circuit ob Output Capacitance Vcb = 5 v, l E = 0, f = 1 mc 8 8 pf These parameters must be measured using pulse techniques, PW = 300 /xsec. Duty Cycle < 2%. 'cES ma y k e usec ' ' n pl°ce of IpRO ^ or c ' rcu it stability calculations. TABLE A. 3 JEDEC registered operating characteristics at 25 C free-air temperature. PARAMETER TEST CONDITIONS 2N929 2N930 UNIT MAX MAX NF Average Noise Figure Vce = 5 v, I c = 10 jiia, R s = 10 kil Noise Bandwidth 10 cps to 15.7 kc 4.0 3.0 db 204 Transistor Characteristics A. la Typical Characteristics Z 1000 S E £ o U 100 10 ilium tttt — "A/ aximum i,. ~~H Vfct =5v •a 1 1 f = kc ii 1 T 1 vtf< A„ 1 V y# en i c T 1 ^ " a - • *\ * . *-^ 1 1 1 1 \K L TJ T\ w inimum h fe Ta = 25°C 01 1 0.01 0.1 1.0 10 lc — Collector Current — ma 100 Fig. A.2 2N929 transistor: small-signal common- emitter forward current transfer ratio vs. collector current. Z. 1000 u ■a a I £ ioo E E o 10 0.01 . _ »i ~T TTTT1 \ Ci - J V f '= 1 kc lot J, =2.s°r I 2-T a=125°C T T 1 1 II Tm ^j* — <- _.. -v., = 75° c 5— i Ta=25°C Vjjf C ^"^C^C . ..I. \ ► 1 f ' ' 1 t at T A = 25°C ■ [ 1 0.1 1.0 10 Collector Current — ma 100 Fig. A. 3 2N930 transistor: small-signal common- emitter forward current transfer ratio vs. collector current. -§10,000 I ft E 1000 _ 100 I 10 y.. = 5 = 1 V- v\ ^ s- — - 125°C = 25°C = -55°C ,.''' ■" ' ^ U Maximum at T A = 25 K "C II Minimum hj b / l|atT A =25°C ' ^V» 1.0 •0.01 •0.1 1.0 -10 l E — Emitter Current — ma Fig. A. 4 2N929 and 2N930 transistors: small- signal common-base input impedance vs. emitter current. E TJ < J 0.1 E E o U !°-°-l, " ' rzr ^ Ma --- at T A -25°C --; V - "i V CB J f - ' kc X n*~ -55"C ---- •-i- 25°C 125°C 01 -0.1 -1.0 , — Emitter Current — ma -1 Fig. A. 5 2N929 and 2N930 transistors: small- signal common-base output admittance vs. emitter current. 205 Transistor Circuit Analysis * ]Q : o > E E o U I0" 4 1 .. 1 1 Ml Ma> imum h rb atT » = 25°C V c .=5v "JO f ke T* = 12 5°C T* = 25 = -5 'C 5°C -0.01 -0.1 -1.0 L — Emitter Current — ma -10 Fig. A. 6 Small-signal common-base reverse voltage transfer ratio vs. emitter current. U E E o U 30 20 10 7 5 4 3 2 f = 30 mc = 25°C V CE = iuv V r[ = 5 V Minimum •atV CE =5 1 1 h V -I 1 1.0 10 l c — Collector Current — ma 100 Fig. A. 7 Small-signal common-emitter forward cur- rent transfer ratio vs. collector current. 16 12 I v CE = 5 V ""' No 10c Ta se Bandwic ps to 15.7 = 25°C Jth kc ? 6 / k Max ', atl imum rsr c = 10 pa // 2 N ?30- ,' \ <? #' ;; \c «\ J 1 10 -o I o 6 & I 52 0.1 1.0 10 R G — Generator Resistance — k fl Fig. A. 8 Average noise figure vs. generator resistance. Mil V CE ' 5 v l c = 1 ma T A = 25°C A i \ / k NF p ) vs f 100 1.0 10 f - 100 Frequency — mc 1.0 0.8 0.6 s O 0.4 E O 0.2 L 1000 Fig. A. 9 Optimum spot noise figure and optimum generator resistance vs. frequency. 206 Transistor Characteristics I c s u L. £ u _• "o o 15 30 45 60 V ce —Collector-Emitter Voltage — v Fig. A. 10 2N929 transistor: common-emitter collector characteristics. 75 e 3 u Lb £ o _e 3 10 8 - j ^ *tt — «i-^ — i — . — ^•Minimum BV cco <-**i< T A =25°C •^"^ .rUS^ \» - v = o.oi ma 1 ,,= = 0.005 ma r l r° 15 30 45 60 V CE — Collector-Emitter Voltage — v Fig. A. 11 2N930 transistor: common-emitter collector characteristics. 75 1000 _o J? c E 3 u -D I O 100 10 0.001 — 1 ~"tf TT It II * N •u ^ ,' *Ai !* 1 -ji c 1>V ill V CE =5v See Note 6 1000 0.01 0.1 1.0 10 100 l c — Collector Current — ma Fig. A. 12 2N929 transistor: static forward current transfer ratio vs. collector current. Note that these parameters were measured using pulse techniques, PW = 300 /isec, Duty Cycle < 2%. •8 i 3 u "2 a i £ 100 10 0.001 " T in T in T III T 111 t •c -A?E ^•*"Zc°C^ ■\> 1 i-c ■u *-* 1 V CE =5v See Note 6 0.01 0.1 1.0 l r — Collector Current — ma 10 100 Fig. A. 13 2N 930 transistor: static forward current transfer ratio vs. collector current. Note that these parameters were measured using pulse techniques, PW = 300 ptsec, Duty Cycle < 2%. 207 Transistor Circuit Analysis 1.2 1.0 J 1 0.8 "o > .? 0.6 E 0.4 V^ Maximum, V, f at l,= 0.5ma, l c = 10ma i 1 ?: ^5 4* »o ■ *H S > ^^ f<^ V ce =5v, l c = 100,ja-^ Vce = 5v , l c = 1 ma 1 See Note 6 1 0.2 -75 -50 -25 25 50 75 100 125 150 T A — Free-Air Temperature — °C Fig. A. 14 2N929 transistor: base-emitter voltage vs. free-air temperature. 1.2 1.0 8.0.8 o ~o > I 0.6 0.4 0.2 >*, Maximum V lE 1, =0.5ma,l c at = 10 ma ^ ^ 0.5 ' '0- *, * ^ "a ^ v CE = 5v, l c = 100 ya ■ See Note 6 1 -75 -50 -25 25 50 75 100 125 19 T A — Free-Air Temperature — °C Fig. A. 15 2N930 transistor: base-emitter voltage vs. free-ai r temperature. 10 « O) o E 1.0 £ 0.1 o U 0.01 * c Maximum V c , =0.5 ma, 1 1 I.01) a * lc - 1 D ma .Sma.lc'J^. In ~ c , =0.05 mo u — u^T — in ■ ) ma OM° 1 = 1 \ia, \ c ■- r- See Note 6 1 10 -75 -50 -25 25 50 75 100 125 150 T A — Free- Air Temperature — °C Fig. A. 16 2N929 transistor: collector-emitter satu- ration voltage vs. free-air temperature. Note that these parameters were measured using pulse tech- niques, PW = 300 ^tsec, Duty Cycle < 2%. 1.0 2 2 E £ 0.1 o U 0.01 ^__ Maximum V ce( , oi1 at .{ Ib = 0.5ma,lc = 10 ma * . ^ .5 ««■'■•"' — 1 u=o ns mo. >c° 1 ma 1 = ' ..n. \r = 100 ua 1, _ , — , - 1, - 0.5 pa, l c - 10 pa See Note 6 -75 -50 -25 25 50 75 100 125 15 T A — Free-Air Temperature — °C Fig. A. 17 2N930 transistor: collector-emitter satu- ration voltage vs. free-air temperature. Note that these parameters were measured using pulse tech- niques, PW = 300^sec, Duty Cycle < 2%. 208 Transistor Characteristics A.2 Types 2N1 1 62 thru 2N1 1 67 Transistors* TABLE A.4 Electrical characteristics, general. (At case temperature of 25°C± 3° except where noted.) Characteristic Symbol Mil Typ Max Ulit Collector Cutoff Current Vcb = BVcbo (max), I B = Vcb = 2V, I E = Vcb = 15V, In = 0, T c = 90°C (2N1 162, 2N1 163) Vcb = 30V, Ik = 0, T r = 90°C (2N1 164 through 2N1 167) IcBO — 3 125 10 10 15 225 20 20 ffiA *•* mA mA Collector-Emitter Breakdown Voltage Ic = 500mA, V SB = (2N1162, 2N1163) (2N1164, 2N1165) (2N1166,2N1167) BVcsa 35 60 75 — — Vdc Vdc Vdc Emitter Cutoff Current V BE = 12V,Ic = Iebo — 0.5 1.2 mA Collector - Emitter Saturation Voltage Ic = 25A,I B =1.6A Vce(sat) — 0.3 1.0 Vdc Base - Emitter Drive Voltage (at Saturation) Ic = 25A,I B =1.6A Vbe — 0.7 1.7 Vdc TABLE A. 5 Electrical characteristics, common emitter. Characteristic Symbol Min Typ Max Unit DC Forward Current Gain Vce = 2V, Ic = 5A Vce = IV, Ic = 25A llFB 15 65 25 125 65 — Frequency Cutoff Vce = -2V, Ic = 2A f.. - 4 - kc 30 25 20 15 =i 10 1 4 1 1 ° 1 .6 1.0 0.8 06 0.4 Ib = 0.2 AMP i 0.2 0.4 0.6 0.8 V CB , COLLECTOR- EMITTER VOLTAGE (VOLTS) Fig. A. 18 Saturation region, common emitter (constant base current). 1.0 30 25 S. 20 15 3 10 ; • i<i . ( 18- . 0.6 ,.0.4 Ib = 0. 2 AMP 1 *°- 1 t — > Vbb = r— 1 r— 25 50 75 100 125 150 BVces, %RATED MAXIMUM COLLECTOR- EM ITTER VOLTAGE (VOLTS) Fig. A. 19 Collector characteristics, common emitter. * MOTOROLA Inc., Semiconductor Products Division. 209 Transistor Circuit Analysis 25 s? 20 - s 5 15 a io / V i , +25-C _50°C f \ +75-C / / f y ''A A y V CE = 2 VOLTS CASE TEMPERATURE °C 25 20 s £ 15 £ 10 0.4 0.6 1.2 1.6 2.0 l B , BASE CURRENT (AMPERES) Fig. A. 20 Collector current vs. base current. V CE = 2V0LTS CAS F TFMPrPlTIID : or . $ / // +7 5»C- / '/ / L -50»C // y+25'C / V J S 2.4 0.2 0.4 0.6 0.8 1.0 1.; V 8E , BASE-EMITTER VOLTAGE (VOLTS) Fig. A. 21 Output current vs. emitter-drive voltage. 100 140 120 100 80 60 40 20 lc = = 1V = 25A —60 —40 -20 20 40 T c , CASE TEMPERATURE («C) Fig. A. 22 hpp vs. temperature. 60 80 100 ff 10 1.0 o o -° 0.1 0.01 »CB — VZOTCES — — 2N1] 62- 2N1 67 20 40 60 80 100 T c , CASE TEMPERATURE CC) Fig. A. 23 \qq vs. temperature. 210 Transistor Characteristics 75 50 25 25 50 75 -60 - 25 AMP. j 5 AMP Vce = 2V 1 -40 -20 20 40 T c , CASE TEMPERATURE (°C) Fig. A. 24 gpj: vs. temperature. 60 80 100 A. 2a Peak Power Derating The peak allowable power is: p (Tj — Ta — fljA Pss) e,c (i) + * CA(t,/t) Cp is a coefficient of power as obtained from the chart. Tj is junction temperature in °C; T A is ambient temperature in °C; 9ic is junction to case thermal resistance in °C/W; C a is case to ambient thermal resistance in °C/W; 0ja is the sum of 0jr + »ca; ti is pulse width; t is the pulse period; (t,/t) is the duty cycle; Pss is a constant power dissipation and Pp is the additional allowable pulse power dissipation above the amount of Pss. The above equation is usable when a heat sink is used which has thermal capacity very much larger than the transistors thermal capacity. The chart is normalized with respect to the thermal time con- stant, which is on the order of 50 milliseconds for these power transistors. (Fig. A. 25.) t.- Mt „i time — ► EXAMPLE Given: Pss = 10W Ta = 40°C Pulse width (ti) = 1 msec Duty Cycle = 20% »ca = 3°C/W 9 JC = 0.8°C/W T Jm .,= 100°C r *« Solution: Enter the graph at U/r = 1 msec/50msec, and Duty Cycle 20%. Find Cp = 5. Solve equation _ 100-40- (3 + 0.8) 10 Pp - 08_+ 3 x 0.2 5 P P = 29 watts in addition to the steady 10 watts resulting in 39 watts peak. Zi 1000 0.1% 100 10 ^0. \% — % DUTY CY( :l: "ABSOLUlt IMUNKtCUKKtNT 0. 5% % 1 2"/o 1 DUTY CYCLE 5% | 1 10% * '1)7 3 bU% 1 1 100% io-* io-» io-» 10- 1 1 10 Vr, PULSE WIDTH / THERMAL TIME CONSTANT Fig. A.25 Normalized peak allowable power. 211 Transistor Circuit Analysis A.3 Types 2N1302, 2N1304, 2N1306, and 2N1308 n-p-n Alloy-Junction Germanium Transistors* Table A .6 JEDEC registered electrical c naracteristics at 25 C free -air tem perature. PARAMETER TEST CONDITIONS 2N1302 2N1304 2N1306 2N1308 UNIT MIN TYP MAX MIN TYP MAX MIN TYP MAX MIN TYP MAX BV CBO Collector-Base Breakdown Voltage l E = 100 jua, l E = 25 - - 25 - - 25 - - 25 - - V BV EBO Emitter- Bast Breakdown Voltage l E = 100 pa, l c =0 25 - - 25 - - 25 - - 25 - - V *V PT Punch Through Voltaget Veb<i = ' » 25 - - 20 - - 15 - - 15 - — V *'cBO Collector Cutoff Current V CB = 25v. l E = - 3 6 - 3 6 - 3 6 - 3 6 t"> 'E80 Emitter Cutoff Current V EB = 25v, l c = - 2 i - 2 6 - 2 6 — 2 6 IM %E Static Forward Current Transfer Ratio V CE = ' ». l c = 10 ma 20 100 - 40 115 200 60 130 300 to 160 - — V CE = 0.35v, l c = 200 ma 10 100 - 15 no - 20 125 - 20 140 — — *v K Base-Emitter Voltage 1, = 0.5 ma, l c =z 10 ma 0.15 0.22 0.40 0.15 0.22 0.35 0.15 0.22 0.35 0.15 0.22 0.35 V * V CE|wt Collector-Emitter Saturation Voltage l B = 0.5 ma. l c = 10 ma - 0.07 0.20 - - - - - - - - — V l B = 0.25 ma, l c = 10 ma - - - - 0.07' 0.20 - - - - - — V l B — 0.17 ma, \q = 10 ma - - - - - - - 0.07 0.20 - — — V l B = 0.13 ma, l c== 10ma - - - - - - - - — — 0.07 0.20 V "ib Small-Signal Common-Base Input Impedance V CB = 5v, f = Ike l E = — I ma - 21 - - 20 - - 28 - - 28 - ohm •Vb Small-Signal Common-Base Reverse Voltage Transfer Ratio V CB = 5», 1= Ike l E — — 1 ma - 5 xlO" 4 - - 5 x 10-+ - - 5 il0-« - - 5 x 10-* - - h ob Small-Signal. Common-Base Output Admittance V CB = 5«, f = lkt l E = — 1 ma - 0.34 - - 0.34 - — 0.34 — — 0.34 _ fimho K Small-Signal Common-Emitter Forward Current Transfer Ratio V CE = 5v, f = lkc l c = 1 ma - 105 - - 120 - - 135 - - 170 - - * f hfb Common-Base Alpha- Cutoff Frequency V CB = 5v, l E = — 1 ma 3 12 - 5 14 - 10 16 - IS 20 - mc *<ob Common-Base Open Circuit Output Capacitance V CB =5». f = 1 mc l E = - 14 20 - 14 20 - 14 20 - 14 20 Pf c ib Common-Base Open-Circuit Input Capacitance »E5=5«. f = lmc i c = " 13 - " - - 13 - - 13 - Pi tV pT is determined by measuring the emitter-base floating potential V EB(| The collector-base voltage, V rl/ Is Increased until V„„ = 1 volt- this value of »C. = (»PT + 1 v). Table A 7 JEDEC registered switching characteristics at 25°C free-air temperature. PARAMETER TEST CONDITIONSft 2N1302 2N1304 2N1306 2N1308 UNIT MIN TYP MAX MIN TYP MAX MIN TYP MAX MIN TYP MAX +d Delay Time l c = 10 mo, Ijjh = 1.3 ma !„,,= 0.7 ma, V^ (offl _ 0.8 v \ — 1 k fi (See Fig. 1) - 0.07 - - 0.07 - - 0.06 - - 0.06 - /tsec +r Rise Time - 0.20 - - 0.20 - - 0.18 - - 0.15 - fliW +. Storage Time - 0.70 - - 0.70 - - 0.64 - - 0.64 — jusec *f Fall Time - 0.40 - - 0.40 - - 0.36 - — 0.34 — fiiK Q,b Stored Base Charge 'Bill = ' m °< 'c = " mo I s " Fi 9- *) - 800 - - 760 - - 720 - - 680 - peb tt/Vollage and current values shown are nominal; exact values vary slightly with device parameters. TABLE A. 8 JEDEC registered operating characteristics at 25°C free-air temperature. PARAMETER TEST CONDITIONS 2N1302 2N1304 2N1306 2N1308 UNIT MIN TYP MAX MIN TYP MAX MIN TYP MAX MIN TYP MAX NF Spot Noise Figure V CB = 5, l E = - 1 ma f = 1 kc, R s = 1 k n - 4 - - 4 - - 3 - - 3 - db 'Indicates JEDEC registered data (typical values excluded). TEXAS INSTRUMENTS Incorporated. 212 Transistor Characteristics A. 3a Typical Characteristics 12 3 4 5 V CE — Collector-Emitter Voltage — v Fig. A. 26 2N 1302 transistor. 12 3 4 5 Vce — Col lector- Emitter Voltage — v Fig. A. 28 2N1306 transistor. "1.4 I .21. o si- 2 I— fo. 3 <J £ .so. a 80.2 o z t a =: 25° c N E Iv s A £ F NP 4 t^ V v V c = 0.35v \z CE = = -0.35v 1 10 100 Mc I ~ Collector Current — ma Fig. A. 30 Normalized static forward current transfer ratio vs. collector current. 1000 50 E40 t 30 3 u o J 20 "o U I 10 -. /J5*£- T A = 25°C o^S- 1 \»2. Jg^rn r \[= 0.20mo_ .= 0.15ma l«= O.lOma - 1 1 """ 1 1 l,= 0.05ma T- l»= 12 3 4 V CE — Collector-Emitter Voltage - Fig. A.27 2N1304 transistor. 50 40 t 30 u | 20 "5 U 1.10 ■N OS ma T A = 25°C WJJt^r \.= °^&- u^J^p- r |.= 0.1 jma — -l B = 0.05 ma E 1 f 1.7 12 3 4 5 V ce — Collector- Emitter Voltage — v Fig. A. 29 2N 1308 transistor. -60 -40 -20 20 40 60 T A — Free-Air Temperature — °C Fig. A. 3 1 Normalized static forward current transfer ratio vs. free-air temperature. 213 Transistor Circuit Analysis 0.7 0.6 I £.0.5 0.4 0.3 I 0.2 0.1 Ta = 25° C o!^ 1 / ^ -0. 0. 5J° . 1 10 100 1000 Nc — Collector Current — ma Fig. A. 32 Base-emitter voltage vs. collector current. -60 -40 -20 20 40 60 8C T A — Free -Air Temperature — °C Fig. A. 33 Base-emitter voltage vs. free-air temperature, > 0.35 1 0.30 0.25 0.20 £ 1 k 0.15 o U 0.10 0.05 T A = 25 °( It fe ■ J * "-I V a. Lb' 7" It II * /J .- C .a "*> Vti_«d \0 P tr* §3 m o ' 0.35 10 100 1000 ll c I —Collector Current — ma Fig. A. 34 Collector-emitter saturation voltage vs. collector current. o > 0.30 .2 0.25 <* 0.20 E 0.15 q 0.10 0.05 ?* D «^ tf >^ \0«*^ \»^ ^=J0 ma, *c* 200 ma , NPN l,= -0.5ma, l c = -lOma, PNP 1 | 1 | == - l t = 0.5ma, l c = lOma, NPN -60 -40 -20 20 40 60 8C T A — Free-Air Temperature — °C Fig. A. 35 Collector-emitter saturation voltage vs. free-air temperature. 214 Trans is tor Character is t ics 100 8: o O o 3 I 10 X /& X x.k .^ .-^ ^ J&<*-i ' /$ y/V 20 30 40 50 60 70 Ta — Free-Air Temperature — °C Fig. A. 36 Collector cutoff current vs. free-air temperature. 80 I « 8. o U i 3 s o E E 6 o 28 24 20 Z 16 12 T»=25 ,, C f = 1 mc .1,-0 Nt^l / "NP 012 34567 89 |V C ,| — Collector-Base Voltage — v Fig. A. 37 Common-base open-circuit output capacitance vs. collector-base voltage. 10 0.35 0.30 £ * 0.25 - 0.20 £ o 0.15 0.10 1 0.05 V ■25 •c fc ' 4 <0 7/* Lb' /* £.« I i II II " >'* {fa in *' \0 $' & «*>' 10 100 1000 Fig. A. 38 Collector-emitter saturation voltage vs. collector current. 0.35 J> 0.30 "5 > .2 0.25 1 2 <2 0.20 | I 0.15 £ 3 °-'° J. I 0.05 _^ tf y/ \0«**^ 1,-10 mo. *c* 200 m° , NPN l,= -0.5ma, l c = -I0ma, PNP | | | | ' l,= 0.5ma, l c » lOma, NPN -60 -40 -20 20 40 60 80 T A — Free-Air Temperature — °C Fig. A. 39 Collector-emitter saturation voltage vs. free-air temperature. 215 Transistor Circuit Analysis A.4 Types 2N1 529A thru 2N1 532A, 2N1 534A thru 2N1 537A and 2N1529 thru 2N1538 Transistors* TABLE A. 9 Electrical characteristics. (At 25°C case temperature unless otherwise specified.) AQL and inspectio levels apply to "MEG-A-LIFE" series (2N1529A thru 2N1532A and 2N1534A thru 2N1537A) only. Thermal resistance, 0J(;!4 0.6°C/W typical, 0.8°C/W maximum. Panneter Symbol ■ 1 Mia Mix Unit Collector-Base Cutoff Current (V CB = 25V) 2N1529A, 2N1534A* (V CB = 40V) 2N1530A, 2N1535A* (V CB = 55V) 2N1531A, 2N1536A* (V CB = 65V) 2N1532A, 2N1537A* (V CB = 80V) 2N1533, 2N1538 ^BOl ■ I 2.0 2.0 2.0 2.0 2.0 mA Collector-Base Cutoff Current (VCB = 2V) All Types (V CB = 1/2 BV CES rating; All Types Tc = +90° C) ICBO ■ 0.2 20 mA Emitter-Base Cutoff Current (V EB = 12V) All Types lEBO 1 0.5 mA Collector-Emitter Breakdown Voltage (Ic = 500 mA, V EB = 0) 2N1529A, 2N1534A* 2N1530A, 2N1535A* 2N1531A, 2N1536A* 2N1532A, 2N1537A* 2N1533, 2N1538 BVCES ■ ■■■■■■■■■■ - volts Collector-Emitter Leakage Current (V BE = IV; V CE @ rated BV C BO) All Types J CEX ■■■■■■■■■l L^hVAVAbIbVi 20 mA Collector-Emitter Breakdown Voltage (I c = 500 mA, I B = 0) 2N1529A, 2N1534A* 2N1530A, 2N1535A* 2N1531A, 2N1536A* 2N1532A, 2N1537A* 2N1533, 2N1538 BVcEO ■ ■ - volts Collector-Base Breakdown Voltage (IC = 20 mA) 2N15.29A, 2N1534A* 2N1530A, 2N1535A* 2N1531A, 2N1536A* 2N1532A, 2N1537A* 2N1533, 2N1538 BV CB0 ■ 100 ■ 120 - volts Current-Gain (V CE = 2V, I C - 3A) 2N1529A - 2N1532A 2N1534A - 2N1537A 2N1529 - 2N1533 2N1534 - 2N1538 h FEl ■ ■ ■ 40 70 40 70 Base-Emitter Drive Voltage (I c = 3A, I B = 300 mA) 2N1529A - 2N1532A 2N1534A - 2N1537A 2N1529 - 2N1533 2N1534 - 2N1538 V BE ^^^^^^^^^B 1.7 1.5 1.7 1.5 volts Collector Saturation Voltage (IC = 3A, I B = 300 mA) 2N1529A - 2N1532A 2N1534A - 2N1537A 2N1529 - 2N1533 2N1534 - 2N1538 v CE(sat) 1 1.5 1.2 1.5 1.2 volts Transconductance (V CE = 2V, I C = 3A) 2N1529A - 2N1532A 2N1534A - 2N1537A 2N1529 - 2N1533 2N1534 - 2N1538 SFE 1 ■ - mhos * Characteristics apply also to corresponding, non-A type numbers ** Each parameter of the "MEG-A-LIFE" series only is guaranteed to an individual AQL of 0. MOTOROLA Inc., Semiconductor Products Division. inspection level n 216 Transistor Characteristics Fig. A.40 Power-temperature derating curve. The maximum continuous power is related to maximum junction temperature, by the thermal resistance factor. For d-c or frequencies be- low 25 cps the transistor must be operated within the constant Pj = V c X l c hyperbolic curve. This curve has a value of 90 watts at case temperatures of 25 C and is watts at 100 C with a linear relation between the two temperatures such that allowable p 100°- T c 0.8 20 40 60 80 T c , CASE TEMPERATURE (°C) 100 A.4a Collector Characteristics at 25°C: Types 2N1529A thru2N1532A and 2N1529 thru 2N1533 Transistors r 3 S 2 500 "■ 4rm, ^^300^--^ 200 100 50 20 l,= 10 mA 0.5 1 1.5 Vce, COLLECTOR-EMITTER VOLTAGE (VOLTS) Fig. A. 41 Saturation region, common emitter, constant base current. 2.0 50 100 150 V CE , % RATED MAXIMUM COLLECTOR-EMITTER VOLTAGE Fig. A. 42 Collector characteristics, common emitter. fc 3 3 2 500 / 300/ 200 \yS 100 50 20 1, = 10 mA 4 - — 150 f 100 ST a. LU ac . 50 ^ 2 o o 1, = 10 mA ) 1 < s » f — ' 0.5 1 1.5 V CE , COLLECTOR-EMITTER VOLTAGE (VOLTS) Fig. A. 43 Saturation region, common emitter, constant base current. 50 100 150 V cs , % RATED MAXIMUM COLLECTOR-EMITTER VOLTAGE Fig. A. 44 Collector characteri sti cs, common emitter. 217 Zo =50$) 1 i I j 20 V I — i PULSE GENERATOR Transistor Circuit Analysis 0.5 M f TEST TRANSISTOR Fig. A.45 Switching time measuring circuit. Also see Table A. 10. Table A.10 See Fig. A.45. Ic (Amps) V (Volts) Ion (mA) R (ohms) td+t, (yusec) t. (/*sec) t, (/<sec) 2N1529A-32A 2N1529-33 3 3 300 65 10 2 5 2N1534A-37A 2N1S34-38 3 3 200 100 8 3 5 0.05 0.10 0.15 0.20 I* BASE CURRENT (AMPS) 0.25 0.30 Fig. A. 46 Collector current vs. base current. 1.0 2.0 V«, EMITTER-BASE VOLTAGE (VOLTS) Fig. A.47 Collector current vs. emitter base voltage. 150 125 : 100 75 V a = 2 1 _2N1 2^ 534A- 1534 — 2N1537 2N1538 * 2N >29A — 529- ZN1532A 2N1533 1.0 2.0 3.0 Ic, COLLECTOR CURRENT (AMPS) 4.0 5.0 Fig. A.48 D-c current gain vs. collector current. 1.0 2.0 Vb, emitter-base voltage ivoltsi Fig. A. 49 Base current vs. emitter base voltage. .218 Transistor Characteristics 100 < 10 E 1.0 - 8 0.1 .01 V c , = V 2 BV CK ALL TYPES 20 40 60 80 T c , CASE TEMPERATURE (°C) Fig. A. 50 1,-q vs. temperature. 100 140 130 120 110 100 90 80 70 1 - 2N1529A-2N1532A -•~--<2ni529-2ni533 " lc = 3 A "V« = 2 V 2N1534A-2N1537A. 2N1534-2N1538 _60 —40 _20 20 T c> CASE TEMPERATURE (°C) 40 60 80 100 Fig. A. 51 hpp vs. temperature. CJ 140 in CM 130 i— 120 UJ =D 110 <c >■ 100 o 90 UJ 80 10 | c - -1ft | V CE = 2V - 2N1529A-2N1532A / 2N1529-2N1533 N1534A-2N1537A 'N15>' J4-2f L_ U53 1 3 _60 —40 —20 20 40 T Cl CASE TEMPERATURE (°C) Fig. A. 52 gpp vs. temperature. 60 80 100 219 Transistor Circuit Ajialysis A. 4b Determination of Allowable Peak Power When a heat sink is used for increased heat transfer and greater thermal capacity, the following equation can be used to determine the allowable pulse power dissipation, desig- nated as Pp. The allowable pulse power plus the steady state power, P gg , gives the peak allowable power dissipation. Al- lowable pulse power is related by the following equation : Tj — T A — 6] a Ps Pp = fJA r SS «,o(l/C P )+0 OA (Vt) where C P = Coefficient of Power (from peak power derating curve) , Tj = Junction Temperature (°C), T A = Ambient Temperature (°C), 0j O = Junction-to-Case Thermal Resistance (°C/W), CA = Case-to-Ambient Thermal Resistance (°C/W), 0JA = #JC + 0C A > (t x /t) = Duty Cycle = Pulse Width/ Pulse Period, P ss = Steady State Power Dissipation, and P P = Allowable Pulse Power Dissipation Above P ss . T = Thermal Time Constant « 50 msec The peak power derating curve is normalized with respect to the thermal time constant, T . The following example shows the application of this equation in conjunction with the peak power derating curve. EXAMPLE: Given : P ss = 10 W, T A = 40°C, tx = 1 msec, (ti/t) = 20%, CA = 3°C/W, 6, c = 0.8°C/W, T 7 max = 100°C Solution : Enter the derating graph at tj / T = 1 msec/ 50 msec, and duty cycle of 20% . Find C P = 5. Substitute this value and the given parameters into the peak pulse power equation. This gives P P = 29 watts. Thus the peak allowable power is P P + P sg , or 39 watts. 1000 0.1% 100 10 0.5 % =3 ( -H %DUTY CY m ^ = :: "' ABSOLUTE NONRECURRENT T 1 ^ 1% "".. ?%, 1 DUTY CYCLl 5% 10% 2U"}i.- — H-- 50% 1 1 100% 10- 10-' 10-' io- : Vr, PULSE WIDTH / THERMAL TIME CONSTANT Fig. A. 53 Pulse power derating curve. Caution: In all cases the peak pulse power should stay within the Safe Operating Area. 220 SUMMARY CHARTS B APPENDIX TABLE 3.1 Conversion from hybrid to hybrid-^ parameters. h ie - r bb < hie ~ r bb' . _ hie ~ r bb' ;y «, = h -h 1 + h t' ~ h '° ~ h "oe "re = tl oe Tce ft i« -Tbb' [3.56] [3.60] [3.61] [3.63] fen 'be ♦. I i 1!U (a) Common-emitter configuration. Hybrid hie Ke h„ Common- base ha l + *« l + A/6 -hfb 1 + Afb h b h,b 1+A /b Common- collector 1-h, -d + A/c) Tee- equivalent r e r b + 1-a «•. (1- -a)r c a 1 - a 1 (1 - a)r c (c) Approximate parameter conversion formulae. (b) Hybrid equivalent circuit. hi, = 2200 fl />,. = 2 x 10-' h,e = 290 h oe = 30 x 10" 6 mhos (d) Typical values for type 2N929 transistor at J c =4 ma, V CE = 12v. Fig. 3.24 Conversion to common-emitter h- parameters. * For convenience in reference, the original table numbers have been retained in this Appendix. 221 Transistor Circuit Analysis v «b V cb 1 I 1 " . (a) Common-base configuration. 0>) Hybrid equivalent circuit. Hybrid Common- emitter ft,. 1 + ftf. h .ftoe i. 1 + ft/. ft/. 1 + ft/. ftoe 1 + A + "/. Common- collector ft/c ft/c ftrc-1 ft/c ftp ft/c 1 + ft/c ft/c ftoc Tee- equivalent r. +(l-a)r 6 (c) Approximate parameter conversion formulae. />,„ = 7.57 Q h rb = 0.268 x 10- 4 />,„ = -0.996 h ob = 0.103x10-* mhos (d) Typical values for type 2N929 transistor. Fig. 3.25 Conversion to common-base h-parameters. rCH "be L (a) Common-collector configuration. hie Ui+. (b) Hybrid equivalent circuit. Hybrid ft/c ftrc Common- emitter ft/. l"ftre=l -d+ft/e) ftoe Common- base h lb l + ft/6 1- t h ib h ob 1 + ft/o 1 l + ft/6 ftob s-i 1 + A /& Tee- equivalent r , r « ° 1 - a * e 'Vi (l-a)r e -1 1- a 1 (l-a)fe (c) Approximate parameter conversion formulae. h lc = 2200 Q h rc = 0.9999^1.0 ft/ e = - 291 h oc = 30 x lO" 6 mhos (d) Typical values for type 2N929 transistor. Fig. 3.26 Conversion to common-collector A-parameters. 222 Summary Charts EO ^ — f • WV — • — OC (a) Tee-equivalent circuit, common-base. tH"-** bo WV— f— •— WV— • — OC l+p E (b) Tee-equivalent circuit, common-emitter. Tee- param- eter r. P Common- emitter A/. + 1 h,.- *r.(U*f.) Common- base - */b *i» -(! + *»>■?* Aft A/6 hi h ob 1 + ft/b (c) Approximate parameter conversion Common- collector 1 + /. /c ftoc , /■,.(! -ft rc ) "fc + 7 -d+A, e ) formulae. a =+0.996 r c = 9.7 Mil r e = 6.667 fl r„ = 260 ft (d) Typical values for type 2N929 transistor. Fig. Table 3.27 Conversion to tee-parameters. 5.1 Single-stage amplifier formulae. •l A -Z> A-parameters (Second subscript omitted) r L * h - 1 h, -R L *, l + h R L *l [-("- j e)] Common-base tee-equivalent circuit (see Fig. 2.31) a*A + (i_a)^ ' i r, ") i r, ' c '' a r c + r b r ' + r » TT K)* 'c 2-r, + r»(l-a), !i<KP *£ « 1 1+ r 1 + R t r e '» + *«■ « 1 'c ««L r t + R t ^ , --(— £)' " Common-emitter tee-equivalent circuit (see Fig. 2.31) Note: r; = r, + R B ; Rg = circuit resistance in emitter 1, Rl r „ r: n »-3 A -• r. + r;(U/» r ' (1+ ^ u = r»+r;(l+/3), *L + 'I « r* l + r ' + " r * i, V'» *L.. /j + l r„(/9 + l) 'X « r d r : !, «L r, /j, rl + M [i, * 1 • r„</3+l)' r;(l + ^)V r„ ; <y Rt, 'l«'i, P » 1 . *l « 'a r; ^«1 Common-collector tee-equivalent circuit (see Fig. 2.31) r , Wt+r.)(/S+l) = r b + (R L+ r.)(/3 + l), r c »(R t +r.)(/3 + l) R 4 +r 6 1 r tf (l + /3) a + p) — l —- 'i = 1 + /5, r. + R t « rj 1 i i • U/3' R, « r c l+/,.+_a_YL l + i»' \* 1 + ^R«. R L '» « 'c 223 Transistor Circuit Analysis Table 6.1 Type No. 2N930. Electrical Parameters Ic = 13 fia l c = 4 ma Unit bib 2100 17 ohm K„' 0.045 x 10" 6 0.056 x lO" 6 mho A/e* 200 370 h,b 1.5 x 10- 2.3 x 10- u ** 420,000 6300 ohm ^oc" 9 X 10" 6 20.8 x lO" 6 mho h lc " -201 -371 \c" 1 1 • Published data ** Derived data using conversion formulae of Chap. 3 Table 6.: » Type No. 2N930. Electrical 'c = 1 ma / c = 2 ma Ic- = 3 ma Parameters 25° C 100° C 25° C i 100° C 25°C 100° C Unit h le 320 460 340 [ 480 360 500 h ib 30 36 20 ] 24 16 20 ohm K b 0.076 x 10" 6 0.086 x 10' 6 0.11 x 10" 6 l 0.13 x 10" 6 0.15 x lO" 6 0.18 x 10" 6 mho hrb 1.8 x 10' 4 2.5 x lO- 2.0x10- 2.8x10- 2.2 x 10- 3.0 x 10- r c = 1/Aob 13.2 x 10 6 ll. 6 x 10 6 9.1 x 10 6 j 7.7 x 10 6 6.7 x 10 6 5.57 x 10 6 ohm h ic = h ib (1 + h te ) 9630 16,650 6820 1 11,600 5780 10,000 ohm h oc = h ob (X +h te ) 24.4 x 10"* 39.6 x 10" 6 37.4 x lO" 6 | 62.5 x 10" 6 54 x lO' 6 90.5 x 10" 6 mho h re =h ib h ob (l +h te )-h rb 5.5 x 10- 11.7 x 10" 4 5.5 x 10- | 12.2 x 10- 6.4 x 10- 15 x 10- r h =/>ie - T e (l+h fe ) 2230 2750 1710 j 2450 1330 1500 ohm r e = h re /h oe 23 1 30 15 19 I 12 1 17 ohm 224 Summary Charts TABLE 7.1 Class A amplifier design formulae. Item Formula (R T * 0) Formula (Rr=0) Vcch = p c Ri S^.x + 2R T l Q 2 0.5 Ri Ri + 2R T BV max — Aj- 2/ 'O(opt) rr ^^max 2 Pc BV max 2 0.5 0.5 BV„ 4R, 2P C V C c /g = Pc 2/ c SK„ ^cc/q = Pc 2/r (7.10a) (7.10b) (7.10c) (7.10d) (7.10e) (7.10f) (7.10g) (7.10h) (7.10i) Transformer coupling to load assumed. TABLE 7.3 Measured parameters on a 2N930 transistor. ; c, , T,= 175°C T, = 25° C Tj = 175°C v (ma) hi. /b ^.) V BE (v) K BE (v) 1 460 2.2 0.515 0.215 2 490 4.1 0.54 0.24 5 500 10 0.56 0.26 10 490 20 0.58 0.28 20 425 47 0.61 0.31 30 380 79 0.635 0.335 40 335 120 0.655 0.355 50 310 161 0.67 0.370 60 290 206 0.68 0.38 70 250 240 0.695 0.395 225 Transistor Circuit Analysis TABLE 7.2 Class B push-pull amplifier, design formulae Item CE r Ri. CE r ' cc cc r r cc p« Formula (R T \0) VccRL 4(R 't + Rl) V 2 y cc n 2 Ri n RL 4 R T + RL cc Formula (K T =0) Vcc 4RL cc n 2 Pr " 2 ( RL V 4\Rt + RL) ^Vcc-BV^Vcc + ^PcRtV, cc - "'max °C ^T" V*r CC n(R T + RL) 0.785 "cc n 2 Pn — = 2.466 4 Vcc = ^ cc nRi For/ c=*'c max , *<; cc Rt + Rl kV cc *RL (per transistor) kn 4 k 2 V 2 cc 4RL (per transistor) Vcc RL (7.12a) (7.12b) (7.12c) (7.12d) (7.12e) (7.12f) (7.12g) (7.12h) (7.12i) (7.12J) (7.12k) 226 c APPENDIX FREQUENCY RESPONSE PLOTTING C.1 Introduction The frequency response of networks used in transistor amplifiers can be represented by the product of a constant term, and frequency sensitive terms of the form listed below: 1 + jaT, jaT 1 + jcoT a 2 T 2 +2j£coT+l, This limitation on the realizable form of practical transfer functions leads to a quick method of plotting those transfer functions. The key is to use logarithmic coordinates, i.e., logarithm of gain vs. logarithm of frequency. On these co- ordinates, the transfer functions are very accurately described by straight line asymptotes. Phase angle in linear coordinates is normally plotted versus the same log frequency as the gain function. C.2 The Asymptotic Plot The utility of logarithmic coordinates is based on the elementary fact that the logarithm of a product of terms is the sum of the logarithms of the individual terms. Thus, the logarithm of the gain of a complex transfer func- tion is the sum of the logarithms of the component terms. Since the variety of component terms is very limited, as listed in C.1, the basis for rapid sketching of the gain function is established. The logarithmic gain is conventionally expressed in db, where Thus, a gain of 2 corresponds to 6 db; a gain of 10 corresponds to 20 db. Attenua- tion is expressed as negative db. For example, an attenuation of j is expressed The asymptotic plotting technique is based on the surprisingly excellent cor- respondence between limiting asymptotes and the actual frequency response curve Elaboration on the actual plotting techniques is carried out by means of a series of problems, gradually increasing in complexity. Asymptotic gain is plotted 227 Transistor Circuit Analysis on logarithm}? coordinates, while phase shift is plotted on linear coordinates versus log frequency. Note how asymptotes are determined in the following examples, by taking the simplified forms of the transfer functions for extreme conditions. PROBLEM C.l Plot the following transfer function on logarithmic coordinates: 10 G(jco) = ;<u(l + ;W10) Solution: At very low frequency, |G(;o))| - 10 jto Express gain in decibels for the low frequency region : 10 db = 20 log 10 20[log 10 10-log 10 co] (C.l) 20[l-log 10& )]. (C.2) Note that if db is plotted versus log 10 co, the resulting curve will be a straight line as in Fig. C.la (dashed line). This line crosses the zero db axis at w = 10, since at that point, (C.2) = 0. The line is an approximation to the actual fre- quency-response curve only at very low frequencies. 20 t •"> ^ Slope of —20 db/decade CO. — i 3 " -20 1 1 10N>s 100 -40 1+/" _ Slope CO >\ x 10 yS \ of -40 db/decade >< (a) 20 3 —20 - 3 db down Exact curve -40 t I -2 -90 o> ° -135 6 db/octave or 20 db/decade Corner frequency \Q^^ 100 10 - r - 12 db/octave, or 40 db/decade CO- (b) Fig. C.l (a) Separate Bode components which are combined to obtain the resultant diagram of Fig. C.l (b). (b) Bode plot of the transfer function G(jco) = -° JCO ( 1+ '?.) Now consider the very high frequency region where — » 1. In this region, |G(/«)| 10 (; "■® 100 2 (C.3) 228 Frequency-Response Plotting Express gain in decibels: db = 20 log 10 ^ = 20 [log 10 100 - log 10 co 2 } CO = 20[2-21og 10& >] - 40[l-log 10 o>] . (C.4) This is again a straight line; i.e., db is a linear function of log 10 o>. This line crosses the zero db axis where log 10 a> = 1, or at o> = 10, by coincidence, at the same point where the low-frequency approximation curve crosses the zero db axis. Note that the slope (-40) of (C.4) is twice the slope (-20) of (C.2). Figure C.lb shows the superposition of the straight line approximations to the very low- and very high-frequency regions of the |G(/w)| curve, as well as the actual curve. As frequency reduces or increases indefinitely, the curve |G(;"o))| approaches the straight lines asymptotically. Hence, this type of dia- gram is often called an asymptotic diagram. Practically, the degree of approxima- tion represented by the two asymptotic portions to the left and right of their common intersection point, is remarkably good. The excellence of the straight line approximation to the actual curve is the key reason for the widespread use of this type of diagram. Figure C.lb also shows the plot of the phase angle plotted on the same frequency scale. PROBLEM C.2 For the transfer function of Prob. C.l with generalized gain and time constant, calculate the gain error at the intersection of the two asymptotes. Solution: Consider the mathematical expression for |G(/'<u)|. Use K and T instead of the numerical values of the previous example. We have K G{jco) = \G(jco)\ jcoil + ja>T)' K When coT » 1 (high frequency) or when coT « 1 (low frequency), the approxima- tion method applies with high accuracy. Where coT = 1, the approximation is poorest. Here, the asymptotic diagram indicates a value (at the intersection of the two asymptotes) of 20 log 10 KT. The exact value is KT KT KT \G(jco)\ coWl + UT) 2 Wl + 1 \ll KT db = 20 log 10 -^- = 20 log 10 KT - 20 log 10 V2 , approximate error value so that error (db) = -20 log l0 \ll = -10 log 10 2 = -10(a301) = -3.01 db. At this poorest point, the actual curve is very close to 3 db below the value indicated by the asymptotes. There ate other interesting points to note from the asymptotic curve of Fig. CVb. Qiipsive thcr slopes of the two sttafgM Uae sections: High-frequency atope = -40 229 Transistor Circuit Analysis Suppose we let frequency o> 2 = 2<o lt a one octave range, and determine the change in decibels over this octave for the two slopes. Therefore, the changes at the low- and high-frequency slopes are, respectively, -20 log I0 ^ = -20 log l0 2 = -6.02 db, -40 log I0 -^ = -40 log 10 2 = - 12.04 db. The slopes of the straight line approximations are 6 db/octave and 12db/octave, respectively, for the low-frequency and high-frequency regions. For a decade band, <u,/<u, = 10, log l0 Wj/wj = 1, and the slopes become 20 db/decade and 40 db/decade, respectively. Therefore, 6 db/octave = 20 db/decade, (c.5) 12 db/octave = 40 db/decade . (^.6) Six db/octave means that gain is changed by a factor of two for a doubling of frequency. Twelve db/octave means that gain is changed by a factor of four for a doubling of frequency. With respect to plotting phase angle, the method is more sophisticated than the one for amplitude. However, it is simple enough for practical purposes. Con- sider the phase shift at very low frequency of the function of (C.l). The approximate low frequency expression is G(;<u) = — i which corresponds to a 90° phase lag. The high-frequency approximation Uto) 2 T indicates a phase lag of 180°. The phase shift versus frequency curve is plotted in Fig. C.lb. At extreme frequencies, actual phase shift approaches these asymptotic values. At aT = 1, phase shift may be found directly from the transfer function (C.l). Thus, G(jca) -: K KT KT y'«u(l + jaT) ;(1 + ;') 1/90° V2 /45° At aT = 1, phase lag is 135°, as shown by Fig. Clb. The phase angle can be estimated directly from the asymptotic gain diagram by applying the following When asymptotic slope is db/octave, phase angle approaches 0°. When asymptotic slope is i 6 db/octave, phase angle approaches i90°. When asymptotic slope is ± 12 db/octave, phase angle approaches 1 180°. When asymptotic slope is ± 6n db/octave, phase angle approaches tntr/2, By means of these rules, combined with an actual calculation of phase angle at certain critical points, phase angle curves are easily plotted. PROBLEM C. 3 Plot the gain and phase characteristic of where 1 + ja>T K= 10. 230 Frequency-Response Plotting Solution: Refer to Fig. C.2. At low frequency, the asymptote is simply K, a horizontal line whose ordinate is given as db = 20 1og 10 K. At high frequency, |G(/&))| = K/oT, which intersects the low-frequency asymp- tote at <oT - 1, or co = 1/T. The slope at high frequency falls off at 20 db/decade, since the high-frequency slope has co to the first power only in the denominator. The low-frequency phase shift is approximately zero, approaching a 90° lag at very high-frequency. At <oT = 1, phase lag may be calculated from G(jco) = K hence, phase lag = 45°. 20 t £ 4 -45 a c -90 db = 201og 1(> K 1 I 10T T Corner frequency) <0 —*■ High frequency asymptote (20 db/decade) Note: Negative phase angle — phase lag Fig. C.2 Plot on Bode coordinates. -20 log l0 -i- - Fig. C.3 Bode plot of transfer function. Note that the phase moves toward -90 and then returns to its asymptote. PROBLEM C.4 Sketch the asymptotic diagram and the approximate phase char- acteristic of the following transfer function: G(; w ) Solution: Refer to Fig. C.3. 1 + jcoT 2 1 +;'<u7", r t >r 2 PROBLEM C.5 Sketch the asymptotic diagram and the approximate phase char- acteristic of the following transfer function: G (, w ) = i±i^, ri >T 2 . 1 + j<oT 2 231 Transistor Circuit Analysis Solution: Refer to Fig. C.4. 20io Kl0 -zr - t 20 1og 10 K~j 2*60 •a 01 -60 Asymptotes l/r^ i/r, i/r 3 /i/r 4 Fig. C.4 Bode plot of transfer function. Note that the phase moves toward +90 and then returns to its 0° asymptote. Fig. C.5 Bode plot of transfer function. Note relative con plexity of phase diagram. PROBLEM C.6 Sketch the asymptotic diagram and the approximate phase char- acteristic of the following transfer function: G(jco)= K (1 + ja)T 2 )(l + jcoT 3 ) (1 + ;o)r,)(l + jcoT t ) Solution: Refer to Fig. C.5. r t >7i> T 3 >T 4 . PROBLEM C.7 Sketch the asymptotic diagram and the approximate phase char- acteristic of the following transfer function: G(/Vu) Solution: Refer to Fig. C.6. K (i + joW + /Vuz;) ' T X >T 2 ■ 12 db/octave 12 db/octave Fig. C.6 Bode plot of transfer function. 12 db/octave Fig. C.7 Bode plot of transfer function. PROBLEM C.8 Sketch the asymptotic diagram and the approximate phase char- 232 Frequency-Response Plotting acteristic of the following transfer function: (jw)(l + JcoT 2 ) Solution: Refer to Fig. C.7. C.3 More Complex Frequency-Response Functions The most general frequency- response functions en- countered in elementary servomechanisms may be represented by proper fractions of the form G( iu ) KMia,) {jo>)"B(jco) The A(j<o) and B(/a>) are polynomials with real coefficients which may be fac- tored into the product of linear and quadratic expressions of the following forms: Linear factors, 1 t jaT- Quadratic factors, -a/T 1 + 2Cja>T + 1. The quadratic factors have complex zeros. The values of £ and T are dependent on the numerical values of the parameters. In general, transfer functions consist of combinations of terms of these types, with the degree of the denominator exceeding the degree of the numerator. Typical forms taken by these functions are listed below: G(i )- 1+ i0iT% G(i )- ^ + ;&>ri) lUW; (I + jvTJO. + jvTJ ,W> ' i ffl (l + jcoTJ (1 + jcoT 3 ) G 3 0V>=— -~ 7 — ^— — — , G 4 (; W )= 1 + iaT > (-^T' + D + jlCcoT, -<o 2 a+jcoT 2 ) G 5 (/«,) = d+W , G t (jo>) = (Lb W(l + W [(-a> 2 r 2 + l) + ;2^ra)][l + ;6>rj (1 + jcoTJ (1 + ja>TJ PROBLEM C.9 Plot the Bode diagram of a quadratic function, using the standard expression G(icu) =— — — ^— -— — . (C.7) -co T + ICioi T + 1 • Solution: As before, determine asymptotes for low and high frequencies. At low frequency, |G(;'a>)| = 1 with zero phase shift. This corresponds to db = 20 log 10 1 = 0. The low frequency gain is thus zero db. At very high frequency, 1 JO(/»)| - with 180° phase shift. Converting gain at high frequency to db, db = -20 log.otuT 2 = -40 log I0 w T . This expression, plotted on suitable coordinates, is a straight line with a nega- tive slope of 40 db/decade (corresponding to 12 db/octave), with phase shift 233 Transistor Circuit Analysis Resonant peak -90 o- -180 Fig. C.8 Bode plot of quadratic function. approaching 180° at very high frequencies. This straight line asymptote inter- sects the zero db axis where -40 \o% m cjT = 0, or <oT « 1 . Thus, a quadratic ex- pression of the form of (C.7) has two asymptotes, the same as for a linear ex- pression, except that at the intersection point of the asymptotes (the corneT frequency), the slope of the asymptotes changes by 40 db/decade. Consider now the region in the vicinity of the corner frequency, where the asymptotic approximation is apt to be least accurate. In (C.7), let a>T = 1. Then G(j(o) -. 2do)T 1 Hi (C.8) This expression indicates a gain of l/2£ and a phase lag of 90°. Plotting the high and low frequency asymptotes, the exact corner frequency point, and sketch- ing in phase shift, the approximate gain curve of Fig. C.8 is quickly drawn. The damping factor £ for the quadratic expression is exactly the same «s the oae for the simple second order servo whose characteristic equation is also a quad- ratic. A zero damping factor means infinite amplitude at the corner or "resonant" frequency. Heavy damping eliminates the resonant p**k. A damping factor greater than unity means that the quadratic expressmen »ay be factored into two linear expressions which can be plotted by methods previously described. Am- plitude and gain functions of the quadratic plotted versus dimensionless frequency <oT as a function of the parameter £, are shown by Pigs. C.9 and CIO, re- spectively. 20 10 -10 -20 -30 -40 o.i b> \aL£=0.2 -\V^£=0.25 \N^\^£=0.3 <r=o.< £=o.« 1 / ) s ).4 = 0.5 /-? ( j,.\ — 1 -a> 2 T 2 + 2Cj(oT+l 1 1 1 0.2 0.3 0.4 0.5 0.6 0.8 1.0 (OT- 5 6 8 10 Fig. C.9 Magnitude of output/input versus dimensionless frequency (oT for various values of £. 234 lit • k a • -o -20 -40 -60 a. -60 c » -100 o -120 -140 -160 -180 0.1 0.2 Frequency-Response Plotting ^^1 --r=o.o5 £>--f =0.15 -\>-r=o.2 r=o.4'v \V U^-i =0.2b ^■f =0.3 r=o. f=0. f = 0. r=i. 5 ^s 6"0 i 0-^ fel^ ill ^ 0.3 0.4 0.5 0.6 0.8 1.0 5 6 8 10 Fig. CIO Phase shift of output/input versus dimension less frequency CjT for various values of £. PROBLEM CIO Sketch the Bode diagram of the expression 1 G(;e>) = -o) 2 T 2 + jcoT + 1 (C.9) Solution: Compare the middle term of the denominator with the standard form 2£ja>T, 2£ = 1 or £ = -§ . Refer to the £ = \ of Fig. C.9. Note that gain is db at coT = 1 for this damping factor. Figure CIO shows the companion phase shift curve. Consider now the plotting of transfer functions containing linear and quad- ratic factors. This may be carried out in a routine manner by remembering that the contributions of each factor of the gain function may be added, since a multi- plying factor becomes an additive term when using logarithmic (decibel) units. Phase shifts an, of course, directly additive. PROBLEM CI 1 Plot the gain and phase characteristics on Bode coordinates of the transfer function G(ja) = K T t >T 2 {1 + jaT t )(X + ioT t ) ' Solution: Proceeding as before to determine the gain characteristic, K (CIO) |G(;o>)| Expressed in db: db = 20 log, |i + /'<yr 1 ||i + ; jr J K = 20 (log 10 K - log l0 Jl. + ja> T t | - log 10 1 1 + jo T 2 | ) . (c. 11) Note that each factor in the transfer function makes its separate contribution to 235 Transistor Circuit Analysis the decibels of attenuation, and in plotting the asymptotic diagram, each term may be considered separately. Thus, as shown by Fig. C.lla, the separate asymptotic contributions of each term are drawn to provide an asymptotic or straight line approximation to the entire curve of Fig. C.llb. This approximation is at its best in regions far from the corner frequencies. Similarly, each term of the transfer function contributes its phase shift component, which may be added, as shown by Fig. C.llc. 20 log 10 K t (a) 3 20 1og 10 K t (b) -=■ o 1 1 T 2 F- \ ^xc ^^ . «. / ^^. db/octave i ^W> 6db/octav>- 6 db/octave ^^.^ T-C^^^ Resultant asymptotic plot^ 1 ^^. 1 2 1 / s\ 12 db/octav Actual resultant ^^V. i curve ^v 1 Vs. t V L. 01 V -o ■- -90 — o> c o S-180 i/r, l/r, ^^^^ Arcton 0)T l ^"--^- Arctan fc)T 2 ^v Combined >«^phase angle Phase lag due to l/(l+;<y7\) = arctan <uT, ; Phase lag due to l/(l+/ft)T a ) = arctan coT 2 ; Over-all phase lag is sum of separate phase lag components. (c) Fig. C.ll (a-b) Bode plot showing how components of asymptotic diagram are added to develop ovei (c) Bode plot showing how separate phase lags of transfer function are added together. ■ II di i a gram. Note in Figs. C.lla- c that corner frequencies for the resultant asymptotic curve occur at co = 1/T X , and co = 1/T 2 . Note further that each successive corner frequency marks the point where the corresponding term of the denominator starts to increase rapidly with frequency, resulting in a sharper slope of the asymptotic line following the corner point. The phase shift curve tends toward the values corresponding to the gain asymptote slopes; i.e., 90° for 6 db/octave, 180° for 12 db/octave, etc. The phase shift curves tend toward these values only when corner frequencies are widely separated. When the corners are less than a few octaves apart, the limiting phase shift values constitute only a crude guide. A more accurate plot would require calculating the exact phase shift at a few well chosen points. Since the corner frequencies represent points of poorest approximation, it is particularly convenient to calculate phase shift at these corners. In practical cases, phase shift is important in limited regions where exact calculations may be made once the approximate phase curve is known. PROBLEM C.12 Plot the Bode curves for the function Kd + joTJ G(jo) r, > t 2 . jco (1 + jco T 2 ) Solution: Refer to Fig. C.12. At very low frequency, K ,„,, M K (C.12) G(;a>) = )<o \G(jco)\ This curve, or an extension thereof, crosses the zero db axis at co = K. At co = 1, db = 20 log, K. This straight line extends indefinitely to the left (toward co = 0, 236 Frequency-Response Plotting on the logarithmic scale). Phase shift at very low frequency tends toward 90° lagging. At co = 1/7", , the term in the numerator starts to increase substantially in magnitude, thereby introducing a leading phase shift component. The increase in <o7i starts to balance the increase in the u> of the denominator (coT 2 still much less than unity), so that the gain curve flattens out. The (1 + jaT^) factor in the numerator creates a positive slope change of 20 db/decade to balance the initial negative slope of 20 db/decade. At o = 1/T 2 , the second denominator factor introduces a second corner frequency, adding a 20 db/decade negative slope to the asymptotic gain curve. Phase shift tends toward the values corre- sponding to the different slopes between the corner frequencies. 6 db/octave db/octave jL 20 log 10 K t -o -a t : 6db/oc 12 db/octave T *» N Asymptotes 6 db/octave 1 l/7\ K l/T. l/T x l/T, -90 Fig. C.12 Bode plot of transfer function PROBLEM C.13 G(;a)) Plot the Bode diagram of the transfer function K Ua>ya + io>T) Fi g. C.13 Bode plot of transfer function. (C.13) Solution: Refer to Fig. C.13. The gain curve has a single corner frequency at «u = l/T. The initial slope is 12 db/octave until the comer frequency, at which point the curve falls at 18 db/octave. Phase shift is 180° lagging at low fre- quencies, approaching 270° at higher frequencies. PROBLEM C.14 Plot the Bode curves for G(/fl>) = K Uo>)(0.1;o + 1) Solution: Refer to Fig. C.14. This shows the Bode plots for K = 1. (C.14) 20 10 -10 I -20 -30 -40 -50 - V K 3- O 90 2 -Ph ise cu rve o 3 <o - Ej (act curve- -»Ss; k Asym >tote -180 l_ s « <o 5 - -270 0.1 10 100 CO- Fig. C.14 Bode diagram, amplitude and phase. 237 Transistor Circuit Analysis PROBLEM C. 15 Plot the Bode curves for K Giico) = {jo){jo + l)(j^ + l\ Solution: Refer to Fig. C.15. This shows the Bode plots for K = 2. 10 -10 ! -20 -30 -40 -50, - E xact ^ curve - V - Phase \ ^*-- Asymptotes - curve " - - - 0.1 13 3- ■90 ■180 -270 100 1 10 Fig. C.15 Bode diagram, amplitude and phase. (C.15) 238 DISTORTION CALCULATION D APPENDIX D.1 Distortion Distortion is a measure of the degree to which a given periodic waveform departs from non-sinusoidality. Implicit in this definition is the presence of a fundamental frequency component, upon which distortion com- ponents <are superimposed. An exact analysis of the distortion in a given periodic waveform is routinely carried out using numerical methods based on Fourier series. By this means, it is possible to achieve any degree of accuracy justified by the given waveform data. Furthermore, specific higher harmonics may be investigated where desired. However, this method of harmonic analysis is very tedious, and much too refined for the simpler problem of determining distortion in the output of a power amplifier. The principal distortion components in power amplifier output are second and third harmonic, and frequently, even the third harmonic component may be neglected. Accordingly, simplified methods of Fourier analysis have been devised to give only the important low-order distortion components. These methods, taking no more than a few minutes to apply to a specific waveform, are summarized in Figs. D.l and D.2. Figure D.l applies when only second harmonic distortion is present, while Fig D.2 applies when the distortion includes second and third harmonic components. Individual distortion components may be designated by the ratios of their individual amplitudes to the fundamental amplitude. Total distortion is usually calculated as follows: where D t represents the amplitude of the i-th harmonic, while D, represents the fundamental amplitude. Using the formulae of Fig. D.2, the following expressions may be derived for the components denoted by y of a distorted waveform: 7o . i [ Xl + x, + 2 (x '+*")], 6 ,r» = ~[*, + x, -x'-x"], jr $ » i-[x,-x,-2x'+2x"]. ® (D.la) (D.lb) (D.lc) (D.ld) X. + Xn — 2 X„ x„ — quiescent value of waveform Fig. D.l Calculation of second har- monic distortion component in an ap- proximately sinusoidal waveform. Higher harmonics are negligible; D = percent distortion. Xl 1 I I l\ I Time — 1 | I --I J I I I I <ur I I I I I I 1 1 1 1 O)t +77 Di (%) - *' + x 2 — -x 2 -h X ' — X x' — X x 100 D, *1 -x 2 - 2(x'- x") ( '° 2( D, =Vl X \ ~ x 2 + x'- *") f 3 + D\ Fig. D.2 Calculation of second and third harmonic distortion components in an approximately sinusoidal wave- form. Higher harmonics are negligi- ble; D = percent distortion. 239 Transistor Circuit Analysis PROBLEM D.l Refer to Fig. D.2. Analyze the wave described by the following parameters for distortion components: x, = 2, x 2 = 0, x q = 1, x' = -, x" = 1 . Also use (D.l). 2 2 Solution: Using direct substitution, the following results are obtained: 2+0- 1-4 D 2 (%) = 1 !=0, 2+1 - 1 2 2 2-0-2(1-1) Z) 3 (%) = ? L. = 0> 2 (2 - + 1 + I) yo= I[2 + + 2(2)] = l, o yi =I[2-0 + i--l] = l, y 2 = l [2 + o- |-i] = o, y 3 = I [2 - - 3 + 1] = 0. 6 The wave exhibits no second or third harmonic distortion components. As a matter of fact, at the points defined in this problem, the wave has the parameters of a pure sinusoid. Test the formula of Fig. D.l for second harmonic distortion: D ' (%) =2i^) xl00 = a This result is of course necessary for consistency. 240 ...U. LIST OF SYMBOLS E APPENDIX A v B BV CBO BV CEO BV CER BV CES BVEBO BV R c c ib C,c Cie C ob c oc c oe E (hfb ( hle h FB hfb h FC "/c h FE h, e current gain small-signal average power gain voltage gain base electrode breakdown voltage, collector to base, emitter open breakdown voltage, collector to emit- ter, base open breakdown voltage, collector to emit- ter, with specified resistance be- tween base and emitter breakdown voltage, collector to emit- ter, with base short-circuited to emitter breakdown voltage, emitter to base, collector open breakdown voltage, reverse collector electrode input capacitance (common-base) input capacitance (common-collector) input capacitance (common-emitter) output capacitance (common-base) output capacitance (common-collector) output capacitance (common-emitter) emitter electrode small-signal short-circuit forward current transfer ratio cut-off fre- quency (common-base) small-signal short-circuit forward current transfer ratio cut-off fre- quency (common-collector) small-signal short-circuit forward current transfer ratio cut-off fre- quency (common-emitter) static value of the forward current transfer ratio (common-base) small-signal short-circuit forward current transfer ratio (common-base) static value of the forward current transfer ratio (common-collector) small-signal short-circuit forward current transfer ratio (common- collector) static value of the forward current transfer ratio (common-emitter) small-signal short-circuit forward current transfer ratio (common- emitter) h IB hib h ic h jc h IE h ie h OB h ob h 0C hoc h OE hre I B •b 'c 'c ICBO 1 CEO ICER ! CEX tcES static value of the input resistance (common-base) small-signal value of the short-cir- cuit input impedance (common-base) static value of the input resistance (common -collector) small-signal value of the short-cir- cuit input impedance (common- collector) static value of the input resistance (common-emitter) small-signal value of the short-cir- cuit input impedance (common- emitter) static value of the open-circuit out- put conductance (common -base) small-signal value of the open-cir- cuit output admittance (common -base) static value of the open-circuit out- put conductance (common-collector) small-signal value of the open-cir- cuit output admittance (common- collector) static value of the open-circuit out- put conductance (common-emitter) small-signal value of the open -cir- cuit output admittance (common- emitter) small-signal value of the open-cir- cuit reverse voltage transfer ratio (common -base) small-signal value of the open-cir- cuit reverse voltage transfer ratio (common -co Hector) small-signal value of the open-cir- cuit reverse voltage transfer ratio (c ommon-emitter) base current (d-c) base current (instantaneous) collector current (d-c) collector current (instantaneous) collector cut-off current (d-c), emit- ter open collector cut-off current (d-c), base open collector cut-off current (d-c), with specified resistance between base and emitter collector current (d-c), with speci- fied circuit between base and emitter collector cut-off current (d-c), with base short-circuited to emitter 241 Transistor Circuit Analysis 'EBO If •f Ir •r I B n BE 'CE Pi Pl Po Rb r b Re R e Rl T emitter current (d-c) emitter current (instantaneous) emitter cut-off current (d-c), collec- tor open forward current (d-c) forward current (instantaneous) reverse current (d-c) reverse current (instantaneous) saturation current region of a device where electrons are the majority carriers region of a device where holes are the majority carriers total power input (d-c or average) to the base electrode with respect to the emitter electrode collector junction dissipation total power input (d-c or average) to the collector electrode with respect to the base electrode total power input (d-c or average) to the collector electrode with respect to the emitter electrode total power input (d-c or average) to the emitter electrode with respect to the base electrode large-signal input power small -signal input power load power large-signal output power small-signal output power total power input (d-c or average) to all electrodes small-signal forward resistance external base resistance base resistance external collector resistance collector resistance external emitter resistance emitter resistance saturation resistance load resistance temperature T A ambient temperature Tq case temperature Tj junction temperature 8 thermal resistance thermal resistance, ambient i-A e i-C Vbb Vbc v bc junction to thermal resistance, junction to case base supply voltage (d-c) base to collector voltage (d-c) base to collector voltage (instantaneous) Vbe base to emitter voltage (d-c) v be base to emitter voltage (instantaneous) Vcb collector to base voltage (d-c) v cb collector to base voltage (instantaneous) V cc collector supply voltage (d-c) V C e collector to emitter voltage (d-c) v ce collector to emitter voltage (instantaneous) ^CE( sat ) collector to emitter saturation volt- age V EB emitter to base voltage (d-c) Veb emitter to base voltage (instantaneous) V EC emitter to collector voltage (d-c) v ec emitter to collector voltage (instantaneous) Vee emitter supply voltage (d-c) V F forward voltage (d-c) vp forward voltage (instantaneous) V CBF d -c open-circuit voltage (floating potential) between the collector and base, with the emitter biased in the reverse direction with respect to the collector. V ECF d-c open-circuit voltage (floating potential) between the emitter and collector, with the base biased in the reverse direction with respect to the collector Vr reverse voltage (d-c) v R reverse voltage (instantaneous) 242 INDEX Acceptor impurity, 4 Amplifier performance: small-signal, 97 common-emitter, 97 common-base circuit, 107 common-collector, 109 formulae, 113 Amplifiers: basic circuits, 17 capacitor-couples, 124 direct-coupled, 121 emitter-follower, 54 multi-stage, 121 performance calculations, 52 Approximation techniques, 92 Audio amplifier, 97 Avalanche effect, 161 i Base spreading resistance, 17 Bias circuits: constant base voltage, 71 temperature sensitivity, 71 general configuration, 79 emitter bias, 83 approximate analysis, 92 Bias compensation, 85 Bias drift, 149 Bias point stability, 69 Black box, 39 Bode diagram, 199 By-pass capacitor, 124, 155 Capacitances, 22 Capacitor coupling, 124 Carriers: majority, 4, 6 minority, 4, 6 Characteristic curves, 24 Collector resistance, 17 Classes of operation, 178 Class A push-pull, 178 Class B push-pull, 180 Collector-base feedback, 81 Collector-base leakage, 67 Common-base: characteristics, 25 circuit, 107, 127 connection, 13, 17, 173 parameters, 47 derivation, 48 power gain, 14 voltage gain, 14 Common-collector: characteristics, 26 circuit, 17, 109, 176 Common-emitter, 17 characteristics, 25 circuit, 17, 97 Complementary transistors: direct-coupled, 155, 156 d-c feedback, 157 Complex impedance, 126 Composite characteristics, 179, 181 Composite load-line, 181 Constant power hyperbola, 164 Conversion formulae, 45, 52-4 Coupling capacitor, 124 Covalent bonds, 2, 8 Crossover distortion, 181 Current feedback, 190, 194 Current gain, 26, 32 Cut-off, 73 Cut-off frequency, 113 Cut-off region, 160 D-c bias, 67 D-c feedback, 145, 157 D-c models, 20 D-c stabilization, 73 Diffusion, 5 Diffusion current, 6 Diode compensation, 85, 95 Direct coupling, 144 stability with temperature, 148 Distortion, 163, 174 Donor impurity, 4 Driving impedance, 121 Dynamic resistance, 8 Ebers-Moll model, 15 Electron-hole pairs, 3, 6 Emitter bias, 83 Emitter-follower, 54, 95, 109, 121 current gain, 112 gain, 56 voltage gain, 1 12 Emitter resistance, 17 Energy gap, 2 Equilibrium, 6 Equivalent circuit, 41 Ebers-Moll model, 15 d-c models, 20 hybrid-77, 16, 21, 62, 114 small-signal, 16 tee-equivalent, 16, 43 Equivalent model, 41 Excitation level, 2 Feedback, 36, 186 Feedback amplifier, 190 Feedback and distortion, 188 Feedback and frequency response, 187 Feedback, types, 190 Forbidden gaps, 2 Forward conductance, 22 Frequency response, 124 asymptotic diagram, 137 effect of emitter resistors, 129 high-frequency response, 132 interstage coupling network, 125 low-frequency, 124 multiple time constants, 136 universal curve, 126 Gain margin, 198 Gain stability, 201 General bias circuit, 74, 79 General bias equation, 74, 77 Generator resistance, 175 H h-parameters, 38 High-frequency behavior, 22 High-frequency circuit, 140 High-frequency performance, 113 Hybrid parameters, 38 conversion to tee parameters, 43 conversion to hybrid-77 parameters, 64 243 Transistor Circuit Analysis Hybrid-77 : calculations, 115 capacitances, 114 circuit, 114 conversion formulae, 64 equivalent circuit, 62 parameters, 114 measurement, 116 I Impedance matching, 138 Impedance variation, 108 Impedance variation with R 6 , 101 Impedance variation with R^ , 101 Incremental output impedance, 33 Input impedance, 97, 121 Instability, 187 Integrator, 202 Intermediate frequency range, 136 Interstage, coupling circuit, 130 Intrinsic resistivity, 3 Junction temperature, 85 Junction transistor, 13 K Kernel, 2 Large-signal operation, 30 Leakage, 67, 68, 90 current, 67 equivalent circuit representation, 68 Load line, 27, 143 Load line curvature, 179 Load line of input circuit, 28 Logarithmic plotting, 131, 137, 140 Low-frequency response, 124 Low-level transistor characteristics, 72 M Majority carriers, 4 Miller effect, 118 Minority carriers, 4 Multiplication effect, 161 Multi-stage amplifier, 121, 197 N Negative feedback, 187 Nonlinearity : nonlinear behavior, 20 nonlinear operation, 30 nonlinear region, 30 Nyquist criterion, 196 Nyquist plot, 198 Nyquist point, 196 Open-loop, 186, 196 Operating limits, 160 Operating point, 26, 67 drift, 67, 69 Operational amplifier, 200 Output admittance, 39 Output impedance, 97 Output transformer, 141 p-n junction, 5 Parameter definition, 39 Positive feedback, 187 Potential barrier, 5 Power amplifier, 160 Power amplifier design equations, 171 Power dissipation, 95 Push-pull amplifier, 178 design formulae, 182 equivalent circuit, 180 Quiescent point, 26 Recombination, 3 Rectifier, 5 circuit, graphical analysis, 11 equation, 7 po we r di s s ipat ion , 11 static equivalent circuit, 16 Reverse leakage, 7, 10 Reverse voltage ratio, 39, 46 Saturation, 36, 73 Saturation region, 160 Saturation resistance, 162 Self-heating, 85 Semiconductor crystal, 2 Single-stage amplifier, 97 Small-signal: amplifier, 97 operation, 30, 38 Stability, 196 Stability factors, 73-76, 81 Stability margin, 198 Static characteristics, 24 Summing circuit, 200 Surface leakage, 10 Switched operation, 35 Switching circuit, 93 Tee-equivalent circuit, 46 Temperature effects, 68, 162, 168 Thermal resistance, 162 Thermal runaway, 18, 89 Thevenin's theorem, 71, 129 Three-stage amplifier, 148 Transconductance, 166 Transformer-coupled amplifier, 141, 165 Transformer coupling, 138 equivalent circuit, 138 frequency response, 140 Transistor: basic circuits, 17 bias, 13 breakdown, 19 current gain a , 13 current gain |3 , 14 description, 12 junction transistor, 12 parameters, 25 Two-stage amplifier, 125, 129, 133 Two-transistor circuit, 34 U Universal curve, 126 V Voltage feedback, 190 Voltage gain, 33 244 SIMON and SCHUSTER TECH OUTLINES COMPLEX VARIABLES by Arthur Hauser, Jr. including 760 solved problems $4.25 #18901 FOURIER ANALYSIS by Hwei P. Hsu including 335 solved problems $4.25 #18904 ENGINEERING MECHANICS by Lane K. Branson including 1095 solved problems $4.50 #18902 PULSE CIRCUITS by C. H. Houpis and J. Lubelfeld including 310 solved problems $3.95 #18905 FEEDBACK AND CONTROL SYSTEMS by Sidney A. Davis including 355 solved problems $3.95 #18903 TRANSISTOR CIRCUIT ANALYSIS by Alfred D. Gronner including 235 solved problems $4.25 #18906 VECTOR ANALYSIS by Hwei P. Hsu including 580 solved problems $3.95 #18907 IN PREPARATION PROBABILITY, STATISTICS, AND STOCHASTIC PROCESSES by Louis Maisel ENGINEERING THERMODYNAMICS by Harwood Mullikin LOGIC DESIGN FOR DIGITAL COMPUTERS by Frank Flanagan and Henry S. Sacks ELECTRONIC CIRCUIT ANALYSIS by Andrew Cohen FLUID MECHANICS by Lloyd Polentz ' DIFFERENTIAL EQUATIONS by Edward Norman GENERAL TOPOLOGY by Paul Slepian ENGINEERING MATHEMATICS by Bernard G. Grunebaum SET THEORY by Paul Slepian UNIVERSITY PHYSICS by Solomon Liverhant and Lane K. Branson