TRANSISTOR
CIRCUIT ANALYSIS
SIMON md SCHUSTER
TECH OUTLINES
ALFRED D. GRONNER
theory and
stepbystep
solutions to
235 problems
TRANSISTOR
CIRCUIT ANALYSIS
ALFRED D. GRONNER
SingerGeneral 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 socalled
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 seniorlevel,
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 largesignal parameters.
Nonlinearities, in particular, are easily investigated by means of the graph
ical methods described.
Chapter 3 provides a thorough coverage of the smallsignal equivalent circuit,
with emphasis on the teeequivalent and hybrid configurations. The hybrid* circuit
is introduced in connection with the highfrequency 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 smallsignal amplifier. Multi
stage amplifiers, together with various feedback circuits, are considered in Chap. 6.
Power amplifiers, both singleended and pushpull, 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 highgain
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 pn Junction 5
1 .4 The Transistor 1 2
1 .5 The EbersMoll 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 DC Models 20
1.10 The HybridJt 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 LargeSignal AC Circuits 30
2.5 Supplementary Problems 37
SMALLSIGNAL EQUIVALENT CIRCUITS
3.1 Introduction 38
3.2 Hybrid Equivalent Circuit 38
3.3 TeeEquivalent Circuit 43
3.4 CommonBase Parameters 47
3.5 Derivation of CommonBase Parameters 48
3.6 Calculation of Amplifier Performance 52
3.7 Hybridjr Equivalent Circuit 62
3.8 Supplementary Problems 66
8
BIAS CIRCUITS AND STABILITY
4.1 Introduction e _
4.2 Leakage Current 67
4.3 TeeEquivalent 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 SelfHeating gg
4.9 Thermal Runaway 8g
4.10 Approximation Techniques 92
4.1 1 Supplementary Problems 96
SINGLESTAGE AMPLIFIERS
5.1 Introduction g
5.2 CommonEmitter Circuit g7
5.3 CommonBase Circuit 107
5.4 CommonCollector Circuit (EmitterFollower) 109
5.5 HighFrequency Performance , 1 o
5.6 Hybrid* Circuit 1 ^
5.7 Supplementary Problems 120
MULTISTAGE 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 CommonBase Connection j 1 73
7.5 CommonCollector Power Amplifier Stage \ 1 76
7.6 PushPull Amplifiers _" "i 78
7.6a Class A PushPull Amplifier 178
7.6b Class B PushPull 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 npn 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 npn AlloyJunction
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 FrequencyResponse 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
Solidstate 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 solidstate 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 electronhole pairs due to thermal agitation in a
germanium crystal shown twodimensionally.
(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 electronhole pair thereby ceasing to exist. At any
given temperature, equilibrium exists where the rate of thermal generation of
electronhole 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 ntype (n for
negative) impurities, and the crystal thus treated becomes an ntype semiconduc
tor. Analogously, impurities that contribute extra holes are called acceptor or
ptype (p for positive) impurities, and the crystal thus treated becomes a ptype
semiconductor. Figure 1.5 shows how the type of impurity determines whether a
semiconductor becomes either an ntype or ptype.
(a)
Fig. 1.5 Effect of impurities on pure germanium crystals, (a) Donor impurity provides
mobile electrons. The positivelycharged 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 Qcm.
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 ntype material,
thereby creating an electric current. Similarly, the principal current in ptype
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 ntype material, electrons are called majority carriers and holes are called
minority carriers. In ptype material, the holes are majority carriers and the
electrons minority carriers. Both ptype and ntype 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 thermallygenerated electronhole pairs lead to a higher
proportion of minority carriers.
1 .3 The pn Junction
If ptype and ntype 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
pn junction or junction diode.
Since both the ptype and ntype 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 pn 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 ptype to ntype material, while the electrons move in the opposite
direction.
Figure 1.6a shows the pn 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 ptype and ntype materials. How
ever, if electronhole pairs are formed by thermal agitation in the ptype mate
rial, electrons will flow across the pn junction aided by the electric field.
Similarly, holes in the ntype 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 pn junction of Fig. 1.7a.
With the polarity shown in Fig. 1.7b, the junction is forwardbiased 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 backbiased (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 pn
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
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+ T — 1 +
+ O]GK}
0_© O_0
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Junction
(a)
— 4 + —
4
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00:00
OO0
OOI0O
+
O0jOO
0O1OO
©@!0
©©!oo
(b)
Q Donor atoms
\~^ Acceptor atoms
+ Holes
— Electrons
• Imperfections, etc.
Fig. 1.6 A pn 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*
©_..© .©
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H
1 —
(b)
p
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& £+ 9.
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H
h
(c)
Fig. 1.7 (a) Unenergized pn junc
tion, (b) The effect on thepn junc
tion of application of forward bias,
(c) When the battery connections
are reversed, the electrons and
holes are drawn away from the
pn junction .
6 Transistor Circuit Analysis
A plausible expression for current flow across a pn junction as a function of
applied voltage may be developed by using relationships from semiconductor
physics. Referring to Figs. 1.7ac, 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 ntype material with
its relatively high concentration of mobile electrons.
2. A diffusion current I dp due to hole flow from the ptype 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 nregion and holes in the pregion. As a result of the flow of
these current components, at the junction the ntype material develops a net
positive charge and the ptype material a net negative charge, which leads to a
potential barrier across it. This barrier limits further diffusion except for the ef
fect of thermallygenerated electronhole pairs on both sides of the junction.
Consequently the remaining two current components are:
3. A current /„„ due to thermallygenerated free electrons in the pregion,
which are accelerated across the junction by the barrier voltage.
4. A current l ep due to thermallygenerated free holes in the nregion, 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 pregion and holes in the nregion.
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 thermallygenerated 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 opencircuit or shortcircuit condi
tions, the net current flow must be zero. Therefore,
The barrier potential adjusts itself until an equilibrium exists between the diffu
sion and thermallygenerated 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 electronhole 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 (/,),
\ lldl.,
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 pn
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 thermallygenerated 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 pn 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 pn
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 electronhole pairs than does germanium, the saturation leakage of silicon
is much less than for germanium. Surface leakage is therefore more important in
silicon pn 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 voltageindependent 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 pn 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 pn 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
Jbtraightiine piecewise linear approximation of Fig. 1.14. The slope of the
gitraightline 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 twoline 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 forwardbiased 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
0.50.134
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 pn
junction diode as a rectifying device. Consider now two pn junctions, /, and /,,
as in Fig. 1.17. Junction / t is forwardbiased so that majority carriers (in this
case, electrons) flow from n to pmaterial. Junction /, is reversebiased, so that
only /, is flowing. If we now combine these junctions and make the pregion 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 pregion are sup
■ plied by I B . Since a is close to but less then unity, Ib is almost entirely trans
ferred to the righthand 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 threelayer 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 npn and pnp cases. The emitterbase junction is forwardbiased and
phe basecollector junction is reversebiased for ordinary amplifier operation. For
JmiP 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 (pnp 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 collectorbase circuit is
ponsiderably higher than the power in the baseemitter circuit. This power gain,
properly exploited, is the basis for the amplifying properties of the transistor.
m Figure 1. 19 shows a pnp transistor in the socalled commonbase 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
rwv — 'rT^r — v ^~i
Fig. 1.17 Juxtaposition of two pn
junctions to make a transistor, (a)
Junction /, is forwardbiased, (b)
junction J 2 is reversebiased, and
(c) an npn junction transistor.
Emitter
Collector
o + + o
Emitter
Collector
o
(a) npn
Fig. 1.18 A junction transistor showing flow of currents in normal amplifier operation.
vii r\j
V E E
'cc
Fig. 1.19 Commonbase connection of a pnp transistor. Lower case letters
represent smallsignal components of voltage and current superimposed on
steady dc 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 commonbase 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 reversebiased 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 commonbase 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
la
(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 EbersMoll Model of the Transistor
We are now in a position to develop a general model of
the transistor applicable to forward and reversedbias conditions, and adaptable
to ac as well as dc conditions. The EbersMoll model, in effect, presents the
transistor static characteristics derived from previously developed concepts in
convenient form. Although the EbersMoll model will later be expanded to cover
highfrequency characteristics, for the moment its use will be confined to low
frequency where capacitive effects are negligible. The lowfrequency 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, [17]
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 =  Qr 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 EbersMoll 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 forwardbiased emitter junction,
and a reversebiased collector junction, the EbersMoll model may be simplified
for this condition. This simplified model is particularly wellsuited 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
npn transistor showing stan
dard polarities of voltages and
currents, (b) Alternate represen
tation of npn transistor.
16
Transistor Circuit Analysis
~lER= a RlcR
a F I EF=~ICF
EO •■
H4
* »
>f
'CR
oc
Fig. 1.22 The EbersMoll 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 smallsignal equivalent circuit is one which depicts the response of the
transistor to smallsignal inputs appearing as small variations about the bias or
operating points. The smallsignal model is basic to the design of smallsignal
ac amplifiers.
The most useful forms of the equivalent circuit are the tee, the hybrid, and
the hybrid7r 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 EbersMoll model with forwardbiased emitter
junction and reversebiased collector junction.
Fig. 1.24 Simplified teemodel with equivalent repre
sentation of emitter diode.
'e
EO
'b
B
Fig. 1.25 Equivalent teecircuit
with leakage current omitted.
For the present, the teeequivalent circuit will be developed. Referring to
Fig. 1. 23, the input diode of the EbersMoll 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
dc bias, it can be neglected in the incremental ac 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 smallsignal
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
commonbase, the commonemitter, and the commoncollector 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 collectorbase junction is reversebiased, and the emitterbase junc
tion is forwardbiased. (Of course in an actual circuit, the polarities of the bias
ing voltage depend on whether the transistor is pnp or npn.)
Figures 1.2728 show the configurations of pnp and npn 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
teecircuit for the commonbase
connection.
fr "t
L 7¥ T T H T
Commonbase connection
l c =al E
(a)
Commonemitter connection
Commoncollector or emitter
follower connection
/c=/3/ B »
a
1a
(b)
1a
(c)
Fig. 1.27 Common circuit configurations for the pnp 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 
Commonbase connection
(a)
Vbb
Commonemitter connection
(b)
Vbb ~ v cc
Common col lector or emitter
follower connection
(c)
Fig. 1.28 Basic transistor circuits using an npn 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 commonemitter and commoncollector connections. Note that^he
commonbase connection, collector current flows with v„ T
V CB =0
(b)
Fig. 1.29 Commonbase (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 collectorbase junction. This leaka~
component ^analogous to diode leakage (Fig. 1.9). Since the emitterbase junc
tion is normally forwardbiased, leakage current here is not significant.
Figure 1.30 ac shows the basic leakage current components. Current I co (as
i is usually called) is the commonbase leakage comment, and is more pre
crrcutted. Analogously. I EO rI EBO applies to the emitterbase junction when it
is reversebiased (not a normal condition) and the collector is opencircuited
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 dc
operating or b ias point of the transistor, and can result in nonlinear operation.
17"'; ^ SC COnnection ^Pencircuited, 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 reversebiased, 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 \la/
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 commonemitter 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 baseemitter 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 collectorbase 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 dc transistor
equivalent circui t, commonbase
con fi gu ra ti on .
(8+l)/ C BO
Fig. 1.34 Simplified dc transistor
equivalent circui t, common emitter
confi guration.
1.9 DC Models
Let us briefly relate the teeequivalent circuit (Figs. 1.22
25) derived from the EbersMoll model to the npn transistor under static (dc)
conditions as shown in Fig. 1.33. The teeequivalent circuit suggests, loosely
speaking, the representation of the transistor as two backtoback junction diodes.
Output current includes two components, leakage current and amplified input
a l E , as was previously explained. The forwardbiased 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 emitterbase characteristic is independent of
collector voltages.
The collector is always reversebiased 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 teeequivalent 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 dc equivalent circuit for the commonemitter configura
tion. Note the leakage component (fi + l)/ CBO , which is the previously derived
result. The basecollector current gain is /8.
PROBLEM 1.30 The common base dc equivalent circuit of an npn transistor
is shown in Fig. 1.35 driving a resistance load R L . Using the equivalent circuit,
calculate the dc 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 Commonbase amplifier, (a) Dc 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 Commonemitter amplifier, (a) Dc 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 = VccRlIc, / 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 TheHybrid7T Equivalent Circuit
The hybridn is an equivalent circuit configuration par
ticularly suited to highfrequency 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 teeequivalent
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 hybridiT circuit is derived for the commonemitter configuration in Fig.
1.38. Only the smallsignal model is considered. The pertinent smallsignal
parameters are
v b.
I Sn ~
dv
BE
(1.21a)
U m
Fig. 1.38 Simplified representation
of hybrid77 equivalent circuit for the
common emitter connection.
(1.21b)
Note that lower case letters and subscripts are used for smallsignal parameters.
These smallsignal relationships lead to the simplified equivalent circuit of Fig.
1.38, in which l CBO and the dc 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 collectorbase junction depletion layer with increasing collectorbase
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 hybridTT
equivalent circuit forthe common
emitter connection.
Oc
O E
Fig. 1.40 Hybrid77 equivalent circuit incorporating capaci
tances for a more accurate representation of highfrequency
characteristics.
Highfrequency 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 highfrequency 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 collectorbase 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 electronhole 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 pn junction which is (a) backbiased
and (b) forwardbiased.
PROBLEM 1.40 In a transistor in the commonbase 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 commonbase con
nection always less than unity?
PROBLEM 1.42 Why is the voltage gain of the commoncollector 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 pnp transistor for the
three common connections.
2
CHAPTER
TRANSISTOR
CIRCUIT ANALYSIS
V CE
BO
o c
Fig. 2.1 Voltoges and currents in
a pnp transistor.
Fig. 2.2 Comm
sistor circu
ibase npn
/?L is a loa
tor.
CC
tran
d
2.1 Characteristic Curves
The static characteristic curves of the transistor define
the steadystate relationships among its input and output currents and voltages.
The curves are easily obtained by means of dc 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 threeterminal 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 commonbase 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 Commonbase (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 commonbase parameters are measured. The collectorbase
voltage and emitter current are the independent variables; the collector current
and baseemitter voltage are the dependent variables (see Fig. 2.3).
Analogous to the commonbase configuration of Fig. 2.2 and its curves of
Fig. 2.3 are families of curves particularly descriptive of the behavior of the com
monemitter and commoncollector configurations. As with the commonbase con
nection, the circuit characteristics of the latter are conveniently described by
static characteristic curves as in Figs. 2.56, respectively. Note that the char
acteristics  although shown to different scales  are essentially identical be
Fig. 2.4 Simple dc circuit for ob
taining the curves of Fig. 2.3.
Shown is a commonbase connec
tion for a pnp transistor. With an
npn 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 Commonemitter (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 dc or lowfrequency 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 vacuumtube
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 commonemitter characteristics of Fig. 2.5, and I B =
20 fxa, and V CE = 20 v, find the collector current and the baseemitter 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 * ^ ^ baseemitter 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 commoncollector configuration using the 2N929 transistor,
V EC = 2 v and l E = 0.6 ma. Find the collector current l c and the baseemitter
voltage drop V BE .
Solution: Use the commoncollector characteristic curves of Fig. 2.6. (Note
that if commoncollector curves are unavailable, commonemitter 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 baseemitter drop is found to be V BB = 0.53 v.
PROBLEM 2.4 The operating point of a commonemitter 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 dc circuit for
obtaining the curves of Fig. 2.5.
"The characteristic curves of the Jesas Instrument 2N929 n^pn 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 commonbase 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 commonemitter curves
are provided by the transistor manufacturer. Therefore use only the commonemitter
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. Baseemitter current increases very rapidly with increasing baseemitter
voltage. The baseemitter current is, in effect, the forward current of a diode.
Excessive baseemitter voltage overheats the baseemitter junction and may burn
out the unit. It is important to control base current and not apply low impedance
voltage sources to the baseemitter 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 ac applications, capacitors are generally used to isolate
dc signals while permitting ac 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 commonemitter 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 voltampere charac
28
Transistor Circuit Analysis
teristic of the batteryresistor combination. The slope of the load line represents
the voltagecurrent 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 commonemitter 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 voltagecurrent 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 commonemitter 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 Commonbase characteristics with superim
posed load line.
PROBLEM 2.7 For a commonbase 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 commonbase configuration. Because of the higher impedance
level of the output collector circuit, voltage gain can be realized.
PROBLEM 2.8 For the commoncollector 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 commoncollector amplifier, (b) Commoncollector 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 commoncollector (emitterfollower) 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 LargeSignal AC Circuits
Refer to the basic commonemitter circuit used as an ac
amplifier in Fig. 2.13. The input coupling capacitor blocks dc signals but per
mits transmission of ac signals. In effect, relatively small ac 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 outputtoinput 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 smallsignal parameters. Smallsignal parameters actually vary somewhat with
the bias point, even in the linear region.
PROBLEM 2.9 Given the commonemitter transistor amplifier of Fig. 2.13, ca
pacitor C presents negligible impedance to ac. Resistance R B is adjusted so that
l B = 15 /za. An input ac current having a 5 /xa peaktopeak (pp) amplitude is im
pressed at the input terminals. Using pointbypoint graphical construction, plot
i B (f). Also repeat for a 30 ^a pp 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 dc characteristics of the
transistor over the fall operating range, including nonlinear regions. As a result,
they vary substantially with signal amplitude. Largesignal behavior is important
in dc 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 Commonemitter output characteristics
with superimposed load line.
PROBLEM 2.10 For the commonemitter 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 dc voltage V L across the load, and the power P L dissipated in the
load resistor.
(d) The input dc 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
Peaktopeak
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) Dc 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 largesignal 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 Commonemitter 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 = 11.65.3 = 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 = 58125=67mw.
PROBLEM 2.11 From Prob. 2.10, determine the dc 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,
Dc 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 dc 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 (dc) 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 commonemitter 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. 2.18ab,
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 commonemitter connection by
definition.
(a)
(b)
Fig. 2.18 (a) Determining transistor smallsignal out
put impedance from the commonemitter 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 commonemitter 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
baseemitter junction.
PROBLEM 2.17 Using the commonemitter 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 145x10'
A/*" 5x10  290 = ^
The static current ratio with V CE constant in the commonemitter 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 commonlyused twotransistor circuit is shown in Fig. 2.19.
Find the quiescent operating point and the overall 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 Commonemitter output characteristics in
the very low current region of the 2N929 transistor.
Consider Fig. 2.20, the commonemitter 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
Wro
: 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. 2.1920.
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 baseemitter voltage, the
base is usually fed a current input. The gain calculated above is in effect the
ac gain to smallsignal 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 commonemitter 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 baseground 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 baseemitter voltage. If the base
emitter voltage is relatively small, the baseground voltage approximately equals
the emitter resistor drop. Thus,
A ~ ^ L
K E
as long as the voltage gain is much higher with R E shortcircuited.
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 commonemitter 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 collectoremitter 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 collectoremitter 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 commonemitter 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
Collectoremitter 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 commonemitter 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 dc voltage V L across the
load and the power P L dissipated in the load resistor, (d) the input dc 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 dc 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 dc 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 teeequivalent 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 smallsignal basis, where essentially linear rela
tionships hold for smallsignal excursions about the operating (Q) point on the
characteristic curves.
Figures 3.1ab show typical transistor input and output characteristics with
smallsignal 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 highlycurved 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 ' 31 T/Pe 2N929 commonemitter 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
SmallSignal Equivalent Circuits
39
it combines both impedance and admittance parameters, known as the bparameters.
The ease of measurement of the Aparameters has contributed to its widespread
.a^pttoa.
A set of Aparameters 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 commonbase, commonemitter, and
commoncollector, has a corresponding set of Aparameters, both for small and
largesignal operation.
The development of the hybrid equivalent circuit is illustrated by the fol
lowing problem.
PROBLEM 3.1 Derive the equivalent commonemitter circuit equations from the
following functional relationships that characterize the families of curves shown
$y Pigs. 3.1ab:
'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 smallsignal 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 smallsignal operation, (3.3) and (3.4) become
': Note the mixed or hybrid nature of the Aparameters in (3.5) through (3.8).
Tim second subscript e is applied to the individual Aparameters, 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 commonemitter connection,
(a) Largesignal parameter (dc) and
(b) smallsignal 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 commonemitter
configuration, (a) Input side and (b)
output side.
instance, it signifies the commonemitter connection. For the commonbase and
commoncollector connections, the subscripts b and c apply, respectively.
Figures 3.2ab illustrate the character of this blackbox approach by black
box representations of the commonemitter circuit for small and largesignal
parameters. As was explained in Chap. 2, the largesignal parameters exhibit
decidedly nonlinear characteristics.
PROBLEM 3.2 Illustrate the physical significance of (3.9) and (3.10) by ref
erence to Figs. 3.1ac. Also establish numerical values for the parameters at
the operating points on the input and output characteristics.
Solution: Consider Figs. 3.1bc in relation to the expression
*'c =^oe^ce +f>fei, [3.9]
and remember that h oe and h te are assumed constant for smallsignal 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 Aparameters can be obtained for
both the commonbase and commoncollector 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.
SmallSignal Equivalent Circuits
41
PROBLEM 3.3 Show how (3.9) and (3.10) may be represented by equivalent
Solution: Figures 3.34 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. 3.5ab, compute the /iparameters 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. 3.12,
'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 commonemitter con
nection. This circuit applies to
smallsignal 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 hparameters 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 smallsignal variations about an operating
point is, of course, immediately adaptable to the analysis of smallsignal ac 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 ac 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 shortcircuit 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 ac load power to transistor
ac 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
smallsignal 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 dc 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 ac 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 ac hybrid model corresponding to the
circuit of Fig. 3.7.
Solution: The first step is to draw an ac 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.
SmallSignal 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 commonemitter 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 = 14,4,1 = 252 x 574 = 145,000.
(e) Input impedance = h ie = 2200 fl.
(f) Output impedance = l/h oe = 33,000 Q. This is the ac impedance seen
looking toward the transistor from R L .
3.3 TeeEquivalent Circuit
The teeequivalent 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 teeparameters directly with high accuracy, in contrast to the ease
with which hparameters may be measured. The best way therefore to determine
teeparameters is to convert from known Aparameters.
The basis for converting between A and teeparameters 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 shortcircuited.
2. Output impedance is measured with input opencircuited.
3. Reverse voltage ratio is measured with input opencircuited.
4. Foward current gain is measured with output sbortcirceited.
For the above conditions, all valid equivalent circuits must yield the same nu
PROBLEM 3.6 Using the commonemitter hybrid and teeequivalent circuits of
Figs. 3.910, 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 teeequivalent circuit
for the common emitter connection.
The parameters r bl r e , rj, and p are
constant only for smallsignal 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 Commonemitter 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 shortcircuited. 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 teeconfiguration.
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 teeequivalent circuit of
Fig. 3.9, the input is opencircuited. 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/3ib08 + 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 shortcircuited for
the forward current gain calculation, :
'c = "le'bi ~T~ = "t»'
SmallSignal 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 hparameters.
PROBLEM 3.7 Using the results of Prob. 3.6, derive formulae for conversion
between tee and Aparameters.
Solution: The equations derived in Prob. 3.6 lead naturally to expressions for
the hparameters in terms of the teeparameters. 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 teeparameters in terms of Aparameters.
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 teeequivalent 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 teecircuit,
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 teeconfiguration are
 33,000 Q,
ice^
2200 n.
With these parameters, the teeequivalent 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 ac measurements on the transistor. With the transistor properly biased,
a small ac voltage is applied to the output with the input circuit open (to ac),'
as in Fig. 3.13. The ratio of the opencircuit 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 basetoemitter 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 teeequivalent circuit from the hybrid model of
Fig. 3.14.
SmallSignal Equivalent Circuits
47
Solution: The Aparameters 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, commonemitter connection.
The teeequivalent 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 overall effect to the
previous 2200 12 base resistance.
\>.
Fig. 3.15 The teemodel derived
from the hybrid circuit
of Fig. 3.14.
3.4 CommonBase Parameters
Since many manufacturers' data sheets list only the
commonemitter 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 commonbase circuit are
v*. » A«> »'• + A* v «b. ( 3  2 °)
PROBLEM 3.11 Determine the commonbase Aparameters 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 highquality 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 commonbase 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 ac techniques, as previously explained, but it is usually more con
venient to compute them from the generally available commonemitter parameters.
3.5 Derivation of CommonBase Parameters
Commonbase parameters may be derived from common
emitter parameters by the following procedure:
1. Redraw the commonemitter 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
commonbase circuit.
PROBLEM 3.12 Using the procedure given above, calculate the commonbase
hybrid parameters of a transistor from known values of the commonemitter hybrid
parameters.
Solution: Refer to Figs. 3.17ab which show the hybrid commonemitter circuit,
and its redrawing, in which the base B is made common to the input and output.
SmallSignal Equivalent Circuits
49
The four quantities to be calculated are repeated below:
1. Input impedance, measured with output shortcircuited.
2. Output impedance, measured with input open.
3. Reverse voltage ratio, measured with input open.
4. Forward current gain measured with output shortcircuited.
Calculate the input impedance of Fig. 3.17b with the collector shortcircuited
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 shortcircuited, 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 (lh 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 commonbase
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 commonbase input impedance is therefore,
hf.it
h* x
l>U
I 'e i h,
'/«
Me
This may be simplified and equated to the hybrid commonbase 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 shortcircuited 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 commonbase 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 commonbase parameters,
V " c = A le+ iif^si(i_A F#)ii
1  1+ ^(lA 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
commonbase configuration.
' B It remains now to determine h tb in terms of the commonemitter hybrid pa
rameters. Refer to the commonbase 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 .
SmallSignal 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 Commonemitter tee
equivalent circuit.
;,;. The above problem concludes the development of formulae for the conversion
Weaminonetnittef hybrid parameters to coauaeabase 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 commonemitter teeequivalent
^Circuit to corresponding commonbase parameters.
(a)
AAAr^ j
P'fd
(b)
^vOO^
r d
fi'e r d fii c r d
(c)
o
P'etd
(d)
P'd
■A/W
»Iution: The commonemitter teeequivalent circuit is shown in Fig. 3.20. It
redrawn in Fig. 3.21 for the commonbase 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 opencircuit 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^mukW^g^M^^^ ■"' . ' "\ ■■■ . : . " .'
The above network is converted to a current source in parallel with a re
sistor to obtain the commonbase teeequivalent 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 opencircuit 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 commonbase teeequivalent configuration is given in Fig. 3.23.
i Figures 3.247 summarize approximate conversion formulae between hybrid
B and teemodels for the commonbase, commonemitter, and commoncollector con
c . , , r . figurations. Those formulae not derived here may be verified using the methods
rig. 3.23 Commonbase 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 smallsignal amplifier performance. This includes the determination of
voltage, current, and power gains, and input and output impedances.
r€T]
'be f ,
i ltd
(a) Commonemitter 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
1A,c
■(1 + A/c)
Tee
equivalent
Tb +
r e
1a
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 commonemitter hparameters.
SmallSignal Equivalent Circuits
53
v eb v cb
i )^J>
(a) Commonbase 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 hparameters.
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
1ft,
1 + h, b
* 1 + h tb ~
1
1 + Afb
Aob
1+A
/6
Tee
equivalent
" 10
1 r ± — = 1
(la)r e
1
1 a
1
(la)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 commoncollector hparameters.
54
Transistor Circuit Analysis
ie
eo 'VAA/ f • — *\AiA* — • — oc
(a) Tee equivalent ci rcuit, commonbase.
la
■b=p' b
BO VW f * VW1 — oc
<■<*=«•<; (1<X)=.
i+)8
(b) Teeequiva 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 teeparamete
CC
2N929 30v
Fig. 3.28 The ac emitterfol lower
amplifier for Prob. 3.14. Choke I.
and capacitors C l and C 2 isolate
ac and dc circuits.
PROBLEM 3.14 Figure 3.28 shows a singlestage, ac transistor amplifier.
The collector characteristics for the type 2N929 transistor are given in Fig. 3.29
Determine:
(a) Ac voltage gain, for a 10 mv input signal.
(b) Input impedance.
(c) Output impedance.
Solution: (a) This is a commoncollector circuit, thus requiring commoncollector
parameters to be used in the analysis. Let us first establish the dc operating
point at which the smallsignal 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 commonemitter 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 commoncollector parameters using the
conversion formulae of Fig. 3.26c:
SmallSignal 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 commoncollector 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 Commoncollector 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 commoncollector 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.
SmallSignal Equivalent Circuits
57
PROBLEM 3.15 For the emitterfollower 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'
■M1V
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 12x10*
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
1h.
(^)
From the previous gain calculation,
1  ^Se. = 1 _ 15.4 x Hr 4 ,
Substituting in (3.37),
(3.36)
[3.34]
(3.37)
i„ = l2x 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 ac 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 commonemitter 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
SmallSignal 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 commonemitter teemodel. 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 teeequivalent circuit of
Fig. 3.34 for Z { and A v .
Solution: The teecircuit is easily solved using conventional twomesh 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 + a0667)(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
SmallSignal 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 baseemitter 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 Hybrid7T 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 lowfre
quency characteristics, in particular, the development of conversion formulae be
tween h and hybrid* parameters. Accordingly, the capacitors which represent
highfrequency effects are momentarily ignored.
Figures 3.36ab show the h and hybrid* models for comparison purposes.
Note that the hybrid* circuit has five independent lowfrequency 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) Hybrid77 equivalent circuit for ™ UB LCM J.18 Given the fiparameters 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 lowfrequency model. Solution. The basic approach is to compare the behavior of the circuits for the
conditions of the output shortcircuited and the input opencircuited; i.e., short
(b)
SmallSignal 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 /iparameter
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 shortcircuit input impedance is calculated. For the
hybrid;7,
Z l = T bb ' + r.q,
and for the hparameter 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 '. (355)
From (3.54), g m is determined:
*. = — ^— . (3.56)
h le ~ 'bb'
A simple approximate expression for the largesignal 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 Aparameter circuit,
Equating these expressions.
r b'« + r 6 ',
v be = /»r« v c
(3.59)
*W
A*
TABLE 3.1 Conversion from hybrid
to hybridTT 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 >
1A,.
= 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 hybridjr 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 l9T bt) ' h lm r bb ' r eo
r A oe ,
Ac  A M * + *'■*" = A oe . (3.63)
r c« "ler 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
hybrid7r 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]
SmallSignal 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 hybrid77 amplifier model. Using the hy
bridw 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
« Wrf 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 Hybnd77 amplifier circuit
Fiq. 3.37 HybridTT 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 fcparameters
for the commonbase connection.
PROBLEM 3.24 For the characteristics shown in Fig. 3.5, find the /iparameters
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 Aparameters 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 Aparameters at the operating point, and (c) r e , r b , and r d . Draw this
circuit, replacing the transistor by its teeequivalent 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 teeequivalent circuit parameters and (b) the input
and output impedances.
PROBLEM 3.32 For Fig. 3.2b, define the small signal /iparameter in physical
terms for the commonemitter 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 vacuumtube 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 highleakage ger
manium transistors than lowleakage 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
versebiased collectorbase junction of a transistor (emitter open) exhibits simi
lar leakage characteristics. Transistor collectorbase 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 npn type has been selected for examples in this
chapter. Figures 4.23 show its commonemitter 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 npn germa
nium transistor vs. junction temperature for the com
monemitter connection.
Fig. 4.3 Collector characteristics of 2N1308 npn
germanium transistor vs. junction temperature for
the commonemitter connection.
'CBO
Fig. 4.4 Commonbase teeequivalent
circuit. The directions of current flow
correspond to an npn transistor.
has a 5 pa maximum leakage at room temperature, 1 ^a is more typical and
assumed here.
4.3 TeeEquivalent Circuit
Representation of Leakage*
The teeequivalent 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 dc 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
collectorbase leakage.
A modified teeequivalent circuit for the commonemitter 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 dc 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 (42)
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 Teeequivalent circuit in the
commonemitter connection.
Icmo m l&solX + /8X
(4.3)
The current gain factor j8 leads to a relatively high leakage current in the com
monemitter circuit.
Since the leakage current occurs with the base open, the leakage is identical
in the commoncollector circuit. The multiplied leakage currents in the common
emitter and commoncollector 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 commonemitter 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 Commonemitter 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 commonemitter
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.
1.50.12 _ 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 basetoemitter 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 baseemitter bias is impractical.
The previous considerations may be examined from a somewhat simpler view
point. The change in baseemitter 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 baseemitter 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 commonemitter connection with superim
posed load line.
V CE , volt — *
Fig. 4.14 Commonemitter 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
threetoone variation in ft is not uncommon; similar variations occur over mili
tary temperature ranges such as 55 c to t85°C.
Fig. 4.15 Commonemitter 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 baseemitter 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 cutoff, 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, lowleakage silicon tran
sistor circuits tolerate a much higher S than relatively highleakage 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 commonbase 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 commonemitter circuits considered will lead
to results directly applicable to the commoncollector circuits, by setting the
collector circuit 'resistance R L equal to zero.
Figure 4.16a shows the general form of the most commonlyused 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 ac input signal frequency.
Note that in practical ac amplifiers, R B is bypassed to avoid gain reduction
due to ac negative feedback. The negative feedback is, of course, the basis for
dc 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 commonemitter amplifier of Fig. 4.16a.
Input ©
(a)
R„
o /WV
, VccR 2
R,+R 2
Fig. 4.16 (a) General bias circuit for commonemitter 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 (cf) 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 = 1370 v.
=3650fi
R 2
= 1380fi
I CBO = 3/Ua at 25 C "=■
Fig. 4.17 Commonemitter 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= ^=^=002 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 collectorbase 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 teecircuit 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 teecircuit 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 :
VccVgE + 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 (1cO
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
collectorbase 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 = dc 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
VccV 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
50.22 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/
50.12 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 straightforward 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 ) ' il (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 dc. 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 temperaturesensitive
characteristics of auxiliary diodes and transistors. Some of these compensation
methods are now illustrated in the following examples.
PROBLEM 4.17 For a commonemitter circuit with an npn 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 SelfHeating
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 smallsignal 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°C25°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 selfheating
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. 4.23. 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
dc 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 dc 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 ia dlcBO dT) dl CBO dT)
Simplifying, and recalling that dl c /dl CB0 = S,
1 = s d_l^± [Vcc _ 2 J iR + R) i (446)
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 . (448)
dT,
Substitute (4.48) into (4.47):
> 0.07 I CBO [V cc  2/ c (Re + Rz.)] • (449)
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 teecircuit.
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
PVVArOVcc =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:
IeIc =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 baseemitter 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 emitterfollower 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
100.6
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 commonbase 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 runaway and its mechanism.
PROBLEM 4.42 In the circuit of Fig. 4.38a, estimate the power dissipation in
the transistor when a 5Kfl resistor is replaced with a 10Kfi resistor.
SINGLESTAGE
AMPLIFIERS
5
CHAPTER
5.1 Introduction
The performance of the singlestage audio amplifier will
be calculated for smallsignal 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 singlestage ampli
fier is energized by a smallsignal ac 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 CommonEmitter Circuit
Figure 5.2 illustrates the commonemitter circuit adapted
to the analysis of small ac signals. Bias resistors are not included, since their
effect on ac 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 ac frequencies in question. Their effect on amplifier bandwidth will be
considered later.
The ac amplifier is most conveniently analyzed using the hybrid equivalent
circuit discussed in Chap. 3. Note that for the commonemitter connection, i, . i b ,
PROBLEM 5.1 The Aparameter equivalent circuit of the commonemitter 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 midfrequencies, we can also speak of input and output
resistances.
Fig. 5.1 Singlestage amplifier con
nected to input generator and output
load resistance.
Vcc
v,f>
O v o
Fig. 5.2 Simplified commonemitter
circuit for smallsignal ac 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 hparameter equivalent
circuit for the commonemitter
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
commonemitter 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 shortcir
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 ).
SingleStage 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 commonemitter 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 commonemitter 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 shortcircuited),
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. 5.23 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 «..
SingleStage 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 commonemitter 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 bypassed. This cir
cuit is most conveniently studied using the teeequivalent circuit of Fig. 5.7b.
1 + /3
" 1 + /S
A^V — • — — VW
'L Vn
(a)
(b)
Fig. 5.7 (a) Commonemitter circuit with feedback resistor in the
emitter circuit, (b) Teeequivalent circuit of (a).
Fig. 5.8 Commonemitter 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).]
SingleStage 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
commonemitter circuit, the following performance parameters:
Z,, Input impedance ,
Z ot Output impedance,
A t , Current gain,
A v , Voltage gain.
Solution: The teeequivalent 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 teeequiva
lent circuit provides convenient formulae in terms of teeparameters, 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 (la) + 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 (lot)
is much greater than r E . Thus,
A,=
1a
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 (la)+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 (la)+r.
(ar c r)ft L
r b [Rl tr c (la)+ rl ( r (/?/. + r,)
re
H)
r. *
r E L
(1  a)
(W
(5.19a)
In terms of the appropriate equivalent teeparameters,
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)
SingleStage 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 commonemitter 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
unbypassed 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 teeparameters:
Ht
R L = 5000fl
O Output
2N929
R E = 93.30
Fig. 5.9 Singlestage 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
SingleStage 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 CommonBase Circuit
The performance parameters of the commonbase circuit,
namely input and output impedances, and current and voltage gains, are derived in
a similar manner to the derivation of the commonemitter 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 (commonbase) is used instead of the subscript e
(commonemitter).
PROBLEM 5.11 Derive formulas for Z,, Z , A,, and A v for the commonbase am
plifier circuit.
Solution: Modifying the previously derived commonemitter 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 commonbase 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 /iparameters 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
CE
as
267 n
R L =<*>
3
1 iv
*L = '</
= 33.3KQ
—
CB
= 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 + 0996x0.27x10 = 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 = <*> —
CB
«*^ 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 selfevident 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 = °°^^
CE an
i CC
d
as 1
Ay =0
asR L =
/ CB
/
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 singlestage 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 commonbase Aparameters ,
Z = 7.57 + 0996 (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 commonbase con
nection. For the teeequivalent circuit, Z, = r e + r b (1  a).
PROBLEM 5.14 For the commonbase connection and the operating point of Prob.
5.12, determine the variation of Z vs. R t .
Singlestage Amplifiers 109
Solution: Use the commonbase Aparameters 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 commonbase connection and the operating point of
Prob. 5.12, determine the variations in A v and A, vs. R L .
Solution: Use the same commonbase hparameters 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 CommonCollector Circuit
(EmitterFollower)
Proceeding as in the previous section, formulae for the
commoncollector singlestage amplifier configuration (emitterfollower) 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 commoncollector configuration.
Solution: It is only necessary to change the subscripts of the Aparameters 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 Aparameter formulae for the commoncollector 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 Aparameters 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).
SingleStage Amplifiers 111
Input impedance obviously depends almost directly on R L in the useful range
of operation. From the teeequivalent 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 hparameters, 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 teeequivalent 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 ^ [531]
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 /iparameters for the commoncollector 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 commoncollector or emitterfollower 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 /iparameters, 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
commoncollector 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 singlestage
audio amplifier.
SingleStage Amplifiers
113
Table 5.1 SingleStage Amplifier Formulae
Aparameters
(Second subscript
omitted)
Common base
teeequivalent
circuit
(see Fig. 2.31)
Commonemitter
teeequivalent
circuit
(see Fig. 2.31)
Note: rj = r. + R B ;
R B = circuit resistance
in emitter
Commoncollector
teeequivalent
circuit
(see Fig. 2.31)
*,■?
V
a*i.+(io)I»
r . + '(> (l«) + r t
Rr. + f»
= r. + r 6 (la),
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 (la + ^)' "
/3 + 1 r/(/3 + 1)
r„(/3+l) r;(l + ) 8)V r„ /
~ Rl
rl«r dl j3»l, Ri«fd
1 +
HH)*^
"('•i^k l *Tf
r t « 'c
5.5 HighFrequency 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 borderline 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 hybridn circuit is easily the most
convenient.
Consider the ftparameter commonbase equivalent circuit with complex forward
current gain. Let h tbo be the lowfrequency value of h tb . The following approxi
mate relationship based on intrinsic solidstate 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 ) cutoff frequency where the absolute value of cur
rent gain declines to 0.707 of its lowfrequency value.
Now determine the complex forward current gain for the commonemitter 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 lowfrequency value (the /3 cutoff frequency), then
h tB =
Af.c
(5.35)
l+j
'*■
/a
Fig. 5.14 Hybrid77 equivalent cir
cui t for commonemi tter connection.
where h, eo is the lowfrequency value of h t9 .
The cutoff 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 highfrequency capability. From (5.34), we see that the common
emitter configuration has a bandwidth smaller by a factor of 1 + h tbo than the
commonbase configuration. (Recall that h tbo is negative and nearly unity, so
that 1 + h !bo is a very small number.)
5.6 Hybrid7T Circuit
The hybrid77 circuit is superior to other highfrequency
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 commonemitter circuit, represented for small signals
by the hybrid77 model of Fig. 5.14. The lowfrequency conversion formulae from
Aparameters are given by Table 3.1, with r b6 » arbitrarily chosen. Table 5.2 pre
sents a concise procedure for direct measurement of the hybrid77 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) Hybrid77 equivalent circuit setup 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 resistancecapacitance combinations with com
plex impedances. The basespreading 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.
SingleStage 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
tancecapacitance 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
hybridTT equivalent circuit.
(b) Expansion of (a).
116
Transistor Circuit Analysis
TABLE 5.2 Direct measurement of Hybrid77 parameters. (Refer to the hybridff equivalent circuit of Fig. 5.14.)
1. Refer to Fig. (a). Capacitors C, and C 2 are effective shortcircuits to ac 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 shortcircuit 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 shortcircuited 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 ac. Resistance R L is high enough
so as not to shunt the collectorbase capacitance too heavily. Using a capacitance
bridge between points B and C, measure C ob , the output capacity for the commonbase
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 socalled 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 highfrequency transistors.
To measure C b ' e , refer to Fig. (c). Capacitance C is a shortcircuit 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 cutoff 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.
SingleStage 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 capacitorresistor 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) Hybrid77 equivalent output circuit, (b) Simplified equivalent of (a),
(c) Final simplified representation.
PROBLEM 5.22 Develop the input impedance network representation of the com
monemitter 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 hybridTV 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 Commonemitter 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 hybridn 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 shortcircuit. 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 Hybrid77 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 lowfrequency 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)
SingleStage 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 (10.0248 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 lowfrequency
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 Commonemitter 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
_,, . , , . , Ti 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 commonemitter 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 commonemitter circuit.
PROBLEM 5.27 Calculate the commonemitter teeequivalent circuit for the
transistor of Prob. 5.25.
PROBLEM 5.28 Find the effect on the teeequivalent 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 commonemitter
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 teeequivalent
circuit.
PROBLEM 5.31 Derive the formulae for Z it Z o! A,, and A v for the common
emitter circuit using (a) Aparameters and (b) teeparameters.
PROBLEM 5.32 Why is the commoncollector circuit useful? What are its main
characteristics?
PROBLEM 5.33 Define /3 cutoff frequency.
PROBLEM 5.34 Find the gain change from to 1 MHz when a transistor with a
/3 of 300 and a /3 cutoff of 10 MHz is used (a) in a commonemitter configuration
and (b) in a commonbase configuration.
PROBLEM 5.35 In the circuit of Fig. 5.22, use a transistor whose hybrid77
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.
MULTISTAGE
AMPLIFIERS
6
CHAPTER
6.1 Introduction
The techniques developed in previous chapters, particu
larly Chap. 5, may be applied to the analysis of multistage amplifiers. There
are two basic analytical approaches:
1. We can replace each transistor by its equivalent circuit or model, and
analyze the resulting multimesh 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 multistage 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 multistage 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 multistage 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 twostage amplifier of Fig. 6.2, calculate the small
signal ac quantities R it R ol and A v . Use the parameters for the 2N930 transis
tor given in Tables 6.12 and Appendix A.
Solution: Figure 6.2 shows an emitterfollower 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 twostage 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 Twostage directcoupled emitterfollower
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 = Ke ~ 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 dc 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 baseemitter drops in
series. These are approximately 0.6 v each for silicon transistors at room tern
MultiStage Amplifiers 123
perature. Thus the potential at R L becomes 11.2  2(0.6) = 10 v. For R L =
2500 fi, the dc 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 = 11.110.50.61 = 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 dc operating (quiescent) levels. We
now have to determine from the transistor curves the smallsignal fiparameters
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 " • [528]
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 ac 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 twostage device well suited for good isolation of source
and load.
6.2 Capacitor Coupling
In previous chapters, coupling methods were essentially
ignored. Coupling and bypass capacitors were assumed to be infinite, thereby
providing zero impedance to ac signals. In actual practice, however, coupling
and bypass capacitors introduce limits to the lowfrequency amplifier response.
These limits are now considered.
Refer to Fig. 6.3, which shows a twostage commonemitter amplifier using
both capacitors for interstage coupling (C, and C 2 ), and bypass capacitors (C E
and C Ej ). The coupling or blocking capacitors prevent undesired dc coupling
between the bias circuits of the separate stages. The bypass capacitors allow
separate adjustment of dc and ac 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 lowfrequency response of the amplifier.
MultiStage 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 Twostage 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 loglog
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 lowfrequency 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
St
h oe  b " b " 3Q xl0 «_ (290)2x10
fi, e + R 8 2200 + 1000
 = 84,400 0.
MultiStage 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 commonemitter circuit. If R Ej
were not bypassed, 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 lowfrequency
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 commonbase 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 lowfrequency
attenuation at co = 500 radians /sec, omitting bypass capacitor, C Ej . Use the
teeequivalent circuit with the following parameters:
= 290, R Ll =R L2 = 5KO,
r e = 6.67fi, RjRia = KailK 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 unbypassed, C 2 is conveniently small. However the secondstage gain
is considerably reduced.
PROBLEM 6.6 Refer to the twostage commonemitter amplifier of Fig. 6.7.
Calculate C 2 so that lowfrequency 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 dc 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 basetoemitter 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 •
MultiStage 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 Twostage commonemitter amplifier with emitter bias
resistors only partially bypassed.
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 lowfrequency 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 bypass 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 twostage commonemitter 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 highfrequency
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 =/?pli? 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)
MultiStage 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
loglog 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 loglog paper appears as shown in Fig. 6.11. This is a
generalized plot of amplifier lowfrequency 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 loglog
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 bypass 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.
MultiStage 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 teeequivalent circuit for determining
Po Ab
solution: Figure 6.12 shows the commonemitter teeequivalent 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 twostage 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 frequencysensitive 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 overall 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 +
MultiStage 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 twostage 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
overall 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 highimpedance 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 twostage 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 opencircuited. 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 opencircuited, 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 opencircuit, I bi is easily calculated from the allresistive network
as/ b2 = 15.4 V gi ia.
At higher frequencies, both capacitors act as shortcircuits. 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
shortcircuit, 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.
MultiStage 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 twostage 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. 6.910. 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 dc resistance; R 2 = sec
ondary dc 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 opencircuited; secondary inductance L 2 can
be measured with the primary opencircuited: 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 highfrequency regions. Consider first the
lowfrequency 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,
MultiStaga 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 lowfrequency
condition.
and may therefore be neglected. The simplified equivalent circuit corresponding
to lowfrequency 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 frequencysensitive term in the denominator
of (6.10) becomes dominant. In this intermediate or midfrequency range,
"i,
nR,
V 0i (R 1 + R 0l ) + n 2 (R 2 + R il )
The equivalent circuit for the midfrequency 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 Interstagecoupling equivalent circuit for highfrequency
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 highfrequency region of the frequencyresponse 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 highfrequency 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 lowresistance 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 seriesreso
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)
MultiStage 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 dc 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 lowfrequency 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 dc 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 dc 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 dc. To compensate for the baseemitter 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
transformercoupled 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 Transformercoupled
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 lowfrequency 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 ac 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 stepdown 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
MultiStage Amplifiers
143
O)=200
9.0 xLx^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 multistage am
PROBLEM 6.16 For the circuit of Fig. 6.26, design the bias circuit to permit a
distortionfree 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, peaktopeak.
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 dc 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
2Vcc = 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 distortionfree 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 peaktopeak, 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 lowfrequency
response or allow lowfrequency phaseshift.
2. Quiescent output voltages provide input bias to subsequent stages, avoid
ing bias networks. This allows higher basecircuit input impedances.
MultiStage 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 threetransistor directcoupled
amplifiers, which may then be cascaded, as desired, using ac 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 twostage directcoupled amplifier with dc 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
bypass capacitor C E prevents ac feedback which would reduce amplifier gain.
OVr
cc
M
ICBO^ + Pi>
+ /V B2
Rb, + r b 2
Rf,  Rr
+ R
Bib
Fig. 6.29 Twostage directcoupled amplifier.
To analyze the circuit of Fig. 6.29, use the techniques of Chaps. 34 to set
up the dc 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 = VccRlJc 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 dc equiv
alent circuit of the twostage 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 distortionfree 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.
MultiStage 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
peaktopeak voltage swing. Allowing a minimum of about 1 v for Vce 2 •
Ve 2 = 12 5.671 = 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 peaktopeak 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 dc 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 multistage 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 Threestage, directcoupled commonemitter
transistor amplifier.
Rxi I c l \iR Ll
R 2
OV,
CC
ov„
Fig. 6.32 Simplified equivalent circuit of threestage
amplifier for bias calculations. (See Fig. 6.31.)
MultiStage Amplifiers
149
56.6K12
10Kfi> R Ei> R
o _^ 3K " 1 L3I
Fig. 6.33 Threestage, direct coup led commonemitter 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 dc 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 :
Vcc1 Ci Rl^V Bi .
Substituting and solving, R L = 17.4 K 0.
Continuing the above with bias calculations,
MultiStage 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 cutoff, V c ,
age may swing from 10 v to 24 v, with a quiescent level of 17 v. The peakto
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 = 001 ^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 multistage directcoupled 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 peaktopeak
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 stepbystep series of calcula
tions of input and output impedances, starting with the last stage. At 1000 Hz
(gj = 6280 rad/sec), all emitter bypass capacitors are essentially shortcircuits.
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 bypassed
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.
MultiStage Amplifiers
153
Now calculate the currents and voltages in successive stages as required to
determine overall 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 overall 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 peaktopeak 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 bypassed, so that the voltage V Ei is a fixed 9 v re
gardless of the ac signal component. A load line for this amplifier stage is
drawn for an effective dc 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 peaktopeak ac voltage swing is 14 v.
Reexamine the peaktopeak 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 peaktopeak 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 ac react
ance. What is the maximum undistorted output swing?
Solution: Since the 1000 U, load does not effect the dc operating point (because
of the blocking capacitor), this point remains the same in Fig. 6.34. The ac
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 peaktopeak.
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 Qac 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.
MultiStage Amplifiers
155
PROBLEM 6.30 Repeat Prob. 6.27, but with 100 fi (see Fig. 6.36) of each emit
ter resistor unbypassed. 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 unbypassed 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 unbypassed.
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
(npn and pnp types with otherwise similar characteristics) has made it easier
to design directcoupled amplifier stages. They overcome the difficult biasing
problems of directcoupled stages using only one transistor type. The following
example shows how easy it is to devise directcoupled 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 baseemitter 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 cutoff 9 v, for a peaktopeak 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 peaktopeak undistorted output is zero.
Previous multistage amplifier calculations relating to performance and tem
perature sensitivity apply to amplifiers using complementary transistors, except
that there are no dc polarity inversions between stages. Therefore effects of
changes in V BE , I CB o, and /8, always add. Ac 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 ac gain and maximum power ab
sorbed by R L .
Solution: Assume, as a crude estimate (which may be reexamined 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.
MultiStage Amplifiers
157
■ high /8, /3>100
Q x , Q 3 : npn silicon transistor
Q 2 : pnp silicon transistor
R Ei =RE 2 =RE 3 =3Kti, 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 Directcoupled 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 cutoff, 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 dc 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,
32
0,: 2N930
r ei = ll £2, r Cl =2 XlO 6
A = fife, =380; h FEl =380atI Cl = l0™*
Fig. 6.38 Twostage di rectcoupled
ampli fier using complementary tran
sistors, and dc 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 dc input impedance to the first transistor may be estimated:
^58 Transistor Circuit Analysis
R,
'lldc
= (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 ac signals, R 3 is in parallel with R L so that effective load resistance
is 31200, 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 peaktopeak, or
2.12 v rms. Maximum undistorted power output P is
p _Vl (2.12)*
f = — = — — = li5 mw.
Rl 2 31
Under quiescent (zero ac) 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 ac input signals.
Note in Fig. 6.38 that if only part of R Ei is bypassed by a capacitor, ac
MultiStage 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 zeroresistance 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 dc 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 smallsignal 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 smallsignal stages. Refer to Fig. 7.1
which shows the transistor output characteristics for the commonemitter connec
tion over their full range of interest. In particular, let us examine the character
istic curves in the highvoltage region.
Saturation^
region
Saturation .
resistance
line
Cutoff ~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 collectorbase 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 electronhole 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 commonbase 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 collectorbase junction with the emitter
opencircuited (BV C bo). and V is the applied junction voltage. In the breakdown
region, the collectorbase current is multiplied by M.
An expression for the effective forward current gain for the commonemitter
connection is easily derived:
a* an h * F l
Solving for h FB ,
h * = _f^_. (7.3)
'""ictir;
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 pnp transistor, and n = 5 for a germanium npn 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 (collectoremitter 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 welldefined limit. Practi
cally, the maximum current is limited by the falloff 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 collectorbase 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 ceni
ter of square sheets of
1/8inch 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*
MK10
MK15
MK20
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 cutoff region. When I c <Icbo, Ie reverses sign,
the emitterbase diode blocks, and transistor action ceases. At I c = 'cboi ^b =
 1 CB0 and I E = 0. The transistor is said to be cutoff. 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 collectorbase voltage, or P c = V CB I C * The locus
of constant power dissipation may be plotted on the commonemitter 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 emitterbase junction is usually negligible compared with dissipation at the
collectorbase junction.
164
Transistor Circuit Analysis
2N1537
Fig. 7
20 36 40 60F max 80 100 120 140
Vce / volt ».
4 Collector characteristics of the commonemitter 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
"^MfcW'',
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»
8025
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 highvoltage 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 dc load line sees only the tran
sistor drop, and extends from V C c = 40 vertically to Q. The ac 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 transformercoupled output.
166
Transistor Circuit Analysis
7.2 Distortion
240
200
80
2N1537A
Vc
;e =
r
2v
2 3
fc , amp
Fig. 7.8 The falloff in h FE at high
Iq is a distortionproducing 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 distortiongenerating 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"
peaktopeak input voltage swing.
In the above instance, the transistor base is driven from a lowimpedance
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. emitterbase voltage V EB .
PROBLEM 7.4 Given a set of transistor commonemitter 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 dc
quiescent power, so that transformer coupling to the load is normally used. The
transformer primary dc resistance leads to a much smaller quiescent power loss.
Figure 7.13 shows a transformercoupled 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 bypassed in power amplifiers due to the awkwardly
large size of the required capacitor. In addition, an unbypassed R B will provide
improved ac 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 dc and ac 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
Baseemitter 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. baseemitter 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) baseemitter 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 commonemitter 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 dc resistance is 10% of the load resistance referred to the primary.
R^ = lon
Fig. 7.16 Transformercoupled power amplifier. (See Prob.7.6.)
Solution: Use the configuration of Fig. 7.16. Note the unbypassed emitter re
sistor R E used for bias stabilization. Now following the procedures of Prob. 7.2,
we establish a Qpoint 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 Qpoint (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 peaktopeak 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 dc and
ac 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 baseemitter
voltage drive.
Power Amplifiers 171
From the saturation point P w we determine V BB at saturation. If we apply a
symmetrical sinusoidal baseemitter 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 = 105 a
at point Q. This is the range of the ac 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.
Peaktopeak 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.) Peaktopeak baseemitter 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 (VccV C s\ I D (VccVcsf
3. Ac 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 dc emitter
ci rcui t resistance
R c = dc 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 CommonBase Connection
The commonbase 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
commonbase csnaeeMoa, leakage ewrent is negligible. The maximum voltage
(BYcboI is gteaie* then for the commonemitter or commoncollector 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 transformercoupled,
distortion is very low. If a transformercoupled 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 Commonbase characteristics of the 2N1537A
transistor at 7) = 80°C and BV CBO = 100 v.
Fig. 7.21 Commonbase trans former coupled output stage,
BV CBO = 100 v.
PROBLEM 7.10 Design a transformercoupled commonbase power output stage
under approximately the conditions of Prob. 7.6. Assume the transformer dc 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 commonemitter 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 8025
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 commonbase 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 commonbase 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 secondharmonic 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 commonemitter circuit, explaining why the commonbase 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 commonemitter connection.
The only significant advantage over the commonemitter 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 commonbase amplifier is a serious disadvantage.
Now consider the same commonbase amplifier circuit, Fig. 7.21, driven from
a sinusoidal voltage source. This could be, for instance, a transformer input with
a sufficient voltage stepdown 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 Emitterbase 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 CommonCollector Power
Amplifier Stage
The commoncollector or emitterfollower circuit has a
characteristic voltage gain of somewhat less than unity. The techniques for the
design of commonemitter amplifiers may be directly transferred to the design of
the commoncollector amplifiers. The load, in effect, becomes R B , with R L = 0,
and the transformer dc resistance now serves as a dc 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 commoncollector power stage to drive an 8 fl
loudspeaker 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 powerstage
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 42watt 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, pp. 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 Emitterbase 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 transformercoupled 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 dc power, is 17.3/77 =
22.5%. This poor efficiency is primarily due to the dc dissipation in the 8 Q
emitter load resistance.
178
Transistor Circuit Analysis
7.6 PushPull Amplifiers
Fig. 7.27 PushPull transistor am
plifier with transformercoupled 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 distortionfree 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 dc cancellation in the output transformer, and greatly
increased efficiency as compared with singleended 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 commonemitter 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 PushPull 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 pushpull 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 pushpull 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 pushpull 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 pushpull 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. 7.67, the following quiescent operating conditions for the sep
arate transistor circuits are established:
I Ct = / c , = 1045 a,
' CE
40
The voltage drive is 0.7 v pp, and bias is adjusted for the required Qpoint.
The design is similar to the design of singleended 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 viceversa.
Power Amplifiers
179
r[ =2 nR L =38 Q
Fig. 7.29 Pushpull power amplifier showing tempera
turecompensated 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 pp. 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 singleended stage.
Also, the supply voltage V cc is unchanged, and I Q is doubled. Thus, efficiency
is the same as for a singleended stage.
We have assumed in Prob. 7.12 that the transistor load lines are perfectly
straight. This is not exactly true for pushpull 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
pushpull 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 pushpull 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 singletransistor circuit of Fig. 7.33, where the
circuit parameters are modified appropriately by factors of two. The equations for
the singletransistor 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 pushpull
Class A amplifier.
2' 6 8
O ^ VAr
vs/v — *
r d
n:l
Fig. 7.33 Simplified singletransistor equivalent circuit of
pushpull 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 PushPull 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 pushpull 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 Pushpull Class B amplifier.
Load = n 2 R L =R' L
Vcc
2V r
Fig. 7.35 Load line for a Class B amplifier,
cutoff for half a cycle.
Figure 7.35 shows the typical loadline 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 pushpull operation.
,t
t
v ec 2 J
£l_z\.
^j  ^
Double amplitude
= 2V CC
Double amplitude
= 2V CC
g. 7.37 Waveforms in a pushpull Class B amplifier.
(a)
Time
(b)
Fig. 7.38 (a) Superimposed 2N1537A transistor characteristics showing distortion in
pushpull output. This distortion is due to an equivalent offset in V B £. (b) Typical
crossover distortion in a Class B pushpull 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 crossover distortion.
To avoid crossover distortion, it is advisable to operate slightly in Class AB
with an added bias, above the cutoff 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 pushpull 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 AB pushpull 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, dc power for Class B approaches
zero as the input signal is reduced, so that in normal use, the average dc power
Power Amplifiers
183
I?T ~ Rs "*" ^C "*" ^£
Ic sin cot
Fig. 7.40 Class B pushpull 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 collectoremitter 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 cV C S=BV max
Fig.
7.41
Load 1
ine characteristics
appl
cabl
e to the
pushpull Class B
circuit.
AAAr
R c =30O
Fig. 7.42 Pushpull 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 crossover distortion, particularly important at smallsignal 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 commonemitter 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 pushpull 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»
10/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
10A;
where the logarithms are to the base 10.
= 201ogl_/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 closedloop gain becomes
infinite. (Closedloop refers to the amplifier with the feedback in the circuit, and
openloop 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 (lj84)» 1fiA
Thus the fractional changes are related by
(l pAJA
Mtb 1 dA (g g)
A lb 1PA 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 highfrequency cutoff 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„
lfcl. + £
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 highfre
quency cutoff is higher for the feedback amplifier than for the openloop 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 lowfrequency cutoff 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 openloop and closed
loop cases.
Solution: First consider the openloop 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 closedloop case,
e, = e + 8e ,
li^Sifliillil9i^Btti^.,,, .,,,,
Solving these equations for e„ in terms of ej and e d ,
e + j9e , "V
 At e + e', L
e„ =
e, +
lpA 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 openloop case,
Feedback
189
(8.14)
And for the closedloop 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 openloop case,
P =
e o<*
M,4.e,j l^e,)
l^e„
ie*r
And for the closedloop 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
eJdjBiMOe, •
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
openloop case. Then the signal to noise ratio for the closedloop amplifier
becomes
\lPA t A t \ Jil*ll 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 closedloop 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
clearcut 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 commonemitter 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 openloop 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 fedback 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 opencircuit 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 fedback 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,
lA 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 commonemitter feedback amplifier. Feedback is
taken across a resistor in series with the load. The fedback voltage is propor
tional to load current; this is current feedback. Figure 8.5b shows the equivalent
circuit of this feedback amplifier. The fedback 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. ^oA. 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 closedloop 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 closedloop 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 shortcircuited. 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,dA 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 widetolerance transistors in precision
However, feedback cannot be applied indiscriminately. Note that from (8.4),
»tt
lA/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 lowfrequency feedback factor can become unity
at some high frequency, in which case, selfsustained 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 closedloop circuit
Fig. 8.6 Openloop 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 openloop 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 openloop 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 = ■
lAfi
[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 openloop 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 openloop 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 openloop 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 selfsustained oscilla
tions occur. Somewhat more generally, if the openloop gain characteristic
encircles* the Nyquist point, the closedloop 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 midfrequency range of amplifier operation, gain is relatively
uniform with frequency, with phase shift approximately zero. It is only in the
low and highfrequency regions of operation that large phaseshifts occur. Thus,
in the design of feedback amplifiers, much attention must be given the low and
highfrequency gain and phaseshift characteristics to avoid instability, even
though normal operation may be required only in the midfrequency region.
There is a practical point to be observed in relating openloop 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 openloop analysis.
PROBLEM 8.6 Consider the simple multistage 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 highfrequency 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 highfrequency
region, C i is essentially a shortcircuit, 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 multistage 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 lowfrequency 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 midfrequency region, K/ = 1. Similarly, for (8.30), as frequency de
creases to the mid frequency region, K h = 1. These limiting high and lowfre
quency conditions must, of course, lead to the same midfrequency gain if our
approximations are valid.
Openloop 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 highfrequency
regions. The loop is clearly unstable.
Note particularly that (for this special case), the high and lowfrequency
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° phaseshift 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 midfrequency 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 phaseshift 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, wellsuited 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 highgain 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 dc 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) zerofrequency forward gain, (c) feedback gain at 1 Hz,
(d) zerofrequency openloop gain, (e) closedloop gain at 1 Hz, (f ) zero
frequency closedloop gain, and (g) frequency at which the closedloop has
dropped 3 db below its initial value at zerofrequency.
PROBLEM 8.23 For the circuit of Fig. 8.9b, determine (a) the closedloop
gain at &> = 200, (b) the change in the closedloop 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 npn
Planar Silicon Transistors*
FOR EXTREMELY LOWLEVEL, LOWNOISE, HIGHGAIN,
SMALLSIGNAL AMPLIFIER APPLICATIONS
• Guaranteed h F E at 10 pa , T A =55°C and 25°C
• Guaranteed LowNoise Characteristics at 10 fia
• Usable at Collector Currents as Low as 1 \ia
• Very High Reliability
• 2N929 and 2N930 Also Are Available to MILS19500/253 (Sig C)
^3  COLLECTOR
ALL DIMENSIONS ARE
IN INCHES
UNLESS OTHERWISE
SPECIFIED
THE COLLECTOR L£ IN ELECTRICAL
CONTACT WITH THE CASE
ALL JEDEC T018 DIMENSIONS
AND NOTES ARE APPLICABLE
Fig. A.l JEDEC registered mechanical data.
TABLE A.l JEDEC registered absolute maximum ratings at 25 C freeair temperature
(unless otherwise noted).
CollectorBase Voltage 45 v
CollectorEmitter Voltage (See Note 1) 45 v
EmitterBase Voltage 5 v
Collector Current 30 ma
Total Device Dissipation at (or below) 25°C FreeAir 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 baseemitter diode is open circuited.
Derate linearly to 175 C freeair 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 freeair temperature (unless otherwise noted).
PARAMETER
TEST CONDITIONS
2N929
2N930
UNIT
MIN
MAX
MIN
MAX
BVceo CollectorEmitter Breakdown Voltage
lc = 10 ma, l B = 0, (See Note 1)
45
45
V
BVebo EmitterBase 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 BaseEmitter Voltage
Ib = 0.5 ma, l c = 10 ma, (See Note 1)
0.6
1.0
0.6
1.0
V
VcEi.t] CollectorEmitter Saturation Voltage
l B = 0.5 ma, l c = 10 ma, (See Note 1)
1.0
1.0
V
. SmallSignal CommonBase
ib Input Impedance
V C b — 5 v, l E = — 1 ma, f = 1 kc
25
32
25
32
ohm
SmallSignal CommonBase
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
SmallSignal CommonBase
n ° b Output Admittance
Vcb = 5 v, l E = l ma, f = 1 kc
1.0
1.0
jLtmho
SmallSignal CommonEmitter
f * Forward Current Transfer Ratio
V C e = 5 v, l c = 1 ma, f = 1 kc
60
350
150
600
1 1 SmallSignal CommonEmitter
1 '*' Forward Current Transfer Ratio
Vce = 5 v, l c = 500 (jua, f = 30 mc
1.0
1.0
 CommonBase OpenCircuit
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 freeair 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: smallsignal 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
2T
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: smallsignal 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 /
latT 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 commonbase 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 commonbase 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 Smallsignal commonbase 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 Smallsignal commonemitter 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 —CollectorEmitter Voltage — v
Fig. A. 10 2N929 transistor: commonemitter
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 — CollectorEmitter Voltage — v
Fig. A. 11 2N930 transistor: commonemitter
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
ic
■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 — FreeAir Temperature — °C
Fig. A. 14 2N929 transistor: baseemitter voltage
vs. freeair 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 — FreeAir Temperature — °C
Fig. A. 15 2N930 transistor: baseemitter voltage
vs. freeai 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: collectoremitter satu
ration voltage vs. freeair 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 — FreeAir Temperature — °C
Fig. A. 17 2N930 transistor: collectoremitter satu
ration voltage vs. freeair 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
CollectorEmitter 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
, +25C
_50°C
f
\ +75C
/
/
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 , BASEEMITTER VOLTAGE (VOLTS)
Fig. A. 21 Output current vs. emitterdrive 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
_ 10040 (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
npn AlloyJunction 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
CollectorBase
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
BaseEmitter 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 CEwt
CollectorEmitter
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
SmallSignal CommonBase
Input Impedance
V CB = 5v,
f = Ike
l E = — I ma

21


20


28


28

ohm
•Vb
SmallSignal CommonBase
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
SmallSignal. CommonBase
Output Admittance
V CB = 5«,
f = lkt
l E = — 1 ma

0.34


0.34

—
0.34
—
—
0.34
_
fimho
K
SmallSignal CommonEmitter
Forward Current
Transfer Ratio
V CE = 5v,
f = lkc
l c = 1 ma

105


120


135


170


* f hfb
CommonBase Alpha
Cutoff Frequency
V CB = 5v,
l E = — 1 ma
3
12

5
14

10
16

IS
20

mc
*<ob
CommonBase Open Circuit
Output Capacitance
V CB =5».
f = 1 mc
l E =

14
20

14
20

14
20

14
20
Pf
c ib
CommonBase OpenCircuit
Input Capacitance
»E5=5«.
f = lmc
i c =
"
13

"


13


13

Pi
tV pT is determined by measuring the emitterbase floating potential V EB( The collectorbase 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 freeair
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 freeair 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 — CollectorEmitter 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 — CollectorEmitter 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 — FreeAir Temperature — °C
Fig. A. 3 1 Normalized static forward current
transfer ratio vs. freeair 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 Baseemitter voltage vs. collector current.
60 40 20 20 40 60 8C
T A — Free Air Temperature — °C
Fig. A. 33 Baseemitter voltage vs. freeair 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 Collectoremitter 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 — FreeAir Temperature — °C
Fig. A. 35 Collectoremitter saturation voltage
vs. freeair 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 — FreeAir Temperature — °C
Fig. A. 36 Collector cutoff current vs. freeair
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 , — CollectorBase Voltage — v
Fig. A. 37 Commonbase opencircuit output
capacitance vs. collectorbase 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 Collectoremitter 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 — FreeAir Temperature — °C
Fig. A. 39 Collectoremitter saturation voltage
vs. freeair 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 "MEGALIFE" 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
CollectorBase 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
CollectorBase Cutoff Current
(VCB = 2V) All Types
(V CB = 1/2 BV CES rating; All Types
Tc = +90° C)
ICBO
■
0.2
20
mA
EmitterBase Cutoff Current
(V EB = 12V) All Types
lEBO
1
0.5
mA
CollectorEmitter Breakdown Voltage
(Ic = 500 mA, V EB = 0)
2N1529A, 2N1534A*
2N1530A, 2N1535A*
2N1531A, 2N1536A*
2N1532A, 2N1537A*
2N1533, 2N1538
BVCES
■
■■■■■■■■■■

volts
CollectorEmitter Leakage Current
(V BE = IV; V CE @ rated BV C BO)
All Types
J CEX
■■■■■■■■■l L^hVAVAbIbVi
20
mA
CollectorEmitter Breakdown Voltage
(I c = 500 mA, I B = 0)
2N1529A, 2N1534A*
2N1530A, 2N1535A*
2N1531A, 2N1536A*
2N1532A, 2N1537A*
2N1533, 2N1538
BVcEO
■
■

volts
CollectorBase Breakdown Voltage
(IC = 20 mA)
2N15.29A, 2N1534A*
2N1530A, 2N1535A*
2N1531A, 2N1536A*
2N1532A, 2N1537A*
2N1533, 2N1538
BV CB0
■ 100
■ 120

volts
CurrentGain
(V CE = 2V, I C  3A)
2N1529A  2N1532A
2N1534A  2N1537A
2N1529  2N1533
2N1534  2N1538
h FEl
■
■
■
40
70
40
70
BaseEmitter 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, nonA type numbers
** Each parameter of the "MEGALIFE" series only is guaranteed to an individual AQL of 0.
MOTOROLA Inc., Semiconductor Products Division.
inspection level n
216
Transistor Characteristics
Fig. A.40 Powertemperature derating curve.
The maximum continuous power is related to
maximum junction temperature, by the thermal
resistance factor. For dc 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, COLLECTOREMITTER VOLTAGE (VOLTS)
Fig. A. 41 Saturation region, common emitter,
constant base current.
2.0
50 100 150
V CE , % RATED MAXIMUM COLLECTOREMITTER 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 , COLLECTOREMITTER VOLTAGE (VOLTS)
Fig. A. 43 Saturation region, common emitter,
constant base current.
50 100 150
V cs , % RATED MAXIMUM COLLECTOREMITTER 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)
2N1529A32A
2N152933
3
3
300
65
10
2
5
2N1534A37A
2N1S3438
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«, EMITTERBASE 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 Dc current gain vs. collector
current.
1.0 2.0
Vb, emitterbase 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 
2N1529A2N1532A
•~<2ni5292ni533
" lc = 3 A
"V« = 2
V
2N1534A2N1537A.
2N15342N1538
_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
 2N1529A2N1532A
/ 2N15292N1533
N1534A2N1537A
'N15>'
J42f
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 = JunctiontoCase Thermal Resistance (°C/W),
CA = CasetoAmbient 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) Commonemitter configuration.
Hybrid
hie
Ke
h„
Common
base
ha
l + *«
l + A/6
hfb
1 + Afb
h b
h,b
1+A
/b
Common
collector
1h,
d + A/c)
Tee
equivalent
r e
r b +
1a
«•.
(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 commonemitter 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) Commonbase 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
ftrc1
ft/c ftp
ft/c
1 + ft/c
ft/c
ftoc
Tee
equivalent
r. +(la)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 commonbase hparameters.
rCH
"be
L
(a) Commoncollector 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
si
1 + A
/&
Tee
equivalent
r , r «
° 1  a
* e 'Vi
(la)r e
1
1 a
1
(la)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 commoncollector Aparameters.
222
Summary Charts
EO ^ — f • WV — • — OC
(a) Teeequivalent circuit, commonbase.
tH"**
bo WV— f— •— WV— • — OC
l+p
E
(b) Teeequivalent circuit, commonemitter.
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 teeparameters.
5.1 Singlestage amplifier formulae.
•l
A Z>
Aparameters
(Second subscript
omitted)
r L * h 
1
h,
R L *,
l + h R L
*l
[(" j e)]
Commonbase
teeequivalent
circuit
(see Fig. 2.31)
a*A + (i_a)^
' i r, ") i r, ' c ''
a r c + r b
r ' + r » TT
K)*
'c
2r, + r»(la),
!i<KP *£ « 1
1+ r 1 + R t
r e
'» + *«■ « 1
'c
««L r t + R t ^ ,
(— £)' "
Commonemitter
teeequivalent
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
Commoncollector
teeequivalent
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 pushpull 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)
^VccBV^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 10log 10 co]
(C.l)
20[llog 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
quencyresponse 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
FrequencyResponse Plotting
Express gain in decibels:
db = 20 log 10 ^ = 20 [log 10 100  log 10 co 2 }
CO
= 20[221og 10& >]
 40[llog 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 lowfrequency 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 highfrequency 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:
Highfrequency 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 highfrequency 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 lowfrequency and highfrequency 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 highfrequency 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
FrequencyResponse 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 lowfrequency asymp
tote at <oT  1, or co = 1/T. The slope at high frequency falls off at 20 db/decade,
since the highfrequency slope has co to the first power only in the denominator.
The lowfrequency phase shift is approximately zero, approaching a 90° lag at
very highfrequency. 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
FrequencyResponse Plotting
acteristic of the following transfer function:
(jw)(l + JcoT 2 )
Solution: Refer to Fig. C.7.
C.3 More Complex
FrequencyResponse 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
FrequencyResponse 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 ^^.^
TC^^^ 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
S180
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 ;
Overall phase lag is sum of separate phase
lag components.
(c)
Fig. C.ll (ab) 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
FrequencyResponse 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 nonsinusoidality. 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 loworder 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 ith 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 14
D 2 (%) = 1 !=0,
2+1  1
2 2
202(11)
Z) 3 (%) = ? L. = 0>
2 (2  + 1 + I)
yo= I[2 + + 2(2)] = l,
o
yi =I[20 + il] = 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
smallsignal 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 shortcircuited to
emitter
breakdown voltage, emitter to base,
collector open
breakdown voltage, reverse
collector electrode
input capacitance (commonbase)
input capacitance (commoncollector)
input capacitance (commonemitter)
output capacitance (commonbase)
output capacitance
(commoncollector)
output capacitance (commonemitter)
emitter electrode
smallsignal shortcircuit forward
current transfer ratio cutoff fre
quency (commonbase)
smallsignal shortcircuit forward
current transfer ratio cutoff fre
quency (commoncollector)
smallsignal shortcircuit forward
current transfer ratio cutoff fre
quency (commonemitter)
static value of the forward current
transfer ratio (commonbase)
smallsignal shortcircuit forward
current transfer ratio (commonbase)
static value of the forward current
transfer ratio (commoncollector)
smallsignal shortcircuit forward
current transfer ratio (common
collector)
static value of the forward current
transfer ratio (commonemitter)
smallsignal shortcircuit 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
(commonbase)
smallsignal value of the shortcir
cuit input impedance (commonbase)
static value of the input resistance
(common collector)
smallsignal value of the shortcir
cuit input impedance (common
collector)
static value of the input resistance
(commonemitter)
smallsignal value of the shortcir
cuit input impedance (common
emitter)
static value of the opencircuit out
put conductance (common base)
smallsignal value of the opencir
cuit output admittance (common base)
static value of the opencircuit out
put conductance (commoncollector)
smallsignal value of the opencir
cuit output admittance (common
collector)
static value of the opencircuit out
put conductance (commonemitter)
smallsignal value of the open cir
cuit output admittance (common
emitter)
smallsignal value of the opencir
cuit reverse voltage transfer ratio
(common base)
smallsignal value of the opencir
cuit reverse voltage transfer ratio
(common co Hector)
smallsignal value of the opencir
cuit reverse voltage transfer ratio
(c ommonemitter)
base current (dc)
base current (instantaneous)
collector current (dc)
collector current (instantaneous)
collector cutoff current (dc), emit
ter open
collector cutoff current (dc), base
open
collector cutoff current (dc), with
specified resistance between base
and emitter
collector current (dc), with speci
fied circuit between base and emitter
collector cutoff current (dc), with
base shortcircuited 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 (dc)
emitter current (instantaneous)
emitter cutoff current (dc), collec
tor open
forward current (dc)
forward current (instantaneous)
reverse current (dc)
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 (dc or average) to
the base electrode with respect to
the emitter electrode
collector junction dissipation
total power input (dc or average) to
the collector electrode with respect
to the base electrode
total power input (dc or average) to
the collector electrode with respect
to the emitter electrode
total power input (dc or average) to
the emitter electrode with respect to
the base electrode
largesignal input power
small signal input power
load power
largesignal output power
smallsignal output power
total power input (dc or average) to
all electrodes
smallsignal 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
iA
e
iC
Vbb
Vbc
v bc
junction to
thermal resistance, junction to case
base supply voltage (dc)
base to collector voltage (dc)
base to collector voltage
(instantaneous)
Vbe base to emitter voltage (dc)
v be base to emitter voltage
(instantaneous)
Vcb collector to base voltage (dc)
v cb collector to base voltage
(instantaneous)
V cc collector supply voltage (dc)
V C e collector to emitter voltage (dc)
v ce collector to emitter voltage
(instantaneous)
^CE( sat ) collector to emitter saturation volt
age
V EB emitter to base voltage (dc)
Veb emitter to base voltage
(instantaneous)
V EC emitter to collector voltage (dc)
v ec emitter to collector voltage
(instantaneous)
Vee emitter supply voltage (dc)
V F forward voltage (dc)
vp forward voltage (instantaneous)
V CBF d c opencircuit voltage (floating
potential) between the collector and
base, with the emitter biased in the
reverse direction with respect to
the collector.
V ECF dc opencircuit voltage (floating
potential) between the emitter and
collector, with the base biased in
the reverse direction with respect
to the collector
Vr reverse voltage (dc)
v R reverse voltage (instantaneous)
242
INDEX
Acceptor impurity, 4
Amplifier performance:
smallsignal, 97
commonemitter, 97
commonbase circuit, 107
commoncollector, 109
formulae, 113
Amplifiers:
basic circuits, 17
capacitorcouples, 124
directcoupled, 121
emitterfollower, 54
multistage, 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
Bypass 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 pushpull, 178
Class B pushpull, 180
Collectorbase feedback, 81
Collectorbase leakage, 67
Commonbase:
characteristics, 25
circuit, 107, 127
connection, 13, 17, 173
parameters, 47
derivation, 48
power gain, 14
voltage gain, 14
Commoncollector:
characteristics, 26
circuit, 17, 109, 176
Commonemitter, 17
characteristics, 25
circuit, 17, 97
Complementary transistors:
directcoupled, 155, 156
dc feedback, 157
Complex impedance, 126
Composite characteristics, 179, 181
Composite loadline, 181
Constant power hyperbola, 164
Conversion formulae, 45, 524
Coupling capacitor, 124
Covalent bonds, 2, 8
Crossover distortion, 181
Current feedback, 190, 194
Current gain, 26, 32
Cutoff, 73
Cutoff frequency, 113
Cutoff region, 160
Dc bias, 67
Dc feedback, 145, 157
Dc models, 20
Dc 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
EbersMoll model, 15
Electronhole pairs, 3, 6
Emitter bias, 83
Emitterfollower, 54, 95, 109, 121
current gain, 112
gain, 56
voltage gain, 1 12
Emitter resistance, 17
Energy gap, 2
Equilibrium, 6
Equivalent circuit, 41
EbersMoll model, 15
dc models, 20
hybrid77, 16, 21, 62, 114
smallsignal, 16
teeequivalent, 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
highfrequency response, 132
interstage coupling network, 125
lowfrequency, 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
hparameters, 38
Highfrequency behavior, 22
Highfrequency circuit, 140
Highfrequency performance, 113
Hybrid parameters, 38
conversion to tee parameters, 43
conversion to hybrid77 parameters, 64
243
Transistor Circuit Analysis
Hybrid77 :
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
Largesignal 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
Lowfrequency response, 124
Lowlevel transistor characteristics, 72
M
Majority carriers, 4
Miller effect, 118
Minority carriers, 4
Multiplication effect, 161
Multistage 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
Openloop, 186, 196
Operating limits, 160
Operating point, 26, 67
drift, 67, 69
Operational amplifier, 200
Output admittance, 39
Output impedance, 97
Output transformer, 141
pn junction, 5
Parameter definition, 39
Positive feedback, 187
Potential barrier, 5
Power amplifier, 160
Power amplifier design equations, 171
Power dissipation, 95
Pushpull 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
Selfheating, 85
Semiconductor crystal, 2
Singlestage amplifier, 97
Smallsignal:
amplifier, 97
operation, 30, 38
Stability, 196
Stability factors, 7376, 81
Stability margin, 198
Static characteristics, 24
Summing circuit, 200
Surface leakage, 10
Switched operation, 35
Switching circuit, 93
Teeequivalent circuit, 46
Temperature effects, 68, 162, 168
Thermal resistance, 162
Thermal runaway, 18, 89
Thevenin's theorem, 71, 129
Threestage amplifier, 148
Transconductance, 166
Transformercoupled 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
Twostage amplifier, 125, 129, 133
Twotransistor circuit, 34
U
Universal curve, 126
V
Voltage feedback, 190
Voltage gain, 33
244
SIMON and SCHUSTER
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