an Sonos EMba eee os reaGreek Symbols « acalpha @ge d-c alpha Bae beta Bae d-e beta Bop Darlington-pair beta Abbreviations and Symbols A voltage gain Ap voltage gain with feedback bandwidth C. — collector-diode capacitance C, — emitter-diode capacitance Ce emitter bypass capacitance C; junction capacitance Ci load capacitance C, — source capacitance fa alpha cutoff frequency fo beta cutoff frequeney fo resonant frequency fr gain-bandwidth product Gm transconductance hu input impedance with out- put shorted hi, reverse voltage gain with input open hy forward current gain with output shorted he output admittance with in- put open hy the ha of a CB circuit hye the hey of a CC circuit hy the ha: of a CE cireuit Bo low-frequency beta AB change in beta At change in current Av change in voltage » amplification factor hy the hi of a CB circuit hic the hu of a C hie the hu of a C he the he: of a CB cireuit Foc the haz of a CC circuit ioe the hay of a CE circuit circuit ‘ireuit hw the his of a CB circuit Ie the his of a C hee the his of a CE circuit Hz hertz (cps) ireuit a total current ty —a-e base current iy total base current Tn d-c base eurrent i, a-e collector current ic total collector current Ic d-e collector current Teno collector-base leakage eur- rent with open emitter Tero collector-emitter leakage current with open base Tco same as Icno Towa) collector saturation eur- rent i, a-¢ emitter current-s Seid — ig — total emitter current Ig d-c emitter current Treat, emitter saturation cur- rent Ty forward current i, — load current Treakage Collector leakage current Ip diode reverse current IV current-voltage K beta sensitivity kHz kilohertz (ke) MHz megahertz (me) n turns ratio, or number of stages Q figure of merit Ri||R2 Rs in parallel with Re Tye ae resistance 7, base spreading resistance rs bulk resistance Re external base resistance “emitter a-c resistance re — unbypassed emitter resistance Ry external emitter resistance ry forward resistance vj a-c junction resistance rz ae load resistance Rr reverse resistance ry plate resistance T. —_ source resistance rs zener resistance Rx cathode resistance Ri dee load resistance Rx||R_ Rx in parallel with R ve total collector-ground voltage ves total collector-base voltage vce total collector-emitter voltage vy feedback voltage vy peak ae voltage v% a-¢ source voltage Vex d-c base-emitter voltage _ Ve d-ccollector-ground voltage Vee collector supply voltage Vex d-ecollector-emitter voltage Ve d-c emitter-ground voltage Vex emitter supply voltage Vx knee voltage Ve reverse voltage Vz — zener voltage_fpomrle f- Vo. rk Mot CL 2 — = 1 Sap tpn esos U fee i : ste c ae fe 3 a ’ om ee Kili RTRANSISTOR CIRCUIT APPROXIMATIONSTRANSISTOR CIRCUIT APPROXIMATIONS Albert Paul Malvino Foothill College Los Altos Hills, California Vice President Time Systems Corporation Mountain View, California McGraw-Hill Book Company New York « St. Louis « San Francisco London « Toronto Sydney « Mexico « Panama‘TOR CIRCUIT APPROXIMATIONS Copyright © 1968 by McGraw-Hill, Inc. All Rights Reserved. Printed in the United States of America. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, ‘mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Library of Congress Catalog Card Number: 68-13521 39846 5 6 78 9 10 (MPMM) 7 43210To my wild Irish roseIt is the mark of an instructed mind to rest satisfied with that degree of precision which the nature of the subject admits, and not to seek exactness where only an approximation of the truth is possible ARISTOTLEPreface When using a transistor, we must realize that we are dealing with a device that is inexact and unreliable as far as its characteristics are concerned. For instance, one important transistor quantity is its current gain ¢ (also designated h.). Typically, the 8 can vary over at least a 2:1 range when we change from one transistor to another of the same type. Because of this variation, it is impossible to predict exactly how a transistor will operate in a specific circuit arrangement. Also, when the surrounding temperature changes, the 8 changes, thereby adding another degree of uncertainty to transistor circuit operation. There is a way out of this predicament. To eliminate the wide variations in transistor circuit performance, we can use negative feedback or some similar technique for trading off some of the gain in exchange for more stable operation. However, when we do this, we are making the operation of the circuit almost independent of the transistor characteristics, In other words, after using enough negative feedback to stabilize the circuit operation, knowing the exact value of @ and other transistor character- istics is no longer important. The point is simply this: exact formulas for transistor circuit analysis are of limited value to most of us because the exact characteristics of a transistor are seldom known. Considering the tolerances of transistor parameters, it is appropriate to use approximations in analyzing various transistor circuits. Most of this book stresses the ideal-transistor approach, that is, an approximation that retains only the most significant features of transistor action, With this approach, the novice can quickly obtain a feeling for how transistor circuits work; oddly enough, this highly sim- plified approach is adequate for much of the transistor work that we encounter. In the later chapters of the book we discuss some of the second-order effects that are discarded in the ideal-transistor approach. To round out viiviii Preface the presentation, the last chapter deals with the h-parameter approach, because this technique, at least in theory, gives exact answers. This book was written for an electronics technician in a junior college or technical institute; many others in the electronics field will find the book useful, especially if they need a practical introduction to transistors. The length of the book makes it suitable for an introductory three-unit course. The prerequisite is a sound knowledge of algebra and basic electricity. I would like to express my thanks to Daniel J. Mindheim of Time Systems Corp. for his advice throughout the writing of this book and to Clifford Burrous of Ames Research Center for his careful review of the final manuscript and for his many excellent suggestions. ALP, MALVINOa Contents Semiconductor Physics 1-1. Germanium and Silicon Atoms 1-2. Germanium and Sili- con Crystals 1-3. Conduction in Pure Germanium and Silicon 1-4. The Hole Concept 1-5. Extrinsic Semiconductors The p-n Junction Diode 2-1. The p-n Junction 2-2. The Rectifying Properties of a p-n Junction 2-3, The 1V Characteristics of a Semiconductor Diode Large-signal Diode Approximations 3-1. The Ideal Diode 3-2. The Second Approximation of a Real Diode 3-3. The Third Approximation of a Real Diode 3-4. Using the Diode Approximations 3-5. Approximating the Reverse Current 3-6. Zener Diodes 3-7. The Second Approxi- mation of a Zener Diode Small-signal Diode Approximations 4-1. The Superposition Theorem 4-2. Superposition in Hlec- tric Circuits 4-3. The A-C Resistance of a Diode 4-4. For- mulas for the A-C Resistance of a Diode 4-5. Applying the Superposition Theorem to Diode Cireuits 4-6. Diode Capaci- tance in the Reverse Region Common-base Approximations 5-1. Terminology and Schematic Symbols 5-2. Biasing the Transistor 5-3. The IV Characteristics of a Common-Base 12 22 AT TA10 Contents Transistor 5-4. The Alpha of a Transistor 5-5. The Ideal Transistor 5-6. Using Superposition to Analyze CB Circuits 5-7. A Complicated A-C Equivalent Circuit 5-8. Formulas for CB Analysis 5-9. Notation for Voltages and Currents Common-emitter Approximations 6-1. The [V Characteristics of the CE Connection 6-2. The Beta of a Transistor 6-3. The Ideal CE Transistor 6-4. Base Bias of a Transistor 6-5. Emitter Bias of a Transistor 6-6. Analyzing CE Transistor Circuits 6-7. The Voltage Gain of an Emitter-biased Stage 6-8. Effects of Source Resistance 6-9. Tube-to-transistor Transformations Common-collector Approximations 7-1. The Basic Idea of the CC Connection 7-2. Derivation of CC Formulas 7-3. The Darlington Pair Large-signal Operation 8-1. The D-C Load Line 8-2. Load-line Interpretation of an A-C Signal 8-3. The A-C Load Line 8-4. Obtaining Maximum Unelipped Signal p-n-p Load Lines 8-6. Load Lines for the CE Connection 8-7. The D-C Load Line for the CC Con- nection 8-8. The A-C Load Line for the Emitter Follower Bias Arrangements 9-1. The Concept of 8 Sensitivity 9-2. Base Bias 9-3. Base Bias with Emitter Feedback 9-4. Base Bias with Collector Feed- back 9-5. Base Bias with Collector and Emitter Feedback 9-6. Emitter Bias with Two Supplies 9-7. Emitter Bias with One Supply 9-8. A Comparison of Sensitivities 9-9. Location of the Ground Point 9-10. Biasing p-n-p Transistors A-C Operation 10-1. CE Operation 10-2. A-C Analysis of CE Circuits 10-3. Emitter Feedback 10-4. CC Operation 10-5. CB Operation 10-6. The Effect of Source Resistance 10-7. Stabilizing the Voltage Gain from Source to Output 10-8. p-n-p Operation 10-9. The Ground Point 10-10. Maximum Signal-handling Capability ill 148 166 202 241Contents iL 12 13 4 Cascading Stages 1-1. RC Coupling M-2. Two-stage Feedback 11-3. Induc- tive Coupling 11-4. Transformer Coupling 11-5. Tuned Am- plifiers 11-6. Direct-coupled Amplifiers Temperature Effects 12-1. Changes in Emitter-junction Resistance 12-2. Changes in 8 12-3, Changesin Vx 12-4, Leakage Current in a Grounded- base Circuit 12-5. Leakage Current in a Grounded-emitter Cir- cuit 12-6. The Stability Factor Frequency Response 13-1. Response of an RC-coupled Amplifier 13-2. Lower Cutoff Frequency of a Typical CB Stage 13-3. Upper Cutoff Frequency of a CB Stage 13-4, Lower Cutoff Frequency of a CE Amplifier 13-5. Transistor Cutoff Frequencies 13-6. Base Spreading Resistance 13-7. Upper-frequency Limit of a CE Stage 13-8. Response of Caseaded Stages h Parameters 14-1, Concept of Parameters 14-2. Input Impedance of a Network 14-3. Voltage Gain Using h Parameters 14-4. The h Parameters of a Transistor 14-5. The h Parameters of the Ideal CB Transistor 14-6. The h Parameters of the Ideal CE Transistor 14-7. Conversion of h Parameters 14-8. Practical Observations on the h Parameters References for Further Reading Answers to Odd-numbered Problems Index xi 286 317 340 366 395, 397 400Semiconductor Physics Throughout most of this book we will be interested in how the transistor is used in typical circuits. We will develop a number of simple approxi- mations to allow a rapid and casy analysis of such circuits. At the outset of our discussion, however, it is important to consider some aspects of atomic theory. In this chapter we discuss semiconductors, free electrons, holes, and doping. This material will make it easier for us to understand how the transistor actually works. 1-1 Germanium and Silicon Atoms In this section we examine the atomic structure of germanium (Ge) and silicon (Si) because these materials are widely used in the fabrication of transistors. Most of us already know that all matter is composed of atoms and that atoms contain a nucleus surrounded by revolving electrons. If we ex- amine an isolated atom of germanium, we find that the nucleus contains 32 protons. When this atom is in a normal state, there are 32 electrons revolving around the nucleus. Further, these revolving electrons distrib- 12 Transistor Cirenit Approximations ute themselves in a definite pattern and occupy what are commonly called shells. Figure 1-1a symbolically illustrates a germanium atom. In the center there is a nucleus with 32 protons. The revolving electrons distribute themselves in different shells, following the pattern of 8,18... Qn? where n is the number of the shell. In other words, there are 2 electrons in the first shell, 8 in the second shell, and 18 in the third shell. The last 4 electrons are in the fourth, or outer, shell. In a similar way we find that an isolated atom of silicon has a nucleus with 14 protons. When this atom is in a normal state, there are 14 revolv- ing electrons. As shown in Fig. 1-1), the first shell contains 2 electrons, and the second shell contains 8. The 4 remaining electrons are in the third, or outer, shell of the silicon atom. 2-8-18-4 2-8-4 (0) (6) {o) (4) Fig. 1-1 (a) Germanium; (b) silicon. Fig. 1-2 Outer shell of (@) germa- nium and () silicon. We are primarily interested in the cater shells of germanium and sili- con atoms; as a result, we will simplify the diagrams of these atoms. In Fig. 1-2a we have shown only the outer shell of the germanium atom. The inner part of this figure is called the germanium core, and it contains the nueleus plus the inner shells, In similar way, Fig. 1-2b shows a silicon core surrounded by an outer shell with four electrons. 1-2 Germanium and Silicon Crystals In the previous section we were discussing isolated atoms of germanium and silicon. Now we discuss how a number of germanium or silicon atoms combine to form a solid piece of germanium or silicon. Let us consider germanium first. The outer shell of an isolated ger- manium atom contains four electrons. This shell is incomplete in the sense that more electrons can be added to it. It is an experimental fact that germanium atoms combine with each other in such a way as to have2 ‘Transistor Circuit Approximations ute themselves in a definite pattern and occupy what are commonly called shells. Figure 1-la symbolically illustrates a germanium atom. In the center there is a nucleus with 32 protons. The revolving electrons distribute themselves in different shells, following the pattern of 2,8, 18... , Qn? where n is the number of the shell. In other words, there are 2 electrons in the first shell, 8 in the second shell, and 18 in the third shell. The last 4 electrons are in the fourth, or outer, shell. Ina similar way we find that an isolated atom of silicon has a nucleus with 14 protons, When this atom is in a normal state, there are 14 revoly- ing electrons. As shown in Fig. 1-1b, the first shell contains 2 electrons, and the second shell contains 8, The 4 remaining electrons are in the third, or outer, shell of the silicon atom. SO ® © 2-8-18-4 ae (a) (o Fig. 1-1 (a) Germanium; ® silicon, Fig. 1-2 Outer shell of (a) germa- nium and (b) silicon. We are primarily interested in the outer shells of germanium and sili- con atoms; as a result, we will simplify the diagrams of these atoms. In Fig. 1-2a we have shown only the outer shell of the germanium atom. The inner part of this figure is called the germanium core, and it contains the nucleus plus the inner shells, In similar way, Fig. 1-2b shows a silicon core surrounded by an outer shell with four electrons. 1-2. Germanium and Silicon Crystals In the previous section we were discussing isolated atoms of germanium and silicon. Now we discuss how a number of germanium or silicon atoms combine to form a solid piece of germanium or silicon. Let us consider germanium first. The outer shell of an isolated ger- manium atom contains four electrons. This shell is incomplete in the sense that more electrons can be added to it. It is an experimental fact that germanium atoms combine with each other in such a way as to haveSemiconductor Physics 3 a total of eight electrons in the outer shell of each atom. In order to accom- plish this, the atoms align themselves in a structure known as a crystal. When we examine a crystal of pure germanium, we find that each atom is surrounded by four neighboring atoms that actually share electrons with the central atom. Figure 1-3 brings this concept out more clearly. ‘The central atom has @ total of eight electrons in its outer shell. Four of these electrons belong to the central atom; an additional four electrons have been added by the neighboring atoms because each of these neighbors actually shares one of its outer-shell electrons with the central atom. In turn, each of the neighboring atoms will have four neighbors; in this way, every atom within the crystal has an outer shell with eight electrons. Fig. 1-3 An atom within a crystal. In a completely analogous way a large number of silicon atoms will join to form a crystal in which each atom has four neighbors that share electrons. Once again, each silicon atom will have a total of eight elec- trons in its outer shell. For our purposes, we will accept as experimental fact that the outer shell of a germanium or silicon atom is incomplete until it contains eight electrons. When there are fewer than eight electrons, an atomic force of attraction exists until enough electrons are added to produce a total of eight electrons in the outer shell. Once there are eight electrons in this outer shell, these electrons are tightly held or bound to the atom and cannot escape from the atom unless some outside force is applied. 1-3 Conduction in Pure Germanium and Silicon How well does a crystal of germanium or silicon conduct an electric eur- rent? To answer this question, consider Fig. 1-4a, in which we have shown4 Transistor Circuit Approximations a crystal of pure germanium connected to a battery. When the tempera- ture is at absolute zero, we find that there is no current in the cireuit. The reason for this is that there are no free electrons within the crystal. All electrons are tightly held within the atoms of the crystal; the electrons in the inner shells are buried well within the individual atoms and cannot contribute to current flow; the electrons in the outer shells are also tightly bound to the atoms and cannot participate in current flow. Therefore, at absolute zero temperature the germanium crystal is an insulator. Pure Metol Germanium Metal t t t == Electron motion Hi th (a) (6) Fig. 1-4 Conduction at (a) absolute zero temperature and (6) higher tempera- tures. As the temperature increases, however, thermal energy is added to the germanium crystal. This energy can actually dislodge some of the outer- shell electrons, thereby making them available for current flow. An elec- tron that has been knocked out of an outer shell is called a free electron. In Fig. 1-4b we have indicated the free electrons by minus signs. These electrons are now free to move, and under the influence of the battery they move to the left, setting up a current flow. The size of this current is usually small because only a few free vleclruns are produced by thermal energy. Of course, if the temperature is increased further, more electrons will be liberated from outer shells, and a larger current will result. At room temperature (around 25°C) we find that the amount of current is too small to consider the germanium a conductor and yet too large to continue calling it an insulator. Therefore, we refer to the germanium crystal as a semiconductor. A silicon erystal behaves much the same as a germanium erystal as far as electric current is concerned. At absolute zero temperature all elec- trons are tightly bound to their atoms, and the silicon erystal acts like an insulator. As the temperature increases, some electrons are knocked out of outer shells and become available for current flow. However, in silicon the outer-shell electrons are more tightly held than in germanium; as aresult, more thermal energy is needed to dislodge an outer-shell clec- tron from a silicon atom than from a germanium atom. Therefore, weSemiconductor Physies 5 note one important difference between silicon and germanium: at the same temperature there are fewer free electrons in a silicon crystal than in a germanium crystal. 1-4 The Hole Concept An interesting concept in discussing semiconductors is that of a hole. To bring this idea out clearly, eonsider Fig, 1-5. A typical atom of germanium (or silicon) is shown with four neighboring atoms. As already indicated, thermal energy can dislodge an electron from the outer shell of an atom. O= Hole Free electron @= Electron Fig. 1-3 Production of a hole. When this happens, the released electron becomes a free electron and is capable of moving to another part of the crystal structure. The departure of this electron produces a vacancy in the outer shell that is called a hole. The hole behaves like a positive charge in the sense that it will attract and capture any electron in the immediate vicinity. In a crystal of pure germanium or silicon, an equal number of holes and free electrons are created by thermal energy. The free electrons drift randomly throughout the crystal structure. Occasionally, a free electron approaches a hole closely enough to be attracted and captured by the hole. When this happens, the hole disappears, and the free electron be- comes a bound electron. The action of a free electron moving into a hole is called recombination. At any one instant in time the following takes place within the crystal: 1. Some free electrons and holes are being generated by means of thermal energy. 2. Other free electrons and holes are recombining.6 ‘Transistor Circuit Approximations 3. Some free electrons and holes exist in an in-between stage; that is, they were previously generated and have not yet recombined. The aver- age amount of time that these free electrons and holes exist before re- combining is called the lifetime. There are actually two components of current possible in a semicon- ductor: the movement of free electrons is one component, and the move- ment of holes is another component. The movement of free electrons was briefly discussed in Sec. 1-3 and is illustrated in Fig. 1-4. The holes also move. To understand how this takes place, consider Fig. 1-6. At the extreme right a single hole is shown. Adjacent to this hole is a bound electron at position A. This bound electron is attracted by the hole and can move into the hole. When this happens, the original hole dis- appears, and a new hole appears at position A. The new hole at position A Electron drift Hole drift Fig. 1-6 Hole movement. Fig. 1-7 Components of current. can now attract. and capture the bound clectron at position B. When this happens, the hole at position A disappears, and a new hole appears at position B. This process can continue with a bound electron moving along the arbitrary path shown by the arrows. In this way, a hole can move from one atom to another. Note that the hole moves opposite to the direction of the bound electrons. In the absence of an electric field the hole movement is a random process; that is, holes move in all directions, so that there is no net cur- rent in any one direction. If, however, an electric field is present within the crystal, a net current takes place in the direction of the field. For instance, in Fig. 1-7 a battery is connected to a pure semiconductor. Thermally produced holes and free electrons are represented by plus and minus signs. Because of the battery polarity, the free electrons move to the left, and the holes move to the right. Of course, there will be some recombination of free electrons and holes. However, free electrons and holes are being continuously produced by thermal energy, so that at anySemiconductor Physics 7 one instant in time there are some free electrons and holes that can par- ticipate in a current flow. Note the two distinct components of current in Fig. 1-7. The drift of free electrons to the left is one component of this current. When the elec- trons reach the left end of the semiconductor, they are collected by the metal plate and enter the positive terminal of the battery. In the mean- time, the negative terminal of the battery injects free electrons into the right end of the semiconductor, thereby maintaining a continuous flow of free electrons. The movement of holes to the right is the second com- ponent of current. When the holes reach the right end of the semicon- ductor, electrons from the battery combine with these holes and become bound electrons. In the meantime, new holes are generated within the crystal by means of thermal energy. In this way, a continuous flow of holes is maintained. There is one more point. Strictly speaking, a hole is not a positive charge; it is simply a vacancy in the outer shell of an atom. Yet, a num- ber of experiments (such as those involving the Hall effect) indicate that holes move and act as though they were positive charges. For this reason we will use a plus sign to represent a hole and will think of it as a positive charge. 1-5 Extrinsic Semiconductors The number of free electrons and holes produced by thermal energy in a pure semiconductor is generally too small to be of any practical use. It is possible, however, to increase the number of free electrons and holes by a process called doping. This simply means that we can add impurity atoms to the germanium or silicon crystal. A semiconductor that has been doped is called an extrinsic semiconductor; a semiconductor that is still pure, or undoped, is called an intrinsic semiconductor. One way of doping pure germanium (or silicon) is to first break down the crystal structure by melting it. A small amount of an impurity ele- ment with five electrons in its outer shell can then be added to the molten germanium. (Examples of elements with five outer-shell electrons are phosphorus, arsenic, antimony.) Assume that we add a small amount of arsenic. The arsenic atoms will diffuse throughout the molten ger- manium. When the germanium is cooled, a solid crystal will form. Once again, we find that each atom within the crystal has four neighboring atoms which share their outer-shell electrons. Examining the crystal on the atomic level, we find that most of the atoms are germanium; occasion- ally we find an arsenic atom. This arsenic atom has taken the place of a germanium atom in the crystalline structure, and, as a result, it has four8 ‘Transistor Cireuit Approximations neighboring atoms. In Fig. 1-8 we see that the central arsenie atom has cight electrons in its outer shell, This atom originally had five electrons in its outer shell. Each neighboring atom is sharing one of its outer-shell electrons with the arsenic atom. Therefore, there is an extra electron which becomes a free electron. It is clear that each arsenic atom in the doped semiconductor produces one free electron. Obviously, by control- ling the amount of arsenic that is added, we can control the number of free electrons in the doped crystal. Note that the production of free electrons by doping is quite different from the thermal production of free electrons and holes. Only free elec- trons have been produced by doping the semiconductor with arsenic, As : Free electron Fig. 1-8 An n-type semiconductor. Fig. 1-9 A p-type semiconductor. a result, there are more free electrons in the doped germanium than there are holes. The only holes present are those produced by thermal energy. A pure semiconductor that has been doped by an element with five electrons in its outer shell is called an n-type semiconductor. The n stands for negative, referring to the excess of free electrons in the doped ger- manium, Since a piece of n-type semiconductor has more free electrons than holes, we commonly speak of the electrons as the majority earriers and the holes as the minority carriers. We can also dope a semiconductor to obtain an excess of holes. In this case, we must use an impurity element with three outer-shell electrons (elements like boron, aluminum, gallium, ete.). If a small amount of this kind of impurity element is added to the molten germanium, then after cooling to obtain a crystal we find that each impurity atom has displaced one of the germanium atoms. Again, we note that most of the atoms in the erystal are germanium. Oceasionally, we find an impurity atom. In Fig. 1-9 we have shown an impurity atom with the usual four neighbors.Semiconductor Physies 9 ‘As shown, there are only seven electrons in the outer shell (three origi- nally in the impurity atom, and four from the neighboring atoms). Thus, there is a hole in the outer shell of the impurity atom. Clearly, by varying the amount of impurity we can control the number of holes in the crystal. A semiconductor doped by impurity atoms having three outer-shell electrons is called a p-lype semiconductor. The p stands for positive, referring to the excess of holes in the erystal. Since the p-type semi- conductor has more holes than free electrons, the holes are the majority carriers, and the free electrons are the minority carriers. Figure 1-10 summarizes the two types of semiconductors. M-type P-type + Mojority carriers ++ ++ ++ ++ +t = Mojority corriers t4e rae + + + |—Minority carriers J— Minority carriers (a) (6) Fig. 1-10 Majority and minority carriers. Doping makes it possible to obtain usable levels of current in semi- conductors. Of course, a single piece of n-type or p-type semiconductor is not much better than a carbon resistor. It is when p-type and n-type materials are brought together that something new and useful results. This is discussed in later chapters. SUMMARY The two most widely used semiconductor materials are germanium and silicon. Isolated atoms of these materials have four electrons in the outer shells. A crystal refers to the geometric structure that results when a large number of germanium or silicon atoms combine to form a solid piece of material. Each atom in the erystal has four neighboring atoms which share electrons to produce a total of eight electrons in the outer shell of each atom. At absolute zero temperature, pure germanium or silicon acts like an insulator because there are no free electrons available for current flow. All eight electrons are tightly held in the outer shell of each atom. As the temperature increases, however, thermal energy dislodges some electrons from outer shells. This results in free electrons and holes, which can drift under the influence of an electric field to produce a current. The size of this current is usually too small to be of any practical use. Generally10 Transistor Cireuit Approximations speaking, pure silicon has fewer free electrons and holes than pure ger- manium at the same temperature. A hole refers to a vacancy in the outer shell of a germanium or sili- con atom and occurs when there are fewer than eight electrons in this shell. A hole will exert a force of attraction on any nearby electron. Move- ment of a free electron into the hole is called recombination. In this case, the hole disappears. On the other hand, when a bound electron moves into the hole, the hole simply moves to a new position. A pure semiconductor is called an intrinsic semiconductor. Adding im- purities to a pure semiconductor is called doping, and we call the doped semiconductor an extrinsic semiconductor. Whereas the n-type semicon- ductor has an excess of free electrons, the p-type semiconductor has an excess of holes. GLOSSARY crystal The internal structure of a solid piece of germanium or silicon. In this structure each atom has four neighboring atoms that share outer-shell electrons. doping Adding impurity atoms to pure germanium or silicon in order to increase the number of free electrons or holes. extrinsic semiconductor Doped germanium or silicon. hole A vaeaney in the outer shell of an atom. It can be produced either by thermal energy or by doping. intrinsic semiconductor Pure germanium or silicon. The only current carriers are the free electrons and holes produced’by thermal energy. lifetime The average amount of time that a free electron or hole exists after being generated but before recombining. n-type semiconductor A semiconductor that has been doped to produce an excess of free electrons. p-type semiconductor A semiconductor that has been doped to produce an excess of holes. recombination The merging of a free electron and a hole. semiconductor Material like germanium or silicon whose electrical prop- erties lie between those of an insulator and a conductor. REVIEW QUESTIONS 1, What does a silicon core refer to? 2. How many electrons are found in the outer shell of an isolated ger- manium atom under normal conditions?Se miconductor Physies 11 . What is a crystal? . Ina erystal of germanium or silicon how many outer-shell electrons are there in each atom? . Why does a pure semiconductor behave like an insulator at absolute zero temperature? . Why do we use the word semiconductor to describe a erystal of pure germanium or silicon? . At the same temperature which conducts better, a erystal of pure silicon or a crystal of pure germanium? What is the reason for this? . What is a hole? Name two ways in which a hole is produced. . What word describes the merging of a free electron and a hole? . Define an intrinsic semiconductor and an extrinsic semiconductor. . To produce an n-type semiconductor we add an impurity element with how many electrons in its outer shell? 2. What are the majority carriers in a p-type semiconductor?2 The p-n Junction Diode To understand how transistors work in different circuits it is first neces- sary to understand how a typical semiconductor diode works. We begin this chapter by joining a piece of p-type material to a piece of n-type material. The result is the junction diode. After discussing why this p-n junction acts like a diode, we obtain the typical ZV characteristics of a semiconductor diode. Our work in this chapter is concerned with under- standing the important features of diode behavior. We need this for our later work in transistors. 2-1 The p-n Junction We know that at room temperature a piece of p-type material has mostly holes produced by doping and only a small number of free electrons pro- duced by thermal energy. Also, a piece of n-type material has mostly free electrons with only a small number of holes produced by thermal energy. In Fig. 2-la we have symbolically shown a piece of p-type material. The plus signs represent the holes, which are free to move around in the crystal. The circled minus signs represent the atoms associated with these 2The p-n Junction Diode 13 holes. Thus, if a hole moves away from the atom associated with it, that atom becomes negatively charged. Note that the negatively charged atom is not free to move around within the crystal. This atom is held in position because it is part of the crystal structure. Therefore, in Fig. 2-La the plus signs represent positive charges that are free to move, and the circled minus signs represent negatively charged atoms that are immobile. Fig. 2-1 Current carriers and im- mobile atoms. (a) p-type and (b) n-type material. + O+ Or D+ O+ O+ |v O+@+ O+ QO @ OOO |= Qe @ (a) = Similarly, in Fig. 2-16 we have a piece of n-type material. Here, the minus signs symbolize the free electrons, and the circled plus signs sym- bolize the atoms associated with these free electrons, Note that if a free electron moves away from its associated atom, it leaves a positively charged atom behind, Once again, this positively charged atom is not free to move; it is embedded at its particular position within the crystal structure. For the moment we are disregarding the minority carriers in both types of material, and therefore we have not shown the small number of free electrons in the p-type material or the small number of holes in the n-type material. Let us remember that there are some thermally produced mi- nority carriers in both materials. It is possible by various manufacturing techniques to produce a single crystal with p-type material on one side and n-type material on the other. Does anything unusual happen at the junction of the p- and n-type ma- terials? To answer this question, consider Fig, 2-2a. At the instant that the junction is formed, the holes are still in the p-type material and the free electrons are still in the n-type material. These charges are free to move; as a result, they do move randomly in all directions. Some of the free electrons and holes at the junction of the two materials will move across the junction and recombine. When this happens, the free charges disappear, since they neutralize each other. This recombination of free charges produces a narrow region at the junction called the depletion region (see Fig. 2-2b). In this region there are essentially no free electrons or holes; there are only positively and negatively charged atoms, which are not free to move. As the depletion region builds up, a difference of potential builds up, simply because there are positively charged atoms on the right side of4 Transistor Cireuit Approximations ill ae region P uv e W 66 élea@ & Sica 6 = boat “a 6641556 5 dielete 6 +e ¥4 eee Seti Sete 680 ej@e © © 9 joe @ © fo) (6) Fig. 2-2 A p-n junction. (@) At the instant of formation; (6) depletion region. the junction and negatively charged atoms on the other side. Eventually, this potential becomes large enough to prevent further electron and hole movement across the junction, To understand why, note in Fig. 2-2b that an electric field exists at the junction; the direction of this field is from right to left. This means that when an electron tries to pass through the junction, repelled by the negative column of ionized atoms on the left and attracted by the positive column of ionized atoms on the right. In other words, a free electron is repelled into the n-type material. Similarly, any holes in the p-type material that try to move across the junction after the depletion region has formed will encounter a repelling force that drives them into p-type material. The difference of potential at the junction is called the barrier potential. We find that at room temperature (about 25°C) the barrier potential is approximately 0.3 voli for germanium and 0.7 volt for silicon. Let us summarize the main points of our discussion: 1. After the p-n junction is initially formed, free electrons and holes move across the junction and recombine. 2. A depletion region appears at the junction, in which there are essen- tially no free charges but only immobile charged atoms. 3. The charged atoms in the depletion region set up a barrier potential that prevents further movement of free electrons and holes across the junction. (This summary of main points disregards the small number of minority carriers in each material, If they are taken into account, the first two steps of the summary are the same, but step 3 is slightly different. At equilibrium, a small number of minority carriers are swept across the junetion by the barrier potential, while an equal number of majority carriers move across the junction in the opposite direction.) 2-2 The Reetifying Properties of a p-n Junction A p-n crystal can be used as a diode because it allows current to flow more easily in one direction than in the other. To understand why this isThe p-n Junction Diode 15 so, consider Fig. 2-3, where we have shown a battery connected to a p-n crystal. For the moment, disregard the minority carriers on either side of the junetion. The barrier potential still tries to retard the movement of holes and free electrons across the junction. However, the battery drives holes and free electrons toward the junction. This means that all the free electrons in the n-type material move en masse to the left; as they move to the left, new free electrons are injected by the battery into the right end of the crystal. Thus, a flow of free electrons is set up in the wire connected to the negative battery terminal. The holes in the p-type material are also driven toward the junction. As they move en masse to the right, new holes are created at the left end of the erystal because bound electrons leave the p-type material and enter the positive terminal of the battery. The important point to grasp here is that the battery potential has overcome the barrier potential and has established a flow of electrons in the external circuit. This flow is sus- tained because the holes and free electrons converging on the junction recombine with each other. To better understand how the current flows in Fig. 2-3, let us follow one electron around the entire circuit. Here is approximately what hap- pens. A free electron leaves the negative terminal of the battery and enters the n-type material. Once in the n-type material, this free elec- tron is driven toward the junction by the battery potential. Somewhere in the vicinity of the junction this electron recombines with one of the holes from the p-type material. When this happens, the free electron be- comes a bound electron. As a bound electron, it can continue drifting to the left by moving into an adjacent hole, and it now travels to the left through the p-type material until it reaches the left end of the crystal. When it leaves the crystal and enters the wire, the bound electron be- comes a free electron, which then flows through the wire into the positive battery terminal. More or less, this is what happens to each of the elec- trons that participate in the total current flow. Note that the current carriers inside the p-n crystal are the holes and the free electrons. However, in the external circuit, that is, in the battery wires, only free electrons are moving. Also note that the conventional cur- rent is in the same direction as the holes. In drawing circuit diagrams we will be using conventional current, as shown in Fig, 2-3. This is a matter of convenience; there are many electronic devices whose schematic sym- bols are based upon conventional current; as a result, it will be easier to understand the meaning of these schematic symbols if we adopt the use of conventional current. Simply remember that conventional current flows in the same direction as the hole flow or opposite to the direction of the electron flow. What happens if we reverse the polarity of the battery, as shown in16 ‘Transistor Circuit Approximations Fig. 2-4? The holes in the p-type material now move to the left, and the free electrons in the n-type material move to the right. The depletion layer widens as these carriers move away from the junction, and momen- tarily there will be a current in the external cireuit. However, this current cannot continue indefinitely because there are no new holes and free elec- trons created by the battery, as in Fig. 2-3. The battery now aids the barrier potential in preventing further movement of majority carriers across the junction. Thus, for practical purposes we can say that after a momentary flow of current, the battery potential aids the barrier poten- tial in preventing majority carriers from moving across the junction. Actually, there is a small amount of current in the circuit of Fig. 2-4. Recall that we have shown only the majority carriers in Figs. 2-3 and 2- Holes —— ——Electrons Conventionot current += I v Fig. 2-3 Current in a diode. Fig. 2-4 Reverse bias. Because of thermal energy, there are also some holes in the n-type material and some free electrons in the p-type material. The battery drives these minority carriers across the junction, and therefore there is a small current in the circuit of Fig. 2-4. Thus, the p-n crystal acts as a diode, since it allows current to flow easily in one direction only. To summarize the major points we note: 1. When the positive terminal of the battery is connected to the p-type material, current flows easily because the battery potential overcomes the barrier potential and drives majority carriers across the junction. The diode is then said to be forward-biased. 2. When the positive terminal of the battery is connected to the n-type material, very little current flows because the battery aids the barrier po- tential in preventing majority carriers from moving across the junction. The only current is the flow of the minority carriers. We call this con- dition reverse bias or back bias.The p-n Junction Diode 7 2-3. The JV Characteristics of a Semiconductor Diode A p-n diode and its schematic symbol are shown in Fig. 2-5. From our discussion of the preceding section we know that current flows easily when the positive terminal of the battery is connected to the p-type material. Easy direction for conventional current selves Ap Anode Cathode Fig. 2-5 Schematic symbol of a diode. This means that conventional current flows easily from the p-type to the n-type material. As a memory aid, note that the triangle of the schematic symbol points in the easy direction of conventional current. In other words, whenever a diode is used in a circuit where the conventional current is trying to flow in the direction of the triangle, the diode offers a low- impedance path. Whenever the conventional current is trying to flow against the triangle, the diode offers a high-impedance path. To get a more accurate idea of diode action, consider the following hypothetical experiment. A battery is connected across a diode as shown in Fig. 2-6a. If the battery is adjusted to 0 volts, we will find that there is Burnout Fig. 2-6 The forward characteristic - of a diode. a oo (a) zero current in the diode. As we increase the battery voltage, current begins to flow in the diode. The current will increase slowly at first, but as we increase the battery voltage toward higher values, the current in- creases significantly. In other words, when the battery voltage is large enough to overcome the barrier potential, current flows easily. Figure 2-60 illustrates the current-voltage relation. The voltage Vx is called the knee voltage; it simply refers to the approximate voltage above which the diode current increases sharply. As an approximation, the knee voltage is equal to the barrier potential. Thus, for germanium diodes the knee voltage is approximately equal to 0.3 volt, and for silicon diodes the knee voltage is approximately equal to 0.7 volt.18 Transistor Cireuit Approximations Of course, there is a limit to the amount of current that the diode can pass without burning out. When we increase the voltage well beyond the knee voltage, we eventually reach a burnout current. The diode will burn out simply because it has a maximum power dissipation. For instance, a typieal semiconduetor diode may have a maximum power rating of 0.5 watt. When the product of voltage and current exceeds 0.5 watt, the diode burns out (assuming d-c voltage and current). What happens if we reverse the battery, as shown in Fig. 2-7a? With the battery reversed, we find that very little current flows. The actual conventional current flows opposite to the direction of i. Also, the actual voltage across the diode is opposite to the polarity of v. When we increase Fig. 2-7 The reverse characteristic of a diode. . | Breakdown point | (0) Burnout (a) the battery voltage, very little current flows, as shown in the reverse characteristic of Fig. 2-76. This is because of the small number of mi- nority carriers that are actually flowing in the diode. Note in Fig. 2-7) that there is a limit to the amount of reverse voltage that we can apply to the diode. When enough reverse voltage is applied, the diode current begins to increase sharply. The approximate voltage where this happens is called the breakdown voltage Vz of the diode (analogous to the peak inverse voltage of a vacuum-tube diode). If we continue to increase the battery voltage, more current flows, until eventually we reach a value of current that burns out the diode. Once again, this burnout is caused by exceeding the diode’s maximum power rating. For instance, if the diode has a power rating of 0.5 watt, and if it breaks down at 100 volts, the maximum current is As long as we keep the current below this value, the diode is not de- stroyed. In other words, it is possible to break down a diode without destroying it, provided that the current is kept below the burnout value. The breakdown phenomenon in a semiconductor diode is caused by either of two effects, zener or avalanche. In essence, what happens isThe p-n Junction Diode 9 that bound electrons are being knocked out of outer shells and are then available for current flow. The breakdown voltage varies from one diode type to another and typically can occur in the range of a few volts to several hundred volts. Forward % Fig. 2-8 The IV characteristic of a diode. Reverse The overall IV characteristic of a typical semiconductor diode is shown in Fig. 2-8 (we have combined the forward and reverse characteristics). To summarize our hypothetical experiment we note: 1, In the forward direction only a few tenths of a volt is needed to obtain significant values of current. 2. In the reverse direction very little current flows below the break- down voltage. 3, Beyond the breakdown voltage, current increases sharply, but the diode is not necessarily destroyed. 4, In either the forward or the reverse direction, the diode can be burned out if its maximum power rating is exceeded. Examrie 2-1 A semiconductor diode has a maximum power rating of 1 watt and a breakdown voltage of 150 volts. If the diode is operating in the break- down region, what is the value of direct current that burns out the diode? Sonvrion P=VI or P 1 T= 5 = 759 = 667 ma EXAMPLE 2-2 The diode of the preceding example is forward-biased. When the voltage across the diode is 2 volts, the diode burns out. What was the value of the current at the burnout point?20 ‘Transistor Cireuit Approximations Sonvrion T= =0.5 amp SUMMARY A p-n junetion diode is a solid crystal with p-type material on one side of the junction and n-type material on the other side. After the junction is formed, majority carriers diffuse across the junction and recombine This produces a depletion region containing immobile charged atoms, which in turn producc a barrier potential. For germanium the barrier potential is about 0.3 volt, and for silicon it is around 0.7 volt. When the positive terminal of a battery is connected to the p side of the crystal, the battery can overcome the barrier potential and establish a current in the external circuit. We call this condition forward bias. When the negative terminal of the battery is connected to the p side of the crystal, the diode is reverse-biased, and only a small amount of current flows. The IV characteristic of a p-n diode indicates that in the forward direction a large current can flow if the diode voltage is greater than the knee voltage. In the reverse direction only a small current flows un- less the diode voltage exceeds the breakdown voltage. At breakdown the diode is not necessarily destroyed unless the maximum power dissipation of the diode is exceeded. GLOSSARY avalanche At. higher reverse voltages minority carriers can attain suf- ficient velocity to dislodge outer-shell electrons, which in turn can gain sufficient: velocity to dislodge more outer-shell electrons, etc., with the result that there is a significant increase in reverse current. barrier potential The voltage across the p-n junction. This voltage is produced by the layer of charged atoms on both sides of the junc- tion. Approximately 0.3 volt for germanium and 0.7 volt for silicon. breakdown voltage (Vz) That value of reverse voltage beyond which there is a significant inerease in reverse current. conventional current Current that flows in the same direction as the holes and opposite to the direction of the electrons. depletion region A region on both sides of a p-n junction. It is relatively empty or depleted of free charges and primarily contains immobile ionized atoms.The p-n Junction Diode 21 forward bias Applying external voltage across a diode with a polarity such that the conventional current is trying to flow in the direction of the diode triangle, that is, from the p-type to the n-type material. knee voltage (Vx) The approximate value of forward voltage across a diode beyond which the forward current increases sharply. reverse bias Applying external voltage across a diode with a polarity such that conventional current is trying to flow against the diode triangle, that is, from the n-type to the p-type material. zener effect. In the back-bias condition outer-shell electrons can be dis- lodged by the electric field set up by the reverse voltage with the result that there is a significant increase in current. REVIEW QUESTIONS 1. What is the depletion region of a diode? 2. What are the approximate values of the barrier potential in germanium and silicon diodes? 3. What effect does the barrier potential have upon the majority carriers that try to move across the junction? 4, What causes the barrier potential? 5, What are the majority carriers in the p-type material? In the n-type material? In the external wires of a battery connected to the diode? 6. What happens to the size of the depletion layer when the diode is back-biased? 7. What is the approximate value of the knee voltage for a germanium diode? For a silicon diode? . Does reverse breakdown immediately destroy the diode? Why? 9. When a diode is in the breakdown region, how are the current carriers produced? ” PROBLEMS 2-1 A diode has a maximum power dissipation of 0.25 watt. What is the maximum direct current allowable in the forward direction if we allow an approximate voltage drop of 1 volt? 2-2 A diode has a breakdown voltage of 150 volts. If the maximum power dissipation is 0.25 watt, what is the approximate value of break- down current that burns out the diode? 2-3 A diode with a maximum power rating of 0.5 watt burns out when the forward current is 400 ma. What was the d-c voltage across the diode at the burnout point?3 Large-signal Diode Approximations Before we attempt to analyze transistor circuits, we must first be able to analyze diode circuits. When we can analyze diode circuits quickly and easily, we will find that transistor circuit analysis is only a step beyond. We now work with the aim of obtaining approximate answers for diode and transistor circuits; this is essential to rapid, easy analysis of such cireuits. In short, we are abandoning the quest for exact answers because such answers are found only after performing difficult and time-consuming calculations. Our intentions are to simplify diodes and transistors as much as possible in order to get at the essential ideas behind the use of these devices. In this chapter we discuss some of the widely used approximations for the rectifier diode and for the zener (or avalanche) diode. These approxi- mations are primarily for cireuits driven by large-signal sources, that is, sources whose voltages are much larger than the diode knee voltage. 3-1 The Ideal Diode As we saw in the last chapter, a typical semiconductor diode has the characteristic shown in Fig. 3-1. This characteristic is too complicated for 22Large-signal Diode Approximations 23 practical circuit analysis; we will therefore approximate it by a simpler graph. The first (and simplest) approximation that we consider is the ideal diode. By this we mean a perfect diode. When we look at Fig. 3-1, it is clear that the real diode has imperfections. For instance, the breakdown Fig. 3-1 Diode characteristic curve. Reverse phenomenon is undesirable in any rectifier diode. Also, the small amount of reverse current below breakdown is undesirable. Finally, in the for- ward direction, we see that a few tenths of a volt are needed before the diode conducts heavily. An ideal or perfect diode would have none of these obvious defects. Figure 3-2a shows the characteristic of an ideal diode. There is no breakdown, no reverse current, and no forward voltage drop. Of course, such a diode cannot be built. It is only a theoretical approximation of a real diode. However, we will find that in well-designed rectifying circuits the behavior of real diodes approaches the behavior of an ideal diode. No forward voltage drop No reverse a current Fig. 3-2 Ideal diode. i Ideal oF = eee Conventional current (a) (a) In using the ideal-diode approximation, it is helpful to visualize the ideal diode as a switch, as shown in Fig, 3-2. Note that these two proper- ties apply to the ideal diode: 1. When conventional current is trying to flow opposite to the direction of the diode triangle, the diode is like an open switch.24 Transistor Cireuit Approximations 2. When conventional current is trying to flow in the same direction as the diode triangle, the diode is like a closed switch. To repeat, the ideal diode does not exist. It is only a convenient and simple approximation of the behavior of a real diode. We will use the ideal-diode approximation whenever we wish to obtain a basic idea of how reetifier cireuits work. Occasionally, the answers obtained by using the ideal-diode approximation are grossly inaccurate. In these cases we will use better approximations to be developed in later sections. As an illustration of using the ideal-diode approximation, consider the simple half-wave rectifier shown in Fig. 3-3a. How can we find the voltage waveform across the 10-kilohm resistor? First, note that during each posi- tive half cycle the generator polarity is plus-minus, as indicated in Fig. 3-3b. Therefore, conventional current is trying to flow in the direction of the diode triangle. This means that the ideal diode is shorted. Hence, each positive half cycle must appear across the 10-kilohm resistor. % Idea! Short 20: eo nn 4 + : Ov og AG “ne |: (a) (a) Open 5 ¢ 5 $ ; 20) ve ike |v 20 ggg t (c) (a) Fig- 3-3 Half-wave rectifier. During each negative half cycle, the source polarity reverses, becoming minus-plus, as shown in Fig. 3-3c. Conventional current now tries to flow against the diode triangle; therefore the diode is open. With an open diode there can be no current through the 10-kilohm resistor; hence, there is no voltage across the resistor. The final waveform is shown in Fig. 3-3d. This is the familiar half-wave- rectified sine wave found in some power supplies. As another example of using the ideal-diode approximation, consider the circuit of Fig. 3-4a. Let us find the waveform of voltage across the diode. Again, the key to analysis is to determine when the ideal diode isLarge-signal Diode Approximations 25 open and when it is shorted. During each positive half cycle the circuit has the form shown in Fig. 3-4b, Conventional current is trying to flow in the direction of the triangle; therefore the diode is shorted. The voltage across a short, regardless of how much current is flowing, must be zero. All the source voltage is dropped across the 10-kilohm resistor. Therefore, throughout each positive half cycle the voltage across the diode is zero. 9 10k lok 20: it t "9 Ideal ; Vy 7 20. - Short | {> ta) (a) 10K fore : - : Law -20. (a) (e) Fig. 3-4 Positive clipper. During each negative half cycle, the circuit has the form shown in Fig. 3-4c. Conventional current tries to flow against the triangle, and therefore the diode is open. With an open diode there can be no current through the 10-kilohm resistor, and hence there can be no voltage drop across the resistor. From Kirchhoff’s voltage law we know that the volt- age across the diode must equal the source voltage. Stated another way, the voltage across the diode must equal the source voltage minus the voltage across the 10-kilohm resistor. With no voltage drop across the 10-kilohm resistor, all the source voltage appears across the diode. Thus, each negative half cycle appears across the diode. The final waveform is shown in Fig. 3-4d. Note that all the positive portions of the source voltage have been removed. We call a cireuit that removes the positive parts of an input signal a positive clipper. If a circuit clips off the negative parts of an input signal, we call it a negative clipper. The circuit of Fig, 3-4a can be made into a negative clipper by reversing the direction of the diode. EXAMPLE 3-1 Sketch the waveform for v in the circuit of Fig. 3-5a.26 ‘Transistor Circuit Approximations SoLurton We must determine when the ideal diode is shorted and when it is open. First, note that as long as the source voltage is less than 10 volts, the battery voltage exceeds the source voltage, and therefore conven- tional current tries to flow against the triangle, as shown in Fig. 3-5b. Hence, the ideal diode is open. With an open diode, no voltage can appear across the 10-kilohm resistor. Therefore, the output voltage 0 must equal the source voltage. 9 30: : Oo) '6 a = buise 3 (e} (a) Fig. 3-5 Positive clipper with 10-volt clipping level. Whenever the source voltage exceeds 10 volts, conventional eurrent will try to flow in the direction of the triangle, and the ideal diode will appear shorted, as in Fig. 3-5¢. With the diode shorted, the output voltage » must equal 10 volts, the value of the battery voltage. We conelude, therefore, that whenever the source voltage is less than 10 volts, the output voltage v follows the waveform of the source volt- age. Whenever the source voltage is greater than 10 volts, the output is held fixed at 10 volts. Because of these conclusions we ean draw the final waveform shown in Fig. 3-5d. We can think of the circuit of Fig. 3-5a as a positive clipper with a clipping level of 10 volts. All parts of the input signal above 10 volts have been clipped off. Note that the clipping level is equal to the battery voltage. In other words, if we change the battery voltage to another value, say 18 volts, the clipping level will be changed to 18 volts.Large-signal Diode Approximations 27 EXaMpLe 3-2 Sketch the output waveform for the circuit of Fig. 3-6a. Sonution By inspection, the diode-battery combination on the left clips off all parts of the input signal that are greater than 10 volts (discussed in Example 3-1). The diode-battery combination on the right is a negative clipper with a clipping level of — 10 volts (the reader should be able to work this out, if in doubt). Thus, all parts of the input signal above +10 volts are clipped off, and all parts below —10 volts are clipped off. The final output waveform is shown in Fig. 3-6b. 9 y 50 7 10] ~10) : -50 (8) Fig. 3-6 Positive- and negative-clipper combination, Note that this is one way of obtaining square waves (approximate) from sine waves. Also note that the clipping action takes place at +10 and —10 volts, regardless of the shape of the input waveform. Any input waveform—triangular, sawtooth, etce.—may be used, and if it exceeds the clipping levels, all positive parts above +10 volts and all negative parts below —10 volts will be clipped off. EXAMPLE 3-3 Sketch the output waveform for the cireuit of Fig. 3-7a. SoLvtTion During each positive half cycle the ideal diode is shorted. Therefore during each of these positive half cycles, the circuit behaves like the simple voltage divider of Fig, 3-75. From this circuit it is clear that the output voltage v must equal one-half the input voltage. Thus, the output will be a triangular wave with a peak value of 25 volts. During each negative half cycle the ideal diode is open, and the circuit becomes equivalent to that shown in Fig. 3-7c. As noted in earlier examples, no current can flow through the 10-kilohm resistor. Therefore, the output voltage must follow the waveform of the input voltage.28 Transistor Circuit Approximations o lok. 10K : 7 7 ft Ow)" He v -50 lok - J ike - 4 | (a) (4) lok i (e) (a) Fig. 3-7 Example 3-3, We can now draw the final waveform of the output voltage, as shown in Fig. 3-7d. 3-2 The Second Approximation of a Real Diode The ideal diode is the simplest but crudest approximation of a real diode. The answers we obtain using the ideal-diode approach provide us with an initial idea of how rectifier circuits operate. To improve our approximation of the real diode we can take into account the forward voltage drop across the real diode. A simple way to do this is to allow a forward voltage drop equal to the knee voltage of the diode. For instance, in Fig. 3-8a we have shown the characteristic of the second approximation of a real diode. In this case our viewpoint is that the diode does not conduct until the voltage across the diode reaches the knee voltage. ‘ v oD Il-e Fig. 3-8 The second approximation. (a) (4)Large-signal Diode Approximations 29 The equivalent circuit for the second approximation is given in Fig. 3-8). There is an ideal diode in series with a battery whose voltage equals the knee voltage of the real diode. Our idea of the circuit action is simply this: whenever conventional current tries to flow in the direction of the triangle, the ideal diode is shorted, and the net voltage across the termi- nals of the combination becomes equal to the knee voltage. In other words, in the second approximation we allow a voltage drop of 0.3 or 0.7 volt across the real diode when it is conducting in the forward direc- tion. (Remember: 0.3 volt for germanium and 0,7 for silicon.) EXxaMPLe 3-4 For the simple half-wave rectifier of Fig. 3-9a, sketch the waveform of the output voltage » by: (a) Replacing the silicon diode by an ideal diode. (t) Replacing the silicon diode by its second approximation. % Silicon 4 Ideal diode 10 @ . ail 10 i t R. / “10: “ i (o) (0) Meo 07, ‘ DA 2nd approximation + @ 8: | {e) (d) Fig, 3-9 Half-wave rectifier. So.urion (a) The ideal-diode approach simply gives us the standard half-wave- rectified sine wave shown in Fig, 3-96. (6) To use the second approximation we visualize the silicon diode as an ideal diode in series with a 0.7-volt battery, as shown in Fig, 3-9c. As the source voltage builds up from 0 to 0.7 volt, conventional current tries to flow against the triangle, so that the diode is open. Thus, as long as the source voltage is less than 0.7 volt, the output voltage re- mains at zero. However, once the source voltage is greater than 0.7 volt,30 ‘Transistor Cireuit Approximations the knee voltage is overcome, and the diode is turned on. The output voltage now begins to build up, following the waveform of the source. At the instant that the source voltage equals the peak voltage of 10 volts, the output must equal 10 volts minus the 0.7-volt drop. Thus, the peak value of output voltage is 9.3 volts. The final output waveform is shown in Fig. 3-9d. Note that this waveform is almost the same as the waveform obtained by the ideal- diode approach (Fig. 3-9b). Exampie 3-5 Use the second approximation to find the output waveform for the circuit of Fig. 3-10a. % R R y 5 veal Y | Os ® “Troe, -5. Silicon [> ~ Of ates { Slay cea (a) (5) to) Fig. 3-10 Positive clipper. SonuTion First, note that an ideal-diode approach would simply give us an output waveform that is positively clipped at the 0-volt level. In using the second approximation of the silicon diode, we visualize the cireuit as shown in Fig. 3-10b. The simplest way to analyze this circuit is to realize that it is a positive clipper with a 0.7-volt clipping level (similar to Example 3-1). Therefore, we conclude that the output waveform is that of Fig. 3-10c. 3-3. The Third Approximation of a Real Diode Still another approximation can be made for a real diode to account more accurately for its forward voltage drop. We recall that the forward char- acteristic of a real diode is not vertical above the knee but actually slopes upward to the right. This means that as more current flows through the real diode, more voltage is dropped across it. Figure 3-1la shows the characteristic of the third approximation of a real diode. By inspection, this characteristic still does not give the exact forward behavior of a real diode, but it does represent a more accurateLarge-signal Diode Approximations 31 model than our first two approximations. Our interpretation of this graph is that the diode begins to conduct above Vx (0.3 or 0.7 volt). Once the diode is conducting, it acts like a resistor, because the change in voltage is directly proportional to the change in current. The value of this resist- ance is called the forward or bulk resistance of the diode, and we will use the notation rz to represent this resistance. i { Av ee | Ar Ideal ety ig. 3-11 The third ximation. : Fig. 3-11 The third approximation. i: SEA (a) (6) The equivalent circuit for the third approximation is shown in Fig. 3-116. Our viewpoint here is that the real diode acts like an ideal diode in series with a battery of value Vx and a resistor of value re. Whenever the external circuit tries to force conventional current in the direction of the triangle, the ideal diode is shorted. The voltage drop across the termi- nals of the combination is then the sum of Vx and the drop across the rp resistor. In other words, in the third approximation it takes at least 0.3 or 0.7 volt even to turn the diode on. The diode then acts like a resistor of value rp, which drops additional voltage depending upon how much current is flowing. A word or two on finding the value of ra is in order. First, if an JV eurve tracer is available, the diode characteristic can be displayed. From this graph we need only select. two points well ahove the knee of the curve where the characteristic is almost linear. The change in voltage between the two points divided by the change in current is the approximate value of rp. For instance, suppose the curve-tracer display of a diode is that shown in Fig. 3-12a. We would select two points well above the knee of the curve. Two such points are shown, To find the bulk resistance rs we would divide the change in voltage by the change in current between the i 50mat----. Fig. 3-12 Estimating the bulk resist- ance rz. v 07 10 {a}32 Transistor Cireuit Approximations two points. That is, OO tos ohma 50(10-) — 10(10-%) ~ 4010-8) ~~ Another approach is to find the approximate value of bulk resistance directly from the manufacturer’s data sheet for the particular diode. On the data sheet one of the quantities normally specified is the forward current at 1 volt. Knowing whether the diode is germanium or silicon, we can subtract the knee voltage from 1 volt and divide this result by the forward current. For instance, suppose that a data sheet for a par- ticular silicon diode indicates a forward current of 50 ma at 1 volt. In effect, we are being given one point on the forward characteristic of the silicon diode, as shown in Fig. 3-12b. By using the third approxima- tion of the diode, we can use the knee as the second point. The change in voltage between the two points is Av = 1—0.7 = 0.3 volt The change in current between these two points is Ai = 50 ma Therefore the bulk resistance is approximately Av The third approximation of a real diode can be used to refine the answers obtained by the simple ideal-diode approach. EXamMPLe 3-6 Sketch the output voltage o for the circuit of Fig. 3-13a by using the third approximation of the germanium diode. This diode has a forward current Ir of 28 ma at 1 volt. Sonurion First, calculate the value of bulk resistance Next, replace the germanium diode by its third approximation, as shown in Fig. 3-13b. From this circuit it is clear that the ideal diode cannot turn on until the source voltage is greater than 0.3 volt. When- ever the source voltage exceeds 0.3 volt, the ideal diode is shorted; the 25- and 75-ohm resistors then form a voltage divider, so that the out-Large-signal Diode Approximations 33 % Germanium Ideal 0.3 25 7 es ’ | 10 : %y Bey ALY Se 727 ; “10: { (a) (a) (ec) Fig. 3-13 Example 3-6. put voltage is a reduced version of the positive half cycle. To find the peak value of the output voltage during the positive half cycle, we can find the peak current in the cireuit and multiply it by 75 ohms. At the instant when the source reaches its peak value of 10 volts, the cur- rent will reach a peak value of _ 10-03 _ i, = Sapp = 97 ma The peak output voltage is vp = 97(10-)(75) = 7.27 volts Whenever the source is less than 0.3 volt, the ideal diode is open, and there is no current through the 75-ohm resistor; therefore the output voltage is zero. The sketch of the total waveform is shown in Fig. 3-13c. 3-4 Using the Diode Approximations The diode approximations discussed so far can considerably shorten the amount of time required to analyze various diode circuits; they also are the basis of transistor circuit analysis. The ideal-diode approxima- tion is by far the simplest and most often used approximation in transistor circuit analysis. Occasionally, we will want to refine our answers by using the second and third approximations. Let us summarize the use of these approximations: 1. In analyzing any rectifier-diode circuit, start with the ideal-diode approach. This yields a basic idea of how the circuit operates and is adequate for most situations. 2. If the answers obtained by the ideal-diode approach indicate that 0.3 or 0.7 volt is significant, reanalyze the circuit using the second approximation.34 ‘Transistor Circuit Approximations 3. If the external resistance in series with the diode is not large com- pared to the bulk resistance of the diode, use the third approximation. 3-5 Approximating the Reverse Current In the reverse direction below breakdown, a small amount of current does flow through the diode. Up to now, we have completely disregarded the reverse current. In some circuits the reverse current can be a problem. Therefore, we need an approximation for a reverse-biased diode operating below the breakdown point. We want as simple an approximation as possible; therefore, we will view the diode as a large resistance whose value equals the reverse voltage Vp divided by the reverse current Iz. For instance, if a manufacturer’s data sheet indicates that a particular diode has a reverse current of 10 4a for a reverse voltage of 50 volts, then we calculate a reverse resistance Rr of Rr= Ta = TOU05 = 5 megohms A point worth making at this time is that silicon diodes have much Jess reverse current than comparable germanium diodes. It is not at all uncommon for a silicon diode to have a reverse resistance that is 1000 or more times larger than that of a comparable germanium diode. For instance, a 1N277 is a germanium diode with a reverse current of 0.25 ma for a reverse voltage of 50 volts. Therefore it has a reverse resistance of ___80 * 0.25(10-) An SG1825 is a comparable silicon diode; however, it has a reverse current of only 10 na at 120 volts. Therefore it has a reverse resistance of ued 20 neces * Todo) ~ Rr = 200 kilohnis R. 12,000 megohms Note that the higher the reverse resistance of a diode, the more closely it approaches the ideal, or perfect, diode. Generally speaking, silicon diodes and transistors are far superior to germanium diodes and transis- tors in this respect. Originally, the germanium diodes and transistors were less expensive than silicon devices. This is no longer true; com- parably priced silicon diodes and transistors are now available. The use of a resistance to represent the diode in the reverse region yields reasonably accurate results in low-frequency circuits, provided that the diode remains below the breakdown voltage. If the diode breaks down, we must adopt a different viewpoint, as discussed in Sec. 3-6.Large-signal Diode Approximations 35 EXAMPLE 3-7 Sketch the waveform of the output voltage v in the circuit of Fig. 3-14a, Neglect the forward voltage drop but take reverse current into account, The diode has a reverse current of 1 za for a reverse voltage of 50 volts. SoLvTIon The reverse resistance is Re = 50 megohms eeu ~ 10-6 During the positive half cycle the diode is shorted, and the output voltage follows the source voltage. % 50M v 50 1S ‘9 IM -50: leis +o] be is +—+ 8 (a) (8) (e) Fig. 3-14 Effect of reverse resistance. During the negative half cycle the diode acts like a 50-megohm re- sistor, as shown in Fig. 3-14. This circuit is a voltage divider, so that the output voltage is about 145 of the source voltage. At the negative peak the source voltage is 50 volts, and the output voltage is about 1 volt. The final waveform is shown in Fig. 3-14c, If the 1-volt level is objec- tionable on the negative half cycle, then either of two changos is possible to remedy the situation. First, the 1-megohm resistor can be reduced to a lower value, like 100 kilohms. In this case, only about }499 of the source voltage will appear at the output during the negative half cycle. Second, we can change diode types to obtain a much larger reverse resist- ance. For instance, we can easily obtain a diode with a reverse resistance of 500 megohms. In this case, the output voltage will be only about 1490 of the source voltage during the negative half cycle. 3-6 Zener Diodes When using the diode as a rectifying device we generally make sure that the driving voltage across the diode does not exceed the breakdown volt-36 Transistor Circuit Approximations age. On data sheets the breakdown voltage Vz is often designated by either PIV (peak inverse voltage) or by BV (reverse breakdown voltage). In typical rectifying and detecting applications the PIV of the diode should be sufficiently large compared to the driving voltage to prevent any possibility of breakdown. We might be tempted to think that diodes are never used in the break- down region; this is not true. There definitely are uses for the breakdown phenomenon, as we shall see shortly. By careful manufacturing techniques the breakdown ean be made very sharp and almost vertical, as shown in Fig. 3-15a. Diodes exhibiting this sharp knee at the breakdown point aro called zener diodes. The voltage Vz is the approximate voltage where the zener diode breaks down. The amount of current flowing in the breakdown region will depend upon the external circuit driving the diode. There is, of course, a maximum value of current at which the zener diode burns out. This value of current is determined by the zener voltage and the maximum power dissipation of the diode. An important point to realize is that the diode is not immedi- ately destroyed just because it has entered the breakdown region. As long as the external circuit connected to the diode limits the diode current to less than the burnout current, the diode will not burn out. i i ee el, + “ v (a) (o) Fig. 3-15 Zener diode. (a) Typical characteristic; (b) ideal characteristic; (c) equivalent circuit. To enable us to analyze zener-diode circuits quickly and easily, we will approximate the graph of Fig. 3-15a by redrawing it as shown in Fig. 3-15). We call this the characteristic of an ideal zener diode. Note that in the forward direction the diode acts like a short. In the reverse direc- tion the diode is open until we reach the breakdown voltage, Beyond the breakdown point the diode voltage remains constant, even though the current can change considerably, depending upon the external cireuit con- nected to the diode. This constant-voltage characteristic is the most use- ful property of a zener diode. Our circuit viewpoint of a diode in the breakdown region is shown in Fig. 3-15c. An ideal zener diode looks like a battery of Vz volts, because an ideal battery is a device whose voltage remains constant even thoughLarge-signal Diode Approximations 37 the current through it changes. Remember we are not saying that the zener diode is a battery; we are only saying that in the breakdown region it acts like a battery. The usual schematic symbol for a zener diode is also shown in Fig. 3-15¢. Memorize the polarity of voltage and the diree- tion of the conventional current; this is essential for the analysis of zener- diode cireuits. Note that in the breakdown region the conventional cur- rent flows against the triangle and the voltage is plus-minus as shown. The breakdown phenomenon is actually a combination of two effects: zener effect and avalanche effect. The zener effect refers to removing bound electrons from outer shells by means of an electric field. In other words, as the reverse voltage is applied to a diode, an electric field ap- pears at the junction. When this field is intense enough, outer-shell elec- trons are dislodged, resulting in a significant increase in reverse current. ‘The avalanche effect is different. In this case, when the diode is reverse biased, minority carriers are flowing. For higher reverse voltages these minority carriers can attain sufficient velocity to knock bound electrons out of their outer shells. These released electrons then attain sufficient velocity to dislodge more bound electrons, ete. The process is well named, since it is suggestive of an avalanche. When a diode breaks down, both zener and avalanche effects are present, although usually one or the other predominates. It has been found experimentally that below 6 volts the zener effect is predominant; above 6 volts the avalanche effect is predominant. Strictly speaking, diodes with breakdown voltages greater than 6 volts should be called avalanche diodes, and sometimes they are so called. However, the gen- eral practice is to refer to diodes exhibiting either effect as zener diodes. EXAMPLE 3-8 Find the current through the zener diode in the circuit of Fig. 3-16a. Use the ideal-zener-diode approximation. SoLurion Note that the zener diode is back-biased by a source that exceeds the breakdown voltage of the diode. Therefore, the zener diode must be in the breakdown region, and we can visualize it as a 30-volt battery, as shown in Fig, 3-16b. The difference of potential across the 2-kilohm 2k Fig. 3-16 Example 3-8. (4)38 Transistor Cireuit Approximations resistor must be the difference of 50 and 30 volts, that is, 20 volts, The current through the 2-kilohm resistor is 20 T = 3999 = 10 ma Since we have a series circuit, 10 ma must also flow through the zener diode. Thus, the zener diode is operating in the breakdown region with a voltage of 30 volts and a current of 10 ma. EXAMPLE 3-9 In Fig. 3-174, use the ideal-sener-diode approximation to find the following: (a) The current through the zener diode when R = 30 kilohms. (b) The current through the zener diode when R = 4 kilohms. Soivtion (a) When & = 30 kilohms, there is more than enough voltage appearing across the diode to cause breakdown, and we can think of the zener diode as a 30-volt battery (Fig. 3-17b). The voltage across the 2-kilohm resistor is still the difference between the source voltage of 50 volts and the zener voltage of 30 volts. There- fore, there is still 20 volts across the 2-kilohm resistor, anda current of 10 ma flows through it, as shown in Fig. 3-17b. This 10 ma of current splits at the junction of the zener diode and the 30-kilohm resistor. 10ma—~— Fig. 3-17 A simple voltage regulator. To find how this current divides, we note that the voltage across the 30-kilohm resistor must be 30 volts, the zener voltage. Therefore, there is 1 ma of current in this resistor. From Kirchhoff’s current law, we know that the current in the zener diode must equal the difference between 10 and 1 ma, that is, 9 ma, as shown in Fig. 3-17b. (6) When R = 4 kilohms, more current must flow through this resistor. With 30 volts across the zener diode, there must also be 30 voltsLarge-signal Diode Approximations 39 across the 4-kilohm resistor, and therefore there is a current in the resistor of The current in the zener diode must be the difference between the input 10 ma and the 7.5 ma flowing in the 4-kilohm resistor. Therefore, the zener-diode current is 2.5 ma, as shown in Fig. 3-17e. Note what has happened. The load resistance R has changed from 30 to 4 kilohms, and yet the voltage across this resistor has been held constant at 30 volts by the zener diode. This is one of the major uses of zener diodes, namely, to hold the voltage across a load resistance con- stant, even though the load changes. In Fig. 3-17a regulation is lost when R is less than 3 kilohms, because for this value, the load current has just reached 10 ma, and the zener current has reached zero; the diode is on the verge of coming out of the breakdown region, and any further reduction in the size of 2 will result in a voltage of less than 30 volis. In general, there is a limit on the minimum value of R. If R is made too small, the zener diode will come out of the breakdown region, and regula- tion will be lost. To ensure that the diode is in the breakdown region, we must have at least a small amount of zener current. Exampte 3-10 Find the value of v in the circuit of Fig. 3-18a. Use the ideal-zener- diode approximation. Also, find the minimum and maximum value of zener-diode current. 2k 40-60 volts | (a) (d) Fig. 3-18 Voltage regulation for a changing source voltage. Sonution The source voltage varies from 40 to 60 volts. The equivalent cir- cuits for both of these conditions are shown in Fig. 3-18) and c. In both cases, note that there is enough source voltage to cause breakdown.40 Transistor Circuit Approximations When the source is at its minimum value of 40 volts, there must be 10 volts across the 2-kilohm resistor, and therefore there is a current of 5 ma flowing in the zener diode. When the source voltage equals 60 volts, there must be 30 volts across the 2-kilohm resistor, and a current of 15 ma flows through the zener diode. In either case, the voltage across the zener-diode terminals is 30 volts. The point of this example is to show how the voltage across the zener diode remains constant in spite of changes in the source voltage. Thus, zener diodes can be used to regulate voltage under conditions of changing source voltage and changing load resistance (Example 3-9). 3-7 The Second Approximation of a Zener Diode Tn the preceding section we discussed the ideal-zener-diode approxi- mation, in which we think of the zener diode as a battery whenever it is in the breakdown region. This simple model is adequate for most trouble- shooting and preliminary design. To improve our analysis of zener-diode circuits we ean take into account the slope of the breakdown characteristic. In other words, for the second approximation we will use the IV characteristie shown in Fig. 3-19a; note that the breakdown region is not vertical but actually has a slope, so that it more closely resembles the breakdown character- istie of a real zener diode. Our interpretation of this not quite vertical breakdown region is simply this: when more current flows through the diode, the voltage does not remain exactly constant but inereases slightly. ='2 Fig. 3-19 Second approximation of oF . zener diode. (a) Characteristic curve; iy “z — (®) equivalent circuit. (a) (o} In other words, once the diode is in the breakdown region, it resembles a resistor as far as changes in voltage and current are concerned. ‘The size of this resistor ean be found by taking the ratio of a change in voltage to a change in current between any two points on the breakdown char-Large-signal Diode Approximations 41 acteristic. That is, naa where Av is the change in voltage between two points, and A7 is the change in current between the same two points. The resistance rz is called the zener resistance and is normally given on the data sheet for the particular diode. The circuit model that we use for the second approximation is shown in Fig. 3-19b. In the breakdown region we think of a zener diode as a battery in series with a resistor. Thus, the voltage across the zener diode is Vz + ire (3-1) where Vz is the voltage right at the knee of the breakdown and #rz is the additional voltage drop across the zener diode produced by the zener resistance and the current. The use of Eq. (3-1) is straightforward. For instance, suppose that a particular zener diode has the following values: Vz = 30 volts, rz = 20 ohms. When the diode just broaks down, there is no current flowing, and therefore the total voltage across the diode is simply 30 volts. When there is 1 ma of current through the diode, the voltage across the diode is the sum of Vz and the additional drop across rz, that is, Va + trz = 30 + 10-9(20) = 30.02 volts If we increase the current from 1 to 2 ma, the voltage across the zener diode becomes Va + irz = 30 + 2(10-*)(20) = 30.04 volts ‘The voltage across the zener diode has progressively changed from 30 to 30.02 to 30.04 volts as we changed the current from 0 to 1 to 2 ma. Thus, we see that all the second approximation does is to take into account the small additional voltage drop across the diode that occurs when we increase the amount of current through the diode. This means that in a voltage regulator using a zener diode (see Examples 3-8 to 3-10) the output voltage of the regulator is not exactly constant but changes slightly when the current through the diode changes. Note in Fig. 3-19a that the forward characteristic of the zener diode shows the knee voltage of 0.7 volt. The reason for this is that all zener diodes are made out of silicon; germanium diodes have too much reverse current below breakdown, which prevents them from having a sharp knee at the breakdown point; the silicon diodes, on the other hand, have such low reverse current below breakdown that a very sharp knee can be produced at the breakdown point.42 Transistor Circuit Approximations SUMMARY The ideal diode is the simplest and most useful approximation of a rectifier diode. There is no breakdown, no reverse current, and no forward voltage drop. Our circuit model for the ideal diode is a switch that is closed whenever conventional current flows in the direction of the tri- angle and open whenever conventional current tries to flow against the triangle. The second and third approximations refine the ideal diode by tak- ing into account the forward voltage drop across a diode. The second approximation allows 0.3 volt for germanium diodes and 0.7 volt for silicon diodes. The third approximation allows for an additional drop by taking the bulk resistance of the diode into account. A real diode does have some reverse current through it when it is back-biased. Below breakdown we think of a diode as being a large resistance Rx. A very important difference between germanium and silicon diodes is that silicon diodes have much larger reverse resistances than comparable germanium diodes. With enough voltage applied to the back-biased diode, the breakdown point is reached. Zener diodes have an extremely sharp knee at the breakdown point and an almost vertical breakdown region. The ideal- zener-diode approximation simply views a zener diode in the breakdown region as a battery with a voltage equal to the breakdown voltage Vz. To refine this simple approximation we can put a small resistance rz in series with the battery to account for the not quite vertical breakdown characteristic. GLOSSARY avalanche A breakdown phenomenon based upon minority carriers dis- lodging outer-shell electrons, which in turn dislodge more outer-shell electrons. bulk resistance The resistance of a diode well above the knee of the forward characteristic. This is the ohmic resistance of the p-type and n-type materials. ideal diode The first approximation of an ordinary rectifier diode. The circuit model is a switch. negative clipper A circuit that removes the negative parts of an input signal. positive clipper A circuit that removes the positive parts of an input signal.Large-signal Diode Approximations 43 zener diode A specially processed silicon diode that has an extremely sharp breakdown point and an almost vertical breakdown region. REVIEW QUESTIONS 1, When visualizing the ideal diode as a switch, what determines whether the switch is closed or open? 2. What is a positive clipper? A negative clipper? How can the clipping level be changed? 3. What is the equivalent circuit for the second approximation of a real diode? What does the IV characteristic look like? 4. Name one approach in obtaining square waves (approximately square) from sine waves. . What is the equivalent circuit for the third approximation of a real diode? What does the IV characteristic look like? 6. What does the forward, or bulk, resistance of a diode refer to? Name two ways of finding the approximate value of rp. 7. Asa general rule, when can we neglect: the knee voltage of a diode? When can we neglect the bulk resistance of the diode? 8. How do we find the approximate value of the reverse resistance of a diode that is not operating in the breakdown region? 9. What is the outstanding advantage that silicon diodes have over germanium diodes? 10. What is a zener diode? What circuit device do we use to represent the ideal zener diode operating in the breakdown region? 11. What does the zener effect refer to? What does avalanche refer to? For breakdown voltages greater than about 6 volts, which effect is dominant? 12. What is one of the major uses of zener diodes? ge PROBLEMS 3-1 In Fig. 3-20a, use the ideal-diode approximation to find the value of direct current J. 10K 500 + os 4] Y Silicon oS i Germanium, (a) (4) Fig. 3-2044 Transistor Circuit Approximations 3-2 In Fig. 3-208, use the ideal-diode approximation to find J. 3-3 In Fig. 3-21a, sketch the waveform of tout. i thet a vin 3K 5 | 20 Ideall + © 30 Yout a) YW “out WO) 7 -20) fj; * il (0) Fig. 3-21 3-4 In Fig. 3-21b, sketch the waveform of vout- 3-5 Sketch the waveform of vou in Fig. 3-22a. 3-6 Sketch the waveform of vout in Fig. 3-22. 3-7 Find the direct current J in Figs. 3-20a and 6 by using the second approximation. 3-8 Instead of using an ideal diode in Fig. 3-21a, use the second approxi- mation of a silicon diode and sketch the waveform of tout. 3-9 In Fig. 3-21b, replace the ideal diode by the second approximation of a silicon diode and sketch the waveform of tout. 3-10 In Fig. 3-22a, use the second approximation of a silicon diode and sketch tout. vin 50; -50- 50- -50:Large-signal Diode Approximations 45 3-11 In Fig. 3-22h, sketch vou using the second approximation of a sili- con diode in the place of the ideal diode. 3-12 In Fig. 3-20a, the silicon diode has a forward current I» of 50 ma at 1 volt. Compute the approximate value of bulk resistance rs and state why it is negligible in this circuit. 3-13 In Fig. 3-20b, the germanium diode has a forward current Ip of 10 ma at 1 volt. Compute the bulk resistance rz and the value of direct current 7. 3-14 In Fig. 3-23a, the silicon diode has a forward current Ir of 20 ma at 1 volt. (a) Compute the bulk resistance ra. () Sketch the waveform of the current i using the ideal-diode approximation. (c) Sketch the waveform of the current 7 using the third approxi- mation of the silicon diode. Yin 500 Yn 100K 10} 30; + f vin i| \ZSiligon f Yo “out co | ie) r tis (a) (6) Fig. 3-23 3-15 In the circuit of Fig. 3-230, sketch the waveform of vou: for the following diodes: (a) A germanium diode with a reverse current of 0.125 ma for a reverse voltage of 50 volts. Neglect forward voltage drop. (0) A silicon diode with a reverse current of 1.25 wa for a reverse volt- age of 50 volts. Neglect forward voltage drop. 3-16 In Fig. 3-24a, find the minimum and maximum current through the zener diode. Use the ideal-zener-diode approach. SK 5K 80 to 120. +50 80 to 120 4250 volts es is a (a) 7) Fig. 3-2446 ‘Transistor Circuit Approximations 3-17 In Fig. 3-24b, use R = 10 kilohms. Find the minimum and maxi- mum value of zener-diode current. 3-18 In Fig. 3-24, find the value of R that causes the zener diode to come out of the breakdown region for the following source voltages: (a) Source voltage equals 120 volts. (b) Source voltage equals 80 volts. 3-19 In Fig. 3-24), what is the maximum power dissipation in the 5-kilohm resistor for the given range of source voltage? What is the maximum power dissipation in the zener diode for any source or load condition? 3-20 In Fig. 3-24a, the zener diode has an rz = 90 ohms. Compute the minimum and maximum voltage appearing across the zener divde for the range of source voltage. Use the second approximation.4. Small-signal Diode Approximations In the previous chapter we discussed large-signal diode approximations, that is, approximations that are suitable whenever the signal driving the diode is larger than the knee voltage. There are times, however, when the driving signal is smaller than the knee voltage; in this case we must. use the small-signal diode approximations that will be developed in this chapter. In addition, we will review the superposition theorem, a theorem of great importance in transistor-cireuit analysis. 4-1 The Superposition Theorem The superposition theorem is widely used in science and engineering, as well as in more liberal subjects like economies and philosophy. To under- stand this important theorem, consider Fig. 4-1a. We have shown a linear system in which several causes are acting simultaneously to produce a net effect. This net effect is the result of all the causes acting together. The superposition theorem tells us that one approach to finding the net effect is the following: determine the individual effect produced by each cause acting by itself; all the individual effects added together then give the net effect. In other words, to find the net effect produced by all causes 4748 Transistor Circuit Approximations Severo! causes acting—>] Linear system |—>-Net effect First cause] Linear system |» First effect simultaneously (a) (4) Second couse—ry Linear system |—> Second effect (c) Fig. 4-1 The superposition theorem. acting simultaneously in Fig. 4-1a, we single out any one cause and de- termine the effect it produces, as in Fig. 4-1b. Next, we select another cause and determine the effect it produces, as in Fig. 4-1c, We continue in this manner, finding the individual effect produced by each cause; finally, we add all the individual effects to obtain the net effect produced by all the causes acting simultaneously. The superposition theorem sounds quite simple, and it is. Whenever many causes are acting together to produce a total effect, we can find this effect by isolating the causes, one at a time, to find the individual effects. Then by adding the individual effects we get the total effect. Remember, however, that we can apply the superposition theorem only to linear systems. A linear system is one in which the effect is directly proportional to the cause; in other words, if a given cause is producing an effect, doubling the size of the cause will double the size of the effect. If a system is nonlinear, there will not be a direct proportion between cause and effect, and therefore we cannot use superposition theorem. The general proof of the superposition theorem is straightforward. Consider Fig. 4-2a. There is an effect y being produced by a cause 2. In a linear system y is directly proportional to z; that is, y= mer (4-1) where m is a constant of proportionality. Now suppose that the input is actually several causes acting simultaneously, as in Fig. 4-26. In other words, suppose that tentutat--- Then by substituting into Eq. (4-1) we get yem@itatiat- +) or y = mx, + ma. + mast +++ (4-2)Small-signal Diode Approximations 49 4—+] Linear system }-H—> y Ay Hata gto Linear system L/L» y {a} (4) Fig. 4-2 General proof of superposition. Examine this equation carefully; the proof of the superposition theorem is implied in it. Note that y is the total effect produced by all causes acting together. Also, on the right-hand side of this equation we have the individual effect produced by each cause acting by itself. In other words, mz; is the effect produced by the first cause; mz. is the effect produced by the second cause, and so on. Equation (4-2) tells us that we can add the individual effects to find the total effect. To summarize the all-important idea of the superposition theorem we note: 1. The system must be linear. 2. The individual effect produced by each cause is found. 8. All the individual effects are added to find the total effect produced by all causes acting simultaneously. 4-2 Superposition in Electric Circuits We have discussed the superposition theorem in its most general terms in order to establish the underlying notion behind this theorem. With the fundamental idea in mind, we now turn our attention to how the superposition theorem is applied to electric circuits. Specifically, wo are interested in circuits in which two or more sources (causes) are acting simultaneously to produce a net voltage or current (effect). To find this voltage or current by superposition we find the component produced by each source and add all the components to obtain the net effect. Of course, the circuit must be linear, meaning that the currents produced are directly proportional to the sources causing them. A simple way to recognize a linear circuit is to determine whether or not the resistors, inductors, and capacitors remain fixed in value as the voltage across them changes. For instance, suppose a circuit contains a 1-kilohm resistor, a I-henry inductor, and a 1-uf capacitor. If the values of these circuit elements remain at 1 kilohm, 1 henry, and 1 uf for different terminal voltages, the circuit is linear, and we can apply the superposition theorem to it. In analyzing electric circuits we find the superposition theorem quite useful in determining the voltage or current in any part, of a linear circuit50 Transistor Cireuit Approximations driven by two or more sources. Let us summarize the superposition theorem as it applies in circuit analysis: 1. Compute the current (or voltage) produced by each source with all other sources reduced to zero. 2. Add the individual currents to find the net current produced by all sources acting simultaneously. In applying step 1, reducing all other sources to zero means that voltage sources are shorted and current sources are opened. In applying step 2, adding the individual currents means algebraic addition; that is, if eur- rents flow in the same direction, the magnitudes are added, but if the currents flow in opposite directions, the magnitudes are subtracted. To bring out the full meaning of the superposition theorem, we will consider several examples. The circuit of Fig. 4-3a is a linear circuit with two voltage sources. What is the value of /r in this cireuit? One way to solve this problem is by applying superposition. We proceed as fol- lows. First, determine how much current is produced by the left battery with the right one shorted, as in Fig. 4-3b. The current I; is the individual current produced by the left battery acting by itself. It is obvious that this current is peace apg abe : he= 10 1010 7 15 = 3 amp to the right Incidentally, note the use of the vertical parallel lines in this equation. This will be our shorthand notation for two resistors in parallel. In (a) Fig. 4-3. Solving a circuit problem with superposition.Small-signal Diode Approximations 51 general, when we have two resistors R, and Hz in parallel, we will indi- cate this by using Rill Re. Next, we find the current produced by the right battery with the left one shorted, as shown in Fig. 4-3c. It is clear that the current Ip out of the battery will split when it reaches junction A. ‘The current. that we are after is Js, From the circuit it is apparent that 30 To = 9 F 10]10 = 2amp and. = & = 1 amp to the left We now find the total current in the original circuit of Fig. 4-3a by algebraically adding the individual currents 7, and J;. Since the currents flow in opposite directions, Ip = 3 — 1 = 2 amp to the right This is the current that will actually flow in the circuit when both batteries are acting together. This same result could have been obtained by using other methods, such as those derived from Kirchhoff’s laws. For instance, we could have written two loop equations and solved them to find Ip. However, this latter approach is usually more difficult. The really important point here is that the superposition theorem offers us an alternate approach in the analysis of multisource circuits. ‘As another example of using the superposition theorem, consider the circuit of Fig. 4-4a. Again, there are two sources. How can we find the % + 1Omv “\ \ . WK = 10 10K -10my (a) (0) lomv t _. -10mv > 0.999me | — t Fig. 4-4 Superposition of d-c and a-c components.52 Transistor Circuit Approximations current tr in the 10-kilohm resistor? We begin by finding the current produced by the battery, as shown in Fig. 4-4b, Obviously, the battery alone will produce a direct current of Next, we find the current produced by the a-c source with d-c source shorted, as shown in Fig. 4-4c. The signal out of the generator is a sine wave with a peak voltage of 10 mv. This voltage will produce a sine wave of current with a peak value of 10-? 104 Ip = = 10-6 =1ya To find the total current in the original circuit of Fig. 4-4a, we com- bine the a-c and the d-c components. The total current waveform is shown in Fig. 4-4d. Note that the current is a fluctuating current that varies from a low of 0.999 ma to a high of 1.001 ma. We will use the superposition theorem frequently to analyze different transistor circuits, especially those circuits containing d-c and a-c sources. In drawing the equivalent circuits for the d-c and a-c sources it is impor- tant to remember the following: 1. Direct current cannot flow through a capacitor; therefore, all capacitors look like open circuits in the d-c equivalent circuit. 2. When a-c signals are involved, capacitors are generally used to couple or bypass the a-c signal. In order to accomplish this, the designer deliberately selects capacitors that are large enough to aprear as short circuits to the a-c signal. Therefore, when we draw the a-c equivalent circuit, we will show all capacitors as short circuits, unless otherwise instructed. As an example of using these two guides, consider Fig. 4-5a. What is the total voltage vy simultaneously produced by the a-c and d- sources? The d-c equivalent circuit is shown in Fig. 4-5b. Since the capacitor is open to direct current, there can be no direct current in either 10-kilohm resistor. Therefore, the entire battery voltage must appear across the output terminals, that is, V = 10 volts. The a-c equivalent circuit is shown in Fig. 4-5c. We will assume that the capacitor is large enough to appear essentially as a short cireuit to the a-c signal. (The reader may verify that the capacitor does look like a very low impedance compared to 10 kilohms by calculating the capacitive reactance at 10 kHz.) Note that as far as the a-c signal is concerned, the circuit acts like a simple voltage divider. Since 10 kilohms is in series with 10 kilohms, the a-c output voltage v is one-half of theSmall-signal Diode Approximations 53 % Ok / b omy. + t =l0mv: 7 %@ ike 10k tomy | t t 10K e ws a omy ©) 2 9.995volts { t (ce) Fig. 4-5 A-c and d-c equivalent cireuits. a-e source voltage. In other words, the output voltage v is a sine wave with a peak value of 5 my. The total voltage vr produced by both sources acting simultaneously is the sum of the a-c and d-c components. We have shown this voltage in Fig. 4-5d. Note carefully that it is a fluctuating voltage with a low of 9.995 volts and high of 10.005 volts. In other words, it is a d-c voltage of 10 volts with a 5-my-peak sine wave superimposed on the d-c level. A word about notation is appropriate at this time. In order to keep the d-c and a-c components distinct in the various formulas to be devel- oped, we will use the following rules: 1. All d-c or fixed quantities will be denoted by capital letters. For instance, to represent a d-c voltage or current we will use V and J, respectively. 2. All a-e or varying quantities will be designated by lowercase letters. Thus, we will use » and i to represent a changing voltage or current. Further refinements in our notation will be introduced as the need arises. Examp_e 4-1 Find the voltage vr appearing across the 10-ohm resistor in Fig. 4-6a. Soiution The d-c equivalent circuit is shown in Fig. 4-65. The d-c voltage appearing across the 10-ohm resistor is approximately 10 mv, because54 ‘Transistor Circuit Approximations the 10-ohm resistor and the 10-kilohm resistor form a voltage divider that divides the a a by a factor of 1000. That is, V= 10 = 10 mv Foo a ed 1000 The a-c equivalent circuit is shown in Fig. 4-6c (the battery and capacitor have been shorted according to the usual rules). Note that % ‘9 10K 10K 10 mv. 90 pow 5 oe =10my v4 t + WY) Xp negligble S10 v, OH EEY 2 foroc fy oe aay (a) ) %9 1Omv 90 90 “10 : my + 9 10K lo y % lo ee eee i (ce) (a) Fig. 4-6 Example 4-1. for practical purposes the 10-ohm resistor in parallel with the 10-kilohm resistor is essentially 10 ohms; therefore we can redraw the a-c equiv- alent circuit, as shown in Fig. 4-6d. We can see that the a-c voltage » must be one-tenth of the a-c source voltage because of the 10:1 voltage divider. In other words, the a-c output voltage v is a sine wave with a peak value of about 1 mv.Small-signal Diode Approximations 38 The total voltage vr produced by both sources acting simultaneously is the sum of the a-c and d-c components; this total voltage is shown in Fig. 4-6e. Note that it has an average value of 10 mv with a 1-my-peak sine wave superimposed on this 10-mv level. Exampie 4-2 The reactance of the capacitor in Fig. 4-7a is very small as far as the a-c signal is concerned. Sketch the voltage vr across the 10-kilohm resistor. SoLution The d-c equivalent circuit is shown in Fig. 4-7b. Since the capacitor is open to direct current, the d-c voltage V is equal to zero. The a-c equivalent circuit is shown in Fig. 4-7c. Since the d-c source has been shorted, the two 10-kilohm resistors are in parallel as far as the a-c signal is concerned. The a-c equivalent circuit can be redrawn as shown in Fig. 4-7d. In this figure it is clear that the a-c voltage 9 eee 10 10K Es lOmv t |» oo -l0mv ‘5K TOK | bm 5K 10K 10K v (a) (eb Fig. 4-7 Example 4-2.56 Transistor Circuit Approximations across the output is one-half of the a-c source voltage. Therefore, the a-c voltage across the output is a sine wave with a peak value of 5 mv. The total voltage vr equals the sum of the d-c and a-c components, and is shown in Fig. 4-7e. In this case, there is no d-c component. In fact, coupling capacitors are generally used for this very purpose, namely, to pass the a-c component but to block the d-c component. 4-3 The A-C Resistance of a Diode In order to apply the superposition theorem to diode and transistor cir- cuit analysis we must discuss how a diode acts as far as a-c signals are concerned. We will see that if the a-c signal is very small, the diode re- sembles a resistance whose value is given by the ratio of the a-c voltage across the diode to the alternating current through the diode. Let us begin our discussion by considering the circuit in Fig. 4-8a. A d-c source is in series with a small a-c source. What happens is quite simple: the d-e source establishes the average voltage across the diode, while the a-c source causes small changes above and below this average voltage. For instance, suppose that the d-c source voltage is 1 volt and the a-c source voltage is a 1-my-peak sine wave. Then, the voltage across the diode will have an average value of 1 volt; in addition, there will be a fluctuation of 1 mv above and below this 1-volt level. ‘9 _0-¢ current + l i d=c operating if Point =Imv —_—+ ov Q-C voltage {a) (2) Fig. 4-8 The a-c resistance of a diode above the knee. How much current flows through the diode? To answer this question, we draw a typical 7V characteristic, as shown in Fig. 4-85. Assume that we have adjusted the battery voltage so that the d-e operating point is well above the knee of the diode characteristic. The average operating point is point A; the d-c voltage and current at this operating point are V and J, respectively. We have shown a small arc signal in Fig. 4-8)Small-signal Diode Approximations 8T along the v axis. This a-c voltage causes the instantaneous current in the diode to change above and below the average value of I. On the positive half cycle the a-c voltage causes the diode current to change from point A to point B, and on the negative half cycle the current changes from A to C. If the a-e voltage is very small, the changes in current will be very small. In fact, notice that for small excursions, the instantaneous oper- ating point of the diode will be moving along a segment of the curve that is almost linear. This means that the change in current is almost directly proportional to the change in voltage. For instance, suppose that a change of 1 my produces a change of 0.1 ma; then a change of 2 mv will cause a change of about 0.2 ma. The almost linear relation between small changes in voltage and cur- rent allows us to think of a diode as a resistance as far as small a-c signals are concerned. The value of this resistance is simply te = eee AG. where Av and Ai represent small changes in voltage and current about the d-c operating point. As an example, suppose that in Fig. 4-8 the changes in voltage and current between points B and C are Av = 1 mv and Ai = 0.1 ma. As far as the arc signal is concerned, the diode appears to be a resistance of Av _ 10-8 rue = 5 = DAGOry = 10 ohms What happens to the a-e resistance of the diode when we change the battery voltage? Suppose we adjust the battery voltage to a smaller value; then the d-e operating point will be lower down on the diode curve, as shown in Fig. 4-9a. Again there is an average voltage V and an average current I flowing through the diode. With the same small a-c voltage as before, the changes in current will be less than before as shown i / 0-c voltage o-ccurrent v )* c current (0) (4) Fig. 4-9 The a-c resistance below the knee.58 ‘Transistor Cireuit Approximations in Fig. 4-9a. Therefore, the a-c resistance of the diode is larger than it was before. As a matter of fact, it is clear that as we reduce the d-c current I through the diode, the a-c resistance rae will increase. We can reverse-bias the diode to obtain the d-c operating point shown in Fig. 4-9b. In this case, the small a-c voltage produces extremely small changes in current. In other words, the a-c resistance of a diode is quite high in the reverse direction. Let us summarize the key points of our discussion: 1. The d-e source establishes the operating point about which the a-c excursions take place. 2. For small a-e excursions the changes in voltage and current are almost linear (directly proportional). : 3. As far as small a-c voltages and currents are concerned, the diode looks like a linear resistance whose value is given by fn = A i at 4, In the forward direction the a-c resistance decreases when the direct current is increased. 4-4 Formulas for the A-C Resistance of a Diode When a diode is turned on hard, that is, when it is operating well above the knee of the diode characteristic, the only opposition to the current is the bulk resistance of the diode (discussed in Chap. 3). In other words, well above the knee of the diode curve, only the ohmic resistance of the p- and n-type material remains to impede the current, as illustrated by Fig. 4-10a. Note that the ohmic resistance of the p- and n-type material is lumped into a single resistance called the bulk resistance rs. Thus, well above the knee we can say that the a-c resistance of a diode is simply equal to the bulk resistance rz. When the operating point of the diode is below the knee of curve, the Pp N P v =o seyhowr= a (a) (4) Fig. 4-10 A-c resistance. (a) Well above the knee; (b) junction and bulk resistance.Small-signal Diode Approximations 59 barrier potential also becomes important in retarding current. We can indicate this junction effect as shown in Fig. 4-10b, where we have shown an additional resistance r;, the junction resistance of the diode. Hence, below the knee of the diode curve, we say that the a-c resistance of the diode is the sum of the bulk resistance and the junction resistance. ‘The value of junction resistance for any particular diode is not usually published on the data sheet for that diode. However, there are some approximations that are widely used for the value of junction resistance. Theoretically, it can be shown that at room temperature a perfect june- tion diode has a junction resistance of = 2Smy ae a where I is the d-e (average) value of current through the diode. For instanee, suppose there is a direct current of 1 ma through the diode. Then the junction resistance is (4-3) _ 25 mv _ 1) = Taw = 25 ohms If the direct current is changed to 2 ma, tho junction resistance is changed to _ 25 mv = = 12.5 ohms 2 ma % Equation (4-3) gives the value of junction resistance for a perfect junction diode. In practice, we find that most junction diodes will fall in the range of 25 mv 50 mv rags T For instance, when there is 1 ma of direct current, the junction resistance of most diodes will fall in the range of 25 to 50 ohms. If the current is changed to 0.1 ma, the junction resistance will then be about 250 to 500 ohms. When the diode is reverse-biased, it simply looks like a large resistance of Re (discussed in Chap. 3). Recall that Rr is the reverse resistance and is calculated by taking the ratio of any reverse voltage to the cor- responding current. The main points of this section are simply this: (4-4) 1. When the diode is reversed-biased, it has an a-c resistance of =Rr 2. When the diode is forward-biased, it has an a-c resistance of tue = tre60 Transistor Circuit Approximations 3. As an approximation for most junction diodes we will use 25 mv 50 mv T frequency. a i ance, we can disregard the capacitance and use the equivalent circuit of Fig. 4-15a. On the other hand, when the frequency is so high that the capacitive reactance is much smaller than the reverse resistance, we can use the equivalent cireuit of Fig. 4-15b. Exampie 4-9 A silicon diode of the alloy type has a reverse current of 1 wa when the reverse voltage is 50 volts. Find the reverse resistance. Also, find the junction capacitance for reverse voltages of 5 and 10 volts. The value of Co is 10 pf.68 ‘Transistor Circuit Approximations Sonvrion First, we find Rr. TE = igs = 50 megobms Next, we get the junction capacitance using Eq. (4-5) since the diode is an alloy type. When Vr = 5 volts When Vz = 10 volts, The a-c equivalent circuits for each value of reverse voltage are shown in Fig. 4-16. These circuits are the way that a diode looks to a small a-c signal. 5OM SOM Somapailene: et 10 $9 Fig. 4-16 Example 4-9. JL 4 if IT 418pf 3.06pf (a) (4) Exampce 4-10 The data sheet for a silicon diode indicates that it is an alloy type and has a capacitance of 50 pf for a reverse voltage of 4 volts. Find the value of Co and then find the value of C; for a reverse voltage of 20 volts. Sonurton (a) First, we substitute Vz = 4 and C; = 50 pf into Eq. (4-5). Co | Cee Care o Vat Vx | Co | 50(10-42) = —22__ ome V4+07 | By solving this equation we get Co = 108 pf. |Small-signal Diode Approximations 69 (6) Now we can find the value of junction capacitance for a reverse voltage of 20 volts. 108(10-7) V20 + 0.7 Note how the junction capacitance has changed from 50 pf (reverse voltage of 4 volts) to 23.8 pf (reverse voltage of 20 volts). In effect, we have a voltage-controlled capacitance. This ability to change the diode capacitance by varying the reverse voltage leads to a number of inter- esting applications in frequency modulation and control. G 23.8 pf SUMMARY A linear circuit is one whose resistors, capacitors, and inductors main- tain constant values even though the voltage across them changes. The superposition theorem allows us to analyze linear circuits contain- ing more than one source. The basic approach in using superposition is to compute the voltage or current produced by one source at a time. The sum of the individual voltages and currents then gives the total effect produced by all sources. Circuits containing a d-c source and an a-c source are of special im- portance in transistor circuit analysis. In applying superposition to these circuits we draw a d-c equivalent and an a-c equivalent circuit. In the d-c equivalent circuit all capacitors are open circuits. In the a-c equiva- lent circuit the capacitors normally look like short circuits. The a-c resistance of a diode is the resistance that the diode presents to a small a-c signal. This resistance depends upon the amount of direct current flowing through the diode. The superposition theorem can be applied to diode circuits provided we follow the modified procedure discussed in Sec. 4-5. The capacitance of a reverse-biased diode appears in shunt with the reverse resistance. Since the depletion layer widens when the reverse volt- age is increased, the diode capacitance decreases. At extremely low fre- quencies the diode acts like a resistance because the capacitance effects are negligible, whereas at very high frequencies the diode acts primarily like a capacitance. GLOSSARY a-c equivalent circuit The circuit used in computing the a-c component. This circuit is obtained by reducing all d-c sources to zero, replacing70 ‘Transistor Cireuit Approximations all diodes by their a-c resistances, and replacing all coupling capaci- tors by short circuits. a-c resistance The resistance that a diode presents to a small a-c signal. The value of this resistance can be found by computing the ratio of a change in voltage to the change in current about the d-c operating point. bulk resistance The ohmic resistance of the p- and n-type material. The bulk resistance is the only resistance well above the knee of the diode curve. d-c equivalent circuit The circuit used in computing the d-c component. This circuit is obtained by reducing all a-c sources to zero and re- placing all capacitors by open circuits. junction capacitance This is the capacitance produced by the depletion layer and the p and n materials on each side of the depletion layer. Junction resistance The effects of barrier potential can be handled by means of a resistance as far as small a-c signals are concerned. This resistance is called the junction resistance and can be approximated by Eqs. (4-3) and (4-4). linear system One in which effects are directly proportional to causes. superposition In a linear system where several causes are acting simul- taneously to produce an effect, this effect can be found by adding the individual effects produced by all the causes considered one at a time. REVIEW QUESTIONS 1. What is the superposition theorem? To what kind of circuits can we apply it? 2. What is a linear circuit? What can be said about the values of resis- tors, capacitors, and inductors in a linear circuit? 3. How do we treat a capacitor when considering the d-c equivalent circuit? How do we usually visualize capacitors when drawing the a-e equivalent circuit? 4. We use capital letters to designate what kind of quantities? Lower- case letters are used for what? 5. How did we define the a-c resistance of a diode? 6. What happens to the size of the a-c resistance r,. when the direct current in the diode is increased? 7. What is the formula for the junction resistance of a diode? How do we find the a-c resistance of the diode when it is forward-biased? Reverse-biased?Small-signal Diode Approximations n 8. Describe our procedure for applying superposition to diode cireuits containing large d-c and small a-c sources. 9. What is our rule for determining whether or not an a-c signal is small? 10. When the reverse voltage is increased, what happens to the width of the depletion layer? And to the junction capacitance? LL. Is the diode capacitance important at low or high frequency? Why? PROBLEMS 4-1 Use the superposition theorem to find the current in the 10-ohm resistor of Fig. 4-17. 20 30 Fig. 4-17, 4-2 Use superposition to find the current in the 20-ohm resistor of Fig. 4-17, 4-3 In Fig. 4-18a, sketch the waveform of current in the 30-kilohm re- sistor. Sketch the waveform of voltage across the 30-kilohm resistor. 9 20mv. 10K Ik 2K ~20mv + © ig 30K Fig. 4-18 4-4 In Fig. 4-18), the a-c source is generating a sine-wave voltage with a peak value of 50 mv. Sketch the waveform of voltage across the 3-kilohm resistor. 4-5 Sketch the voltage waveform across the 1-kilohm resistor of Fig. 4-19a, The reactance of the capacitor is very small.2 Transistor Circuit Approximations 9 9 20mv 20myv i t “2 : Omy: 9k Ik 20mv § q (a) Fig. 4-19 4-6 Sketch the voltage waveform across the 40-kilohm resistor of Fig. 4-195. The reactance of the capacitor to the a-c signal is very small. 4-7 A diode has a current of 1 ma flowing through it when the voltage across the diode is 0.71 volt. If the voltage is increased to 0.715 volts, the current becomes 1.1 ma. Find the a-e resistance of the diode. 4-8 A germanium diode has a forward current of 55 ma at 1 volt. What is the approximate value of bulk resistance rp? 4-9 A silicon diode has a forward current of 80 ma at 1 volt. What is the approximate value of bulk resistance? 4-10 A diode has a bulk resistance of 2 ohms. Calculate the junction resistance and the a-c resistance for the following direct currents: 0.1, 0.5, 1, and 5 ma. Use Eq. (4-3). 4-11 What is the a-e resistance of the silicon diode of Prob. 4-9 when a direct current of 0.75 ma flows through the diode? 4-12 In Fig. 4-20a find the following: (a) ‘The approximate direct current in the diode. () The approximate peak value of the alternating current in the diode. 4,13 If the d-e supply in Fig. 4-20a is increased to 60 volts, what will the direct current become? What will the peak current become? % 9 % Lorge Lorge : ig capacitor 2mv. copacitor = ro t - 1K “Oy EO "9 (o} (o) Fig. 4-20Small-signal Diode Approximations 3 4-14 In Fig. 4-20b, what is the direct current approximately? The approximate peak value of the alternating current? 4-15 In Fig, 4-20a, what happens to the alternating current in diode if the 30-kilohm resistor is changed to a 60-kilohm resistor? 4-16 In Fig. 4-206, how much a-c voltage appears across the diode when the 30-kilohm resistor is changed to a 15-kilohm resistor? 4-17 In Fig. 4-21, find: (a) The direct current in each diode. () The peak value of the a-c voltage v across the output terminals. +100 9 looK 100K Tvolt 10K IK t bhevyn—eww. — lvolt fae tt % Y IK y eaeay Fig. 4-21 4-18 A reverse voltage of 25 volts is applied to a grown type of silicon diode. What is the junction capacitance if the diode has a Co = 75 pf? What is the capacitive reactance at 1 kHz? At 1 MHz?Common-base Approximations In its physical appearance the transistor is nothing more than two back- to-back diodes; however, because the spacing between these diodes is so small, a new phenomenon takes place in a transistor. This phenomenon makes it possible for us to obtain one of the most important effects in electronics, namely, amplification. In this chapter we study the common-base connection of a transistor. After discussing the basic idea behind transistor action, we develop a transistor approximation called the ideal transistor, which makes it possible to analyze transistor circuits rapidly and easily. Even though the ideal transistor is only a simple approximation of an actual transistor, we find in practice that the ideal transistor is adequate for most everyday needs; it is quite useful for troubleshooting and initial design of transistor circuits. ‘There are more exact methods than the ideal transistor; we discuss these in later chapters after the basic idea of the transistor has become fixed. 5-1 Terminology and Schematic Symbols A transistor is made by growing, alloying, or diffusing pieces of p-type and n-type materials together. A p-n-p transistor is made by placing 4Common-base Approximations 15 n-type material between two pieces of p-type material, as symbolized in Fig. 5-la. The larger region of p-type material is called the collector, and the other region of p-type material is called the emitter. The region in the middle is called the base. Note in Fig. 5-1a that the transistor is like two back-to-back diodes. Bose Bose Emitier 1 Collector Emitter + collector — Pp lw P ny |P No -~. (a) () Fig. 5-1 (a) Structure of a p-n-p transistor; (b) structure of an n-p-n transistor. Of course, we can put a piece of p-type material between two pieces of n-type material, as indicated in Fig. 5-1b. This would be an n-p-n transistor. The relative sizes of the emitter, base, and collector are not accurately shown in Fig. 5-1. Of special importance is the fact that the base region is actually very thin. The reason for this will be brought out shortly. Also, the actual shape of the transistor can be different from the sym- bolic sketch of Fig. 5-1; the important idea is that no matter what the actual shape of the pieces, there is a piece of material between two regions of the opposite type of material. In Fig. 5-2 and h we have shown the schematic symbols commonly used to represent transistors (sometimes the circle is omitted). An arrow- head is placed on the emitter but not on the collector. The direction of Emitter o— P |v} Pf Collector Emitter | WV |P) MW j-*Collector Base Bose Emitter Collector Emitter To Bose Bose (a) (4) Fig. 5-2 Schematic symbols. (a) p-n-p; (b) n-p-n.6 Transistor Circuit Approximations the arrowhead, like the triangle in a diode, points in the easy direction of conventional current, that is, from the p- to the n-type material. For instance, in Fig. 5-2a the p-n diode on the left conducts conventional current easily from the emitter to the base; in the schematic symbol, therefore, we show an. arrowhead from the emitter to the base. Similarly, in Fig. 5-2b con- ventional current would flow easily from base to emitter, and we therefore show an arrowhead from base to emitter. The collector-base part. of a transistor also forms a diode. In Fig. 5-2a the easy direction of conventional current is from collector to base. Even though it is not customary to show an arrowhead on the collector, it is helpful to viswalize an arrowhead pointing from the collector into the base in Fig. 5-2a. In Fig. 5-26 we can visualize an arrowhead pointing from the base to the collector. Thus, whenever we see the schematic symbol of transistor, we can visualize an arrowhead on the collector pointing in the same direction as the emitter arrowhead. 5-2 Biasing the Trar.sistor What happens when we apply voltages to the transistor? Let us consider an n-p-n transistor driven by two d-c supplies, as shown in Fig. 5-3a. Note that the base is grounded; this configuration is called the grounded- base or common-base connection (the base is common to both loops). As already indicated, the transistor is like two back-to-back diodes. For con- venience, we will call the diode on the left, formed by the emitter and the base, the emitter diode. The diode on the right, formed by the collector and the base, will be called the collector diode. E (4 \Qdr 4 NYC A IB ee a ee 1 Fig. 5-3 Biasing. (a) Both diodes off; (b) both diodes on. By inspection of Fig. 5-3a, both the emitter and the collector diodes are back-biased, since the batteries try to force conventional current against the direction of the arrowheads (visualize an arrowhead out of the collector). Therefore, only a small reverse current flows in each diode.Common-base Approximations rT Suppose that we reverse both batteries as indicated in Fig, 5-3b. What happens now? In this case, both diodes are forward-biased, and therefore a large current can flow in each diode. In Fig. 5-3a and 6 nothing of any consequence is taking place. Either both diodes are off, or both diodes arc on. These results are of little im- portance because we can obtain the same results by using two separate p-n diodes instead of a transistor. What makes the transistor different? Consider Fig. 5-4. In this case, the emitter diode is forward-biased, and the collector diode is back-biased. Our instinct tells us that there should be a large emitter current and a small collector current. This is not what happens! What actually happens Fig. 5-4 Emitter diode forward- biased and collector diode reversed- biased. is that there is a large emitter current and an almost equally large col- lector current. This unexpected phenomenon makes the transistor the important device that it is. Instead of having a small collector current be- cause the collector diode is back-biased, we have a large collector current. Here is the reason for the large collector current. In Fig. 5-5a (equiva- lent to Fig. 5-4) there is an excess of free electrons in the n-type emitter region. Since the emitter diode is forward-biased, these free electrons move to the right and enter the base region. Because the base region is deliberately made very thin and is very lightly doped, most of these free electrons do not recombine; instead, they diffuse into the collector-base depletion region, where they are swept across the junction into the col- wv oP oW NP WN Emitter Emitter electrons i Collector ae electron electrons tt Fig. 5-5 Emitter electrons captured by the collector.18 Transistor Cireuit Approximations lector. They are then attracted and collected by the positive terminal of the collector supply, as shown in Fig. 5-5b. Thus we see that the emitter injects free electrons into the base. Be- cause the base is extremely thin and lightly doped, these electrons pass into the collector region, where they are immediately attracted by the positive terminal of the collector supply. Almost all the free electrons that enter the base pass through to the collector. To summarize the circuit action for Pig. 5-4: 1. Under normal conditions we forward-bias the emitter diode and back- bias the collector diode. 2. The size of the collector current is almost equal to the size of the emitter current (ic © is). 3. The base current is very small and equals the difference of the emitter current and the collector current (is = ig — ic). 5-3 The IV Characteristics of a Common-base Transistor In order to obtain a fuller idea of how the currents and voltages are related in a common-base (CB) connection, let us consider a hypothetical experiment. A transistor is connected as shown in Fig. 5-Ga. Note that the emitter diode is forward-biased and the collector diode is back-biased. What is the relation between the emitter current and the emitter volt- age? The current-voltage relation in the emitter diode depends to some extent upon the value of the collector voltage. Suppose that arbitrarily we set the collector supply to 1 volt. In the emitter circuit here is what we would find. When we increase the emitter hattery supply, the emitter current will increase slowly at first; however, after vg has reached a few tenths of a volt, the emitter current will increase significantly with a further increase in voltage. The graph for ig vs. ves is shown in Fig. 5-6. Note that this graph is (a) Fig. 5-6 (a) Obtaining IV curves; (b) emitter-diode curve.Common-base Approximations 9 the typical graph of a semiconductor diode. The emitter current is small until reaching the knee voltage, and then it turns up sharply above the knee. As observed in our discussions of diodes, the knee voltage is about 0.3 volt for germanium and 0.7 volt for silicon. What effect does collector voltage have upon emitter current? If we increased the collector voltage from 1 to 10 volts, the relation between emitter current and voltage would change slightly, as indicated by the dashed curve of Fig. 5-6b. This suggests that there is a feedback effect. from the collector to the emitter; however, this feedback effect is small, as implied by the small separation between the two graphs. As a first approximation we can definitely say that the emitter-base part of a transistor acts like an ordinary semiconductor diode. Now, we turn our attention to the collector circuit. How is collector current related to collector voltage? From our discussion of Sec. 5-2 we know that almost all the electrons leaving the emitter pass through the base into the collector. In other words, collector current almost equals emitter current. In order to get a graph of ic vs. vex we must specify a particular value of emitter current. Arbitrarily, suppose that the emitter supply is adjusted to produce an emitter current of 1 ma. In the collector circuit here is what we would find. When we increase the collector volt- age, the collector current will increase only slightly because almost all the emitter electrons are captured by the collector; increasing the collector voltage cannot significantly increase the collector current because it is more or less fixed by the size of the emitter current. Figure 5-7a graphically illustrates what takes place in the collector circuit. Note that the collector current remains fixed at about 1 ma even though the collector voltage is increased. Of course, there is a limit; when the breakdown voltage of the collector diodo is exceeded, a significant increase in current can take place. Normally, the transistor is operated below this breakdown voltage. The graph of Fig. 5-7a below the breakdown point suggests the con- cept of an ideal current source. An ideal current source is simply a hypo- thetical device whose current is independent of the voltage across it. In Io c 4 Breakdown I region jg*imo Ima| Yee a a —~ tee Fig. 5-7 Collector curves.80 ‘Transistor Circuit Approximations other words, below breakdown the collector current remains at about 1 ma even though the collector voltage is being increased. If the emitter current is changed, the collector current will change. For instance, if we inerease the emitter supply so that 2 ma of emitter current flows, we will find that the collector current almost equals 2 ma. Further, when we increase the collector voltage, very little increase occurs in the collector current until we reach the breakdown voltage, as indi- cated by Fig. 5-7b. Again note that the collector acts like an ideal current source below the breakdown voltage. ‘That is, below the breakdown point the collector current remains at about 2 ma even though we increase the collector voltage. The graphs of Fig. 5-7 suggest that the collector diode is a controlled current source for any collector voltage between zero and the breakdown voltage. In other words, the collector current is controlled by the emitter current. When we change the emitter current to a new value, the collector current will change to this new value. Further, changing collector voltage has no significant effect on collector current. If we continue to obtain graphs like those of Fig. 5-7 for new values of emitter current, we can construct the typical composite characteristic shown in Fig. 5-8. This characteristic shows the relation of ic to ves below the breakdown voltage. Note that for iz = 0, there is a small amount of collector current, This small amount of current is the reverse current of the back-biased collector diode. Note also that for any value of emitter current it is necessary to reduce the collector voltage to slightly less than zero in order to shut off the collector current. te Sma ig250 4mofp ama Fig. 5-8 Overall collector character- 3ma 73m istic, 2mo e22ma Ima ee me eo) rn) Let us summarize the key points of our discussion: 1. The emitter diode acts almost like an ordinary semiconductor diode. 2. Below breakdown the collector diode acts almost like an ideal cur- rent source. The value of this current source is controlled by, and is al- most equal to, the emitter current (ig © i).Common-base Approximations al 3. The collector diode will break down if its breakdown voltage is ex- ceeded. Normally a transistor is operated well below the breakdown voltage, 5-4 The Alpha of a Transistor One of the important transistor quantities is what is called the a of a transistor. Actually, there are both a d-c « and an a-c a. By definition, the d-c a is — ie - ae ~ 5 (5-1) For instance, in Fig. 5-9a there is an emitter current of 1 ma and a col- lector current of 0.98 ma. According to Eq. (5-1), the d-e a is nothing more than the ratio of these currents. That is, The ag, is actually a measure of the quality of a transistor. Ordinarily, the higher the aae, the better the transistor, in the sense that the collector current more closely equals the emitter current. For a perfect transistor the collector current equals the emitter current, and therefore the age equals unity. In practice we find that almost all transistors have an aac in the range of 0.95 to 0.999 . . . . In other words, the collector current is usually no lower than about 95 percent of the emitter current and often much closer to 98 or 99 percent of the emitter current. Ima 0.98ma 105mo 1.028ma a Ty i QB = toozme 0.022ma it | i (a) (4) Fig. 5-9 D-c and a-c @. There is also an a-c @ for a transistor. This refers to the ratio of a change in collector current to the corresponding change in emitter cur- rent. For example, suppose that we change the emitter supply Vee in Fig. 5-9a and that we then have currents of 1.05 and 1.028 ma, as shown82 Transistor Circuit Approximations in Fig. 5-9b. By definition, the a-c a is Aic = Ste 5-2 "= Aig ae where Ac is a small change in collector current and Aig is a small change in emitter current. In our example the change in collector current is Aic = 1.028 — 0.98 ma = 0.048 ma and the corresponding change in emitter current is Aig = 1.05 — 1 ma = 0.05 ma Hence, we calculate an a of _ 0,048 ma “* 0.05 ma = 0.96 Often the values of the d-c and a-c a’s are almost equal. In general, we find in practice that the a-c @ also is typically in the range of 0.95 to 0.999 .... 5-5 The Ideal Transistor In Chap. 3 we idealized the diode by simple approximations that re- tained the essential features of the diode while discarding the less im- portant qualities. These approximations allowed us to analyze diode cir- cuits easily and rapidly. We now wish to approximate the transistor in a similar way. First, eonsider the emitter-base part of a transistor. We know that a change in collector voltage causes only a slight feedback into the emitter circuit. As an ideal approximation of the transistor we disregard this small amount of interaction between collector and emitter and say that for practical purposes the emitter diode acts like a typical semiconductor diode with the characteristic shown in Fig. 5-10a. a te forall colecion |) 00 een fate voltages 3ma e= 3100 2ma}——— = 2ma 4 a tae o> “a (a) (0) Fig. 5-10 Ideal transistor. (a) Emitter curve; () collector curves.Common-base Approximations 83 Next, consider the collector-base part of the transistor. We can idealize the characteristics of the collector by using the graphs of Fig. 5-10d. Note that there is no breakdown. Each curve is perfectly horizontal, and collector current equals emitter current. Of course, this is an ideal approxi- mation; nevertheless, it does represent the most important features of transistor action, namely, that the collector is a controlled current souree. The circuit interpretation of the ideal n-p-n transistor is shown in Fig. 5-Lla. We treat the emitter diode as being just that, a diode. However, the collector diode is represented by a current source (the circle with the arrow through it is the most common schematic symbol for a current source). Remember that whenever we encounter the schematic symbol of the current source as shown in Fig. 5-11a, we understand by definition that we have a device whose current is independent of the voltage across it. In Fig. 5-11a the current flowing in the collector is equal to ig. e ‘o Be oe fe e «€ le Nay: = ye Ideal Ideal (a) (4) Fig. 5-11 Equivalent circuits for ideal transistor. Analogously, the ideal p-n-p transistor is represented by a diode and a current source of opposite polarity from the n-p-n, as shown in Fig. 5-11b. Here.are the key ideas for the ideal-iransistor approximation. As long as the emitter diode is forward-biased and the collector diode is back- biased, then: 1. The emitter diode acts like an ordinary semiconductor diode. 2. The collector diode acts like an ideal current source whose value equals the emitter current (i¢ = iz). EXampie 5-1 Find the value of collector current i¢ in the circuit of Fig. 5-12a. Sotution First, we visualize the circuit as shown in Fig. 5-12b, where we have replaced the transistor by an ideal transistor. In this circuit it is clear that the emitter diode is forward-biased. We know that only a few tenths of a volt appear across the emitter diode; therefore, we can say84 ‘Transistor Circuit Approximations that practically the entire 10 volts from the supply appears across the 10-kilohm resistor. Thus, the emitter current is ip Oe ma ‘s = 707008) ~ Since the collector and emitter currents are almost equal, the collector current must equal about 1 ma. a bc ’ © (e) Fig. 5-12 Examples 5-1 and 5-2. EXAMPLE 5-2 Find the collector-base voltage ver in the circuit of Fig. 5-12c. SoLution First, find the emitter current. It is approximately : 10 ts = S070 = 0.5 ma The collector current is therefore about 0.5 ma. This 0.5 ma flows from right to left through the 10-kilohm resistor and produces a voltage of 0.5 (10-*)(10) (108) = 5 volts To find ves we apply Kirchhoff’s voltage law to the collector loop circuit. In other words, v¢p must equal the supply voltage minus the voltage drop across the 10-kilohm resistor. That is, Yr = 25 — 5 = 20 voltsCommon-base Approximations 85 This type of problem is important in troubleshooting because one of the first checks that should be made on a questionable circuit is to measure the collector-ground voltage. EXAMPLE 5-3 In the circuit of Fig. 5-13a, what value of R, causes ves = 10? Sonurion The schematic drawing of Fig. 5-18a is the usual way of indicating power-supply voltages. The other terminal (not shown) on each power supply is grounded. For instance, the 10-kilohm resistor is connected to the negative terminal of a d-c supply, and it is understood by defini- tion that the positive terminal of this supply is grounded even though it is not shown. -10 +30 -30 +20 fee te 10K he | { Re Ke | & ie + t “ce eed (6) Fig. 5-13 Examples 5-3 and 5-4, By inspection of the circuit the emitter diode is forward-biased, and therefore we have 10 T0(10") =1ma Since the collector current approximately equals the emitter current, there is about 1 ma flowing through the load resistor R;. It should be clear from the circuit that in order to have tee = 10 there must be a voltage drop of 20 volts across Rz. To find the size of Rz, that produces this 20-volt drop we use Ohm’s law. 20 _ on ys Ry = joa = 20 kilohms Exampue 5-4 For the circuit of Fig, 5-130, find the value of Rx that causes ves to equal 10 volts.86 Transistor Cireuit Approximations Sonurion In order for ves to equal 10 volts there must be a voltage drop of 10 volts across the 10-kilohm load resistor. In order to have this 10-volt drop there must be a collector current of pee tea *c = 70704) ~ The emitter current controls the collector current; thus, the emitter current must be made equal to 1 ma. This can be done by choosing a value of Rz that sets up 1 ma of emitter current. Using Ohm’s law, we get Rez = ~~, = 30 kilohms 10-3 EXAMPLE 5-5 In Fig. 5-14a, find the collector-base voltage by the following approaches: (a) Neglect the voltage drop across the emitter diode. (b) Allow 0.7-volt drop across the emitter diode. SoLution (a) First, note that we are using a p-n-p transistor, instead of an n-p-n. Transistor action is essentially the same in both types of tran- sistors except that the holes are the majority carriers in the p-n-p transistor. We still must forward-bias the emitter diode and buck-bius the collector diode to obtain normal transistor action; therefore, we connect the d-c sources as shown in Fig. 5-14a. In this circuit it is clear that the emitter diode is forward-biased and that the emitter current is approximately jee oe te eae == iodo ~ *™ The collector current is therefore almost equal to 1 ma and produces a voltage drop of about 5 volts across the 5-kilohm resistor with the polarity shown. From Kirehhoft’s law we know that the collector-base voltage is the difference between the collector supply and the drop across the 5-kilohm resistor. That is, vcs = 10 — 5 = 5 voltsCommon-base Approximations 87 This is the magnitude of the collector-base voltage. The collector diode of a p-n-p transistor is back-biased by the negative d-c source; thus, the actual voltage across the collector-base part of the transistor is negative, as shown in Fig. 5-14a. f +20 -20 IOK : 'e "ge | 40k Rs t G 10 (a) (0) Fig. 5-14 Examples 5-5 and 5-6. (b) If we want a more accurate answer, we can allow 0.7 volt drop across the emitter diode; this means that the voltage across the 10-kilohm resistor in the emitter circuit will be 10 volts minus 0.7 volt. Therefore, the emitter current must be 10 — 0.7 “10(105) 0.93 ma The voltage drop across the 5-kilolhim load resistor is 0.93(10-#)(5)(108) = 4.65 volts and the magnitude of the collector-base voltage is von = 10 — 4.65 = 5.35 volts EXAMPLE 5-6 In Fig. 5-140, what value of R, produces a collector-base voltage of —10 volts? Sonution By inspection of the emitter circuit we see that the emitter current is approximately Seon te = 700%) ~ .5 ma88 Transistor Circuit Approximations In order to have a collector-base voltage of —10 volts, there must be a voltage drop of 10 volts across R,. Therefore, Rz must equal 10 Re = Tao = 20 kilohms 5-6 Using Superposition to Analyze CB Circuits In the preceding section and in Examples 5-1 to 5-6 we analyzed circuits containing only d-c sources. Now we wish to analyze circuits that con- tain both d-e and a-e sources. Reeall that in Chap. 4 we discussed the use of superposition in ana- lyzing diode circuits containing large d-c sources and small a-c sourees. In analyzing these circuits we found the d-e component using the dc equivalent circuit and the a-c component using the a-c equivalent. cir- cuit, and then we added these two components to find the total voltage or current that we wanted. When we analyze CB transistor circuits, we will use the same basic approach, namely, we will first find the d-e component and then the a-e component, and finally we will add these components. In our early work in transistors we will concentrate upon the ideal- transistor approach because this simple approach will be adequate for most of our needs in the areas of troubleshooting and initial design. Later on, we will discuss more exact methods. For the moment, then, let us make sure that we understand how to use the ideal-transistor approach in d-e and a-e circuit analysis. To begin with, in the d-e equivalent circuit we view the ideal transistor as nothing more than a rectifier diode in the emitter and a current source in the collector. The d-c equivalent circuit for the n-p-n transistor is shown in Fig. 5-15a and that for the p-n-p transistor in Fig. 5-15b. In finding the direct current that flows in the emitter diode, we ean use any of the large-signal diode approximations developed in Chap. 3. In other words, we can use an ideal diode for the emitter diode, or if more accu- racy is required, we can use the second or third approximation of a diode. Once we have found the d-e emitter current, we immediately know that this same value of d-c collector current is flowing. As far as a small a-c signal is concerned, the emitter diode does not rectify; instead, it looks like a resistance (the a-c resistance of the diode discussed in Chap. 4). Therefore, we visualize the a-c equivalent circuit as a resistor and a current source, as shown in Fig. 5-15c. (This equivalent circuit is also valid for the p-n-p transistor.) As an example of using superposition to analyze a CB circuit driven byCommon-base Approximations 39 a large d-c source and a small a-¢ source, consider the cireuit of Fig. 5-16a. The d-c sources set up direct currents in the transistor; the small a-e source causes the transistor currents to fluctuate slightly. In order to find the total voltage vp appearing from the collector to ground, we first find the d-c component by drawing the d-c equivalent circuit as in Fig. 5-16b. It is clear in this circuit that the emitter diode is forward-biased. A one ey ke & & & Y “BO! mae Idea! Ideal la} (4) ’e fe) Fig. 5-15. Ideal equivalent circuits, (a) D-c equivalent circuit for n-p-n; (b) dec equivalent circuit for p-n-p; (c) a-c equivalent circuit for both. By using an ideal diode for the emitter diode, we can see that the d-c emitter current is about 1 ma. (If a more accurate answer were required, we would allow 0.3 or 0.7 volt for the emitter-base voltage, depending on whether the transistor is germanium or silicon.) With 1 ma of emitter current there is about 1 ma of collector current, which produces a volt- age drop of 5 volts across the 5-kilohm resistor. Therefore, the d-c volt- age Vc from collector to ground is 15 volts (the power-supply voltage of 20 volts less the drop across the 5-kilohm resistor). Next, we draw the a-c equivalent circuit as shown in Fig. 5-16c. ‘The coupling capacitor is shown asa short, and both d-c supplies are shorted to ground. The emitter diode acts like a resistance whose value is given by the sum of the junction resistance and bulk resistance. We will assume that the junetion resistance is much larger than the bulk resistance, so that In the d-c analysis we found that the emitter current Iz was about 1 ma.90 Transistor Circuit Approximations -10 +20 ~10 +20 fe vs 10K 5K lok: t, 5K | Sw t + ~5my Yp M215 Or Ideal 1) (0) (a) 4 oe | # Smv a . As loK 25 ne SK He (ce) Y 16. 15. 4 f (0) Fig. 5-16 Applying superposition and the ideal-transistor approximation. Thus, mv : Tee = 25 ohms Ima We can now find the amount of alternating current flowing in the emitter. By inspection of Fig. 5-16c we can see that the a-c emitter current equals the souree voltage divided by the a-c resistance of the emitter diode. That is, the current in the emitter is a sine wave with a peak value of i, = Smv ” = 35 ohms = 0.2 maCommon-base Approximations o1 The a-¢ collector current is approximately equal to the a-c emitter cur- rent. Therefore, the a-c voltage appearing from colleetor to ground must be sine wave with a peak value of vp = 0.2(10-%)(5)(10%) = 1 volt The total voltage vr is the sum of the d-c and a-¢ components and is shown in Fig. 5-16d. In other words, there is an average voltage of 15 volts appearing from collector to ground. Superimposed on this average voltage, there is a sinusoidal variation from 14 to 16 volts Let us summarize the use of superposition in transistor circuits. As with diode circuits, we use a modified form of the superposition theorem. Our approach is: 1. Draw or visualize the d-c equivalent circuit by shorting all a-c sourees and opening all capacitors. The transistor is replaced by a recti- fier diode in the emitter and by a current source in the collector. Compute the desired d-c component of current or voltage. 2, Draw or visualize the a-e equivalent cireuit by shorting all d-e sources and eapacitors (we are assuming that capacitors are large enough to look like very low impedances to the a-c signal), Compute the desired a-c component of current or voltage using the a-c resistance of the emitter diode and a current souree for the collector. 3. Add the d-c and a-c components to obtain the total voltage on current. 5-7 A Complicated A-C Equivalent Circuit The simple a-e equivalent circuit of a transistor shown in Fig. 5-L5¢ will lead to errors in some analysis problems because this model of the tran- istor is incomplete. A very accurate a-c equivalent circuit for a tran- tor is shown in Fig, 5-17a. This rather complicated circuit does take into account a number of transistor effects that we are neglecting for the moment. For instance, there is a capacitance across cach p-n junetion in the transistor; these capacitances are designated by C, and C., the emitter and collector capacitance, respectively. There is also a resistance r/ across the collector current source. This resistance is usually several megohms and is equal to the inverse slope of the collector current-voltage characteristics. There is a generator, designated by u,v, which accounts for the small amount of feedback that takes place from the collector to the emitter. Recall that this feedback has the effect of producing a slight shift in the emitter-diode characteristic (see Fig. 5-60).92 Transistor Cireuit Approximations e & > o (a) (9) Fig. 5-17 Transistor a-c equivalent circuits. (a) Complicated; (b) idealized. The @’ quantity is almost identical to the a discussed in Sec. 5-4. For practical purposes a’ is almost equal to unity. The resistance rj is called the base spreading resistance; it is caused by the ohmic resistance of the base region. Finally, the resistance r/ is the junction resistance of the emitter diode. At room temperature the perfect junction diode has a junction resistance given by _ 25 my Te where ris the resistance seen by a small a-c signal and Ix is the direct current flowing into the emitter diode. Clearly, the equivalent circuit of Fig. 5-17a is much too complicated for practical circuit analysis. Various methods (h parameters, + param- eters, ete.) have been evolved in an attempt to simplify the accurate model of Fig. 5-17a. In our work, especially at the beginning, we want as simple a model of the transistor as possible without losing the essential features. Therefore, in our ideal transistor the following approximations have been made: re (5-3) 1. We have neglected C. and C. because at low frequencies the react- ances produced by these capacitances are negligible. Later on when discussing the high-frequency limits of a transistor, we will consider these capacitances.Common-base Approximations 93 2. We have neglected r! because it is on the order of megohms. 3. Since a’ is normally between 0.95 and 0.999, we have rounded off this value to unity. 4. The quantity u. is typically between 10-* and 10-4, small enough to neglect in a first approximation. 5. The base spreading resistance rj is small enough to neglect in a first approximation. The only quantities that remain are r/ and a current source in the collector, as shown in Fig. 5-17b. Admittedly, we have disregarded many of the effects in a transistor; yet we will find that as a first approxi- mation, the ideal-transistor equivalent circuit is adequate in most initial analysis and design. EXxamPLe 5-7 Use the ideal-transistor approximation to find the total voltage vr across the 10-kilohm resistor in Fig. 5-18a. SoLvution First, we visualize the d-e equivalent circuit; it is clear that the coupling capacitor on the output side is open to direct current. There- fore, there is no d-c component across the 10-kilohm resistor. -10 +30 20K lok Ws & ¢ Smy - , io 10K vy (a) a * & c 7 5 220k Se ve Six Sox - t “05 (d) te) Fig. 5-18 Example 5-7.4 Transistor Circuit Approximations Next, we short the d-c supplies and the coupling capacitors, and we replace the transistor by its ideal a-c equivalent cireuit to get the a-c equivalent cireuit of Fig. 5-18. The a-c emitter current equals the a-c voltage across the emitter divided by the emitter resistance r’. To find rt, we can use Eq. (5-3). , _25mv (ements where Ix is the d-c emitter current. By inspection of the original circuit, Fig. 5-18a, the d-c emitter current is 10 Is = v0(08y = 0.5 ma Now we can find rj. r= 25M _ 50 ohms ma Now that we have the value of the a-c emitter resistance r!, we can find the a-c emitter current. In Fig. 5-18) with a 5-my-peak sine wave voltage appearing across the emitter resistance, we will get an emitter current that is a sine wave with a peak value of Ip = 0.1 ma 5, = 5(10-8) 50 The a-e collector current will therefore be a sine wave with a peak of 0.1 ma. This collector current flows through two 10-kilohm resistors in parallel. Hence, the voltage appearing across the output will be a sine wave with a peak value of 0.1 ma times 5 kilohms (two 10-kilohm resistors in parallel). That is, %» = 0.1(10-%)(5)(108) = 0.5 volt The total voltage appearing across the output is simply the a-c com- ponent shown in Fig. 5-18c. Remember that the d-c component was blocked by the coupling capacitor. Another point worth mentioning is that we used Eq. (5-3) to find 4; this equation is for a perfect diode. In practice, we find that most tran- sistors have emitter resistances that fall within a 2:1 range, that is, 25 my 50 mv age > rj, (5-12) ts Ts Equation (5-12) has great practical value. We know that the value of r is theoretically given by 25 mv/In. In practice, however, most: transistors have emitter resistances that fall within the range of 25 to 50 mv/Ig. As we change from one transistor to another of the same type, the value of ri will change. These possible changes in r) mean that. the voltage gain can change unless we somehow eliminate the effect of a change in rf. One standard method used in practice to eliminate the effects of r! is to use a source resistance r, that is much larger than 7. Thus, in Eq. (5-10) if r, is small compared to 7, changes in r/ will change the value of the voltage gain only slightly. Making r, much larger than 1, is called swamping out the emitter diode. Let us summarize the important formulas for CB transistor operation (Fig. 5-20a). For d-c operation we have y, e provided that Vex >> Vax Ve = Vee — Ick For a-c operation we have Ie Your “te Your in Pout TL when 1, > 1% % Te These formulas are very useful approximations for the analysis of CB transistor circuits. In spite of the fact that we used an ideal transistor in100 Transistor Circuit Approximations deriving these formulas, we will find that they are adequate for most initial analysis and design of CB transistor circuits. EXxampe 5-9 Find the approximate voltage gain from source to output for the circuit of Fig. 5-21. Soiution The a-c load resistance r, seen by the collector is 10 kilohms in shunt with 1 megohm. Therefore, rr = 10(10*)||10° = 10 kilohms The theoretical value of ri is = 25 ohms Iz ima y= 25 mv _ 25 mv We can expect the actual r, to be somewhere in the range of 25 to 50 ohms. Note that r, is 1000 ohms and is much larger than the value of r!. -20 +20 bd Hs mS “out Fig. 5-21 Example 5-9. Thus, r, is swamped out because its value is negligible in Eq. (5-10). Using Eq. (5-12), we get Yous _ rp _ 10,000 _ a 7 = 1,000 = 1° If we had used the more accurate formula given by Eq. (5-10) we would have calculated as follows: When 7! = 25 ohms S pee Pout NO When 7, = 50 ohms va = 050 9.52Common-base Approximations 101 The point is simply this: even though 7, can change over a 2:1 range, the use of a large r, has swamped out the effect of changes in rf so that the voltage gain only changes from 9.75 to 9.52 when r, changes from 25 to 50 ohms. This swamping-out technique is widely used in industry whenever it is desired to have an almost fixed value of voltage gain from source to output; temperature changes, aging of the transistor, or for that matter changing the transistor has virtually no effect on voltage gain once r, has been swamped out. In fact, once the emitter diode has been swamped out, there is little point in even using more exact tran- sistor approaches (h parameters, r parameters, ete.), because the voltage gain is then almost independent of the transistor characteristics. Exampte 5-10 In the circuit of Fig. 5-22 find the approximate a-c output voltage Pout: Sotution By inspection of the circuit there is a d-c emitter current of Ine = 0.5 ma 10 30108) ‘The a-c resistance of the emitter diode is in the vicinity of , 25mv _ 25 my goles Oma = 50 ohms Since r, is much larger than r/, (1000 compared to 50), we can use Eq. (5-12) to find the approximate voltage gain from source to output. Yout __ 10(102)||30(10%) _ 7.5(108) Uneaite 10° 10° -10 +20 20k 10k wn 'Omv ~) 30K Yout Fig. 8-22 Example 5-10.102 Transistor Cireuit Approximations Note that the a-c load resistance seen by the collector is 10 kilohms in parallel with 30 kilohms. By the product-over-sum rule we get 7.5 kil- ohms for the value of rz. To find the output voltage, simply realize that the voltage gain from source to output is 7.5, meaning that the source voltage is amplified by a factor of 7.5. That is, Your = 7.5v, = 7.5(10 mv) = 75 mv rms Exampie 5-11 In Fig. 5-23 find the following: (a) The d-c collector-to-ground voltage Ve. (6) The a-c output voltage vou. Soiution (a) The magnitude of the d-c voltage from collector to ground is given by Eq. (5-7) Ve = Veo ~ Ick Vee is the magnitude of the collector-supply voltage, which in this case is 20 volts. The d-c collector current: I¢ equals the d-c emitter current, which by inspection of Fig. 5-23 is simply 20 Ig 20005 = 1lma Since there is about 1 ma of collector current through a d-c load of 10 kilohms, there is a voltage drop of about 10 volts across this d-c load. Therefore, the d-c voltage from the collector to ground is —10 volts. (0) To find the a-c output voltage v4. we first need to find the volt- +20 -20 Se Fig. 5-23 Example 5-11 IM VoutCommon-base Approximations 103 age gain from source to load. Since r, = 0, we can use Eq, (5-11): Your _ TL Yin 7 The a-c load resistance r, is 10 kilohms in shunt with 1 megohm, which is essentially 10 kilohms. The theoretical value of ri is 25 mv _ 25 mv Ts mae 25 ohms n= The practical range of rf is therefore 25 to 50 ohms. Hence, the mini- mum value of voltage gain is Your _ 10(10*) a a 00 ae and the maximum value is out _ 1010) __ 4g tin 25 Hence, the voltage gain of the circuit in Fig. 5-23 is somewhere in the range of 200 to 400. The output voltage is simply the input voltage times the gain. With 1 my rms input, the output voltage must be in the range of 200 to 400 mv rms. EXAMpLe 5-12 In the circuit of Fig. 5-24 find the following: (a) The voltage gain from the source to the output. (b) The voltage gain from the emitter to the output. () The approximate value of tin. +20 -20 Fig. 5-24 Example 5-12.104 ‘Transistor Circuit Approximations Sonution (a) The voltage gain from source to output is approximately oat ay Fz — 10(10*) _ =r > tor ~ 10 (b) The voltage gain from the emitter to the output is wt _ 10(108) _ perro 400 (For convenience, we used the theoretical value of 25 ohms for r/.) (c) To find vin, we can first find ty. and then divide by 400. The output voltage is Vout = 10v, = 10(2 mv) = 20 mv rms Therefore, the input voltage vi, appearing across the emitter diode is _ 20 mv rms _ fee ty = Tg = 0.05 mv rms = 50 wv rms 5-9 Notation for Voltages and Currents In Chap. 4 we indicated that whenever possible we would use capital letters to represent d-c or fixed quantities and lowercase letters for a-c or varying quantities. To this basic rule we now wish to add the following: 1. When a current or voltage in a transistor is a d-c or fixed quantity, we will use capital letters on both the quantity and its subscript. For instance, to represent the d-c emitter current we use Iz (both the quantity I and its subscript E are eapitalized). As another example, to represent the d-c collector-to-base voltage we use Vex. 2. When the current or voltage is an a-c quantity obtained from an a-c equivalent circuit, we will use lowercase letters for both the quantity and its subscript. For instance, the a-c current in the emitter is desig- nated by i, (both the quantity 7 and its subseript e are lowercase letters). As another example, to represent the a-c voltage from collector to base we would use v.. 3. When we wish to represent the total voltage or current, that is, the sum of both the d-c and a-c components, we will use a hybrid notation, with a lowercase letter for the quantity and a capital letter for its sub- script. For instance, to represent the total current into the emitter, we use ig. To represent the total collector-to-base voltage we use ven. Whenever possible, we will follow the above rules of notation; this will make it much easier for us to understand the meaning of various transistor formulas.Common-base Approximations 105 SUMMARY A transistor is like two back-to-back diodes. There are two basic types, the n-p-n and the p-n-p transistor. For normal transistor operation the emitter diode is forward-biased, and the collector diode is back-biased. Because of the thin and lightly doped base region most of the carriers pass from the emitter to the col- lector. As a result, the collector current is almost equal to the emitter current. The small base current is the difference of the emitter and col- lector currents. The IV characteristics of a transistor indicate that the emitter diode acts essentially like a diode, whereas the collector diode acts like a current source whose value equals the emitter current. If the reverse voltage across the collector diode becomes too large, breakdown can occur. The collector diode is normally operated well below the breakdown voltage. The ideal transistor is an approximation of any real transistor. In the ideal transistor we disregard a number of effects in order to obtain a simple model for preliminary transistor cireuit analysis. In analyzing CB circuits we use the concept of the ideal transistor in conjunction with the superposition theorem. In the d-e equivalent circuit, all a-c sources are shorted and all capacitors are opened. The direct currents and voltages can then be found by using the large-signal diode approximations of Chap. 3. In the a-c equivalent circuit, all d-e sources and all coupling capacitors are shorted. ‘The a-c emitter current can then be found by using the various small-signal diode approx- imations of Chap. 4. OF special importance is the fact that the voltage gain from source to output can be made almost independent of the transistor characteris- ties by swamping out the emitter diode. In other words, when the source resistance is much larger than the emitter-diode resistance, the voltage gain from source to output is approximately equal to the a-c load resist~ ance divided by the source resistance. GLOSSARY alpha (a) This a-c quantity is defined as the ratio of a change in collector current to the corresponding change in emitter current. There is also a d-c a, designated age, defined as the total collector current divided by the total emitter current. base spreading resistance (1) This is the bulk or ohmic resistance of that part of the base region in which the base current flows.106 Transistor Circuit Approximations collector capacitance (C.) This is the depletion-layer capacitance of the back-biased collector-base junction. collector resistance (r) ‘This is the resistance that appears across the collector-current source. The value of this resistance is quite high, usually several megohms. common base (CB) One of the ways of connecting a transistor. In a CB connection the base is at a-c ground. coupling capacitor A capacitor whose purpose is to block the d-e com- ponent and pass the a-c component. Coupling capacitors are deliber- ately chosen large enough in size to offer very little reactance to the lowest frequency that is to be passed. emitter resistance (r1) The effective junction resistance of the emitter diode as seen by a small a-c signal. As a theoretical guide, this resistance equals 25 mv divided by the value of the direct eurrent Ig in the emitter diode. swamping With respect to the emitter diode, this means making value of 7! negligible as far as voltage gain is concerned. REVIEW QUESTIONS 1, What are the two kinds of transistors? Name the different parts of a transistor. 2. Under normal conditions, is the emitter diode forward- or back-biased? Is the collector diode forward- or back-biased? 3. When a transistor is operating under normal conditions, the collector current is almost equal to the emitter current. Why is this so? 4, Is the base current large or small compared to the emitter current? How can we find the base current knowing the emitter and collector current? 5. The emitter-base part of a transistor acts essentially like what kind of device? The collector-base part of transistor acts like what kind of device? 6. What is an ideal current source? 7. Define the two types of a. What is the a of an ideal transistor? 8. What is the formula for the theoretical value of the emitter resistance r,? What is the range of rj for most transistors? 9. What does swamping out the emitter diode mean? PROBLEMS 5-1 Ina transistor the collector current is 4.9 ma, and the emitter cur- rent is 5 ma. What is the ag. for this condition?Common-base Approximations 107 When the emitter current is increased to 10 ma, the collector current becomes 9.7 ma. What is the value of aa: for this new condition? 5-2 When the emitter current changes from 2 to 2.25 ma, the collector current changes from 1.95 to 2.195 ma. (a) Find aay for an emitter current of 2 ma. (®) Find a. 5-3 In Fig. 5-25a, find the approximate emitter current. Also, find the voltage that appears from collector to base. 5-4 In Fig, 5-256, find the following: (a) The emitter current and the collector-base voltage when Rz = 40 kilohms. (0) The emitter current and the collector-base voltage when Rx = 20 kilohms. 5-5 In Fig. 5-25b, what value of Rg produces a collector-base voltage of 15 volts? 15K fe 20K (2) 10K fh Fig. 5-25 5-6 In Fig. 5-25c, find the following: (a) The emitter current and the collector-base voltage when Vgzg = 10 volts. (b) The emitter current and the collector-base voltage when Vez = 40 volts. 5-7 In Fig. 5-25c, how much does the collector-base voltage change when Vez is changed from 10 to 11 volts?108 ‘Transistor Circuit Approximations 5-8 In Fig. 5-25d, find the following: (a) The collector-base voltage when R;, = 5 kilohms. (6) The collector-base voltage when Ry, = 10 kilohms. 5-9 What value of Rz in Fig. 5-25d causes the collector-base voltage to equal 7.5 volts? 5-10 Find the theoretical value of emitter resistance using Eq. (5-3) for the following d-c emitter currents: 0.01, 0.05, 0.1, 0.5, and 1 ma. 5-11 Sketch the total voltage vr from collector to ground in Fig. 5-26a. Show the d-e and a-c components. Use Eq. (5-3). -15 +15 ic 20K JOK mv ' ica m) ¥s ‘ (a) +30 -30 ke 15K 75K Imy t -4 kK U) yy IM (6) a =Common-base Approximations 109 5-12 In Fig. 5-26b, sketch the total voltage vy that appears from col- lector to ground. Use Eq. (5-3). 5-13 In Fig. 5-27a, find: (a) The d-c voltage from collector to ground. (0) The a-c voltage vous (approximately). 5-14 If the voltage gain from source to output in Fig. 5-27 is to be approximately 7.5, what size should r, be? 5-15 In Fig. 5-27, what is approximate value of tou? (Neglect rf.) -20 +20 40K 30K 750. 1 1 iT 1 | Inv OY) ms 10K Yout (0) +20 -20 40K 30K 500 5k Inv e@ rms 5K Yout : ~ (6) i Fig. 5-27ne Transistor Cireuit Approximations -20 +20 20K loK Otolk lok Yout (a) +25 -25 i Ik [and 0 to 25K Yout we (o) Fig. 5-28 5-16 In Fig. 5-28a, what is the maximum and minimum value of You that can be obtained by changing r,? Use 25 mv/Tg to find the value of r!. 5-17 In Fig. 5-286, approximately what are the minimum and maximum.Common-emitter Approximations In this chapter we will discuss the common-emitter connection, un- doubtedly the most widely used of the three basic transistor connections. As we did with the CB connection, we will idealize the common-emitter conneetion in order to obtain a first approximation for the behavior of a transistor. This approximation is adequate for most initial analysis and design; furthermore, the first approximation allows us to become familiar with the essential features of the transistor before we attempt a more thorough analysis. 6-1 The IV Characteristics of the Common-emitter Connection A transistor can be connected with the emitter grounded instead of the base, as shown in Fig. 6-1a; this connection is called a common-emitter or grounded-emitter connection. ‘The base-emitter diode is forward-biased, and the collector diode is back-biased. How are the voltages and currents related in a common-emitter (CE) connection? If we were to run a typical experiment in the laboratory, here is what we would find. First, imagine that the collector supply is nm2 ‘Transistor Circuit Approximations adjusted to 1 volt. If we now vary the base supply Vuz, different values of base current iy will flow. Specifically, when we increase Vgzg from 0 volts toward higher positive voltages, the base current will increase slowly at first; after reaching a few tenths of a volt, however, the base current will increase sharply, as shown in Fig. 6-16. Note how this curve (solid line) rises slowly until we reach the knee voltage and then turns up steeply beyond the knee voltage; this is the usual characteristic for a diode, and we certainly expect this kind of graph because the base-emitter part of a transistor is essentially a diode. As before, the knee voltages are around 0.3 volt for a germanium transistor and 0.7 volt for a silicon transistor. ip Yee? A ft tze"10 h % % toe 0) Fig. 6-1 (a) Obtaining CE curves; (6) base-diode curves. The collector voltage docs have some influence on the shape of the base characteristic. Assume that we change Vee to 10 volts. When we return to the base circuit, we find that varying Vre from 0 volts toward higher voltages now results in a different base characteristic, as shown by the dashed curve of Fig. 6-10. The idea here is simply that there is a small amount of feedback from the collector to the base. For a first approximation, however, we can disregard the small gap between the two curves of Fig, 6-1b and say that the base-emitter part of a transistor acts like a diode. We will refer to the base-emitter diode simply as the base diode. To find the collector characteristics, we can fix the base current at some value and then measure the collector current for different collector voltages. Specifically, in Fig. 6-la imagine that we adjust and hold the base current at exactly 0.01 ma. If we now vary the collector voltage, different values of collector current occur. Figure 6-2a shows a typical collector curve. Note that when we inerease the collector-emitter volt- age veg from 0 volts upward, the collector current ic increases sharplyCommon-emitter Approximations us at first; however, after vgz reaches a few tenths of a volt, the collector current levels off, becoming almost constant. When we continue to in- crease veg, the collector current i¢ remains almost fixed until a break- down occurs. (This breakdown is the usual avalanche effect that takes place at higher reverse voltages across a diode.) Normally, the transistor is operated well below the breakdown point. In Fig. 6-2a the collector current above the first knee is about 1 ma, compared to a base current of 0.01 ma. Thus, the collector current is about 100 times larger than the base current. Of course, we are hypothe- sizing a typical experiment. In practice, we find that the collector current may be anywhere in the range of 20 to 200 times larger than the base current. The important point here is that the collector current is much larger than the base current; this is related to the fact that almost all the emitter carriers reach the collector, and only a few of the emitter carriers actually reach the external base terminal. be Breakdown region ie ig70.02ma “ce (ce) Fig. 6-2 Collector curves. Suppose that we change the base current to a new value like 0.02 ma, In the collector circuit we will find that as we increase the collector volt- age from 0 volts upward, the collector current will increase very sharply at first but then will reach an almost constant value, as shown in Fig. 6-2b. Once again, if we increase the collector voltage too much, we find a breakdown point, where the collector current suddenly inereases. Above the first knee the collector current remains almost constant up to the breakdown voltage. Note that the collector current in Fig. 6-2b is around 2 ma, compared to a base current of 0.02 ma.14 Transistor Circuit Approximations For each new value of base current that we use, we will get a new collector curve. When we plot several collector curves on a single set of axes, we get the typical CE characteristics shown in Fig. 6-2c. (We have omitted the breakdown region for convenience.) Note that even when there is zero base current, there is a small amount of collector current caused by the reverse current in the collector diode. Note also that each time we change the base current, there is a new value of collector current that is significantly larger than the base current. Here are the most important features of Figs. 6-1b and 6-2c: 1. The base diode acts almost like an ordinary diode. 2. Above the knee and below breakdown the collector current is almost constant and is controlled by the base current. 6-2 The Beta of a Transistor In analyzing the common-emitter connection we will be using d-c and a-e equivalent circuits. One of the important quantities that we will use in the d-c equivalent circuit of a transistor is the d-c beta, written Bae. This is defined as (6-1) where Bu. is the d-c 8 of a tran’ Tc is the d-c collector current Iz is the d-c base current When we analyze the a-c operation of transistor cireuits, we will need a quantity called the a-c 8, written simply as 8. This is defined as for a fixed veg 7 (6-2) 4 where @ is the a-c 6 of a transistor 4. is the small-signal a-c collector current 4, is the small-signal a-e base current As an example of using Eqs. (6-1) and (6-2), consider the circuit shown in Fig. 6-3a, In the base circuit there is a d-e and a-c source. The d-c source sets up a d-e base current Ip. This d-e base current in turn produces a d-c collector current Zc. In addition to these d-c com- ponents, there are also a-c components. There is an alternating current in the base of %, which in turn produces an alternating current in the collector of i.. For the sake of illustration, suppose that the total currents in the base and the collector are the waveforms shown in Fig, 6-3b and c.Common-emitter Approximations fhe. us eee 6 Sauna a \, oe 3e¥e 2.09ma a ee as 2mo oo (ee cL igh mo es € / Zp ai : as ke, , te a Ae @ yi te) 6s!” ae we Sor ee A* Fig. 6-3 D-c and a-c B. FF By inspection of Fig. 6-3b the d-c component of base current I is 0.02 ma. The a-c component #, which is superimposed on this d-c level, is a sine wave with a peak value of 0.001 ma. By inspection of Fig, 6-3c the d-c component of collector current Ie is 2 ma; the a-c component i, is a sine wave with a peak value of 0.09 ma. Now we can calculate the two kinds of 6 as follows: to 2ma _ Bac = T, = 9.02 ma ~ 100 and i 0.09 ma _ -% 7 0001 ma ~ In general, the d-c and a-c f’s are not exactly equal; however, they usually are close to each other in value, and at times we will treat these two 6’s as equal. Typically, the 4’s of transistors show a wide variation from about 20 to 200 or more. The f is a very unstable quantity; it changes with temperature, with the d-c operating point, and from one transistor to another. 6-3 The Ideal CE Transistor For our first approximation we will climinate all but the most essential features of the CE connection. For instance, in the base diode we know6 Transistor Cireuit Approximations that there is some feedback from collector to base as illustrated by the gap between the curves of Fig. 6-1b. We will disregard this feedback and treat the base diode as an ordinary semiconductor diode with the characteristics shown in Fig. 6-4a. As far as the collector diode is con- cerned we will: 1. Disregard the breakdown phenomenon. 2. Disregard the small collector current that occurs for ig = 0 (see Fig. 6-2c). 3. Eliminate the slight upward slope in the collector curves. 4. Reduce the knee voltages to zero. 5. Make the d-c and a-c §’s equal, that is, Ba, — 8. ‘a e — All collector voltages Y % % "aE cE (a) (0) Fig. 6-4 Ideal characteristic curves. Figure 6-4b summarizes these various approximations; for any value of base current, the collector current is 8 times larger. In other words, as long as the collector diode is back-biased, the collector current is controlled by the base current and is 8 times larger than the base current. This control that the base current has over the collector current holds for both direct and alternating currents. For instance, if the d-c and a-c 6’s equal 100, then 1 ma of d-c base current will produce 100 ma of d-e collector current; further, 0.02 ma of a-c base current will produce 2 ma of acc collector current. We will call this first approximation the ideal CE transistor. EXxampie 6-1 In Fig. 6-5, the ideal CE transistor has a 8 of 100. Find the d-c collector current Ic and the d-c collector-ground voltage Vc. Sonution First, find the d-c base current I.Common-emitter Approximations uz Fig. 6-5 Examples 6-1 and 6-2. We neglected Vz because it is only around 0.3 or 0.7 volt, depending upon the material used. Next, find the collector current, which is 8 times the base current. Te = 61y = 100(10 wa) = 1000 wa — 1 ma The collector-ground voltage is simply the supply voltage minus the voltage drop across the 5-kilohm load resistor. Thus, Vo = 10 — Ich, = 10 — 10-*(5)(10") = 5 volts Exampie 6-2 If the transistor of Fig. 6-5 is made of silicon, find I¢ and Ve by allowing for the base-diode voltage drop. Use a 8 of 100. SonutIon 10 — 0.7 Te = Joe = 9.3 wo Tc = BIn = 100(9.3 na) = 930 pa = 0.93 ma Vo = 10 — 0.93(10-8)(5)(108) = 5.35 volts it EXxampie 6-3 In Fig. 6-6 the ideal transistor has a 8 of 50. What value of Hz will produce a collector-ground voltage of 10 volts? SoLurtion In order for Vc to equal 10 volts, there must be a voltage drop of 10 volts across the 10-kilohm load resistor; the collector current, ‘| 5 Fig. 6-6 Examples 6-3 and 6-4. iad e)eayus ‘Transistor Circuit Approximations therefore, must be 1 ma. For a 8 of 50 we compute a base current of _ Tc _1ma Pea igi oo We now must find the correct size of Rp that will produce a base current of 20 ya. Using Ohm’s law, we get oo ~ 2000-8) ~ = 20 wa Re 1 megohm We disregard the small voltage drop across the base diode; therefore, all 20 volts appears across the Re resistor. EXxAmpPie 6-4 Suppose that the transistor of Fig. 6-6 is replaced by another tran- sistor with a 8 of 75. If Re = 1 megohm, what will the new value of Veer be? SonvTion The base current is still 20 Ts = 766 = 20 na The collector current for the new transistor is different. Ic = Ble = 75(20 pa) = 1.5 ma With this new value of collector current the voltage from collector to ground becomes Ve = 20 — 1.5(10-%)(10)(10*) = 5 volts 6-4 Base Bias of a Transistor We already know that for normal operation the emitter diode is forward- biased and the collector diode is back-biased. When an a-c signal is in- jected into a transistor, there is the possibility that the signal may swing the emitter diode into the back-biased condition or perhaps swing the collector diode into the forward-biased condition. In either case, normal transistor action is lost, and clipping of the a-c signal results (see Chap. 8). In order to prevent this possibility, it is common practice to set up a suitable d-c operating point in collector current and voltage; the a-c sig- nal then causes excursions from this operating point, and as long as the a-c signal is not too large, the emitter diode remains forward-biased andCommon-emitter Approximations 9 the collector diode remains back-biased throughout the entire a-c eycle. Setting up this d-e operating point in collector current and voltage is commonly called biasing the transistor. In this section we discuss one of the ways to bias a transistor in the CE connection. Figure 6-7 illustrates what we will call base bias. Specific examples of this type of bias were examined in Examples 6-1 to 6-4. Now we analyze the general circuit of Fig. 6-7 to determine the bias formulas. ‘As shown in Fig. 6-7, the collector-emitter voltage Vor is simply desig- nated V¢ and the base-emitter voltage Vez is simply written as Vz. We already know that Vc is equal to the power-supply voltage minus the voltage drop across the load resistor Rz. That is, Vo = Vee — IcRx (6-3) where Ve is the d-e voltage from collector to ground Vce is the collector-supply voltage Ic is the d-c collector current R,, is the d-c load resistance seen from the collector Mec Fig. 6-7 _ Base bias. The d-c collector current, of course, equals the 8 of the transistor times the base current. That is, Ic = Biz where the d-e 8 will be used if there is a difference between the two 8 values. By inspection of Fig. 6-7 we can see that the base current Iz is Veo — Va Re or for those circuits where the drop of a few tenths volt across the base diode is negligible, we have In= wee when Veo >> Va (6-4)120 Transistor Circuit Approximations ‘The base-biased circuit of Fig, 6-7 is one of the simplest ways to bias a transistor. We set the base current Ip by choosing an appropriate size for Rp. This base current controls the size of the collector current I¢, which in conjunction with Rr determines the collector voltage Vc. For instance, in Fig. 6-8a, the base current is Vee _ 30 In =z Rs = qos = 30 va The £ is given as 50, so that we immediately can calculate a collector current of Io = Bl = 50(30 ya) = 1.5 ma This collector current flows through an Rz of 10 kilohms and produces a voltage drop of 15 volts. Therefore, the collector-ground voltage is Ve = Veo — IcRk1 = 30 — 15 = 15 volts Thus, the d-e operating point is set at Ic = 1.5 ma and Vo = 15 volts. When an a-c signal is coupled into the stage, it causes excursions about. this operating point, As long as the a-c signal is not: too large, the emitter diode remains forward-biased, and the collector diode remains back- biased throughout the entire a-c cycle. The base-biased cireuit of Fig. 6-7 is actually the worst possible way to bias a transistor from the standpoint of a stable operating point. The reason that the base-biased circuit is so poor can be understood by re- ferring again to Fig. 6-8a. We have found that for this circuit with the given f of 50 we have a d-c operating point of Ic = 1.5 ma Ve — 15 volts Fig. 6-8 Effect of 8 change in a base-biased circuit.Common-emitter Approximations 121 Suppose that for some reason or other the 8 of the transistor changes from 50 to 100 (a change of this size can occur when replacing transis- tors or when the ambient temperature changes over a large range). The new circuit is shown in Fig. 6-8b. In this circuit the base current is still the same. Vee 30 In = Rr = ji = 30 va But the collector current is now Te = 100(30 ya) = 3 ma This 3 ma of collector produces a collector voltage of Vo = Veo — Icky, = 30 — 3(10-4)(10)(10%) = 0 volts Therefore, the collector diode is no longer back-biased. If an a-c signal were coupled into the base, the signal at the collector would be clipped off on the negative-going half cycle. (Clipping is discussed further in Chap. 8.) Let us summarize base bias: 1. The base current is essentially fixed by the value of Vec and Rp. 2. The collector current is determined by Ip and 8. Because f varies widely from one transistor to another and with temperature change, the collector current also varies widely with these changes. 3. Base bias is the worst way to bias a transistor from the standpoint of stability of the operating point. 6-5 Emitter Bias of a Transistor Now that we have seen the worst way to bias a transistor, let us look at one of the best ways. The circuit of Fig. 6-9a illustrates what we will call emitter bias. The same circuit drawn in a more practical form is shown in Fig, 6-96. The emitter-biased circuit of Fig. 6-9 is quite popular and is widely used whenever two power supplies (positive and negative) are available. Its popularity stems from the fact that the collector current is essentially independent of the 8. In fact, we will show that the collector current is vy, lo To prove this result we first write the Kirchhoff voltage equation around the loop containing Rz and Rp. If we start at the emitter and proceed in a clockwise direction, we get IpRe — Vex + Iske + Vaz =0oo EF OL wy aur ve 122 \ Transistor Circuit Approximations “ee By transposing Vx and Vaz we obtain IpRs + InRs = Vex — Vaz (6-5) To simplify this expression recall that the emitter current is approxi- mately equal to the collector current. Hence, 8 equals ale Tn Ts or toa If we substitute this expression for Zz into Eq. (6-5), we obtain Ins + tke = Vex — Vor u vas Next, we factor Iz to get le ! u gy \ te(Re + 2) = va -(For ) i Finally, we can divide to obtain. _ Vex — Vox ** Rs + Rs/8 Equation (6-6) is almost an exact expression for the d-c emitter current. A simpler and more useful expression for Iz can be obtained by realizing I (6-6)Common-emitter Approximations 123 that in most circuits Vex is negligible compared to Vz (a few tenths of a volt compared to many volts). Also, in most circuits it is possible to choose Rs so that Rz/8 is negligible compared to Rz. Therefore, an approximate expression for Jz that applies to most practical circuits is (6-7) provided that — (the usual case) Vexr>Ve and a Vins Vs(bo Me Rze>— (the usual case) ~ 8 Oy FR 3 Fete Equation (6-7) tells us that the d-c emitter current in the emitter- biased circuit of Fig. 6-9 is essentially equal to the emitter-supply volt- age Vg divided by the emitter resistance Rg. For instance, in Fig. 6-9 if Vex = 10 volts and Rr = 10 kilohms, the emitter current Is is approxi- mately 1 ma, and, of course, the collector current will be essentially 1 ma. Note that Eq. (6-7) does not contain 8; in other words, the emitter current is independent of 8. Herein lies the tremendous advantage of emitter bias over base bias; even though the @ varies with temperature or transistor change, the amount of emitter current in the emitter-biased circuit re- mains essentially fixed and equal to Vre/Rs. ‘A convenient way to remember emitter bias is to realize that it is actually a form of the CB connection, at least as far as the d-c equivalent circuit is concerned. Within the restrictions of Eq. (6-7) the emitter- biased circuit of Fig. 6-10a acts essentially the same as the CB circuit of Fig. 6-10b; the circuit of Fig. 6-10a has simply been redrawn, and Rs has OMe" nom Me ek 4 (a) ae Fig. 6-10 Simplified viewpoint of emitter bias. Ne V oe Vee #8, =e es “Vee 7 Bebe M57 a812 Transistor Circuit Approximations been shown as negligibly small. Thus, when we encounter the emitter- biased circuit of Fig. 6-10a, we can think of it as being like the CB circuit of Fig. 6-10b as far as its d-c equivalent circuit is concerned. The advantage of this viewpoint is that we already know from our earlier studies of the CB circuit that Tyme VE which is identical to Eq. (6-7). ‘The other formula of interest here is for the collector-ground voltage Ve. This voltage is equal to the collector supply minus the voltage drop across the load resistor. That is, Vo = Veo — IcRt (6-8) Equations (6-7) and (6-8) are very useful in both analysis and design. Basically, they give us the d-c operating point of an emitter-biased tran- sistor. These equations are summarized in Fig. 6-11. When a p-n-p transistor is used, the emitter-bias arrangement is quite similar, and the biasing formulas are identical. Figure 6-116 summarizes Me lee fe aS (o aSte a Me ek Re eRe We= oleh = We = Vogl “Yee + lee (a) (6) Fig. 6-11 Emitter-bias circuits and formulas. the biasing of a p-n-p transistor. Note that the d-c voltage from the col- lector to ground is negative and its magnitude is given by Eq. (6-8). Note also that in a p-n-p circuit all d-c voltages and currents are in opposite directions from a similar n-p-n circuit. Therefore, all the d-c formulas that we develop for the n-p-n circuit also apply to the p-n-p circuit as far as magnitudes are concerned.Common-emitter Approximations 125 - EXaMpie 6-5 For the circuit of Fig. 6-12 the germanium transistor has a 6 of 50. Find: (a) The approximate value of Iz. (b) A more exact value of Iz by using Eq. (6-6). Sonurion (a) We know that in an emitter-biased circuit almost all the emitter- supply voltage appears across the emitter resistor Rg. Hence, the emitter current is Ver 20 T= = 300103 = ima (®) When we use the more exact formula, Eq. (6-6), we get ee * 20(108) ¥ 10(10%) 750 ~ Te Note that the more exact answer of 0.975 ma is only 2.5 percent lower than the approximate answer of 1 ma. A small error of this size occurs in any well-designed emitter-biased circuit. B=50 Fig. 6-12 Examples 6-5 and 6-6. paninen EXAMPLE 6-6 Suppose that we change transistors in the circuit of Fig. 6-12 and that the new 8 is 100. Find: (a) The approximate value of Zs. (b) The value of I by using Eq. (6-6).126 ‘Transistor Circuit Approximations Sonution Once again, we note that for an emitter-biased circuit essentially all the emitter-supply voltage is dropped across the emitter resistor. Hence, faa 20 ea Te = Be = Day = 1 When we use Eq. (6-6), we get 20 — 0.3 19.7 Tz = 0.98 ma * 20010) + 10(108)/100 ~ 20.1010) Note carefully that even though # has changed from 50 to 100, the emit- ter current has changed only from 0.975 ma (Example 6-5) to 0.98 ma. ‘This always occurs in a well-designed circuit because the designer makes sure that R»/é is much smaller than Rg for the typical 6 range of the transistor being used. Examp.e 6-7 In troubleshooting circuits one of the first measurements that should be made is to check the d-c collector-ground voltage Ve. In trouble- shooting the circuit of Fig. 6-13a, what should the approximate value of Ve be? Souution First, note that as far as the d-c operation of the circuit is concerned, all capacitors are open circuits. Hence, the circuit of Fig. 6-130 has Fig. 6-13 Example 6-7.Common-emitter Approximations 127 the d-c equivalent cireuit shown in Fig, 6-13b. It is now obvious that the emitter current is Ve 30 | = Re = 30c0y ~ 1 ™ Next, we can see that the collector voltage should equal the collector supply voltage (30 volts) minus the voltage drop across the load resistor. Thus, Ve = Vee — IeRr = 30 — 10-*(104) = 20 volts Iz In troubleshooting a circuit like that shown in Fig. 6-13a the first important measurement to make is to check that the collector voltage is truly around 20 volts. This simple test verifies that three resistors, two power supplies, and one transistor are working properly as far as d-c voltages and currents are concerned. 6-6 Analyzing CE Transistor Circuits In analyzing the complete operation of CE circuits driven by large d-c sources and small a-c sources we again will be using a modified form of the superposition theorem. We use the d-c equivalent circuit to find the d-c operating point of the transistor. We can then use the a-c equiv- alent circuit to find how the a-e signal is amplified or modified by the transistor circuit. In the d-e equivalent cireuit we will view the ideal transistor as a rectifier diode in the base circuit and a current source in the collector circuit. Figure 6-14 illustrates the d-c equivalent: eirenit. (Far the p-n-p transistor we reverse the diode and the current source.) In the a-c equivalent circuit we think of the ideal transistor as a resistance in the base and a current source in the collector, as shown in Fig. 6-15. Note that the a-c resistance looking into the base is Br’, where r! is theoretically given by 25 mv/Ig. The concept of a-c resistance looking into the base is quite similar to that developed for the CB con- nection in Chap. 5. Recall that the d-c sources establish the d-c operating —t Sy a = ——_ Ideal ¥ Bla Fig. 6-14 Ideal d-c equivalent circuit of an n-p-n CE transistor.128 ‘Transistor Circuit Approximations ic —_% bo bo 7 Bip Br deol Fig. 6-15 Ideal a-c equivalent circuit of an n-p-n CE transistor. point on both the collector and base characteristics; the a-c signal then causes small excursions from the operating point. For instance, in Fig. 6-16 the operating point on the base characteristic is shown at point Q. Tn is the d-c base current at point Q. When an a-c signal is applied to the base diode, there are excursions from the operating point. For a typical ac signal the excursions might take place between points A and B. If the a-e signal is small, the changes in voltage and current are almost directly proportional because only a small part of the curve is being used. Because the changes are almost linear, the base diode looks like an ordi- nary resistor to the a-c signal. ‘The value of resistance seen by the a-c signal looking into the base shows a wide variation from one transistor to another. As a theoretical approximation, this a-c resistance is te = 25 my 7 Tn (6-9) where I, is the value of the d-c base current. Thus, when Ip = 10 ya, the arc resistance is around _ 25 mv re = TO ya = 2.5 kilohms Equation (6-9) should be easy to remember because it is analogous to Eq. (5-3). In Eq. (5-3) we saw that the a-c resistance looking into the 8 d-c operating Fig. 6-16 A-c resistance of the base Point diode.Common-emitter Approximations 129 emitter is 25 mv/Iz. In Eq. (6-9) we see that the a-c resistance looking into the base is 25 mv/Tn. Both of these equations can be derived by apply- ing calculus to an exponential equation relating current and voltage across a p-n junction; we will accept these results without derivation. Equation (6-9) can be expressed in a more useful form. We can rearrange it as follows: my _ 25 mv Is Ic/8 or (6-10) The significance of this result is D ; it says that the a-c resistance looking into the base of a transistor is 6 times larger than the a-c resistance looking into the emitter of a transistor. This makes sense because we know that the base current is smaller than the emitter current by a factor of 8; therefore, we should expect the resistance looking into the base to be Jarger by a factor of 8. The equivalent circuits of Figs. 6-14 and 6-15 should be memorized because they enable us to analyze CE emitter circuits quickly and easily. Of course, these equivalent circuits are for the ideal transistor; neverthe- less, as pointed out before, the ideal-transistor approach is adequate for many of our needs in transistor circuit analysis and design. Examp.e 6-8 Sketch the waveform of the total voltage vc appearing from collector to ground in the circuit of Fig. 6-17a. Use a 8 of 50. Sonurion First, draw or visualize the d-c equivalent circuit as shown in Fig. 6-17b. In this circuit it is immediately clear that the d-c base current is Veco _ 20 _ In ee = Tye = 20 no The d-c collector current is 8 times larger than the base current. Ie = 6In = 50(20 wa) = 1 ma The d-c collector-ground voltage is simply the collector-supply voltage minus the voltage drop across the 10-kilohm load resistor. Ve = Vee — Ick, = 20 — 10-*(104) = 10 volts Next, we analyze a-c operation by means of the a-c equivalent cir- cuit, which is shown in Fig, 6-17c. All d-c sources have been shorted; all coupling capacitors are assumed to have a low enough reactance to130 ‘Transistor Cireuit Approximations +20 +20 ideo! Bre {e) (a) Fig. 6-17 Example 6-8. act like short circuits as far as the a-c signal is concerned. Looking into the base, the a-c signal sees a resistance of Br, = p> BY ~ 59 25 MY _ 1950 ohms - ma The source voltage is a I-mv-peak sine wave. This voltage appears directly across the base diode and produces a sine wave of base current, with a peak value of 0.001 = 959 — 08 va Topeak) = B The a-c collector current is 6 times the base current; the collector current, therefore, is a sine wave with a peak value of tetpeak) = Bid(penky = 50(0.8 wa) = 40 pa In Fig. 6-17c it is clear that this a-c collector current flows through the 10-kilohm resistor and develops a sine-wave voltage with a peak value of Yeqpeak) = 40(10-8)(104) = 0.4 voltCommon-emitter Approximations 1BL Now that we have both the d-c and a-c components, we ean super- pose them and obtain the total waveform shown in Fig. 6-17d. Note two points about this waveform: 1. The a-c component is an amplified version of the source voltage. The voltage gain is the output voltage of 0.4 volt divided by the source voltage of 1 mv. Thus, the voltage gain is 400. 2. The output a-c signal is 180° out of phase with the source signal. The reason for this phase inversion is easily understood by referring to Fig. 6-17a. During the positive half cycle of source voltage, the a-c base current will aid the d-c base current, which means that the total base current is increased. This implies that the total collector current is inereased, which in turn means a larger voltage drop across the load resistor. Therefore, the total collector voltage (which equals the supply voltage minus the drop across the load) must be smaller during the positive half eyele of source voltage. Thus, we have established that the total collector voltage is decreasing during the positive half cycle of source voltage, which immediately implies phase inversion. (For the reader familiar with vacuum-tube circuits, the CE transistor stage inverts the signal in the same way that a common-cathode tube connec- tion inverts the signal at its grid.) Exampie 6-9 The cireuit of Fig. 6-18a is driven by a 1-mv-rms signal. If the transistor has a 6 of 100, what is the rms value of the output voltage? Sotution When we visualize the d-c equivalent circui it is clear that the d-c component is blocked by the coupling capacitor on the output side. In other words, only the a-c component reaches the final 10-kilohm load. We still must find the d-c emitter current because we will use it to find the a-c resistance of the base diode. It is clear that the circuit of Fig. 6-18a is emitter-biased. Using Eq. (6-7), we find that Vex __ 20 Re 30(10) Ig= =1ma The a-c equivalent circuit is shown in Fig. 6-18. Especially note that the emitier is at a-c ground because the bypass capacitor looks like a short to the a-c emitter current. The a-c resistance looking into the base diode is simply 25 mv t= vo 25 = 251 e Br, 100 7." = 2500 ohms132 Transistor Cireuit Approximations Ale Yout (a) Be (5) Fig. 6-18 Example 6-9. In Fig. 6-18, note that the a-c source voltage appears across the base diode; therefore, the a-c base current has an rms value of _ % _ 0.001 = be = BB9 7 OF we The a-c collector current has an rms value of it, = Bly = 100(0.4 wa) = 40 pa Finally, we see that as far as the a-e signal is concerned, the 10-kilohm resistors appear in parallel, and therefore the a-c load resistance is 5 kilohms. Thus, the output voltage has an rms value of Yue = 40(10-%) (5) (108) = 200 mv Note that the voltage gain from source to output is Pout _ 200 mv _ ong v, lmv 6-7 The Voltage Gain of an Emitter-biased Stage In this section we will find the voltage-gain formula for the emitter biased circuit shown in Fig. 6-19a. This circuit is one of the widely used types of transistor amplifier, and therefore it will be worthwhile to have a formula for the voltage gain from input to output, that is, a formula for vous/tin. Note that the transistor amplifier of Fig. 6-19a is analogous to the vacuum-tube amplifier of Fig. 6-196. In the vacuum-tube circuit, biasCommon-emitter Approximations 133 + Lop Vee (a) (s) Fig. 6-19 (a) Common-emitter amplifier; (6) common-cathode amplifier. voltage is developed across the cathode resistor Rx. Many readers have studied vacuum-tube circuits and recall that the purpose of the capacitor Cx is to prevent excessive degeneration. In a similar way, the emitter of the transistor is bypassed by ground by means of Cz. The purpose of Cr is similar to that of Cx in a vacuum-tube circuit: it prevents excessive degeneration from taking place. How can we find the formula for the voltage gain of the tran amplifier of Fig. 6-19a? First, we draw the a-c equivalent circuit of 6-20. We are, as usual, assuming that the amplifier is operating in normal frequency range, where all capacitors are large enough to act like short circuits to the a-e signal. With the idcal-transistor approximation the base diode looks like a resistance of Ar’. Also, in the collector cireuit the parallel combination of R,, and R is simply a net resistance of rz. To find the voltage gain we proceed as follows. First, the a-c output Aig “FAR Be Fig. 6-20 A-c equivalent circuit of CE amplifier.134 ‘Transistor Circuit Approximations voltage is Your = Biern The a-c input voltage is = irl Therefore, the voltage gain from input to output is simply Your _ Bier, Yin Br or (6-11) where r;, is the a-c load resistance seen by the collector and r! is the a-c resistance of the emitter diode. Note carefully that the voltage gain of the ideal CE stage shown in Fig. 6-19a is exactly the same as the voltage gain from the emitter to the collector for a CB stage, Eq. (5-11). Equation (6-11) is very useful in analyzing emitter-biased amplifiers. As usual, rf has a theoretical value of 25 mv/Tz, which ean be used in preliminary analysis. If desired, we can allow for the typical range in rf by using 25 mv — , _ 50 mv Te Te Exampie 6-10 In the circuit of Fig. 6-21 assume that all capacitors are a-c shorts. For a 8 of 100, find: (a) The voltage gain from the base to the output. (0) ‘The output voltage for an input voltage of 1 mv rms. So.ution (a) The a-c load resistance in the collector circuit is 10 kilohms in parallel with 10 kilohms, which means that 7, = 5 kilohms. To find rf we can use r= Zo mv 25 mV _ 95 ohms Te ima (It should be apparent that the approximate value of d-c emitter eur- rent is about 1 ma for this emitter-biased circuit. If not, review Sec. 6-5.) Therefore, the voltage gain isCommon-emitter Approximations 135 Fig. 6-21 Examples 6-10 to 6-12. If we wish to allow for the range in r!, we can observe that 7 will be between 25 and 50 ohms; the voltage gain, therefore, will be between 100 and 200. (6) To find the output voltage for an input voltage of 1 mv, we need only multiply by the voltage gain. Thus, the rms output is Yout = 200 v%, = 200(1 mv) = 200 my when rj is 25 ohms. At the other extreme, Yous = 100(1 my) = 100 my when 7, is 50 ohms. EXamp.e 6-11 In the preceding example, what would happen to the voltage gain if the final 10-kilohm output resistor were changed to 30 kilohms? SoLution ‘The only change here is in the value of the a-c load resistance rz. This resistance becomes 10 kilohms in parallel with 30 kilohms, which equals 7.5 kilohms (by the usual product-over-sum method). Thus, for an rf of 25 ohms the voltage gain becomes Exampie 6-12 In the circuit of Fig. 6-21, what happens to the voltage gain when the 20-Kilohm emitter resistor is changed to a 40-kilohm resistor?136 ‘Transistor Cireuit Approximations Sotution First, notice that the d-c emitter current will change to ze _ 20 ~~ 40(108) With 0.5 ma of d-c emitter current the value of rf lies in the range of 50 to 100 ohms. Therefore, the voltage gain lies in the range of 50 to 100. = 0.5 ma 6-8 Effects of Source Resistance In this section we want to find out what effect the source resistance 7, has on the voltage gain of a typical amplifier like that of Fig. 6-22a. In this circuit it should be immediately clear that some of the source voltage will be lost across r.; only part of the source voltage v, actually appears across the base diode of the transistor. To find out how much effect r, has, we first draw the a-c equivalent cireuit for the input circuit, as shown in Fig, 6-22b. The input voltage via Mec ‘8 6 @» Fg Bre vin es RellBre via hewn (2) (ec) Fig. 6-22 Loss of a-c signal across the source resistance.Common-emitter Approximations 137 is the voltage that appears across the shunt combination of Rp and Br’, These two resistors can be lumped into a single resistor 2p||6rt, as illus- trated in Fig. 6-22¢. The circuit of Fig. 6-22c is obviously a voltage divider, 50 that vin is simply _ Rall, re + (RallBr) * Yin (6-12) Note in Eq. (6-12) that when r, is very small compared to Rz|lBr!, the expression becomes dine. form K (Rallfr’) (6-13) In other words, when the source resistance 7, is very small, almost no voltage is dropped across it; almost all the source voltage appears across the base diode. In practice, we can encounter either situation. An amplifier may be driven from a very low-impedance source, in which case we use Eq. (6-13). On the other hand, if the source resistance r, is not negligible, we must use Eq. (6-12) to determine how much of the source voltage actually appears across the base diode As far as the voltage gain from base to output is concerned, this is still given by Your _ Th Yin Thus, given the source voltage v,, we first find out how much of this voltage appears across the base diode. Then, to find the output voltage we simply multiply this voltage vi. by the voltage gain. Exampie 6-13 In Fig. 6-23, the transistor has a 8 of 50. Find the output voltage. Soiution We first must find how much of the 1-mv source voltage actually appears across the base diode. To do this, we must know the value of rl, Using 25 mv/Ix for r', we get p25MY _ 5 25 mv Te 0.5 ma Br, 2500 ohms Next, we obtain the parallel resistance of Ry and Bri. Ral|8ri = 10,000)|2500 = 2000 ohms‘Transistor Circuit Approximations 138 30K Yout An] Imw rms == bypass Fig. 6-23 Example 6-13. Since 50 ohms of source resistance is much smaller than 2000 ohms, we can conclude that almost all the 1-mv source voltage appears across the base diode. The voltage gain is a4 " Your _ Tz _ 10,000//30,000 _ Te t= ag 180 Vin And the output voltage is Your = 150(1 mv) = 150 mv EXAMPLE 6-14 Suppose that the source resistance of the preceding example is changed from 50 ohms to 3 kilohms. Find the new output voltage. SoLution The voltage gain from base to output remains the same because rz/r, is unchanged. What does change is the amount of voltage actu- ally appearing across the base diode. We already know from Example 6-13 that the parallel combination of Rp and Br! is 2000 ohms. Hence, by using Eq. (6-12) we ean find the amount cf voltage that actually reaches the base diode. ____Rallért _ 2000 o Pe = TF (alia) ~ 3000 + 2000 | MY = 0-4 mv Thus, the output voltage is 150 times the actual base voltage. Your = 150(0.4 mv) = 60 mvCommon-emitter Approximations 139 6-9 Tube-to-transistor Transformations Many of us already have a knowledge of vacuum-tube circuits; it seems worthwhile, therefore, to indicate that many of the formulas for vacuum- tube cireuits can be easily transformed into formulas for analogous transistor circuits. In this section we show that a transistor has a gm, a, and an ry and that the gain formulas for various tube cireuits apply to those transistor circuits which are the direct counterparts of the tube cireuits. In Fig. 6-24a we have shown the a-c equivalent circuit of a vacuum tube (biasing not shown). There is an a-c plate current iow and an a-c grid voltage of vj. Recall that the gm of a tube is defined as a Yin for zero load resistance gn = Basically, the g», is the ratio of output current to input voltage; it tells us how effective changes in the grid voltage are in changing the output current. a i Va ; ( : 7 (0) Fig. 6-24 Tube-transistor analogy. Tout In Fig. 6-24b, we have shown the a-c equivalent circuit for a transistor. The output current is 6%. The input voltage at the base is Yin = Br, Therefore, the g», of a transistor is = fut Oe 9m = yer or aE (6-14) "140 ‘Transistor Circuit Approximations Next, recall that the r, of tube is the dynamic, or arc, resistance looking back into the plate. With respect to the tube curves, r, is the inverse slope of the plate characteristics, The collector characteristics of a transistor are quite similar to pentode characteristics. For the ideal transistor, the IV characteristics are hori- zontal; this means that the rp is infinite for an ideal transistor. Finally, recall that the » equals the product of gm and rp. Because rp is infinite, the w of an ideal transistor is infinite. What we can do with these results is simply the following: whenever we encounter a transistor circuit whose a-c configuration is the counter- part of a well-known tube circuit, we can transform the tube formula by replacing gm by 1/r! and by replacing r, and u by ©. That is, we trans- form the tube formula into a transistor formula by using the following transformations: 2 Im ty © ad As an example, consider the circuit of Fig. 6-25a. This is a well-known vacuum-tube amplifier and has a voltage gain of Yout Vin Gut L for pentodes where rz, is the a-c load resistance seen by the plate. In this case, rz = RilR + Epo Vee Ry A 8 fe Ideal 2 ir Yin wy wa vin Fo ee = = = = = = tee (a) (o) Fig. 6-25 Obtaining transistor gain formula from analogous vacuum-tube circuit.Common-emitter Approximations 141 The transistor cireuit of Fig. 6-25) has the same a-c configuration as the tube cireuit. Therefore, we can find the voltage gain of the transistor cireuit by using the g» of an ideal transistor. That is, th This result is identical to Eq. (6-11), which we derived in a completely different way. The point here is simply that there is an analogy between the vacuum tube and the transistor. In a sense, the real transistor is like a pentode because it has a gm, a very high ry, and a very high w. (Ideally, the r, and 4 are infinite.) One big difference, however, between the vacuum tube and the transistor is that the CE transistor has an input resistance of Br, whereas the vacuum tube has an input resistance that approaches infinity. Nevertheless, as far as voltage gain from input to output is concerned, we can transform tube formulas into transistor formulas when analogous circuits are involved. EXxampie 6-15 The vacuum-tube circuit of Vig. 6-26a is the well-known cathode follower. It has a voltage gain of approximately unity. Find the voltage gain of the transistor circuit shown in Fig. 6-26b. * £66 Mec Yin 7 a Fig. 6-26 Example 6-15. Sotution The transistor cireuit of Fig. 6-266 is the transistor counterpart of the cathode follower; it is called an emitter follower. To find the voltage- gain formula for the emitter follower we will transform the vacuum-142 Transistor Circuit Approximations tube gain formula. The voltage gain of a cathode-follower is Gnlex Vin 1+ gmitx for a pentode Since the gn of the ideal transistor is 1/r', and since Rg takes the place of Rx, we get Vout Ra/r, vin 1+ Re/rh This expression can be rearranged to obtain Vout 1 in +7 /Re % when po <1 In most emitter-follower circuits the condition rf/Re < 1 is easily satis- fied; therefore, the emitter follower has a voltage gain of approximately unity. We intend to study the emitter follower in more detail in the next chapter. SUMMARY The IV characteristics of the CE connection are similar to those of the CB connection. The base-emitter part of the transistor acts very much like an ordinary diode, while the collector resembles a current source. The collector current is much larger than the base current. A figure of merit for a transistor is its 8, which is the ratio of its collector current to its hase current. The d-c 8 involves direct currents, whereas the a-c 8 involves alternating currents. Usually, these two 6’s are close to each other in value. The Variation in 8 for various transistor types is roughly from 20 to 200 or more. For our initial work in transistor circuit analysis we use a simple approximation called the ideal transistor. In the CE connection we treat the base diode as an ordinary:diode and the collector diode as a current source with a value of Bin. Biasing a transistor refers to setting a d-e operating point in order to prevent an a-c signal from taking the emitter diode out of forward bias or the collector diode out of back bias. This is necessary to avoid clipping the a-c signal. From the standpoint of bias stability, base bias is the worst way to bias a transistor and emitter bias is the best way. When the emitter of a CE stage is at a-c ground, the voltage gain from base to collector is simply 7z/r! (assuming an ideal transistor). Unlike the grid of a vacuum tube, which normally offers a very highCommon-emitter Approximations 143 resistance to an a-e signal, the base of a transistor looks like a relatively low resistance of Br’, As long as the source resistance is very small, almost all the source voltage arrives at the base diode; when the source resist- ance is large, however, only part of the source voltage appears across the base diode. The transistor is analogous to a pentode in its a-c operation, For an ideal transistor the gm = 1/rl, rp = @, and w = . Any voltage-gain formula for a vacuum-tube circuit can be transformed to get the voltage gain of a corresponding transistor circuit. GLOSSARY a-c resistance The resistance seen by an a-c signal. When an a-e signal is driving the base of a transistor, it sees a resistance of ér!. But when it drives the emitter, it sees a resistance of 1’. base bias A biasing arrangement in which the d-c base current remains essentially fixed even though the f of the transistor changes. bias Setting up direct currents and voltages in a transistor in order to prevent clipping of the a-c signal. beta (8) The ratio of collector current to base current. The d-c 6 refers to the ratio of direct currents, whereas the a-c 8 refers to the ratio of alternating currents. The two 6’s are often quite close in value, and in our ideal-transistor approximation they are equal. emitter bias A biasing arrangement in which the d-c emitter current re- mains essentially fixed even though the 8 changes. gm Ina vacuum tube this is the ratio of a-c plate current to a-c grid voltage under the condition of fixed plate voltage. In a transistor it is the ratio of a-c collector current to a-c base voltage under the condition of fixed collector voltage. For an ideal transistor, gn = 1/r!. » Ina vacuum tube this is the ratio of a-c plate voltage to a-c grid voltage under the condition of fixed plate current. 7, In a vacuum tube this is the ratio of the a-c plate voltage to the a-e plate current under the condition of fixed grid voltage. REVIEW QUESTIONS 1. How are Ba, and 8 defined? 2. What is the typical range of 6 for different transistors? 3. In a base-biased circuit, which remains almost fixed when the 6 changes, the collector current or the base current?4 ‘Transistor Cireuit Approximations 4. What is one of the reasons that base bias is a very poor way to set a d-c operating point? 5. In an emitter-biased circuit, which remains almost fixed when 8 changes, the collector current or the base current? 6. Why is it usually necessary to set up a d-c operating point before applying an a-c signal to the transistor? 7. What is the d-c equivalent circuit for an ideal transistor? And the a-c equivalent circuit? 8. What is the formula for the resistance seen by an a-c signal when look- ing into the base diode? 9. What is the voltage-gain formula for a typical CE circuit like that shown in Fig. 6-19a? 10. If the source driving a CE circuit has a significant amount of source resistance, how do we take this into account? 11. What is the value of the g, of an ideal transistor? PROBLEMS 6-1 A transistor has a d-e base current of 0.02 ma and a d-c collector current of 1.5 ma. What is the Ba: for this condition? 6-2 The a-c base current in a transistor is a sine wave with a peak value of I ya. The 6 of the transistor is 200. Describe the waveform of collector current. 6-3 What is the base current in a transistor if the 8 equals 125 and the collector current equals 2 ma? 6-4 In Fig. 6-27a, the transistor has a Ba, of 50. Find the collector current and the collector-ground voltage, Neglect the voltage drop across the base diode. 6-5 Repeat Prob. 6-4, but allow 0.7 volt for the voltage across the base diode. 6-6 In Fig. 6-275, the transistor has a Ba. of 100. What size should Ry be in order to have a collector-ground voltage of 5 volts? Neglect Vaz. 6-7 If the Sze is 75 in Fig. 6-27, what size should Rz be in order to have a collector-ground voltage of 10 volts? Neglect. Vaz. 6-8 In Fig. 6-27d, an adjustable resistor is used in the base cireuit so that V¢ can be adjusted to 15 volts when different transistors are used in the circuit; however, there is a limit on the range of Bae that can be compensated. In order to be able to adjust Vc to 15 volts, in what range must Buc lie? 6-9 In Fig. 6-28a, the Bae is 100. What is the collector-ground voltage? Suppose that we change transistors and that the new Ba. is 150. What is the new value of collector-ground voltage?Common-emitter Approximations 145 +10 +10 +20 i” 5k Fe 10K (0) = (Oe +20 +30 uy iM A, 10K iM Y= 15 volts (eh = (a) si Fig. 6-27 +30 2M lok (a) ee146 ‘Transistor Circuit Approximations 6-10 In Fig, 6-288, find the following: (a) The approximate value of Ip using Eq. (6-7). (6) The approximate value of Vo. 6-11 In Fig. 6-286, find the value of Iz using Eq. (6-6) and a 8 of 100. 6-12 In Fig. 6-288, find the approximate value of base current by using 28 of 100. Also, what is the value of the d-c voltage from base to ground? 6-13 In Fig. 6-29, compute the following: (a) The approximate value of emitter current. (0) The approximate voltage from collector to ground. -30 Fig. 6-29 % 40k J 6-14 In order to have a collector-ground voltage of —5 volts in Fig. 6-29, what size should Rg be approximately? 6-15 In Fig. 6-29, what value of Rx will cause the collector voltage to become approximately 0 volts? 6-16 In Fig. 6-30a, the transistor has a 6 of 75. Sketch the output waveform, using the theoretical value of 25 mv/Ts to find 1. 6-17 Repeat Prob. 6-16 but use a 2:1 range for r/; that is, compute ri by using +20 25 mv ,_ 50 mv Tego eon aelg 6-18 The transistor of Fig. 6-300 has a 8 of 80. Find the following: (a) The dc voltage from collector to ground. (6) The size of the output signal tou (use 25 mv/Tz). 6-19 In Fig. 6-30, what will happen to the d-c voltage from collector to ground and to the output voltage tou if Re is changed from 20 to 40 kilohms?Common-emitter Approximations (a) Fig. 6-30 6-20 In the circuit of Fig. 6-31a, the @ of the transistor is 100, and r, is 20 kilohms. Using 25 mv/Tz, find the value of vout. +20 -15 20K 15K +h |B=100 30) - Jge7s DOKS Vout ie an 20k a “40K “3 = SOK 45K — 3mv, = => oo ‘i rms +15 = (0) = (0) Fig. 6-31 6-21 Use a @ of 75 for the transistor of Fig. 6-31b and use the theoretical value of 25 mv/Ix for rj. What is the maximum and minimum output voltage possible in this circuit? 6-22 A typical pentode like a 6AU6 has a g», of 5000 pmhos. What is the gm of an ideal transistor when the d-c emitter current is 1 ma? What must the value of d-c emitter current be in order to have a g,, of 500 umhos?Common-collector Approximations We have now studied two basic transistor connections, namely, the common-base (CB) and the common-emitter (CE). Here we discuss the common-collector (CC), the third basic transistor connection. As we will see, the main advantage of the CC connection is that it transforms a given value of load resistance to a higher value. 7-1 The Basic Idea of the CC Connection Figure 7-1 illustrates a CC stage. For a typical design, the cireuit action is as follows. The input signal ¥;, is coupled through the capacitor into the base of the transistor. This signal causes the instantaneous value of the emitter current to inerease on the positive half cycle and to decrease on the negative half cycle. This changing emitter current produces a voltage across the emitter resistor Rx as shown, and the signal is then coupled to the final load resistor R. For a typical CC stage, the output signal is almost equal to the input signal. For instance, if the source signal is a I-my-peak sine wave, the output will be almost a 1-mv sine wave. Note that the output signal is in phase with the input signal because on 48Common-collector Approximations 149 the positive half cycle of the input signal more base current flows; this increase in base current causes an increase in emitter current, which in turn produces a positive-going signal across Rr. Mee Fig. 7-1 The emitter follower. % Ve rq (bose) In discussing the CC stage, we will use rz to represent the a-c load resistance seen by the emitter. In Fig. 7-1, the a-c load resistance seen by the emitter is re = Rel|R (7-1) For instance, in Fig. 7-1 if Re = 10 kilohms and R& = 30 kilohms, we would calculate an a-c load of rz, = 10 kilohms||30 kilohms = 7.5 kilohms As we will show in the next section, the voltage gain of the CC stage of Fig. 7-1 is Pout 1 where r; is the emitter junction resistance and r, is the a-c load resistance seen by the emitter. Note in this expression that if r, is much greater than ri, then r{/rz is a small number, much less than unity, and the voltage gain becomes (7-2) Your 1 when 1) rz (7-3) Vin In many CC circuits, r’ is much smaller than rz, so that for a first approxi- mation we can say that the voltage gain of such circuits is essentially unity. In other words, whatever signal is coupled into the base appears across the final load resistor R. A natural question to ask at this point is: Why use a CC stage if the voltage gain is unity or less? Now we come to the main reason for150 ‘Transistor Cireuit Approximations using a CC circuit. This circuit has the advantage that the a-c input resistance looking into the base of the transistor is much higher than the a-c load resistance seen by the emitter. In fact, we will show that the resistance looking into the base is Tin(oasey & Ary when r) & rz (7-4) This equation tells us that the load rz seen by the emitter is stepped up to a new value of gr, at the base. For instance, if r, = 1 kilohm and 8 = 100, then looking into the base, the input resistance is Tingosse) = 100(1 kilohm) = 100 kilohms Thus, we see that the CC circuit is similar to a transformer in that it can be used to transform, or step up, the value of a load resistance; however, it is quite different from a transformer, because the output signal is almost equal to the input signal. The common collector is sometimes called an emitter follower because the emitter signal follows the signal at the base. For those readers who are familiar with the cathode follower, it is worth mentioning that the emitter follower is the analogous transistor circuit. To summarize our discussion of the emitter follower, we note: 1. It has a voltage gain of unity or less. 2. It is primarily used to transform a load resistance to a much higher value. 7-2 Derivation of CC Formulas In this section we want lu prove Eqs. (7-2) to (7-4). Before we do this, however, let us find the important formulas that pin down the d-e oper- ation of the emitter follower of Fig. 7-1. As usual, we first draw the d-c equivalent circuit by opening all capacitors and shorting all a-c sourees. The d-c equivalent circuit is shown in Fig. 7-2a. This circuit is really a form of the emitter-biased circuit discussed in the last chapter. The only difference here is that there is no resistance in the collector circuit. In other words, we ean visualize the d-c equivalent circuit as indicated in Fig. 7-2b, where we have shown an R; of 0 ohms. The formulas for emitter bias derived in Sec. 6-5 still apply. They are Ip Vee — Von * Re + RalB Ve = Voc — IcRy, As usual, we can simplify the first equation by noting that Vaz is usually much smaller than Vgg and that R»/é is usually much smaller than Rg.Common-collector Approximations 151 the Fig. 7-2 Emitter bias of emitter follower. fe £| Re S Vee (a) Therefore, the first equation becomes simply Vee Re This equation says that the d-c emitter current is approximately equal to the emitter supply voltage divided by the value of the emitter resistor. Note also that with an Rz, of zero, the d-c voltage from collector to ground becomes Ip (7-5) Ve = Vee Now let us prove the a-c formulas given in the preceding section. When we draw the a-c equivalent cireuit for Fig. 7-1, we short all d-c sources and capacitors, to obtain the cireuit shown in Fig. 7-3a. By inspection, the a-c load resistance seen by the emitter is the parallel combination of Rz and R. If we let rz represent this a-c load, we can redraw the cireuit as in Fig. 7-3b. Next, to get approximate formulas for the a-c operation of the emitter follower we will replace the transistor by its ideal a-c equivalent circuit, as shown in Fig. 7-3c. To find the voltage gain of the emitter follower we need only write expressions for toue and ti, and take the ratio of these expressions. For instance, it is clear that the a-c output voltage must equal the a-c emitter current times the a-c load resistance. That is, Your = UerL Because the emitter current is almost equal to the collector current, we can write Vout Stern = Biot, The input voltage v;, is the a-c voltage from base to ground. By inspec- tion of Fig. 7-3c, vin equals the a-c voltage across the base diode plus the152 Transistor Circuit Approximations Fig. 7-3. Deriving gain and impedance formulas. voltage across rz. That is, Yin = BBM + err, dz, 80 that we can rewrite this expression as tin & tor! + Bist, (7-6) As already indicated, i, The voltage gain is tout/Vin. Your Biot, tn bBre + Brn or re By dividing numerator and denominator by rz and rearranging we get Vout 1 vin + re/re It is immediately apparent from this result that when r/ is much smaller than rz, the voltage gain becomes (7-7) Vout tinCommon-collector Approximations 153 Another a-c formula to prove is that the input resistance looking into the base is approximately 8 times the a-c load seen by the emitter. We do this as follows. The loading effect of the base can be found by taking the ratio of the input voltage to the input base current. That is, ee a ingoaw) = Recall that we have already found an expression for tix. By using Eq. (7-6) we have irl + Bisry, Fincoase t Br + Bre Or we can write Tinbase) = G(re + rz) (7-8) Often, 7 is much smaller than rz, and this equation reduces to Fin(base) & BFL (7-9) Thus, we see that to a first approximation the input resistance looking into the base of an emitter follower is B times the a-c load seen by emitter. In other words, the emitter follower transforms a load resistance r, to a higher value of Br; when viewed from the base. Remember that the formulas we have derived are not exact formulas because we used an ideal transistor, in addition to other approximations, However, the formulas we have are adequate for preliminary analysis and design, and they certainly do describe the fundamental operation of the emitter follower to a good first approximation. One more point. The total input resistance of a CC stage is actually the parallel combination of the base resistor and the input: resistance looking into the base. This can be understood by referring to Fig. 7-4. As +e 1 eRelR Yout Vin eh 2 Vout rig(bose} =Blve+ 7) s rip stage) =a Irn (base) -Ve rig(stoged —rin{bose) © Fig. 7-4 Summary of emitter follower.154 ‘Transistor Cireuit Approximations far as the source is concerned, it must provide current for both the base of the transistor and the base resistor. In other words, the source sees a total resistance of Tin(stage) = Rallrincoase) (7-10) The formulas for a-c operation are summarized in Fig. 7-4. As usual, if a p-n-p transistor is used instead of an n-p-n, the formulas are identical because the a-c equivalent circuit is the same for both types of transistors. Exampip 7-1 For the emitter follower shown in Fig. 7-5 find: (a) The input resistance looking into the base. (b) The input resistance looking into the stage. Sotution (a) Using Eq. (7-9), we can find the approximate input resistance looking into the base. Tingoase) & Arr, Note that r, is 20 kilohms in parallel with 1 kilohm. rz = 20,000|/1000 = 950 ohms With a 8 of 50, we get Tin(oasey = 50(950) = 47.5 kilohms +20 | R=50 Fig. 7-5 Examples 7-1 and 7-2. Yn Tt ©) TOOK = 20K IK. i = 4 > ‘This is approximately the value of input resistance looking into the base. If we like, we can obtain a more accurate answer by using Eq. (7-8), which takes rf into account. According to this equation, Tingbacey = BC, + 11) The theoretical value of r/ is 25 MY _ 95 ohms lma je Zoimy °" Te rCommon-collector Approximations 155 (The d-c emitter current Jz equals 1 ma because the emitter supply voltage of 20 volts divided by the emitter resistance of 20 kilohms yields 1 ma.) Now, we compute the more accurate value of rincbase)- Tingoasey = B(r, + rr) = 50(25 + 950) = 48.7 kilohms (Note that taking rj into account does not change the resistance very much.) (6) Using Eq. (7-10), we can find the total input resistance of the stage. It is Tingstage) = Rell Pingoasey & 100 kilohms||47.5 kilohms .2 kilohms ExampLe 7-2 In the circuit of Fig. 7-5, suppose that vj, is a 3-mv-peak sine wave. Find the output voltage by using Eq. (7-7). Sonution We already know from the preceding example that r, = 25 ohms and that r, = 950 ohms. Hence, the voltage gain is Pout _ 1 1 tin 1+ rit, 1425650 Your = 0.975(3 mv) = 2.92 mv = 0.975 Therefore, Thus, the output voltage is a sine wave with a peak value of 2.92 my. Obviously, this is quite close to 3 my, and for a first approximation, we can say that the output voltage essentially equals the input voltage. Exampie 7-3 In Fig. 7-6, the 8 can be between 50 and 150. Find the following: (a) The voltage gain. (6) The input resistance looking into the base. (c) The input resistance of the entire stage. +30 ae 50< B< 150 Fig. 7-6 Example 7-3. © \ | | 100K 40K2 200. r -20156 ‘Transistor Circuit Approximations SoLution (a) First, let us find the theoretical value of ri. " = > = 50 ohms 5 ma Next, we find rz. 71, = 40,000||200 = 200 ohms Now, by using Eq. (7-7) we get ee vin T+ ri/rz 1 + 5% 00 (b) The input resistance at the base will depend upon the 8. For the minimum 8 of 50, we have Tinbasey = B(r, + rr) = 50(50 + 200) = 12.5 kilohms And for the maximum 6 of 150, we get Tin(oase) = 150(50 + 200) = 37.5 kilohms (c) The minimum input resistance of the stage occurs for the mini- mum 8 of 50. Using Eq. (7-10), we obtain Tingstage) = Rallrincvasey = 100,0001}12,500 = 11.1 kilohms The maximum input resistance of the stage is Tin(stazey = 100,000!|37,500 = 27.3 kilohms EXAMPLE 7-4 Suppose that we drive the emitter follower of Example 7-3 with a source that has an internal resistance of 10 kilohms, as shown in Fig. 7-7a. Find vou: for a 8 of 150. Soturion We already found in Example 7-3 that the input resistance of the stage is 27.3 kilohms when 8 = 150. Therefore, we can visualize the input side of the transistor circuit as illustrated in Fig. 7-76. Since we have a voltage divider, some of the source signal is lost across the 10-kilohm resistor. The actual input voltage to the base of the tran- sistor is a sine wave with a peak value of 27.3 Yinweaid = 194-973 50 mv = 36.6 mv We already know from the preceding example that the emitter fol- lower has a voltage gain of 0.8 from base to output. Therefore, the finalCommon-collector Approximations 187 “5 50my 10K t "5 Vout 10K Yout 5Omv peok (A>) 15 273K oy, | 8m sine wove | ' (2) (ch Fig. 7-7 Example 7-4 output signal is a sine wave with a peak value of Youtwresk) = 0.8(36.6 mv) = 29.3 mv A sketch of the output waveform is shown in Fig. 7-7e. Thus, we see that if the source resistance is large enough, some of the source voltage ean be lost before reaching the base of the transistor. EXxaMpLe 7-5 This particular example is included for the reader who is already familiar with the cathode-follower vacuum-tube circuit. In Fig. 7-8a we have the conventional cathode-follower circuit. Recall that the voltage gain of this circuit is Pout 1 vin LA / gare where rz = Rx||R. when » and ry are very high In Fig. 7-8) we have the conventional emitter follower. Notice that the cathode follower and emitter follower are almost identical in form. ‘The only major difference is in the d-c operation, where it is necessary to use the Ver supply to forward-bias the emitter diode. As far as a-c158 Transistor Circuit Approximations +40 +e Yin " x “out out ay 4 be if O] 3 BH gr ip fH fa | : ~ = ~he (a) — a. wy Rg al, Sk Ae a : le og (c) . Fig. 7-8 Example 7-5. equivalent cireuits are concerned, the cathode and emitter followers are identical in form, as indicated by Fig. 7-8¢ and d. Transform the voltage-gain formula for the cathode follower into the voltage-gain formula for the emitter follower by the methods of Sec. 6-9. SoLvtion The g» of an ideal transistor is By substituting this value into the cathode-follower formula we get os nese ee + /gmrn V+ ri/ry, This equation is identical to Eq. (7-7), which gives the voltage gain of the emitter follower. Thus, we have found a transistor formula by using the tube-transistor transformation of Sec. 6-9. This approach to finding transistor formulas is quite useful sometimes because it can save considerable time. Also,Common-collector Approximations 159 it can be used to check the results of a direct derivation of the transistor formula. For instance, in this example, we have checked the formula for the emitter-follower voltage gain that was derived earlier. 7-3 The Darlington Pair As indicated in Sec. 6-2, the 8 of different transistors typically falls in the range of about 20 to 200. There may be times when we need a 8 that is much higher than 200. One way to get this is to connect two transistors together as shown in Fig. 7-9a. This particular connection is called a Darlington pair. In Fig. 7-9a, note that the base current in the first transistor produces a collector current of i, in this transistor. Since the emitter and collector currents are almost equal, we see that the emitter current of this first transistor is approximately i, Note that this emitter current drives the base of the second transistor; therefore, the collector current of the second transistor equals 8%, (assuming identical transistors). As a result, the emitter current of the second transistor is approximately equal to 6%. {Po | xB | 8% ici ) 6% {8% 128% (a) (5) {e) Fig. 7-9 ‘The Darlington pair. Transistor manufacturers sometimes put two transistors connected as a Darlington pair inside a single transistor housing, as shown in Fig. 7-9b. Thus, the Darlington pair acts like a single transistor (Fig. 7-9c) with an effective 8 of Bop = B* for equal 8 transistors where pp is the effective 8 of the Darlington pair. Of course, if the two transistors do not have equal 4’s, we must use the product of the ’s.160 Transistor Cireuit Approximations That is, Bor = 8182 The important point here is that the Darlington pair can give us ex- tremely high 6’s. For instance, if each transistor has a B of 100, the Darlington pair has an effective 8 of Bop = 100(100) = 10,000 Even for A’s as low as 20 the Darlington pair has an effective beta of Bop = 20(20) = 400 Exampie 7-6 For the circuit of Fig. 7-10 find the input resistance looking into the entire stage. Each transistor has a 8 of 50. SoLutton The effective 8 of the Darlington pair is Bop = 50(50) = 2500 The a-c load seen by the second emitter is r, = 20,000||500 = 500 ohms Using Eq. (7-9) we find that the resistance looking into the base of the first transistor is Tincoasey = 2500(500) = 1.25 megohms +20 Fig. 7-10 Example 7-6. 20K Sen -20 zCommon-collector Approximations 161 The input resistance of the stage is the parallel combination of Rx and Tinvasey- Thus, Tingstoge) = L megohm||1.25 megohm = 550 kilohms EXAMPLE 7-7 Find the voltage gain for the circuits shown in Fig. 7-11a and 6. Sotution In Fig. 7-lla, we recognize this as a simple CE amplifier with a voltage gain of Pout 7 _ 10,000 Gina te a20, = 400 Note that r, in Fig. 7-11a consists only of the 10-kilohm resistor in the collector circuit; also, the theoretical value of r/ is 25 ohms, obtained in the usual manner. +20 +20 TOK 10K Vout 500 aE Yout Yin look 20K = Yin 100K 20K == -20 20 (0) (0) Fig. 7-11 Example 7-7. In Fig. 7-116, there is a 500-ohm resistor loading the stage. The new value of rz, becomes 71, = 10,000|]500 ~ 500 ohms and the voltage gain becomes Yous 9 500 _ wm = BB = 20 Obviously, coupling into the 500-ohm resistor has reduced the gain considerably (from around 400 to about 20). This kind of effect, where162 Transistor Circuit Approximations the gain is reduced by coupling the signal into an additional resistor, is called loading down the stage. One way to avoid the loss in gain is to step up the 500-ohm load to a much higher value, as illustrated in the next example. EXampLe 7-8 In Fig. 7-12 we have cascaded the CE stage of the preceding example with an emitter follower using a Darlington pair. Find the following: (a) The voltage gain for the first stage. (b) The voltage gain for the total circuit. Sonution (a) Looking into the second stage we have an input resistance of Yinistece) = 550 kilohms. (This result was worked out in Example 7-6.) ‘The effective r, seen by the collector of the first stage is rz, = 10 kilohms||550 kilohms & 10 kilohms In other words, the collector sees 10 kilohms in parallel with the 550- kilohm input resistance of the second stage. For practical purposes, ? 10K 5=50 @ 2 % h 100K 1 T ee 20k =] “in(stoge) 20K> 500: 1 I = -20 Fig. 7-12 Example 7-8. this is 10 kilohms. Thus, the second stage hardly loads the first stage at all. Also, the voltage gain of the first stage goes back up to around 400, the result found in Example 7-7. (6) The voltage gain of the entire circuit is simply the voltage gain of the first stage times the voltage gain of the second stage. Since theCommon-collector Approximations 163 emitter follower has a voltage gain of approximately unity, we see that the overall gain is around 400. SUMMARY The common collector is also called an emitter follower. ‘The voltage gain of the emitter follower is usually around unity but can be less, depending upon the relation between rand rz. The main reason for using the emitter follower is to increase impedances. As we have seen, a load of rz can effectively be transformed to a higher value of Br, by using the emitter follower. ‘The Darlington pair is a compound connection of two transistors; it has an effective @ equal to the product of the individual 6's. GLOSSARY common collector (CC) One of the basic ways to connect a transistor. The input signal is applied to the base; the output is taken from the emitter. The CC is sometimes referred to as a grounded-collector cireuit because the collector is at a-c ground. Darlington pair A connection of two transistors in such a way that the transistors act like a single equivalent transistor with an effective 6 equal to the product of the individual 6’s. emitter follower Another name for the common-collector circuit. The name is quite descriptive of the circuit because the emitter signal follows the signal applied to the base. loading down a stage Reducing the voltage gain of a stage by connecting the output of the stage into another stage or resistor. REVIEW QUESTIONS 1, What is the approximate value for the voltage gain of the emitter follower? What is the approximate value of input resistance looking into the base? 2. In order for the voltage gain of the emitter follower to approximately equal unity, should r/ be large or small compared to rz? 3. What is the main reason for using an emitter follower? 4, For the emitter follower discussed in this chapter, how can we find the approximate value of the d-c emitter current?164 Transistor Circuit Approximations 5. What is a Darlington pair? Show the schematic for it. 6. What does loading down a stage mean? PROBLEMS 7-1 An emitter follower is loaded by an a-c resistance of 1 kilohm. If the transistor has a 8 of 50, what is the a-c input resistance looking into base? 7-2 A transistor is used in an emitter follower that is loaded by an rz of 500. If the transistor can have a 8 anywhere in the range of 50 to 200, what is the lowest value of a-c input resistance looking into the base? And the highest? 7-3 In Fig, 7-13, what is the approximate output voltage when the input voltage equals a 3-my-peak sine wave? 7-4 In Fig. 7-13, what is the a-c input resistance looking into the base? What is the a-c input resistance of the stage? +15 +10 0, [ Yin i 7-5 Suppose that the 2-kilohm resistor is actually a variable resistance in Fig. 7-13. For what value of resistance will the voltage gain of the emitter follower drop in half? (Use the theoretical value of 1.) 7-6 A Darlington pair has transistors whose 6’s equal 75. What is the effective 8 of the Darlington pair? 7-1 The 6 of the first transistor in a Darlington pair is twice as large as the @ of the second transistor. The Darlington pair has an effective 8 of 10,000. Find the individual 6’s. 7-8 In Fig. 7-14 the transistors each have a 6 of 80. Find the approxi- mate values of: Q | ony 50K 20K’ 2k Yout ae i oqdi -10 -30 Fig. 7-13 Fig. 7-14Common-collector Approximations 165 (a) The a-e resistance looking into the base. (b) The a-c resistance looking into the entire stage. (c) The output voltage. 7-9 Prove that when a third transistor is added to a Darlington connec- tion the effective 8 is approximately 8°. (Use ideal transistors.) +30 50K B=100 100K a Yin look ae KS “out + 2 SOK aE 50k = “15 Fig. 7-15 7-10 For the two-stage circuit shown in Fig. 7-15 find the following: (a) The a-c input resistance of the second stage (including the 100-kil- ohm base resistor). (b) The overall voltage gain of the two-stage circuit.Large-signal Operation So far we have discussed only small-signal operation of the transistor, that is, operation in which the a-c currents and voltages are less than about 10 percent of the d-c currents and voltages, Now our attention shifts to large-signal operation. Here we are inter- ested in a-c signals that cause large changes in the total currents and voltages in transistor circuits. In this chapter, we discuss several concepts that are helpful in dealing with large signals, concepts like the d-c load line, the a-c load line, and maximum signal capability. 8-1 The D-C Load Line Suppose that we have a common-base circuit like that shown in Fig. 8-1a. How is the collector current related to the collector voltage? From our discussion in Chap. 5 we know that ve = Veo — icky (8-1) In other words, the collector voltage equals the collector-supply voltage minus the voltage across the load resistor. 166Large-signal Operation 167 Equation (8-1) tells us that the collector voltage ve is a function of the collector current zc. When the collector current changes, we get different values of collector voltage. For instance, suppose that we adjust the emitter resistor 2s to a value of 20 kilohms. Then, the emitter current is approximately pe Veg 200 =p, ~ 30,000 ~ 1 ™ This 1 ma of emitter current sets up a collector current of about 1 ma, and therefore the collector voltage is te = Veo — ict, = 20 — 0.001(5000) = 15 volts Thus, when ig = 1 ma, v = 15 volts. If we like, we ean plot this pair of values as shown in Fig. 8-16. The plotted point is at A. -20 +20 Ideal % I s] Fig. 8-1 Obtaining the d-c load line. (a) Circuit; (6) load line. If we were to change the emitter resistor Rg, this would change the collector current and voltage to another set of values. For instance, if we make 2 = 10 kilohms, we get an emitter current of py gm ee * = 10,000 ~~ ™ The collector current therefore changes to about 2 ma, and we get a collector voltage of ve = 20 — 0.002(5000) = 10 volts Therefore, the new d-c operating point is ic = 2 ma, ve = 10 volts. Again, we can plot this pair of values to obtain point B in Fig. 8-1b. Suppose we change Rx until there is 3 ma of collector current. Under168 Transistor Cireuit Approximations this condition, we get a collector voltage of ve = 20 — 0.003(5000) = 5 volts When we plot this pair of values, we obtain point C in Fig, 8-1b. Suppose that we make Ry = 5 kilohms. Then, there is about 4 ma of collector current, which produces a collector voltage of ve = 20 — 0.004(5000) = 0 volts After plotting ic = 4 ma, ve = 0, we get the upper end of the line shown in Fig. 8-1. Note that this point is called the saturation point. The reason for this name is that 4 ma is the largest value of collector current that we can get in the circuit of Fig. 8-1a. At this point, the transistor is said to be saturated because the collector diode is no longer back- biased. As we already know, we get normal transistor action only when the collector diode is back-biased. If we were to increase the emitter current above 4 ma, the collector current would not increase, since the collector current is controlled by the emitter current only when the collector diode is back-biased (see See. 5-2 if in doubt). Here is an important idea: if we continued to plot all the possible pairs of collector current and voltage, we would find that these points lie on the line shown in Fig. 8-1b. This line is called the d-c load line; it is the locus of all possible operating points. [The d-c load line is simply a graph of Eq. (8-1), which is a linear equation.] The saturation point on the d-c load line is one of the limits on the values of ve and ic in the circuit of Fig. 8-la. Another limit on these values occurs at the lower end of the d-c load line. In Fig. 8-16 this lower end of the load line is called the cutoff point. The reason for this name is simply that at cutoff there is no collector current. One way to produce cutoff is to make Rz = ©, thatis, to open the emitter resistor. Under this condition, there is no emitter current and therefore no collector current. -20 +20 -20 +20 Ome. : Amo Fe, {<5k Fey 2081 5K Fig. 8-2 (a) Cutoff; (b) saturation. + + 20 0Large-signal Operation 169 (In a real transistor there is a small reverse current through the back- biased collector diode, but we neglect this in our ideal-transistor approxi- mation.) With no collector current, the collector voltage equals the col- lector supply voltage. In other words, with no current in the 5-kilohm resistor, there is no voltage drop across this resistor, and all the supply voltage appears from collector to ground. In the future, we should remember that the d-c load line is drawn between two points, the cutoff point and the saturation point. At the cutoff point there is no collector current, and all the supply voltage ap- pears across the collector diode, as shown in Fig, 8-2a. At the saturation point there is zero voltage across the collector diode, and all the supply voltage is dropped across the load resistor Az, as illustrated in Fig, 8-2b. In fact, we can generalize our results for any CB circuit of the form given in Fig. 8-3a. In this circuit, saturation occurs when all the supply voltage is dropped across R; (Fig. 8-36). Therefore, the saturation value ee +e Vie Hee Yee Shh Fe A, fe lac Sa, + + Xe 0 Ideo! deo! (a) Vee Mec 0 Fe | A + Ideal Yeo (e) (a) Fig. 8-3 (a) Circuit; (6) saturation; (¢) cutoff; (d) d-c load line.170 ‘Transistor Cireuit Approximations of collector current is Ve Towsat) = he (8-2) At cutoff there is no collector current, and all the supply voltage appears across the collector diode (Fig. 8-3c). The d-c load line must pass through the saturation and cutoff points. Thus, we can show the d-c load line as in Fig. 8-3d. Whatever the values of collector current and voltage are, they must plot as a point somewhere along the d-c load line. As long as the collector current is between zero and the saturation value given by Eq. (8-2), we can find the approximate operating point by using SIy w VEE . Tos In & Ra (8-3) and Vo = Veo — Tok (8-4) EXAMPLE 8-1 For the circuit of Fig. 8-4a, draw the d-c load line and the d-c oper- ating point. SoLution By inspection of the circuit, the cutoff voltage equals the collector supply voltage of 30 volts. Also, the saturation value of collector current is Towa = =3ma -30 +30 dc operating point 20K lok % (a) (6) Fig. 8-4 Example 8-1.Large-signal Operation i The d-c load line can now be drawn as shown in Fig. 8-4b. The actual operating point of the transistor lies somewhere along this d-c load line, and it is easily found by Eqs. (8-3) and (8-4). and. Ve = 30 — 1.5(10-*)(10)(108) = 15 volts EXAMPLE 8-2 For the circuit of Fig. 8-5a, draw the d-c load line and locate the d-c operating point. Soivrion Note that this cireuit is similar to that of Fig. 8-4a; the only differ- ence is that we are using an emitter supply of —50 volts instead of —30 volts. First, realize that this does not change the d-e load line at all because it still passes through a cutoff of 30 volts and a satu- ration point of 3 ma. -50 +30 ‘ 3ma: 2.5ma 20K 1oK % 5 30 . Fig. 8-5 Example 8-2. The only change that takes place is in the d-c operating point. It is clear that the emitter current is The corresponding collector voltage is Ve = 30 — 2.5(10-*)(10)(10*) = 5 volts172 Transistor Cireuit Approximations Thus, the d-c operating point shifts to a new location, as indicated in Fig. 8-5b. Exampe 8-3 Draw the d-c load line and find the d-c operating point for the circuit of Fig. 8-6a. Sotution This circuit is the same as that of Fig. 8-4a except that the collector supply voltage is now 50 volts instead of 30 volts. This causes the d-c load line to change because it passes through a cutoff point whose volt- age is 50 volts. Also, the saturation current changes and becomes 20K 10K (a) () Fig. 8-6 Example 8-3. The new d-c load line is shown in Fig. 8-6b. The actual d-c oper- ating point is located at and Ve = Vee — Ick = 50 — 1.5(10-*)(10) (10°) EXaMPpLe 8-4 In Fig. 8-7a, draw the d-c load line and show the d-c operating point.Large-signal Operation 173 Sotution This circuit is the same as that of Fig. 8-4a except that we are using a load resistance of 5 kilohms instead of 10 kilohms. It should be clear that the cutoff point has a voltage of 30 volts, the collector supply voltage. Also, the saturation point has a current of 30, Toa = 5999 = 6 ma -30 +30 20K SK Fig. 8-7 Example 8-4. The d-c load line is shown in Fig. 8-76. The actual d-c operating point lies at 30 Tos Te yg = 1.5 ma and Ve = 30 — 1.5(10-%)(5)(10*) = 22.5 volts 8-2 Load-line Interpretation of an A-C Signal In Fig. 8-8a we have a CB circuit driven by d-e and a-c sources. We studied circuits of this type in Chap. 5; recall that voltage gain of such a circuit is th ® nr Vout In Fig, 8-8a it is obvious that the d-c emitter current is around 1 ma, and therefore the theoretical value of rj is about 25 ohms. Since r, equals 1000 ohms, we can say that the emitter diode is swamped out (Sec. 5-8).174 Transistor Circuit Approximations Thus, the voltage gain from source to output is approximately Pout — 10,000 _ x, — 1000 ~ 10 Since v, is shown as a 0.5-volt-peak sine wave in Fig. 8-8a, the a-c output voltage must be a 5-volt-peak sine wave. The d-c output voltage is clearly equal to Ve = 20 — 0.001(10,000) = 10 volts Hence, the total output waveform (sum of d-c and a-c components) is approximately that shown in Fig. 8-8b. It has an average or d-c value of -20 +20 Ys 20K 10K 05 IK 1 oa + Oo) is % (a) (0) Fig. 8-8 Load-line interpretation of a-c signal. 10 volts; there is sine wave with a 5-volt peak superimposed on this 10-volt level, so that the total instantaneous voltage swings from a low of 5 volts to a high of 15 volts. How is all this related to the load-line concept? First, we know that the d-c load line passes through a cutoff point with a voltage of 20 volts,Large-signal Operation 175 the collector supply voltage. Also, we know that the saturation point on the d-c load line has a current of Veo _ 20 _ 7 ~ 10,000 ~ 2 ™ Teva The actual d-c operating point is on the load line and is found by using Eqs. (8-3) and (8-4). eeeeea Vga ieee eoO aaa To In = 77 = 5999 = 1 ma Ve = Veo — Ick = 20 — 0.001(10,000) = 10 volts Hence, we can draw the d-c load line as in operating point of the transistor is located at Q. When the a-e signal is present, it causes excursions or changes to take place above and below point Q. These changes must take place along the d-c load line because this line is the graph of all possible pairs of ic and vc, as discussed in Sec. 8-1. We found earlier that for the circuit of Fig. 8-82, the total voltage at the collector swings sinusoidally from 5 to 15 volts, as shown in Fig. 8-8b. On the load line of Fig. 8-8c this means that the instantaneous operating point moves sinusoidally between points A and B. A comment worth making at this time concerns the maximum signal that we get from the circuit before clipping occurs. Examine Fig. 8-8c and note that the maximum excursion on the positive peak is limited to the cutoff point on the d-c load line; on the negative peak the maximum excursion is limited by the saturation point. Therefore, in this particular circuit we can obtain an excursion of 10 volts on either side of the d-c oper- ating point before clipping occurs. Thus, if we were to increase the source voltage, we would find that clipping would occur when the output signal reached a value of 20 volts peak to peak. 8-8c. The d-c or average Exampue 8-5 In the circuit of Fig. 8-9, find the following: (a) The d-c load line and d-c operating point. (6) The maximum peak-to-peak unclipped signal. (c) The approximate value of source voltage that causes clipping to occur. Soution (a) Note that this circuit is the same as the cireuit we have been discussing (Fig. 8-8a), except that the emitter resistor Rx is 40 kilohms instead of 20 kilohms. Since the cutoff and saturation points on the d-e176 ‘Transistor Circuit Approximations load line depend only on Veo and Rz, the d-c load line will be the same as before, and is shown in Fig. 8-9. Of course, the d-c operating point will be different from before. The new values of collector current and voltage are fee Van eee ; Te [a = EE = Fogg = 05 mM Ve = Veo — ToRz = 20 — 5(10-4)(10)(10%) = 15 volts (®) It is immediately clear in Fig. 8-95 that the positive swing can be from 15 to 20 volts but no more. On the negative swing, the output -20 +20 Me 40K’ loK IK 0.5mo (a) (2) Fig. 8-9 Example 8-5. signal can go from 15 volts down to 0 volts. Therefore, as the source signal is increased, clipping first occurs on the positive half cycle. As the source is increased further, clipping eventually takes place on the negative half cycle. The limiting factor, of course, is the cutoff that occurs on the positive half cycle. Therefore, the maximum unclipped signal that we can get is 10 volts peak to peak. () Clearly the emitter diode is swamped out, so that the approxi- mate voltage gain is Therefore, clipping occurs when the source voltage is greater than = 1 volt p-p A 10 volts p—p Fi 10Large-signal Operation ut 8-3 The A-C Load Line Suppose that the a-c load seen by the collector is different from the d-c load. What changes will this make in our load-line interpretation of an a-c signal? Consider Fig. 8-10a. It is immediately clear that circuit has a d-e load line with a cutoff voltage of 20 volts and a saturation current of 2 ma. It is also clear that the d-c collector current is about 1 ma and the col- lector voltage is 10 volts. Therefore, we can draw the d-c load line and plot the d-c operating point as shown in Fig. 8-100. The a-c signal causes changes to take place in the collector current and voltage. However, the excursions from point @ no longer follow the d-c load line; instead the changes in current and voltage are along a new line, called the a-c load line. Basically, the reason for this new line is that the d-c load line only takes the d-c load resistance into account. We therefore would not expect a-c excursions along a line that does not account for the a-c load resistance. To better understand why there is a new line called the a-c load line, -20 +20 %5 20K 10K 05 Nn : 1K \ V7 i — | GIG 7 1OK a 0.5mo peck We - ee ee ag dc load tine - a 10K c : : 1 75 10 2515 2% () te) (0-¢ load line Fig. 8-10 A-c load line. (a) Cireuit; (6) load lines; () a-c load seen by collector.178 Transistor Circuit Approximations consider the following discussion. In the circuit of Fig. 8-10a, the 0.5-volt- peak source produces a sine wave of current in the emitter. The peak value of this emitter current is simply 05. = 1000 te = 0.5 ma Since collector and emitter currents are almost equal, the a-c collector current is also a sine wave with a 0.5-ma peak value. The a-c equivalent circuit for the collector circuit is shown in Fig. 8-10c. From this circuit it is clear that the a-c load resistance is only 5 kilohms, and therefore the peak voltage at the collector is Ye = 0.0005(5000) = 2.5 volts Now, here is the crucial point. The peak excursion from point @ in Fig, 8-10b must show a change of 0.5 ma and 2.5 volts. By inspection of Fig. 8-10b such a change is impossible along the d-c load line. When we plot these changes, we actually get two new points A and B, representing the positive and negative voltage peaks. Thus, when the a-c signal varies sinusoidally, excursions occur along the a-c load line between A and B. Here are a few more interesting differences between the a-c and d-c load lines. As usual, if the a-c signal is large enough, clipping occurs. On the positive-going voltage swing, clipping occurs sooner on the a-c load line than on the d-c load line. In other words, cutoff on the d-c load line occurs at 20 volts; this, however, is no longer important as far as clipping is concerned. Since the actual changes in collector current and voltage take place along the a-c load line, we must use the a-c load line to find the maximum possible swings in either direction. The cutoff voltage on the a-c load line is only 15 volts, because the maximum possible current change from the operating point is 1 ma. This means that the maximum positive voltage change from the operating point is 0.001(5000) = 5 volts Thus, we have shown the cutoff voltage at 15 volts in Fig. 8-10b. The saturation point on the a-c load line is at 3 ma, because the maxi- mum negative voltage change from the operating point is 10 volts. This means that the maximum possible current change from the operating point is 10 5000 = 2 ma Therefore, we have shown the saturation current at 3 ma (1 ma oper- ating current plus the 2 ma swing).Large-signal Operation 179 Let us extend the results of the foregoing discussion to the general CB circuit of Fig. 8-1la. The d-c load line passes through a cutoff volt- age of Veo and a saturation current of Voc/Rz. The operating point Q has a d-c current Ic and a d-c voltage Vc which ean be found by Eqs. (8-3) and (8-4). Vee +e Ae (a) Fig. 8-11 Summary of load lines. When the a-c signal is coupled into the emitter, changes in collector current and voltage take place along the a-c load line. The cutoff point on this a-c load line has a voltage of Ve + Jor: In other words, the cutoff voltage must equal the d-c voltage at the operating point plus the maximum positive voltage change, which is Icrz. Also, the saturation current on the a-c load line equals the direct current at operating point plus tho maximum positive current swing, which is Ve/rz. The two kinds of load lines with their cutoff and saturation points are summarized in Fig. 8-110. Especially note on the a-c load line that: 1. The maximum positive swing of a-c collector voltage is limited to Tort. 2. The maximum negative swing is limited to Vo. In other words, the peak-signal-handling capacity is limited to Igrz or Ve, whichever is smaller. EXxampe 8-6 Draw the d-c and a-c load lines for the CB circuit of Fig. 8-12a. Sonution Clearly, the d-c load line passes through a cutoff voltage of 30 volts and a saturation current of 1 ma. The d-c operating point is Ic = 0.5 ma and Ve = 15 volts.180 ‘Transistor Circuit Approximations -20 +30 (0-¢ load line Co) 60 i go gre load line + oe ears) (a (0) Fig. 8-12 Example 8-6. ‘The a-c load resistance seen by the collector is rz = 30(10*)||60(10°) = 20 kilohms The cutoff voltage on the a-c load line is Vo + Tern = 15 + 0.5(10-*)(20) (102) = 25 volts The saturation current on the a-c load line is 15 30005 = 1.25 ma Te + VE = 0.5(10-%) + TL The d-c and a-c load lines are given in Fig. 8-12b. Note that clipping will occur if the a-c signal tries to exceed 10 volts on the positive volt- age swing and 15 volts on the negative voltage swing. EXAMPLE 8-7 In the circuit of Fig. 8-13a, what is the maximum unclipped out- put voltage? What approximate value of source voltage just. causes clipping? Souurion By inspection of the circuit, we see that the d-c load line has a cutoff voltage of 20 volts, a saturation current of 2 ma, and a d-e operating point of Ic = 1 ma, Ve = 10 volts. The a-c load seen by the collector is 11, = 10,000|[30,000 = 7.5 kilohms Therefore, the cutoff voltage on the a-c load line is Ve + Tort = 10 + (10-*)(7.5)(10%) = 17.5 voltsLarge-signal Operation 181 and a saturation current of 10 Taconite ae Te + YE = 10-2 + TL The d-c and a-c load lines are drawn in Fig. 8-130. It is clear that clipping first occurs on the positive-going excursion. The maximum a-c signal is limited to 7.5 volts peak. -20 +20 20K 10K & a 2.33ma Ik 3OKS ryt 2ma al Ima g bes = cs ii (a) 10 175 20 (a) Fig. 8-13, Example 8-7. The easiest way to find the value of source signal that just causes clipping is to compute the voltage gain from source to output and then divide 7.5 volts peak by this gain. The source-to-output gain is 10,000/130,000 _ 1000 ML DU, 75 Therefore, the source voltage that just causes clipping to occur is Tor, _ 7.5 75 7357 1 volt peak Examp.e 8-8 What is the maximum peak-to-peak signal that can be obtained from the circuit of Fig. 8-14? Souvtion The d-c operating point is at Ic =0.333 ma = and = Ve & 10 volts182 Transistor Circuit Approximations 90k 60K Fig. 8-14 Example 88. 30K: Yout ‘The a-e load resistance is 71 = 60,000|/30,000 = 20 kilohms Therefore, the maximum positive swing is Tors, = 0.333(10-4)(20) (102) = 6.66 volts and the maximum negative swing is Vo = 10 volts Clipping occurs first on the positive-going excursion. Hence, the maximum peak output signal is 6.66 volts, or 13.3 volts peak to peak. 8-4 Obtaining Maximum Unclipped Signal In this section we discuss how to obtain the largest possible unclipped signal by biasing the transistor at the optimum operating point. Specifically, consider the circuit of Fig. 8-15a. We already know that the maximum positive swing of collector voltage is limited to Jer, and the maximum negative swing can be no more than Ve. In order to obtain the largest swing possible before clipping occurs, the a-c load line must be located so that clipping occurs for equal excursions in either direction. That is, for maximum unelipped output voltage, Ve = Teri This optimum bias condition is depicted by the load lines of Fig. 8-15). The a-c load line has been positioned to allow equal excursions above and below the d-c operating point. This particular position of the a-c load line yields the largest unclipped output signal because moving it either way along the d-c load line causes premature clipping. In order to obtain a useful relation between supply voltages and load183 Large-signal Operation Vee Mee ars) 4 2k a-Clond line i. RZ Vout Yec/P i Optimum d-c operating point ‘le rcs d-c lood line = = Ee % % 2% Wee (0) {a} Fig. 8-15 Biasing to obtain maximum signal swing. resistance, observe that the stage of Fig. 8-15a has a d-c voltage of Ve = Veo ~ Icky We have already said that for maximum unclipped signal we must have the condition Ve = Tern By equating these expressions we get Ler, = Vee — IcR After solving for Ic, we have (8-5) Veo To = = for maximum unclipped signal where Ic is the d-c collector current at the optimum operating point Vcc is the collector supply voltage R_ is the d-c load resistance 71 is the a-c load resistance Equation (8-5) is quite useful; it tells us how to bias a stage to get the largest possible unclipped signal. Also notice that when a stage is biased at the optimum operating point, (8-6) the maximum available peak-to-peak output is Umarie-») = 2Ve = 2lert EXAMPLE 8-9 In Fig. 8-16, what value should Rz be in order to obtain maximum unclipped signal?184 Transistor Circuit Approximations Sonution It is obvious that R, = 10 kilohms and r, = 5 kilohms. Therefore, the correct amount of d-c collector current is In order to have 2 ma of collector current, the emitter current must be set at approximately 2 ma. Thus Vex _ 10 : ee Sal Re= Te 0.002 5 kilohms -10 430 Re SK : ie kG a mal ‘o-¢ load line 500 ey 2ma % = a 7 = " ma lOnaeetcOnseeusOs es Fig. 8-16 Example 8-9. Fig. 8-17 Example 8-10. ExaMeie 8-10 Draw the d-c and a-c load lines for the preceding example. Sonurion For optimum bias we know that the collector current is set at 2 ma. This means that the d-c collector voltage is Ve = 30 — 0.002(10,000) = 10 volts The cutoff voltage on the a-c load line is Vo + Ter: = 10 + 0.002(5000) = 20 volts The d-c and a-c load lines are given in Fig. 8-17. Notice that the possible voltage swing in either direction is 10 volts. Therefore, we are at the optimum bias condition as far as maximum signal-handling eapa- bility is concerned.Large-signal Operation 185 Exampre 8-11 In Fig. 8-18 the coil has essentially zero resistance. At the driving frequency, the reactance of the coil is so high that it acts like an open circuit to the a-e signal. The capacitors, as usual, appear as short cir- cuits to the a-c signal. Determine whether or not the stage is at opti- mum bias. Sorution The d-c load resistance Rx is zero. The a-c load resistance is simply 10 kilohms, because the coil appears open to the a-c signal. Therefore, the optimum value of d-c collector current is Vee 30 000 =3ma Fig. 8-18 Example 8-11. ‘The actual value of d-c collector current in Fig. 8-18 is Te = Ie = 5095 =15 ma Therefore, the stage is not at optimum bias. As it now stands, the maximum excursion on the positive half cycle is Terr, = 1,5(10-*)(10)(10*) = 15 volts and the maximum negative excursion is Ve = 30 volts In order for the stage to have a maximum signal-handling capability, the d-c collector current must be increased to 3 ma. This is easily accom- plished by making 2g equal to 10 kilohms. Once this is done, the stage can then deliver 60 volts peak-to-peak output before clipping.186 Transistor Circuit Approximations 8-5 p-n-p Load Lines When a p-n-p transistor is used instead of an n-p-n, the d-e voltages and currents are reversed from those of the n-p-n transistor. How does this affect the load lines discussed so far? As already indicated, all d-c formulas are exactly the same for p-n-p and n-p-n transistors if we use the magniludes of voltages and currents. In other words, for the circuit of Fig, 8-19a, the emitter current is still given by Ver de oR and the collector voltage is still given by Ve = Vee — IcRr where V¢ is the magnitude of the collector-ground voltage and Vcc is the magnitude of the collector supply voltage. te Nee +3000 -15 20K = 5K % Eee 75 5 (d} (e) Fig. 8-19 _p-n-p operation. As an example, in the circuit of Fig. 8-19b, there is an emitter current of and the collector voltage is Ve = 15 — 0.0015(5000) = 7.5 volts The polarity of the voltage is minus-plus, as shown in Fig. 8-19. The corresponding d-c load line and operating point are shown in Fig. 8-19c.Large-signal Operation 187 Thus, there is no difference in the construction of load lines for p-n-p and n-p-n transistors if we use magnitudes of voltages and currents. How- ever, there is another approach used in load-line construction that should be mentioned at this time. Consider the circuit shown in Fig. 8-20a; the direction of the d-c collector current and the polarity of the d-c collector voltage have been reversed from those shown in Fig. 8-19a. The true polarity of collector voltage and the true direction of collector current in a p-n-p transistor are opposite those shown in Fig. 8-20a. Neverthe- less, if we insist on using the direction and polarity shown in Fig. 8-20a, we must change our load-line drawing. The correct load line for the cir- cuit is shown in Fig, 8-200. We have deliberately shown the load line in the third quadrant to reflect the reversal in current direction and voltage polarity. Yee “eg te Le Pes 2. |S a a + Veo % A (a) (o) Fig. 8-20 Third-quadrant load line. The use of a third-quadrant load line to compensate for a reversed current and polarity in Fig. 8-20a may seem to be an unnecessary compli- cation, but there is a good reason for it in some applications. For instance, a typical transistor curve tracer uses the current direction and voltage polarity of an n-p-n transistor as a reference and the IV characteristics and the load line of the n-p-n transistor appear in the first quadrant. However, when a p-n-p transistor is used, the curve tracer displays the IV characteristics and load line in the third quadrant because the cur- rents and voltages are reversed in p-n-p transistor. In any event, either approach can be used in dealing with load lines. Throughout this book we will use the first method, that is, we will use magnitudes of voltage and current. This means that all d-c formulas and load lines developed for the n-p-n transistor will be the same for the p-n-p transistor. (Of course, any a-c formulas are identical for either transistor type, because each has the same a-c equivalent circuit.)FL.( 2) = Lath 188 L ‘Transistor Circuit Approximations he OC (et) = Yor Te 0 Exampie 8-12 Draw the d-e and a-c load lines for the circuit of Fig. 8 21a. Use magnitudes of voltages and currents. Soxurion The d-c load line must pass through a cutoff voltage of 30 volts and a saturation current of 3 ma. The d-c operating point is located at Ic = 1 ma and Ve = 20 volts. +15 30 e ISK 10K 366mo 3mo ima 2 30K 20 275 30 (0) (6) Fig. 8-21 Example 8-12. The a-c load resistance seen by the collector is obviously 7.5 kilohms. Therefore, the cutoff point on the a-c load line is Ve + Icrz = 20 + 0.001(7500) = 27.5 volts and the saturation current is Vo _ 20 Ie+ ae 0.001 + 7500 = 3.66 ma The d-c and a-c load lines are shown in Fig. 8-210. 8-6 Load Lines for the CE Connection When a CE connection is used, the load-line construction is the same as discussed for the CB connection. For instance, in Fig. 8-22a.and b we have the base-biased and emitter-biased CE connections. The a-c signal drives the base, and the amplified output is taken from the collector. To draw the d-c load line we merely draw a line through a cutoff volt- age of Vcc and a saturation current of Vcc/Rz. In finding Ic and Ve weLarge-signal Operation 189 use either the base-bias or emitter-bias formulas, depending upon which cirouit is being analyzed. In either ease, we plot the values of I¢ and Ve, thereby locating the d-c operating point, as in Fig. 8-22c, The a-c load line is also drawn in the usual way; that is, it passes through the d-c operating point and has a cutoff voltage of Vo ++ Ierz; the upper end of the a-c load line passes through a saturation current of To + Ve/ri, as shown in Fig. 8-22c, Mec +e ‘a-¢ load line lot Veo/Rs d=¢ operating point i d=¢ load line % Werle) (ce) Yeo Fig. 8-22 Load lines for CE connections. As observed with the CB circuit, clipping occurs when the a-c signal is too large. By inspection of Fig. 8-22, it is clear that clipping oceurs if the positive peak of the a-c signal tries to exceed Ierz; clipping also occurs if the negative peak of the a-c signal tries to exceed Vc. Thus, the maxi- mum peak-to-peak unclipped signal for either CE circuit is Qer, or 2Veo whichever is smaller. : Of course, the a-c load line can be located so that equal excursions occur in the positive and negative directions by making the d-e collector190 ‘Transistor Circuit Approximations current equal to Veo r+ Ry In this case, Ier, = Ve, so that the maximum unclipped peak-to-peak signal is simply 2V¢ (or 2[crz). All the foregoing results for the CE connections are the same as for the CB circuit. Ig = (8-7) EXaMpie 8-13 Analyze the CE circuit of Fig. 8-23a by: (a) Drawing the d-c load line and operating point. (6) Drawing the a-c load line. (c) Finding the maximum peak-to-peak unclipped output that can be obtained from this circuit. Souution (a) The cutoff voltage on the d-c load line is 25 volts, and the satura- tion current is 25 Toss ~ spgigy = 1-25 ma The d-c operating point is Io In 05 = 0.8 ma and Ve = 25 — 0.8(10-%)(20)(10%) = 9 volts The d-c load line and operating point are drawn in Fig. 8-23b. HEE 20K 50K ‘20K == -16 (a) {a} Fig. 8-23. Example 8-13.Large-signal Operation 191 (b) The a-c load resistance seen by the collector is rr = Rx||R = 20(10*)||10(10*) = 6.67 kilohms Therefore, the cutoff voltage on the a-c load line is Ve + ert = 9 + 0.8(10-4)(6,67)(10*) = 14.3 volts and the saturation current on the a-c load line is Io + i = 0.8(10-8) + = 2.15 ma L ceeOae 6.67 (105) ‘The a-c load line can now be drawn as in Fig. 8-23b. (c) Clipping occurs for 2Zcr;, or 2V'c, whichever is smaller. 2Ier, = 2(0.0008) (6670) = 10.7 volts p-p 2Vo = 2(9) = 18 volts p-p Hence, we see that the largest peak-to-peak unclipped output is about 10.7 volts. Exampie 8-14 For the preceding example, change the emitter resistor as needed to permit maximum signal-handling capability. What is the largest peak- to-peak signal available for this condition? SotuTion The optimum value of d-c collector current is found by using Eq. (8-7). v, Ie = Vee ea y 0.987 ma 7+ Rr 6.67(10°) + 2010" To get this value of d-c collector current we need to change Rg to a new value of eeVeg eG eee Re = = ggg7G0- = 17 Kilohms The maximum unclipped signal under this condition is 2V¢, or 2Zerz, since these quantities are now equal. Hence, the maximum unclipped signal is 2Ter, = 2(0.937)(10-%) (6.67) (108) = 12.5 volts p-p 8-7 The D-C Load Line for the CC Connection The CC circuit (emitter follower) must be handled in a different way from the CB and CE circuits as far as load lines are concerned. In Fig.192 Transistor Circuit Approximations 8-24a, we have a typical CC circuit. How are the d-c and a-c load lines drawn for this circuit? In Fig. 8-246, we have shown the d-c equivalent circuit. Because the collector-ground voltage is fixed at Vcc, we will now use the collector- emitter voltage vex in our load-line graphs. By applying Kirchhoft’s voltage law around the loop including the power supplies, the emitter resistor, and the transistor, we get ves + iss ~ Vez ~ Veo =0 After solving for vce, we have vox = Vax + Veo — isk (8-8) This is the equation we use for plotting the d-c load line. The cutoff voltage can be found by noting that at cutoff ig = 0. Therefore, at cutoff ves = Vex + Vee In other words, at cutoff there is no current through the emitter resistor Rg. With no current through this resistor, there is no voltage drop across it, and the net voltage across the collector-emitter terminals is Veo + Var. At saturation, the voltage across the collector-emitter terminals drops to essentially zero, that is, vcr © 0. Thus, at saturation, the current in the Mec Mec (ce) Yee Voc Fig. 8-24 D-c load line of emitter follower.Large-signal Operation 193 transistor is found as follows: ver = Vax + Veo — isRe 0 = Vez + Vee — iske Ver + Veo trea = ane Rg ae (8-9) The d-c load line is shown in Fig. 8-24c. Note that the d-c operating point of the emitter follower is located at a current of Is and a collector- emitter voltage of approximately Voc. The value of Ig is found in the usual way. That is, mw Vee Ig Ra (8-10) This formula assumes that Vez >> Vee and that Rs > Rs/8. These con- ditions are usually satisfied in any well-designed emitter follower. If necessary, the more accurate formula for Iz can be used, that is, Ver — Var Re + Rs/8 The collector-emitter voltage at the d-c operating point is approximately Voc because the emitter voltage is usually only a few tenths of a volt with respect to ground. The actual voltage across the collector-emitter terminals equals In= Ver = Vee — Ve (8-11) where Vz is the voltage from the emitter to ground. It is easily shown that Ve = —Vox — IpR Von is a few tenths of a volt; IpRp is also small, being only a few tenths of a volt in a well-designed emitter follower. As a result, Vz is usually much smaller than Ve¢ and Eq. (8-11) simplifies to Ves = Vee (8-12) Let us summarize the important points of our discussion. 1. The d-c load line of an emitter follower is a graph of ver = Vex + Veo — isRe 2. The d-c load line is a line passing through a cutoff voltage of Vee + Voc and a saturation current of (Vex + Vec)/Re. 3. The d-c operating point of a typical emitter follower is located at Ina VEE = 2 and = Ves = VeeTransistor Circuit Approximations 194 EXAMPLE 8-15 Show the d-c load line and operating point for the emitter follower of Fig. 8-25a. Souvtion Referring to Fig. 8-24e, it is clear that the cutoff voltage is Ves + Veo = 20 + 30 = 50 volts (In using this formula, note that Vex = 20 volts, not —20, because we are using magnitudes of voltages.) +30 a — Smo © 10K 10K 10K 0 2ma L le 4 = = S = 30 397 (2) (4) Fig. 8-25 Example 8-15. The saturation current is Ves+Vee_ 50) _ Re ~ 10,000 ~ 5a The d-c operating point is at 20 Ig 10,000 = 2ma and Vex = Vee = 30 volts ‘The d-c load line and operating point are shown in Fig. 8-25b. 8-8 The A-C Load Line for the Emitter Follower What is the a-c load line of a typical emitter follower like that of Fig. 8-264Large-signal Operation 195, First, realize that the d-c load resistance seen by the emitter is Rr, but the a-c load resistance is ry = RelR In other words, when a-c current leaves the emitter, it sees two paths that it can flow through. In effect, we lump these two paths into a single resistance designated rz. With an a-c signal present, excursions take place above and below the d-c operating point. With a d-c current of Zz, it is clear that the maxi- mum positive voltage swing is Ter Thus, the cutoff voltage on the a-c load line must be located at approximately Veo + Inrn Also, the saturation current on the a-c load line is approximately Tee Tr The a-c load line is shown in Fig. 8-26. As we observed with the CB and CE circuits, clipping occurs when the a-c signal is too large. Note Mee JE he at iet ae dc operating point A)in Sh Re RS Mout ke f a aa T oF “he ele acter (a) (d} Fig. 8-26 A-c load line of emitter follower. that the maximum positive voltage change from the operating point is Lert and the maximum negative voltage change is approximately Vee196 Transistor Cireuit Approximations Also, note that the optimum bias point from the standpoint of largest unclipped signal occurs when both swings are equal, that is, Ter, = Voc Therefore, the optimum d-c emitter current should be set at EXxampie 8-16 For the emitter follower of Fig. 8-274, locate the d-c operating point and draw the a-c load line. For what peak-to-peak swing does clipping occur? SoLution It is clear that the d-c current is Vea enn SOlcn le a =1ma and that Ve = Veo = 25 volts The a-c load resistance seen by the emitter is 11, = Rall = 30,000||10,000 = 7500 ohms We can now find the cutoff voltage on the a-c load line. Veo + Isr = 25 + 0.001(7500) = 32.5 volts Also, the saturation current on the a-c load line is fe 433mo 50K 30K lok Imo 30 = 25 325 (0) (0) Fig. 8-27 Example 8-16.Large-signal Operation 197 We now draw the a-c load line as shown in Fig. 8-27b. From this load line, it is obvious that clipping occurs for a positive swing of 32.5 — 25 = 7.5 volts Therefore, the maximum peak-to-peak unclipped signal is 15 volts. SUMMARY ‘The d-c load line is a locus or collection of all the possible d-c operating points of the transistor. The actual d-c operating point of the transistor must lie somewhere along the d-c load line. When an a-c signal is coupled into a transistor, the currents and volt- ages in the transistor fluctuate. These fluctuations, or changes, take place along the a-c load line. The maximum signal-handling capability of a transistor circuit is found by using the a-c load line. The excursions from the d-c operating point are limited by either the cutoff or the saturation of the transistor. By means of the a-¢ load line we can see which of these excursions is the limiting factor. Optimum bias, in the sense used in this chapter, means locating the a-c load line so that equal excursions above and below the d-c operating point can occur. By doing this, the circuit has maximum signal-handling capability for the given supply voltages. GLOSSARY cutoff In this chapter, cutoff simply means no collector or emitter current. d-c load line A graph of all the possible d-c operating points of a tran- sistor. magnitude The size of a voltage or current without regard to the actual polarity, or direction. mazimum unclipped signal The largest signal output that we can get from a transistor circuit without causing saturation or cutoff to occur. saturation This refers to having essentially zero volts across the collector diode of a transistor. REVIEW QUESTIONS 1. Where are the cutoff and saturation points on a loud line? 2. For a CB circuit, how do you calculate the saturation current on the d-c load line?198 Transistor Cireuit Approximations 3. Under what condition is the a-c load line the same as the d-c load line? 4. Why is the cutoff voltage on the a-c load line different from that on the d-c load line (assume r, # Rx)? 5. For a CB or CE circuit, what are the formulas for the maximum posi- tive and maximum negative voltage swings? 6. When a CB or CE stage is operating at optimum bias, what is the largest peak-to-peak signal that can be obtained? 7. What are the two basic ways of showing the d-c load line for a p-n-p transistor? 8. In an emitter follower, what is the maximum possible a-c swing in the positive direction? In the negative direction? PROBLEMS 8-1 In Fig. 8-28a, draw the d-e load line and show the operating point for the following values of Rx: (a) Ry = 1 kilohm. (0) Ry = 10 kilohms. () Ri = 20 kilohms. 8-2 In Fig. 8-28b, draw the d-c load line and operating point for the following values of Vee: (a) Vee = 10 volts. (6) Vee = 20 volts. (0) Vee = 40 volts. -20 #30 -20 +c -20 +30 20K A “40K lox, Ae 20K (a) (0) (c) Fig. 8-28 8-3 Draw the d-c load line and operating point for the circuit of Fig. 8-28¢ for the following values of Re: (a) Re = 40 kilohms. (6) Re = 20 kilohms.Large-signal Operation 199 8-4 In Fig. 8-29a, draw the d-c load line and operating point. Also, show the points along the d-c load line that represent peak excursions for the given L-volt source signal. -30 +30 -20 +30 Ms Ivolt. 30K 15K 20K 20K ’ 2k Ik at volt wis Ws (a) (0) Fig. 8-29 8-5 In Fig. 8-29, if the ac signal is too large, clipping will occur. Does clipping first occur on the positive or on the negative swing? What is the approximate value of source voltage that just causes clipping to occur? 8-6 Draw the a-c load line for the circuit of Fig. 8-30a. What are the maximum possible voltage excursions in both directions? What is the largest unclipped peak-to-peak voltage that can be obtained from this circuit? T ‘t so Le : cig 4 ae . (a) {a) Fig. 8-30 8-7 What approximate size should the emitter resistor Rz of Fig. 8-30a be to have equal excursions above and below the d-c operating point? 8-8 In the cireuit of Fig. 8-306, what are the maximum positive and negative swings before cutoff or saturation is reached? What approximate size should Rz be to have the largest signal-handling capability?200 Transistor Cireuit Approximations 8-9 Draw the d-c and a-c load lines for the circuit in Fig. 8-31. Ik Fig. 8-31 % ee 8-10 What is the peak-to-peak value of the largest unclipped signal available from the circuit of Fig. 8-31 with the given bias conditions? If the emitter resistor were changed so as to produce the optimum operating point, what would the largest peak-to-peak output become? 8-11 Draw the d-c and a-c load lines for the CE cireuit of Fig. 8-32a. What will happen to the d-c and a-c load lines if the 8 of the transistor changes to 50? +30 o ae 40k ia 1 i 20K. (a) (ah B75 § ee Fig. 8-32 8-12 Draw the a-c load line for the circuit of Fig. 8-32b. What happens to the location of the a-c load line if the 8 changes to 50? To 100? 8-13 Inorder to have optimum bias in the CE circuit of Fig. 8-32a, what size should the base resistor be? 8-14 Draw the d-c and a-c load lines for the emitter follower of Fig. 8-384,Large-signal Operation 201 +20 +20 Fig. 8-33 8-15 In the emitter follower of Fig. 8-33a, for what values of a-c output voltage does clipping occur on each excursion? 8-16 Draw the d-c and a-c load lines for the emitter follower of Fig. 8-33b. What is the largest unclipped peak-to-peak signal that can be obtained from this circuit?Bias Arrangements Up to now, we have discussed base bias and emitter bias. Recall that as far as setting and holding a d-c operating point is concerned, base bias is the worst way to bias, and emitter bias is the best way. In between these two extremes there are a number of widely used bias arrangements. In this chapter we will study six of the most common ways to bias a transistor, emphasizing those concepts and formulas which are most use- ful in practice. 9-1 The Concept of § Sensitivity The d-c operating point in some biasing arrangements shows a heavy dependence upon the exact value of 6, whereas in others there is almost no dependence upon the 8 value. Generally speaking, it is far more de- sirable to have a biasing circuit in which the value of 8 does not matter; in this way, a change in the 8 value will not disturb the desired d-c oper. ating point. (Changes in 8 can occur when the temperature changes or when the transistor is replaced.) In this section we wish to discuss the concept of 8 sensitivity, that is, 202Bias Arrangements 203 the influence that the 8 value has on the d-c operating point. To begin our discussion, consider the cireuit of Fig. 9-1a. The base current in this circuit is clearly et Veo _ 20 Veo — Var Voc _ 20 _ © ches 7 tee In Fig. 9-1a, the 8 is given as 100; therefore, the collector current is Ic = BIn = 100(20 ya) = 2 ma The collector voltage with respect to ground equals the supply voltage minus the drop across the load resistor. That is, Ve = 20 — 2(10-%)(5)(10%) = 10 volts Also, note that the cutoff voltage on the d-c load line is 20 volts, and the saturation current is 4 ma. Thus, we can draw the d-c load line and operating point as shown in Fig. 9-1b. 2040 Fig. 9-1 Base-bias sensitivity. From our earlier work with this base-bias circuit, we know that if for some reason the 8 changes, this will change the d-c operating point. For instance, suppose that the 8 changes from 100 to 110. We can write this as 8+ 46 = 110 where 8 is the original value of 8 and 4@ is the change in the 8 value. Thus, in this case, the change A8 in 8 is 10. The percent change in 6 is defined as the change divided by the original value. That is, AB Pereent change in 6 = Therefore, when 8 changes from 100 to 110, we say that the percent204 Transistor Cireuit Approximations change in @ is 48 _ 10 37 107 0.1 = 10% We can also speak of the percent change in the collector current. In the circuit of Fig. 9-14, when the 8 changes from 100 to 110, the base current remains essentially the same. Ine 90 ya as before Re The collector current, however, now becomes Te + Ale = 110(20 pa) = 2.2 ma where Ic is the original value of collector current and Alc is the change in collector current. In this particular case, the change in collector current is 0.2 ma; therefore, the percent change in collector current is Ale _ 0.2 ma Ic ~ 2ma = 0.1 = 10% Thus, we have seen a 10 percent change in 6 cause a 10 percent change in collector current. In other words, Alc _ 48 crags for base bias ‘This is quite bad; it means that the d-c operating point shifts signifi- cantly when the 8 changes. ‘There are other ways to bias a transistor, and we will find that the percent change in collector current ean be less than the percent change in f for these circuits. In general, the percent change in collector current is Ale _ x 48 Tc 8 where K is a constant of proportionality. Equation (9-1) tells us that the percent change in the collector current equals a constant times the per- cent change in 6. We will call K the 8 sensitivity because the value of K indicates how sensitive the collector current is to changes in 8. For the base-bias cireuit of Fig. 9-1a, we have seen that K equals unity. In the remainder of this chapter we will discuss bias arrangements in which K is less than unity. In general, the sensitivity K is between 0 and 1, with the least sensitive bias arrangements having K values approaching 0. For instance, a good biasing circuit can have a K as low as 0.01. In a (9-1)Bias Arrangements 205 circuit like this, a change of 10 percent in 8 would cause a change of only 0.1 percent in collector current. 9-2 Base Bias In this section we summarize the important formulas for the base-bias circuit of Fig. 9-2a. First, note that the saturation value of collector cur- rent occurs when the collector-emitter voltage is approximately zero. Thus, in Fig. 9-2a it is clear that Vi Tota = (9-2) This is the maximum value of d-c collector current for the base-bias cir- cuit. For normal operation the actual d-c current must be less than this value. The actual d-c collector current equals Ic = 6In (9-3) The base current, as already indicated, equals i aie a Yee (0-4) The collector-ground voltage Ve simply equals Ve = Vee — Ick (9-5) Finally, we have seen that for a base-biased circuit the sensitivity is K=1 (9-6) Figure 9-2a summarizes the important biasing formulas. Me to 4 © ts00) 2 e/A, Re i 49 Veo/Fy le=Blp {elsat)h--—- + VeVeeleh Ra OSlelsatih~~——)-- > fy 7) 7 Fig. 9-2 Base bias.206 Transistor Cireuit Approximations Note that if we substitute Eq. (9-4) into (9-3), we get Ic = B= or ~ Vee a Rn/B 7) It is helpful to graph this equation for I¢ vs. Rz, as shown in Fig. 9-2b. ‘This graph tells us that when Rz is less than 6R;, the transistor is satu- rated; therefore, for normal operation Rp must be greater than BR,. As shown in the graph, when Ry = 26Rz, the d-c collector current is one- half of the saturation value. EXAMPLE 9-1 In Fig. 9-3a, what value of 8 just causes saturation? Sonurton By inspection of the graph in Fig. 9-2b we can see that saturation occurs when Ry = BR, With Re = 50 kilohms and R, = 1 kilohm, we compute a 8 of g— Re _ 50010) _ Ri 10 Thus, if the 8 is more than 50, the transistor will saturate. Fig. 9-3 Examples 9-1 to 9-3. Exampie 9-2 For the circuit of Fig. 9-3b, find the following: (a) The saturation value of collector current. (b) The value of Rg that produces a collector current equal to one- half the saturation value.Bias Arrangements 207 SoLution (a) The saturation value of collector current is Vee 25 Lorene) = %, = 3000 72m (b) From the base-bias characteristic of Fig. 9-2b we see that the collector current is one-half the saturation value when Rg = 26R, = 2(75)(5000) = 750 kilohms EXAMPLE 9-3 Yor the circuit of Fig, 9-36, let Rr = 1 megohm and find the value of Ic and Ve. SoLvurion Using Eq. (9-7), we find that ~ Vee Ico R/b The collector-ground voltage is Ve = Veo — IcR1, = 25 — 1.88(10-8)(5)(108) = 15.6 volts 9-3 Base Bias with Emitter Feedback One way to reduce the sensitivity of collector current to changes in 8 is to use the circuit shown in Fig. 9-4a. The additional resistor in the emitter circuit causes degeneration to take place, thereby stabilizing the d-e operating point against changes in 8. To understand the stabilizing action, assume that the 8 of the transistor increases for some reason. This increase in will cause an increase in the emitter current, which, in turn, causes the voltage across Rg to increase. However, now that the voltage from the emitter to ground has increased, the voltage across the base resistor Rp will decrease, thereby reducing base current, which, in turn, partially compensates for the increase in B. Some of the quantities that are useful in analysis and design are the following. First, the saturated value of collector current, which is found as follows. At saturation, vce is essentially zero, and the supply voltage is distributed across R, and Ry. Since base current is much smaller than collector current, we have Z Town = 15 (9-8)208 ‘Transistor Circuit Approximations Aplsat) eeAR+R) Le MecMFgthy/B) Yo Yeo~ hh eebhthhe Fe Yor = Vo-Ve K=1/4BRe/Re) ae Zo ot) % 0.5ée(sat) Fg BR BlRe+ 2h) (d) Fig. 9-4 Base bias with emitter feedback. In the normal operating region of the transistor the collector current is less than the saturation value given by Eq. (9-8). We can find the formula for the actual collector current as follows. First, we write a voltage equation including the supply voltage, the drop across Rp, the Var drop, and the drop across Rz. That is, Veo = IsRs + Vax + [eRe Since Ip = Ic/8 and Ig = Ic, we can simplify to get Veo = 42 Rs + Vox + IoRs Solving for Zc, we obtain Tom Veo = Vox * Re + Ra/B As before, we take advantage of the fact that usually Vaz Voc. Hence we simply have 7 Veo = Re + Ra/8 Remember that this formula is valid as long as the transistor is not saturated, that is, Jc must be less than Ica. The formulas for voltages are for Vaz « Vee (9-9) Ve = Veo — IcRi (9-10) and Ve = IgRz =IcRe (9-11)Bias Arrangements By means of calculus, the sensitivity is found to be 1 K= TE GRE/Rs Ca Figure 9-4a summarizes the important biasing formulas. As before, we can show a sketch of Ic vs. Rs (Fig. 9-40). This graph tells us that when Rs is less than BRz, the transistor is saturated. For normal operation, Rs must be greater than 8Rz. In fact, note that if Rp = 6(Rz + 2Rz), the value of Ic equals one-half the saturation value. Exampie 9-4 For the circuit of Fig. 9-5a, find the saturation value of collector current and the actual value. Sonution The saturation value is Vee _ 20 Ret+Rz, 3000 The actual value of collector current is oe Vece 20 a Te = Re+ Ra/B ~ 1000 + 30001057100 ~ > ™* Toceat) = = 6.67 ma EXampie 9-5 For the circuit of Fig. 9-5a, find Vc, Vz, and K. SoLution ‘We found in Example 9-4 that the colleetor current in this cireuit is Ic = 5 ma. Therefore, the collector-ground voltage is Ve = 20 — 5(10-*)(2)(10*) = 10 volts +20 +30 300K: 2K Re 5K Fig. 9-3 Examples 9-4 and 9-6, 100 100 1K 1OK210 ‘Transistor Circuit Approximations and the emitter-ground voltage is Vz = IsRe = 5(10-4)(108) = 5 volts Using Eq. (9-12), we can find the sensitivity to changes in 8. K = 0.75 1 ~ T+ 100(10%)7300(10°) A sensitivity of 0.75 means that the circuit is less susceptible to changes in 6 than the ordinary base-bias circuit without feedback. For instance, if the 8 were to change 10 percent, the collector current in the circuit of Fig. 9-5a would change about 7.5 percent. EXAMPLE 9-6 In Fig. 9-5b, use a 8 of 100 and find the following: (a) The value of R» that just causes saturation. (6) The value of Re that produces a collector current of one-half of the saturation value. (c) The sensitivity K at one-half saturation value. So.urion (a) In the design characteristic of Fig. 9-4b we can see that satu- ration occurs when Re = BRz = 100(5000) = 500 kilohms ‘Thus, for normal operation Rx must be greater than 500 kilohms. (6) Using the characteristic of Fig. 9-4b, we get Ry = A(Rz + 2Rz) = 100(10,000 + 10,000) = 2 megohms (c) The sensitivity is fe 1 pee ae K = Ti Re] Re ~ TF TOOT ACOH ~ 9-687 9-4 Base Bias with Collector Feedback Another common bias arrangement found in practice is the circuit shown in Fig. 9-6a. It is like base bias, except that the base resistor is returned to the collector instead of the Vcc supply. Because the voltage for the base resistor Rz is derived from the collector, there is a negative-feedback effect that tends to stabilize the collector current against changes in 8. To understand the stabilizing action, assume that @ increases. This will increase collector current, which then causes the collector voltage to drop.Bias Arrangements 211 However, with less collector voltage applied to Rz, the base current will decrease and partially compensate for the original change in 8. The important d-c operating formulas can be found as follows. First, we know that saturation occurs when cx is essentially zero, Under this condition, all the supply voltage appears across R;. Note that the total current in A, is the sum of collector and base current. Because the base current is much smaller than the collector current, however, we can neg- lect it and get a saturation value of iz Tota = RE (9-13) This represents the maximum value of collector current. To find the actual collector current, we need to write some voltage equations. As usual, the collector voltage equals the supply voltage minus the drop across Ry. That is, Ve = Vee — Ue + In)Ri = Veo — IcRi From the circuit of Fig. 9-64, we note that the collector-ground voltage must also equal Ve = InRn + Vaz Therefore, we can equate these last two expressions to obtain Isha + Vaz = Veo — IcRt Since Iz = Ic/8, we can rewrite this equation as I gks + Vor = Veo — Ichi Zelsat) Yor /R, Le™ Voc MitRe/B) Totla Yo Veo~ LR | K = A+B RRQ) Rp (a) (a) Fig. 9-6 Base bias with collector feedback.212 ‘Transistor Circuit Approximations Solving this equation for Ic, we get the desired result. Veo — Var Rit Rs/8 Again, we note that Vzz is usually small enough to neglect. Hence, our final practical formula for the collector current becomes Io = = Vee = Ic= Rr + alo for Vez K Vee (9-14) The circuit of Fig, 9-6a has a sensitivity of less than unity. By means of calculus, the sensitivity is found to be 1 K=,— pp -15) 4 6Ri/Re (9-15) An alternate expression for the sensitivity is Roe (9-16) Tevsat) Equation (9-16) is quite useful. It tells us how the sensitivity is related to the operating point. For instance, the limits on Ic/Ieat are Ic Tees where the lower limit represents the cutoff condition and the upper limit represents saturation. Therefore, according to Eq. (9-16), the sensitivity at cutoff is 1; at saturation the sensitivity becomes 0. Thus, with the cir- cuit of Fig. 9-6a, the lowest sensitivities occur when the transistor is al- most saturated. The biasing formulas are summarized in Fig. 9-Ga. The sketch of Ic vs. Re is given in Fig. 9-6. This characteristic tells us that the circuit of Fig. 9-6a does not saturate sharply, like the two pre- ceding bias arrangements. As Rg is made smaller, the circuit approaches saturation, but it never quite saturates until Ra = 0. Note also that when Re = BRz, the collector current is one-half of the saturation value. EXAMPLE 9-7 The transistor in Fig. 9-7a has a 8 of 50. Find the approximate value of collector current and voltage. Sonvrion The collector current is ao ara ear aeaesese 20 oO T= BF Ra/B ~ T0®¥ 10750 ~ 3000 ~ 97 ™Bias Arrangements 213 The collector voltage is Ve = Veo — IcRi = 20 — 6.67(10-*)(10*) & 13.3 volts EXAMPLE 9-8 What is the value of K for the circuit of Fig. 9-7a? SoLution We can use either Hq. (9-15) or (9-16). If we use Eq. (9-15), we find 1 K = resoaoy tooo = 9-997 Or, we ean use Bq. (9-16), to find K=1- 1 = 0.667 +20 +30 kK 2k look Ro Fig. 9-7 Examples 9-7 to 9-10. 8 =50 )B=50 (a) (0) ExaMpLe 9-9 Suppose that Ay in Fig. 9-7b is equal to 10 kilohms. Find the value of Ic/Tevat and the value of K. SoLuTIon We can use Eq. (9-13) to find Icon. Veo _ 30 _ Tow = = a999 = 1 ma With Eq. (9-14) we can find Ic. Vee 4 = 13.6 ma to= RF Ry/B ~ 2000 + 10,0 Therefore, 50 Te — 13.6 ma Te(sae) 15 ma = 0.907 A value of 0.907 means that the transistor is near the saturation point.24 Transistor Circuit Approximations To find the sensitivity, we use Eq. (9-15) or (9-16). In this case, it is easier to use Eq. (9-16) because we have the value of Ic/Tcieat worked out. To Tevsaty K=1- = 1 — 0.907 = 0.093 This is a low value of sensitivity; therefore, the circuit of Fig. 9-7b is only slightly susceptible to changes in 8. For instance, a 10 percent change in 6 will only cause a change of 0.93 percent in collector current; however, the transistor is almost at the saturation point, so that the signal-handling capability of the circuit is limited. Exampie 9-10 Using a 6 of 50, find the following for the circuit of Fig. 9-7b: (a) The value of Rp that produces a collector current equal to one- half the saturation value. (0) The sensitivity K at one-half saturation current. SoLution (a) Referring to Fig. 9-6b, we see that collector current equals one- half saturation current when Re = BR, = 50(2000) = 100 kilohms (6) With the collector current at one-half the saturation value, we can find the sensitivity easily by using Eq. (9-16). Te K=1— Leveat) =1-05=05 9-5 Base Bias with Collector and Emitter Feedback There is one more variation of base bias that we want to discuss. The circuit of Fig. 9-Sa uses both collector and emitter feedback in an attempt to lower the sensitivity to changes in g. In this circuit, an increase in 8 results in a larger emitter voltage and a smaller collector voltage. This means that the voltage across Ry is reduced, causing the base current to become smaller, thereby partially offsetting the increase in 8. To find the saturation current, we note that when veg is approximately zero, the supply voltage is distributed across Rz and Re. Since Is is muchBias Arrangements 215, smaller than Jc, and since Ic is almost equal to Iz, we conclude that Ve Tews = RT (9-17) The actual collector current Ic can be found by writing a voltage equation. Note that Vee = Uc + In)Ri + InRe + Var + Inke With our usual approximations, this equation becomes Veo x FoR +! Ry + Vox + Toke After solving for Ig, we have Tp= Veo — Vax °* Rs + Ri + Ra/B tee Ufsatl™ YoeRe+R) Vette fe2 oR Hy /8) ch (ee lofe Le Yog= ee a Kev / yeeuneravne) Le(sat} Saturation = OSJo(sat) BUAEtA) (a) (4) Fig. 9-8 Base bias with collector and emitter feedback. When Vaz is negligible compared to Vcc, this becomes lox a ea for Var Vee (9-18) The collector-ground voltage Vo is approximately Wee vic tae (9-19) and the emitter-ground voltage is VexlcRe (9-20)216 ‘Transistor Circuit Approximations A calculus derivation will show that the sensitivity is given by 1 fq niece 21 T+ ORs + Ri)/Re ae A useful alternate formula for sensitivity is Roe (9-22) Teva The biasing formulas are summarized in Fig. 9-8a. Finally, the sketch of Zc vs. Rx is shown in Fig. 9-8b. Again note that this graph implies a soft saturation, that is, the circuit does not com- pletely saturate until Rp is zero. EXxampLe 9-11 For the cireuit of Fig. 9-9a, find the following: (a) Tecat)- (b) Ve and Veg for a B of 100. (c) The percent change in Jc when 6 changes 10 percent. SoutTion (a) At saturation, the supply voltage is distributed across Rz and Re. Hence, 20 Tews = 19,000 + 10,000 = (6) To find the collector eurrent we use Eq. (9-18). Te 20 © = 108 + 104 + 5(10°) /100 With 0.8 ma of collector current, there must be a drop of 8 volts across the 10-kilohm collector resistor. Therefore, Ve = 20 — 8 = 12 volts lma = 0.8 ma The voltage from emitter to ground is simply Vx = 0.8(10-*)(10)(10*) = 8 volts The voltage from the collector to the emitter is the difference of Ve and Vz. Thus, Ves = Vc — Vg = 12 — 8 = 4 volts (©) We can find the sensitivity by using Eq. (9-22). Io _ , _0.8ma_ Tees a trcuiantida K=1-Bias Arrangements 217 +20 +20 iok 10K 500K 2M. Fig. 9-9 Examples 9-11 and 9-12. 2100 100 1OK 10K (a) (d) This value of sensitivity is fairly good; it means that a 10 percent change in 8 causes a change of about 2 percent in collector current. Examp.e 9-12 When we change the base resistor of the preceding example from 500 kilohms to 2 megohms, we obtain the circuit of Fig. 9-9b. For this new cireuit, find cgay, Vex, and K. So.vution The saturation current remains at 1 ma because Rz, Rez, and Voc are the same as before. The new collector current is To wap— Ve OC" Ke + Ri + Ra/B 0 + 108+ 2009) /100 Note that this collector current is one-half of saturation current. The sensitivity of the new circuit is also different. Using Eq. (9-22), we find that =0f ma To _ _0.5ma_ Ke=1 0.5 ~ Teens T ma Note the degradation of sensitivity that has occurred by inereasing the size of the base resistor. In the preceding example, Rx = 500 kilohms, and K = 0.2. In this example, Ry = 2 megohms, and K = 0.5. 9-6 Emitter Bias with Two Supplies We have now discussed four variations of base bias. In all these circuits, the value of Rs is important in setting up the desired collector current.218 ‘Transistor Circuit Approximations For the simplest form of base bias (no feedback) the sensitivity equals unity; this represents the worst way to bias as far as stability of the d-c operating point is concerned. When feedback is added to the base-bias cireuit, the sensitivity is reduced. We saw that the last two forms of base bias had sensitivities of Tecan Clearly, by almost saturating the transistor, we can get very low values of K; however, we pay the price of limited signal-handling capability when we do this because the possible a-c signal swing is limited. What we really want is a bias circuit that shows almost no dependence upon the value of 8, that is, a bias arrangement whose K is low no matter what the ratio of Zc to eau. The emitter-bias circuit discussed in Chap. 6 is such an arrangement. In Fig. 9-10a, we have shown the d-c equivalent circuit for an emitter- biased cireuit. Recall that the voltage from emitter to ground is approxi- mately zero, provided that the V gg and IpRp voltages are small enough to neglect. (As already indicated, the designer deliberately makes sure that these conditions are met.) Therefore, at saturation the voltage from collector to ground is almost zero, and we have a current of Ki Tew = Re (9-23) Tn Chap. 6 we showed that (9-24) The key to a good design, that is, one in which the collector current is only slightly dependent upon the transistor characteristics, is to make Vex > Voz and Ra Re PR By doing this, the collector current is essentially given by Ves Re This last equation tells us that to a first approximation the value of Ie depends only upon Vex and Rg; it does not depend upon the exact value of 8. In other words, by ensuring that Re 3 Re/B, we are freeing the col- lector current from a heavy dependence upon the value of B. This, of course, is most desirable from the standpoint of a stable d-e operating point. c= (9-25)Bias Arrangements 219 The expression for the collector-ground voltage is Ve = Veo — IcRy (9-26) As already indicated, the emitter-ground voltage is approximately zero for a well-designed circuit. If necessary, this low value of emitter-ground voltage can be found by using Ve = —(Var + InRs) = — (Yee +Tc ) (9-27) For example, suppose that a germanium transistor is used and that 8 = 100 and Re = 10 kilohms. For 1 ma of collector current, we calculate 10" Ve 2 10° ® (03 10-755 ) = —0.4 volt Thus, the emitter is slightly negative with respect to ground. Mec For Rg << BAe ond Yar K Vee Jdlsat) ® Yoe/R, Re ‘“ lec (A, iE ft K=\/U+BRe/Rg) — Ielsailh Pa KBR ond Vag << Veg 0.5/Gsat)| Fg Fe Ae + let y, pitt p : a Yee Veo “Yoo (o) (5) Fig. 9-10 Emitter bias with two supplies. With calculus, the sensitivity can be derived. 1 = 28 ¥ ORe/Ra (9-28) As an example of using this equation, suppose that Rg = zg; then the sensitivity is For a8 = 100, K = 0.01. For 8 as low as 20, K = 0.05. Thus, we can get extremely low values of sensitivity using emitter-biased circuits.220 Transistor Cireuit Approximations Of course, if Ra is made too large, the sensitivity may become ob- jectionable. For instance, suppose that Re = 10Rx. Then, 1 Fe etc eee ee ~ 1+8Re/l0Rze 1+ 8/10 For a 8 of 100, we get a K = pray, = 0.081 Or, if the @ is as low as 20, we get K=—4,- = 0.838 Tre2040 cece: Under this condition, the circuit is becoming too sensitive to changes in B; we are, in fact, defeating the whole purpose of the emitter-biased circuit if we make Rz too large. To avoid this, we simply need to satisfy the general rule already given for a well-designed emitter-bias circuit. That is, Rs Rr> B By satisfying this inequality, we justify the use of Eq. (9-25) and ensure a low value of sensitivity. A sketch of Ic vs. Rx is shown in Fig. 9-10b. Note that Rx is used instead of Ry because in a well-designed emitter-bias circuit, the collector current is controlled by Rx instead of Rx. Exampie 9-13 For the circuit of Fig. 9-11, find the approximate value of (a) Tear) and Ic. (6) Ve, Ve, and Ver. (© K for a 8 of 50. Sovtion (a) Using Eq. (9-23), we get I Moca ial Coe) = "Ri 1010") With Eq. (9-25), we find thatBias Arrangements 221 +20 Fig. 9-11 Example 9-13. Silicon B=50 10K 20K -10 (0) The collector voltage with respect to ground is Ve = Veo — IeR, = 20 — 0.5(10-4)(10)(108) = 15 volts The approximate value of the emitter-ground voltage is zero. If we like, we can use Eq. (9-27) to get a more accurate value. Ve=— (Ye +Te 7) = - (0. 7 + 0.0005 oe) = 0.8 volt The collector-emitter voltage is Ves =Ve-Va= 5 — (-0.8) = 15.8 volts For practical purposes, we can neglect the emitter voltage with respect to ground and say that Vex = 15 volts. (©) To find the sensitivity, we use Eq. (9-28). 1 1 K = TE BRe/Re ~ 1+ 108/108 R 0.01 EXxampie 9-14 In Fig. 9-12, find the value of Rg that produces a collector current of one-half the saturation value. Also, to ensure a low sensitivity select a value of Rz that is }o8Re. Use a 8 of 100. Soivution The characteristic of Fig. 9-106 tells us that I¢ equals 147 eax) When = 220 5000 = 6.67 kilohms222 Transistor Circuit Approximations The value of Ry must be chosen so as to satisfy Ry» Be or Re X6Re 8-100 Fig. 9-12 Example 9-14, % fe Since Re = 6.67 kilohms and we are given a 6 of 100, we must ensure that Rs < 667 kilohms If we make Rs equal to 1406Rz, we get Re = 1%0(667 kilohms) = 33.3 kilohms In a practical design, we would use the nearest standard values, that: is, Re = 68 kilohms and Rp = 33 kilohms. Also, note that in a practical design we would take the 6 spread into account. In other words, for the transistor type that is being used, the 8 may be specified as between 50 and 150. In this case, we would use the worst case, which is a 8 of 50. If Rp is to remain less than }498Rz for all transistors of the given type, we must reduce it to about 16.6 kilohms (15 kilohms for a standard value). 9-7 Emitter Bias with One Supply The emitter bias circuit studied in the preceding section is an excellent low-sensitivity circuit, and is used whenever two power supplies are avail- able. Note that both a positive and a negative supply are required. There are many oceasions, however, when only a single power supply is avail-Bias Arrangements 223 able. In this case, the modified form of emitter bias shown in Fig. 9-13a ean be used. At saturation, the collector-emitter voltage drops to approximately zero, and all the supply voltage is distributed across R, and Rg. Since the collector and emitter currents are essentially equal, we can write Tevet) (9-29) When the cireuit of Fig. 9-13a is well designed, it operates as follows. Resistors 21 and Rs form a voltage divider, so that the voltage from base to ground is approximately ~ Ra Ve= Ret Mee oao) By inspection of the circuit we can see that, Ve =Var+ Ve Because Vez is only a few tenths of a volt, it is usually negligible, and we can write VazVe In other words, to a first approximation, the voltage across Rr is approxi- mately equal to the voltage across 2s. Therefore, the emitter current is Ve l= R, For lla KBRe and Voc <& Vo Yoo MRetRe) Fa Yee Fas “he leh, 7 flee =leAe fc KeV/BRAR IA f Jelsat}- RVR BRe .Sielsat}> a A, R, Hyp i Fe fe (oy (a) Fig. 9-13 Emitter bias with one supply.224 Transistor Cireuit Approximations Finally, since the collector current almost equals the emitter current, we can write Ve . To= Re (9-31) Equation (9-31) is a simple approximation for the d-c collector current that flows in a cireuit like that shown in Fig. 9-13a. It tells us that the collector current equals the ratio of the base voltage (developed by the voltage divider) to the emitter resistance Rz. The collector-ground voltage is simply Ve = Vae — IeRt and the emitter-ground voltage is Va = InRs = Icke Calculus can be used to prove that the sensitivity is K (9-32) 1 ~ T+ BRe/(Ril/Ra) where || is the parallel resistance of R and Ro. Finally, a sketch of Ie vs. R1/R2 is shown in Fig. 9-135. All the foregoing results have assumed a well-designed circuit. What is a well-designed circuit? To answer this question, we must make a more accurate derivation for the collector current. The easiest way to do this is to replace the voltage divider in the base circuit by its Thévenin equiva- lent circuit, as shown in Fig. 9-14. Summing voltages around the base loop, we get Be ks Veo = In(Rills) + Ver + IsRe Fig. 9-14 Thévenin equivalent cir- cuit. Re A+R teBias Arrangements 225 If we substitute Ic & Ig and Ic/8 = Iz, we can solve for Io, to get — Vecks/(Ri + Rs) — Re + (RillR2)/B Equation (9-33) is more accurate than Eq. (9-31); however, the whole point of the circuit is that it should be insensitive to changes in 8. To accomplish this, it is good design practice to make fuls Ie (9-33) Rr>— Also, when possible, V zg is made i. by ensuring that Ry ier Veo > Vaz (The left member of this inequality is simply Vn, the voltage from base to ground.) To summarize these important results, refer to Fig. 9-15. In a circuit satisfying the two inequalities just given, we can say the following. The We Fig. 9-13 Emitter bias with one sup- ply. AR amount of base current Iz is small compared to the current J flowing down through the voltage divider. Because of this, the voltage divider is lightly loaded by the base of the transistor. In other words, Ry Ve25— Sp MV e=h+Rk, © Further, since Vag is small, almost all the base-ground voltage appears across the emitter resistor. That is, Ve2Vs226 Transistor Circuit Approximations Finally, the collector current approximately equals the emitter current, so that Ve To = Rs The circuit of Fig. 9-15 is basically a form of emitter bias because we set the emitter current to an approximately fixed value of Vg/Rz. As far as sensitivity is concerned, this circuit can have a very low sensitivity. An inspection of Eq. (9-32) immediately shows that to have a low sensi- tivity, we need only make R||R: comparable in size to Rs. For instance, for R,||R2 = Re we get a sensitivity of 1 1 “T+ BRe/(RiiR2) 1+8 For a 8 of 100, K = 0.01. Thus, as long as the parallel combination of Ry and Rz is not too large compared to Rz, we will have a low sensitivity. K EXampLe 9-15 For the cireuit of Fig. 9-16a, find the approximate value of collector current. SoLution By inspection of the circuit, we see that the voltage divider delivers a voltage of about 10 volts to the base. Most of the 10 volts appears across Rg because of the small Vag drop. Therefore, the collector cur- rent is 10 Io 10,000 ~ 1 ma +30 +20 20K 10K lok 2k Fig. 9-16 Examples 9-15 and 9-16. 10K 10K 0K akBias Arrangements 227 EXampie 9-16 For the circuit of Fig. 9-16), find the following: (a) Toca and Ie. (6) Ve, Ve, and Ver. Sonurion (a) Saturation occurs when the collector-emitter voltage is approxi- mately zero. Thus, Vegi 2 20) Teva) © Re + Ry ~ 6000 ~ 3.33 ma To find the actual collector current, we note that the voltage divider in the base develops about 10 volts from base to ground. Most of this 10 volts appears across the 4-kilohm emitter resistor. Therefore, the approximate value of collector current is Te = yp0y = 25 ma (b) The collector-ground voltage is Ve = Vee — IcRy, = 20 — 2.5(10-5)(2)(108) = 15 volts and the emitter-ground voltage is Ve =Va ~10 volts The collector-emitter voltage is the difference of Ve and Vx. Vee = Vc — Vg = 15 — 10 = 5 volts Examp.e 9-17 Find the collector current and the sensitivity of the circuit shown in Fig. 9-17a. Use a 8 of 50. Sotvtion The voltage divider develops about 5 volts from base to ground. Therefore, the emitter-ground voltage is about 5 volts, and the col- lector current is approximately Tow ye = On c= Re 5000 The sensitivity can be found by using Eq. (9-3 K Be 1 = ~ T+ BRe/(RilR2) a gz 3 & £ 3 s228 ‘Transistor Circuit Approximations +20, +20, ‘30K 2K a 2K A=50 Fig. 9-17 Examples 9-17 and 9-18. 0K ‘SK TOK ‘SK = (a) ~ a (d) c EXAMPLE 9-18 Find the value of R1 in Fig. 9-17b that produces an Ic equal to 34love: Sonurion We can easily find the correct value of R1 by referring to Fig. 9-13b. To get an Ic of 47 oan, we need an Ry/R; ratio of Ry _, , 2R, _, , 4000 _ Rett REA 1+ go99 = 18 Ry = 1.8R2 = 1.8(10,000) = 18 kilohms 9-8 A Comparison of Sensitivities We have discussed six common ways of biasing a transistor. Generally speaking, the base-bias arrangements show much more sensitivity to changes in @ than the emitter-bias arrangements. As already indicated, a low value of K is desirable if a stable d-c operating point is required. In order to compare the various circuits with each other, look at Table 9-1. The various sensitivities have been calculated on the basis that all circuits are operated at one-half the saturation current. That is, in all circuits, To = 0.5L c¢x1) Further, in those circuits having additional degrees of freedom, arbitrary conditions have been imposed. For instance, for emitter bias with twoBias Arrangements 229 supplies, Rp has been arbitrarily set equal to Rx, and a 6 of 50 has been used. Table 9-1 shows the overwhelming superiority of emitter-biased cir- cuits over base-biased circuits insofar as stability of the operating point is concerned. This is quite important in class A amplifiers if we are to avoid clipping due to a shift in the d-c operating point. It is also impor- tant in d-e amplifiers, where a d-c shift caused by 8 changes is indis- tinguishable from an actual signal. Table 9-1 Comparison of Beta Sensitivities* Circuit Conditions Base bias Base bias with emitter feedback Base bias with collector feedback 0.5 ias with collector and emitter feedback 0.5 Emitter bias with two supplies Base Ry = Rg, B = 50 Ril: = Re, B = 50 Emitter bias with one supply * All circuits compared at Io = O.5Te(auy- 9-9 Location of the Ground Point So far we have shown the ground point at a typical point in the biasing arrangement. There will be times, however, when we want the ground point at some other place in the circuit. To show that we can move the ground point, consider the simple base- bias circuit of Fig. 9-180. We already know that the base current is fixed in this circuit by the value of Vec and Rx. The collector current simply equals 8 times the base current. ‘The circuit of Fig. 9-18a can be redrawn as in Fig, 9-180. Realize that ground is only a reference point; the biasing of the transistor does not depend upon having a ground point. For instance, we can remove the ground altogether, as shown in Fig. 9-18c. In this case, the base diode is still forward-biased, the collector diode is still back-biased, and the value of collector current is exactly the same as before. At times, we will want to locate the ground point on the end of the collector supply, as shown in Fig, 9-18d. Again note that the ground point230 ‘Transistor Circuit Approximations has no effect on the collector current; the transistor js still biased to the same Veg, Ie point as before. We can redraw Fig. 9-18d as shown in Fig. 9-18¢. This circuit has exactly the same value of collector current. as the original circuit of Fig. 9-18a. The collector-emitter voltages in each circuit are also equal. In effect, all we have done is to locate the ground, or reference, point at another position in the circuit. “Yee c) — A Nog (a) (e) Fig. 9-18 Changing the ground point on a base-bias circuit. Of the six biasing arrangements, five use a single supply. Up to now, we have grounded the negative end of this supply for n-p-n transistors. We can ground the other end of this supply as we have just done for the base-biased cireuit. When this is done, the five single-supply bias circuits appear as shown in Fig. 9-19. In all these circuits, we can find the d-c col- lector current by using the formulas developed in earlier sections. The collector-ground voltage is different from before because we have moved the ground point. By inspection of the circuits in Fig. 9-19 it is clear that the collector voltage in each circuit is negative with respect toBias Arrangements 231 a) (ey Fig. 9-19 Changing the ground point on all single-supply circuits. ground, and the magnitude is given by Ve = Ick, (9-84) EXxaMpLe 9-19 Find the collector current and the collector-ground voltage for the circuit of Fig. 9-20a. SoLutron ‘The problem is similar to one we have already solved. In Example 9-7 we analyzed the circuit shown in Fig. 9-7a. Note that the only difference between igs. 9-7a and 9-20a is the location of the ground point.232 ‘Transistor Circuit Approximations The collector current is the same in both circuits. We found in Ex- ample 9-7 that the collector current is To = 6.67 ma With a different ground point, however, the collector-ground voltage is different from before. Clearly, the circuit of Fig. 9-20a has a voltage drop across the 1-kilohm resistor of Vo = IcRz = 6.67(10-)(10*) = 6.67 volts This is the magnitude of the voltage from collector to ground. Because of the polarity, we would measure — 6.67 volts from collector to ground. Fig. 9-20 Examples 9-19 and 9-20. -30 (2) EXAMPLE 9-20 Find the value of the collector current and the collector-ground volt- age in Fig. 9-20. Soiution ‘The voltage divider in the base will develop about 10 volts across the 25-kilohm resistor. Almost all this 10 volts appears across the 10-kilohm resistor. Therefore, the emitter current is about 1 ma. ‘The collector current is essentially 1 ma. This current flows down through the 5-kilohm resistor and establishes a 5-volt drop. Therefore, the collector-ground voltage is about —5 volts.Bias Arrangements 233 9-10 Biasing p-n-p Transistors As we said before, when we use a p-n-p transistor instead of an n-p-n, we must reverse the polarity of all d-c sources. This, in turn, means that all d-c voltages and currents in the p-n-p circuit will be reversed. There is no need to have a separate group of formulas for p-n-p circuits; we can use the formulas developed for n-p-n circuits if we deal with magnitudes of voltages and currents. For instance, in Fig. 9-21a, we have a base-biased circuit with collector feedback. Since a p-n-p transistor is used instead of an n-p-n, we make the collector supply voltage negative instead of positive. As a result, the base current and collector current flow in opposite directions from those of a similar n-p-n circuit. Also, the collector voltage is negative with respect to ground instead of positive. We can use the formulas given in Fig. 9-6a, provided we use magnitudes of voltages and currents. Fig. 9-21 Biasing with p-n-p transis- tors. As a numerical example, consider the circuit of Fig. 9-216. The magni- tude of the collector current is easily found. siete codiaeiastenneana0raneiy Ri ¥ Rs/8 10" + 10*/50 The magnitude of the collector voltage is Ve = Veo — IcRs, = 20 — 0.667(10-*)(104) = 13.3 volts Note that we have used magnitudes throughout. In substituting for Vee, we use the magnitude, which is 20, not —20, volts. Also, the magnitude of the collector-ground voltage is 13.3 volts; the actual voltage is — 13.3 volts with respect to ground. Ie = 0.667 ma234 ‘Transistor Circuit Approximations Thus, in analyzing the d-c operation of p-n-p circuits, we will use mag- nitudes of voltages and currents. After obtaining a magnitude of eurrent or voltage, we will add the direction or polarity to this value. This ap- proach avoids the unnecessary complication of having separate formulas for n-p-n and p-n-p citeuits. SUMMARY The sensitivity K of a biasing arrangement tells us how sensitive the d-c operating point is to changes in 6. As we have seen, base bias with no feedback has a sensitivity of unity, which means (hal the percent change in collector current equals the percent change in 8. By adding feedback to the simple base-bias circuit, we can reduce the sensitivity to some extent. The two forms of base bias with collector feedback (Secs. 9-4 and 9-5) have very low sensitivity provided that the transistor is biased near saturation; this, of course, limits the signal- handling capability of the transistor circuit. ‘The emitter-bias circuits are undoubtedly the best way to bias when stability of the operating point is the prime consideration. GLOSSARY base bias A biasing arrangement in which the d-c base current remains essentially fixed even though the 6 changes. class A amplifier An amplifier in which collector current flows through- out the a-c cycle without the transistor saturating or cutting off. emitter bias A biasing arrangement in which the d-c emitter current re- mains essentially fixed even though the 8 changes. negative feedback In this chapter, this refers to feeding a signal back from either the collector or the emitter to the base to partially offset a change in 8. sensitivity (K) The constant of proportionality between percent changes in d-c collector current and 8. REVIEW QUESTIONS 1. In a simple base-biased circuit without feedback, does the base cur- rent or collector current remain fixed when the 8 changes? 2. How is the 8 sensitivity defined? 3. The sensitivity K must lie between what two values?Bias Arrangements 235 4, What is the sensitivity of a base-biased circuit without feedback? 5. Describe the circuit arrangements of the four forms of base bias dis- cussed in this chapter. 6. In the four forms of base bias, should the value of Rg be small or large to obtain a low sensitivity? 7. In a base-biased circuit using collector feedback, should the transis- tor be operated near saturation or cutoff to get a low sensitivity? 8. What is the approximate value of Vce for any circuit in which the transistor is saturated? 9. In an emitter-biased circuit with two supplies, the d-c emitter current is approximately equal to Vzx/Rz, provided that two conditions are satisfied. What are these conditions? 10. In the emitter-biased circuit with one supply, how do we find the approximate value of the base-ground voltage? And how do we get the approximate value of collector current? 11. In emitter-biased circuits, does the sensi when we inerease Rs (or Ril|R2)? ity increase or decrease PROBLEMS 9-1 A biasing arrangement has a sensitivity K equal to 0.35. If the 6 changes 7 percent, how much will the d-c collector current change? 9-2 When the 8 changes 12 percent, the d-c collector current changes 2 percent, What is the value of sensitivity K? 9-3 In Fig, 9-22a, find the approximate values of base current, collector current, and collector-ground voltage. 9-4 In Fig. 9-22b, what is the approximate value of Rp that just causes saturation? What value of Rg sets the d-c collector current to one-half the saturation value? 9-5 In Fig. 9-22b, Rp = 5 megohms. If 6 changes from 75 to 100, what is the new value of collector current and voltage? sHeRE 10K Fig. 9-22 B=100 (a) (b)236 Transistor Circuit Approximations 9-6 For the circuit shown in Fig. 9-23a, find the following: (a) The collector current Ic. (b) The collector-ground voltage Vc. (c) The collector-emitter voltage Vos. 9-7 What is the value of sensitivity for the cireuit in Fig. 9-23a? If g changes from 150 to 160, what is the percent change in collector current? +15 1oM 20« B= 150 5K (a) : 9-8 The circuit of Fig. 9-236 is to be biased so that Ie equals 14I crsy- For a 8 of 100, what approximate size should Rs be? With this value of Ra, what is the sensitivity of the circuit? 9-9 In the circuit of Fig. 9-24a, what is the value of collector current? Of collector-ground voltage? Of base current? 9-10 In Fig. 9-24a, find the sensitivity of the circuit to changes in B. +20 40K 1M B=100 7) {a (e} Fig. 9-24 Fig. 9-23 +20 30k fe 60KBias Arrangements 237 9-11 The transistor of Fig. 9-24 can have a 8 anywhere in the range of 50 to 150. The base resistance Ry consists of a fixed 100 kilohms in series with a 1-megohm rheostat. Find the following: (a) The smallest possible collector current. (b) The largest possible value of collector current (c) The worst-case sensitivity. (d) The smallest value of collector-ground voltage. 9-12 The 8 in Fig. 9-24c has a value of 100. What is the approximate size of Rg that produces a d-c collector current of one-half the saturation value? If this value of Rp is used in the circuit, what will the value of the d-c collector current become if 8 changes from 100 to 50? 9-13 In the circuit of Fig. 9-25a, find the value of: (a) The collector-ground voltage. (b) The emitter-ground voltage. (c) The collector-emitter voltage. 9-14 What is the sensitivity of the circuit in Fig. 9-250? +30 420 20k 40k IM Re B=100 B=50 10K 10k lo) (4) Fig. 9-25 9-15 In Fig. 9-250, find the following: (@) Teceat. (0) Ic for an Ry of 1 megohm. (c) The value of Rs that sets up an Ic of }4I cies 9-16 In the emitter-biased cireuit shown in Fig. 9-26a, find the approxi- mate values of Iz, Ve, Ve, and Vee. 9-17 What is the sensitivity of the cireuit in Fig. 9-26a?238 Transistor Cireuit Approximations +20 25 50k Silicon coal B= 100 50 fa 1S lout vin fa 1 Vout (0) () fg fp % Sout %eYout CY) Hin, Yin (o) (2) vin® Ry He vost Vin AUR, 2 “out ’3 a 2 i (e) ") Fig. 10-2 A-c equivalent circuits. be the same. By using the ideal-transistor approach, we can redraw the a-c equivalent circuit as shown in Fig. 10-3. From this it is clear that = 08r Also, the magnitude of out is simply Vous = Bixre Dividing vu. by vin, we get the magnitude of the voltage gain. Vout _ Bistr Yin BP,A-C Operation 245 or Vout _ TE _ Pea (10-1) In speaking of the input resistance, we must distinguish between the resistance looking directly into the base and the resistance seen by the bo @ é é 8 22 Yo 6 “ bs Fg bose) Fig. 10-3. Input resistance. source. By inspection of Fig. 10-3, we can see that the input resistance looking into the base is simply Tincoasey = Bre (10-2) The total input resistance seen by the source is tm = RallBr, (10-3) The cireuit of Fig. 10-2/ is actually in the same form as those of Fig. 10-24, b, and e. The only difference is that Rj||R: takes the place of Rp. The voltage gain is still given by rz/r’, and the resistance looking into the base is still 8r!. The resistance seen by the source, however, becomes Tin = Ry||Rali6r, (10-4) ‘The two remaining circuits (Fig. 10-2c and d) have the resistor Rz con- nected from the collector back to the base. What effect does this have on the voltage gain and input resistance seen by the source? To answer this question, we will replace the transistor by its ideal approximation, as shown in Fig. 10-4. The alternating current in Rs must be the difference of the current source and the load current. That is, the current in Re equals Bip — th246 Transistor Circuit Approximations Fig. 10-4 A-c equivalent circuit for collector feedback. Next, we cau wrile a voltage equation around the outside loop that contains rz, Rp, and dn. Vin + Vout — Ra(Bix — ix) = 0 or Vin + Your — Rabo + Rair = 0 (10-5) By inspection of Fig. 10-4 we can see that Yin = bGre or Pein =e Also, note that Your = inh or iy, == Vout se Tr. If we substitute these expressions for #, and i, into Eq. (10-5) and sim- plify, we get Pin + Your — Fe oy Pe Fe oy =0 After factoring and solving for doue/vin, We have Your Ra/m — 1 Yin Re/re +1 In most, practical circuits, Ry must be much larger than rf or rz, to avoid saturating the transistor (see Secs. 9-4 and 9-5). Because of this, Eq. (10-6) simplifies to (10-6) Your — Ra/r — 1 Ral _ ty Yin Ra/rp +1” Raft, 7A-C Operation 247 This final result simply tells us that the value of Ra is so large in most circuits that it has a negligible effect on the voltage gain. What is the input a-e resistance seen by the source in Fig. 10-2c and d? One way to find this input resistance is as follows. Consider Fig. 10-5. ‘The input resistance seen by the source is Pin Th. = Tin Note carefully that tin equals the sum of the alternating current in the base resistor Rz plus the current in the base diode. That is, tin = in | te If the right end of Rg were at a-c ground, the input resistance would be the parallel combination of Rs and Sr;. However, the right end of Rs is not at a-c ground, because there is a signal on the collector. To find the Pr ety Fig. 10-5 Deriving the input re- ( : * sistance. uy a effective resistance that Ry presents to the source, we note that the cur- rent iy must equal the voltage across Rs divided by the value of Rs. That is, = Yin — (Pour) Re for A> 1 (10-7)248 ‘Transistor Cireuit Approximations Equation (10-7) tells us that as far as the source is concerned, the effective resistance of Rz is given by Re/A. In other words, we can re- draw the circuit of Fig. 10-5 as shown in Fig. 10-6. The source sees an effective input resistance of ria = 2 hart 0-8) The a-c equivalent circuit shown in Fig. 10-6 is quite accurate for most typical circuits of the form originally given in Fig. 10-1c and d. It only assumes that Hs is much greater than r, and rz, a condition that almost always exists if saturation is to be avoided. 7 = Fig. 10-6 Input resistance of collec- Vid Re fp! Y, 8: P io Y) Be She De o tor-feedback circuit, 4 To summarize, we have seen that all CE circuits shown in Fig. 10-1 are quite similar as far as a-c operation is concerned. They all have a voltage gain from base to collector of approximately rz/r’. The notes at the bottom of Fig. 10-1 summarize the important a-c formulas. Examp.e 10-1 The £ in Fig. 10-7a is 50. Find the following: (a) The d-c collector-to-ground voltage. (6) The voltage gain from base to collector. (c) The input resistance of the entire stage. Sonurion (a) The d-c base current is about 10 ya; therefore, the d-c collector current is 0.5 ma. This produces a 5-volt drop across the 10-kilohm resistor, so that Ve = 10 — 5 = 5 volts (6) The voltage gain from base to collector is Vout _ Th Mm 7 The theoretical value of r! is 25 my _ 25 mv . Ts 0.5 ma = 50 ohmsA-C Operation 249 The a-e load resistance is ry, = 104/10* = 5 kilohms Therefore, the voltage gain from base to collector is Your _ 5000 vin 50 = 100 If we allow for the spread of 2:1 in the value of ri, we get a voltage gain between 50 and 100. (c) The input resistance is = Roller n By using the theoretical value of r’, we get rin = 10°||50(50) ~ 2500 ohms If we allow for the 2:1 spread in r{, we get an input resistance between 2500 and 5000 ohms. +10 10K Yout vine 10K (a) Fig. 10-7 Examples 10-1 and 10-2. EXAMpie 10-2 In Fig. 10-76, find the voltage gain for: (a) rf = 25 mv/Ts. ()) 1 = 50 mv/Tz. Souutt0n (a) In Fig. 10-7, 8 is given as 100. This bias circuit was studied in Chap. 9. Veo 20 — Re + Ra/B ~ 10" + 2009/1900 ~ 9-667 ma250 Transistor Cireuit Approximations We can now find ri. , _ _25my * = 6.667 ma ~ 37-5 ohms The voltage gain is Pout = Te _ eee = 107 (b) If we use r} = 50 mv/Iz, we have an rj of 75 ohms. Hence, the voltage gain becomes 53.5. Examrte 10-3 The circuit of Fig. 10-8 has a 8 of 75. If vi, = 5 mv rms, find the approximate value of vout. SoLution Banani oo tana 15 a © Ri + Rx/B 20010) + 1.5(10%)/75 Hence, we can compute the theoretical value of r!. I = 0.375 ma r= gap = 66.7 ohms +15 20K wh “out Fig. 10-8 Examples 10-3 and 10-4, Vine 10K The voltage gain is tout _ Te _ 20(108)||10(108) _ Ga re cee eOT, = 100 Therefore, tout = 100vin = 100(5 my) = 500 my rms (Note that had we used r = 50 mv/Tz, we would have obtained an output voltage of 250 my rms.)A-C Operation 251 Exampe 10-4 What is the input resistance seen by the source in the preceding example? Sonution The source sees a resistance of L 5400» ) rin =F llr = ||75(66.7) = 3.75 kilohms Exampie 10-5 Find the voltage gain and the input resistance of the cireuit shown in Fig. 10-9. The transistor has a 8 of 200, and rY is given by 25 mv/Iz. SoLution First, we need the approximate value of Ig. Recall that this circuit is a single-supply emitter-biased circuit. By inspection, we see that the #30 Fig. 10-9 Example 10-5. d-c voltage developed from base to ground is about 10 volts. Hence, the d-c emitter-ground voltage is about 10 volts, and the emitter current is 10 Ta = 5599 = 2 ma The value of rf isTransistor Cireuit Approximations The input resistance of the circuit is Tin = Ril R2||8r, = 50(10*)|[25(10*)||200(12.5) ~ 2.18 kilohms 10-3 Emitter Feedback Recall that in Sec. 5-8 we discussed the a-c operation of a CB circuit. In that section we showed that the voltage gain from source to output could be stabilized against variations in r” by swamping the emitter diode. Swamping simply means using a source resistance that is much larger Vee fa Yout co] R 'e {a} Neo AK Fe Yout “ine R © (ce) Fig. 10-10 (a) Emitter-feedback circuits, to be visualizedA-C Operation 253 than the rf of the transistor. In this way, the exact value of 7! is un- important as far as the gain from source to output is concerned. A similar technique is possible in CE circuits, We have just seen that the voltage gain of a CE cireuit is given by r;,/rl, Because the value of r, varies with temperature, as well as with the particular transistor used, the voltage gain will vary along with r!. To stabilize the gain, we need to swamp out the emitter diode. A widely used method for swamping out the emitter diode is shown in the cireuits of Fig. 10-10. In each of the six biasing circuits a resistor rs Veo Yeo Yine Notes 4 Mout 4 All have a voltage gain of =; re when 1e>> 1! in “er 2. All have an input resistance looking into the base of ip (D086) = BU Yet ret) = B re when 7>> re Circuits 0,4, and e, have on input resistance of Fin Re MI rin (base) 4. Circuits ¢ and ¢ have an input resistance of Fg Nt Fin {bose} A Circuit F has on input resistence of Fin = AWA ll fig( base) as combination of emitter follower and CB amplifier.254 Transistor Cireuit Approximations has been added between the emitter and the a-c ground point. As far as d-c operation is concerned, the value of rg is lumped into the total d-c resistance Rg seen by the emitter (Fig. 10-10b). The d-c voltages and currents can be found by the methods of Chap. 9. The a-c operation of the cireuits shown in Fig. 10-10 will be different from before, because the emitter is no longer at a-c ground. When an a-c signal is coupled into the base of the transistor, the emitter signal tends to follow this input signal (similar to the emitter-follower action). The a-c voltage developed across the rg resistor sets up an a-c emitter current. The a-e collector current, which is almost equal to this emitter current, then develops an a-e voltage across the rz seen by the collector. (For those familiar with vacuum-tube circuits, the use of an uuby- passed emitter resistance is analogous to leaving part of the cathode re- sistance unbypassed. This results in degeneration, which stabilizes the voltage gain of the circuit.) A simple qualitative viewpoint of the circuits shown in Fig. 10-10 is just this: we can visualize these circuits as a combination of an emitter follower and a CB amplifier. The input signal is applied to the base; a signal is then developed across rg because of the emitter-follower action of the circuit. When the circuit is well designed, the a-c voltage across rz almost equals the a-c input voltage to the base. Since almost all the input voltage appears across 7, the circuit now acts like a CB amplifier whose emitter diode has been swamped out. In other words, the a-c voltage across 7g sets up an a-c emitter current; the a-c collector current is al- most equal to this emitter current; this is completely analogous to a CB circuit whose emitter diode has been swamped. To find the voltage gain from base to collector for the circuits shown in Fig. 10-10, consider the simplified a-c equivalent circuit of Fig. 10-11a. This simplified circuit is applicable to all the circuits of Fig. 10-10, in- —_ — eee a 6 4 Be { By Yn é % Wy % ‘e = = = = = = | a (a) (o) Fig. 10-11 Deriving the voltage gain of an emitter-feedback circuit.A-C Operation 255 cluding those in Figs. 10-10c and d, as long as Ra is large enough to neglect (discussed in Sec. 10-2). The transistor in Fig. 10-11a can be replaced by its ideal approxi- mation, as shown in Fig. 10-115. If we write a voltage equation around the base loop, we get Vin = OBIE + ire = Br, + Bis = 148 (r, + rz) The magnitude of the a-c output voltage developed across r, is simply Pout = tery = Birt Dividing vou: by vin, we obtain Your TL if Wn SE Te na Usually the circuit designer deliberately makes rg much greater than r! in order to swamp out the variations in 7’. In fact, this is the whole point of using the rg resistor. Under this condition, the voltage gain from base to collector becomes for rz >” (10-10) Equation (10-10) is very important. Whenever we analyze circuits like those shown in Fig. 10-10, we can easily find the voltage gain to a close approximation by calculating the ratio of the r, to rp (the unbypassed part of the emitter resistance). The condition rz >> r! is usually satisfied; if there is any doubt, we can easily find the approximate value of rf and use the more accurate formula given in Eq. (10-9). As an example, suppose that the a-c load resistance seen by the col- lector is 10 kilohms. If the d-c emitter current is 1 ma, the value of r! is in the range of 25 to 50 ohms. Hence, we can select an rz that is 10 or more times greater than r). For an rg of 1 kilohm, we get a voltage gain from base to collector of Youe te _ 10,000 _ Tae rs = 000 This voltage gain is quite stable; that is, the exact value of r has little effect upon the voltage gain because it has been swamped out. The a-c input resistance of the circuits shown in Fig. 10-10 is also im- portant. The various a-c equivalent circuits are shown in Fig. 10-12. Note that in each of these variations the input resistance looking directly into the base is simply Ars. The reason for this can be understood by256 Transistor Circuit Approximations referring again to Fig. 10-11b. In this figure it is clear that Yin = BBr, + tere & H8(r, + re) or 2 = B(r, + re) is The quantity vj, divided by % is nothing more than the input resistance looking into the base. Hence, Tingbasey = B(re + re) Once again we observe that the whole point of using the rs resistor is to swamp out the value of r!. Under this condition, we get Fin(base) = Gre for rz > ry (10-11) Thus, to a first approximation, the input resistance looking into the base is 8 times larger than the value of rz. The actual input resistance seen by the source is the parallel combi- nation of bias resistors and the input resistance looking into the base. Figure 10-12 illustrates the various input circuits. The a-c equivalent cir- cuit of Fig. 10-12a (applicable to Fig. 10-10a, b, and e) has an input re- sistance of Ra||ére This is what the source actually sees. Circuits 2,0,€ Circuit # Yin Vout vin fg SB Bp 7 AR, & Be {a} Circuits ¢ onda in “out fe BES BOCY « 4 Fig. 10-12 The input resistance of emitter-feedback circuits.A-C Operation 257 The a-c equivalent circuit of Fig. 10-12b (applicable to Fig. 10-10/) has an a-c input resistance of Ri||Rol|6rz In the a-c equivalent circuit of Fig, 10-12c (applicable to Fig. 10-10c and d), the Rp resistor appears as a resistance of Rz/A as far as the source is concerned. ‘Therefore, the source sees an a-c input resistance of 2 lore where A is the voltage gain from base to collector. A summary of the important formulas for voltage gain and input re- sistance is given at the bottom of Fig. 10-10; these will be quite useful for future reference. Exampie 10-6 The transistor in Fig. 10-13a has a 6 of 100. Find the voltage gain and the input resistance. Sonution ‘The voltage gain is Your TE _ 10#||3(104) _ (evga re dec DOD acre Yout 100K (a) Fig. 10-13 Examples 10-6 and 10-7.258 ‘Transistor Circuit Approximations The input resistance consists of the resistance looking into the base in parallel with Rs/A. ria = 85 Frise = “40 |j100(600) = 36.4 kilohms In both of the calculations we have assumed that rz >> 1. To check the validity of this, lot us find Zz and then r!. In Chap. 9 we saw that a base-biased circuit with collector and emitter feedback has a d-c emitter current of Po Vee = 40 eRe +R, + Ra/B 108+ 10° + 200%)/10% Therefore, r{ is in the range of 25 to 50 ohms. Obviously, rz is much greater than 7. =1ma Exampte 10-7 ‘The circuit of Fig. 10-13b has an adjustable rz. Find the approxi- mate range of voltage gain and input resistance. Use a 8 of 100. SoLution By inspection, r,, is 10 kilohms in parallel with 100 kilohms. By the product-over-sum rule, we get an rz of 9.1 kilohms. When rs is ad- justed to its maximum value of 2 kilohms, the voltage gain is mini- mum and equals Yout TL _ 9100 On the other hand, when rz is turned down to zero, there is no emitter feedback, and the voltage gain is given by rz/ri. A calculation will show that I = 1 ma, and so rf will be between 25 and 50 ohms. The maximum possible gain occurs if rf is as low as 25 ohms. This maximum gain is 9100 ORR 364 Vout Thus, we have a variable gain that lies between 4.55 and 364, de- pending upon the rz adjusment. Since the @ is 100, looking into the base with rz at maximum, we get Tingbase) & 8g = 100(2000) = 200 kilohms This 200 kilohms is in parallel with 80 kilohms and 40 kilohms when viewed from the source. Hence, in = 80(10*)||40(10*)||200(108) = 23.5 kilohmsA-C Operation 259 On the other hand, when rg is at its minimum value, the input re- sistance looking into the base is only Tincoase) & Br, = 100(25) = .5 kilohms The source sees 2.5 kilohms in parallel with 80 and 40 kilohms. For practical purposes, the source sees approximately 2.5 kilohms. There- fore, the range of input resistance as rg is varied from minimum to maximum in 2.5 to 23.5 kilohms. 10-4 CC Operation In the CC connection (emitter follower), we couple the input signal into the base and take an output signal from the emitter. The collector is deliberately placed at a-c ground. Because of this, there is no need to use a d-c load resistance R,. In other words, we can connect the collector directly to the Vec supply. Of the six biasing arrangements studied in Chap. 9, there are only three distinct forms that we can use for emitter-follower circuits when Rz = 0. These forms are shown in Fig. 10-14. The d-c operation of these circuits is straightforward. In all cases, R, = 0. Therefore, in using the formulas developed in Chap. 9, we sim- ply set Rz, equal to zero. For instance, the circuit of Fig. 10-14a is a base-biased circuit with emitter feedback; its d-c operation is summarized by Fig. 9-4. Likewise, the two-supply emitter-bias circuit of Fig. 10-146 has direct. currents and voltages that we can find by using the formulas of Fig. 9-10. Finally, we can analyze the d-c operation of the single-supply emitter-bias circuit of Fig. 10-14¢ by using the formulas of Fig. 9-13. The circuits shown in Fig. 10-14 are emitter followers. To find the a-c equivalent circuit, we short all d-c supplies and capacitors. When we do this, we find that all these circuits reduce to the a-c equivalent circuit of Fig. 10-15. In Chap. 7, we discussed the emitter follower using the a-c equivalent circuit of Fig. 10-15. We saw that the voltage gain is Your 1 ae Teen 1 for rz > rh (10-12) The input resistance looking into the base is Tincoasey & BC", + 2) = Br, for rz > ri (10-13) For the circuits of Figs. 10-14a and b the input resistance seen by the source is tin = Ra|lrincvase) (10-14)260 Transistor Circuit Approximations Vee ined “out fs fe R © (6) ee Notes 1 thee 2 All hove a voltage gain of You» 1 oy i in = Te7n, | when 1? Te 3. Circuits @ and 4 hove on input resistance of Fin FR llB 1, 4 Circuit ¢ hos on input resistance of Fin = ANA B Fig. 10-14 Emitter-follower circuits. Yout Fig. 10-15 A-c equivalent of an emit- AL ter follower.A-C Operation 261 whereas the circuit of Fig. 10-14c has an input resistance of Tin = Bi||Rollrincasey (10-15) These various resuls are summarized at the bottom of Fig. 10-14 and should be useful for future reference. Recall the main idea behind the emitter follower: the circuit is used to increase the impedance level; the value of load resistance rz, seen by the emitter is stepped up by a factor of 8 when seen from the base. Exampte 10-8 Caleulate the approximate input resistance of the cmitter follower in Fig. 10-16a. Use a 8 of between 100 and 200. Sonurion The a-c load resistance seen by the emitter is ry = Rz||R = 10,000||500 = 500 The resistance looking into the base is Tincbasey & Br, + rz) If we were to calculate the d-c collector current, we would find that it is 1 ma or more, depending upon the value of 8; therefore, r/ is in the range of 25 to 50 ohms. As a result, rz >> rf (500 ohms compared to 25 or 50 ohms). Hence, we can say that Tingbase) & Bry, When 8 = 100, Tinbase) & 100(500) = 50 kilohms Also, when 8 = 200, Tin(basey = 200(500) = 100 kilohms The source sees 1 megohm in parallel with rinwase). Since 1 megohm is much larger than 50 to 100 kilohms, we can say that the source sees approximately 50 to 100 kilohms of resistance. Note carefully the significance of this result: a 500-ohm load on the output side has been transformed so that it appears as 50 to 100 kilohms when seen by the source. This is the whole point of the emitter follower. Exampte 10-9 Find the voltage gain and the input resistance of the emitter follower in Fig. 10-16). Use a 8 of 200 and an ri = 25 mv/Iz.262 Transistor Circuit Approximations +20 Vino 300<8< 200 Yine] B= 100 Yout “out WOK $500 200K $20K $50 + + i. 4 = (a) 4b)” Fig. 10-16 Examples 10-8 and 10-9. Sonution The emitter sees an a-c load of 20 kilohms in parallel with 50 ohms. Therefore, r; is approximately 50 ohms. Because this is so low, we will need to take r/ into account. This means we need to first find Iz. The circuit is a two-supply emitter-biased connection. We already know that in this kind of cireuit almost all the emitter supply voltage appears across Rx. Therefore, With 1 ma of emitter current, we find that 25 mv _ 25 my Ts Ima = 25 ohms The voltage gain of the circuit is You _ 1 1 Mm T+r/rn 1+ 2560 The resistance looking into the base is Tinos) = B(r, + 7x) = 20025 + 50) = 15 kilohms = 0.667 This 15 kilohms of resistance is in parallel with the 200-kilohm base-to- ground resistor. For practical purposes, the source sees approximately 15 kilohms. Again note how a relatively low value of load resistance has been trans- formed to a much higher value of resistance when viewed from the source.A-C Operation 263 Also, note that at times, the voltage gain can be quite a bit less than unity, as in this case; r{ was not negligibly small compared to rz. 10-5 CB Operation Ina CB connection, we couple the input signal into the emitter and take the output from the collector. The base is held at a-c ground, and as a result, the circuit is sometimes called a grounded-base cireuit. The base does not have to be physically grounded. In other words, there may be a d-c voltage from base to ground; however, in this case, we must use a bypass capacitor to ensure that the base is at a-c ground. When we examine the six biasing arrangements of Chap. 9, we find that two of these have the emitter connected directly to ground. Since the input signal is coupled into the emitter, we cannot use these two circuits for CB operation. The four remaining bias arrangements are shown in Fig, 10-17, along with coupling and bypass capacitors. In each circuit, observe that the input signal is coupled into the emitter and the output signal is taken from the collector. In each case, the base is at a-c ground. The d-c operation of these circuits is straightforward. All capacitors appear open to d-c voltage; therefore, these circuits revert to the stand- ard biasing arrangements of Chap. 9. We can analyze the d-c operation of Fig. 10-17a by using the formulas of Fig. 9-4. Similarly, Fig. 10-176 is analyzed by using Fig. 9-8, Fig. 10-17c by using Fig. 9-10, and Fig. 10-174 by using Fig. 9-13. By shorting ali d-c supplies and capacitors, we get the a-c equivalent circuits. The circuits of Fig. 10-17a, c, and d have the same a-c equivalent circuit, which is shown in Fig. 10-18a. We found the voltage gain of this cirouit in Chap. 5. It is (10-16) Remember that this is the voltage gain from emitter to collector. [When there is source resistance, the voltage gain from source to collector be- comes rz/(r, + r,). This was discussed in Sec. 5-8.] By inspection of Fig. 10-18, it is clear that the input resistance of the circuit is Rz in parallel with r{. That is, Tin & Rallr, =r, when Re > rh (10-17) Usually, in a CB cireuit, Rg is much greater than r/, so that the input re- sistance of the circuit is quite low, being equal to the r’ of the transistor. The CB circuit of Fig. 10-17b has an a-c equivalent circuit that is264 Transistor Circuit Approximations Notes 12 RUR 2. All have a voltage gain of ¥ ‘out Yin 3. All hove input resistance of Tin % All re Ere. when Ae>> rg Fig. 10-17 Common-base circuits.A-C Operation 265 e é aaa Yn Ideal Yost in Idec Yout % see on fe Ste (Pye Sa Sh ‘a) to) Fig. 10-18 A-c equivalent cireuit for CB amplifiers. slightly different from the othor circuits in Fig. 10-17. Its equivalent cireuit is shown in Fig. 10-18). The collector sees an a-c load of three resistors in parallel. That is, ry = Ral|Ri|R However, we observed in Chap. 9 that 2, must be much larger than Ry if we are to avoid saturation. Therefore, in most practical circuits, Rg is large enough to neglect, and we can say that the a-c load resistance is simply r= Rik The circuit of Fig. 10-176 has essentially the same voltage gain and input resistance as the other circuits in this figure. All the results are summarized by the notes at the bottom of Fig. 10-17. ‘There is an important result worth remembering. By using the ideal- transistor approximation, we have seen that the voltage gains of CB and CE cireuits are equal, being given by rz/r!. One important difference between these two connections is the input resistance. A CB circuit has an input resistance looking into the base of about r/, whereas the CE cir- cuit has an input resistance of Bri. Because of its higher resistance, the CE circuit is used more often; nevertheless, the CB circuit does have some use, especially at higher frequency. This is discussed in the chapter on frequency response. Examp.e 10-10 A 2-my-rms signal drives the emitter of the p-n-p transistor shown in Fig. 10-19. Find the approximate value of vou: and the input resist- ance looking into the emitter. Sorvtion As we already know, when a p-n-p transistor is used instead of an n-p-n, all d-c currents and voltages are reversed. The a-c operation,266 ‘Transistor Circuit Approximations however, is identical for either type of transistor because the same a-c equivalent circuit applies to cach. The voltage gain of the circuit is approximately Vout Vin To find rl, we first must find I, the d-c emitter current. Recall the d-c operation of the single-supply emitter-bias circuit. The voltage divider ¥ ° Fig. 10-19 Example 10-10. in the base develops a voltage of about 10 volts. Almost all this 10 volts appears across the emitter resistor, so that With 0.5 ma of d-c emitter current, the value of rf is in the range of 50 to 100 ohms. Hence, for r; = 50 ohms, Yous _ 10,000||10,000 _ se Om 100 If r! is as high as 100 ohms, the voltage gain becomes 50. Thus, the voltage gain is between 50 and 100. Since the input signal is 2 mv rms at the emitter, the output signal will be between 100 and 200 mv rms. The input resistance is the parallel combination of Rx and r’. Obvi- ously, the 20-kilohm value of Rz is so much larger than ri that we have Tin & 7, = 50 to 100 ohmsA-C Operation 267 10-6 The Effect of Source Resistance Up to this point, we have been discussing the voltage gain from the input terminal of the transistor to the output. In this section, we want to study the effect that source resistance has on the overall voltage gain from source to output. Consider Fig. 10-20a. We have shown a source with a resistance of r.. The amplifier inside the box can be a CE, CC, or CB circuit. The voltage gain of the amplifier is A, where A is the ratio of sat to vim Note care- fully that vin is the voltage appearing across the input of the amplifier. This input voltage does not equal the source voltage », because some signal is lost across the source resistor r.. LG 10K © A 3 Ay © 3 i Se r i i rin (a) (o) Fig. 10-20 The effect of source impedance. How large is the actual signal appearing across the input terminals of the amplifier? The easiest way to find this is to apply the voltage-divider theorem. That is, igre ee tia rete Waa) (10-18) The use of this equation is straightforward. For instance, suppose the source has a resistance of 10 kilohms and the amplifier has an input re- sistance of 10 kilohms, as shown in Fig. 10-200. Then the input voltage is In other words, the input voltage equals one-half of the source voltage when ry = rin. It should be clear from Eq. (10-18) that if the source resistance is very small compared to the input resistance of the amplifier, almost all the source signal will appear across the amplifier input, and very little signal is lost across the source resistance. But when the source resistance be-268 Transistor Circuit Approximations comes comparable to the input resistance of the amplifier, a significant part of the source signal is lost across the source resistance. Exampte 10-11 ‘The transistor of Fig. 10-21a has a @ of 100. Find the value of vin and Yout. SoLution The input resistance looking into the base of the transistor (not in- cluding the bias resistors) is Tintbasey & Brg = 100(500) = 50 kilohms The biasing resistors (50 kilohms and 25 kilohms) are in parallel with Tina. Therefore, the input resistance of the amplifier is tim = Ral|Rel|rinesase) = 50(10*) |[25(10*)|/50(10*) = 12.5 kilohms Now we can find how much of the source signal actually reaches the input of the amplifier. Using Eq. (10-18), we get _ 12,500 ~ 5000 + 12,500 We can find vou by first finding the voltage gain A. Since part of the emitter resistance is unbypassed, we know that the approximate gain is Yin 20 mv = 14.3 mv rms 15 +20 10k 1oK =O = 100 20my (0) rms Fig. 10-21 Examples 10-11 and 10-12.A-C Operation 269 Therefore, Vout = 15(14.3 mv) = 214 mv rms Note that a significant part of the source signal is lost across the source resistor r,. If the amplifier were driven by a very low impedance source (like 50 ohms), almost all the source signal would appear across the input, terminals of the amplifier. EXxampLe 10-12 The transistor of Fig. 10-21b has a 6 of 100 and an ri of 100 ohms. Find the value of tou. SoLvution This is a CB stage with an input resistance of approximately 1/3 therefore, = 100 ohms Tin & The source voltage is 20 mv rms. Since the source resistance is 100 ohms, only one-half of the source voltage reaches the emitter terminal. That is, Tin 100 Ee = Tor e100 20 BY = 10 mv tin = The voltage gain of the amplifier is 10,000)!30,000 _ ae Te 10,000180,000 75 Yin 7% 100 Your Therefore, the output voltage is Your = 75(10 my) = 750 mv rms 10-7 Stabilizing the Voltage Gain from Source to Output In this section we discuss the conditions that are necessary to ensure a stable or fixed voltage gain from the source to the output in spite of changes in the transistor characteristics. We already know that Your = Adin where A is the voltage gain from input to output. We have just seen that the actual input voltage at the amplifier input is Tin + Tin Vin = Us270 Transistor Circuit Approximations By substituting this equation into the preceding one, we get Tin Pour = A ne" or Your Tin fe hee A Te. + Tin aed Equation (10-19) tells us how to find the voltage gain from source to output; it contains voltage gain A of the amplifier and the effect of the voltage divider formed by r, and rin. In transistor work, it is often desirable to have a fixed voltage gain from source to output, a gain that remains constant even though the transistor characteristics change. How can we get this fixed value of voltage gain? If the amplifier uses a CE circuit, the usual approach in getting stable voltage gain from source to output is to: 1, Fix A by using emitter feedback. 2. Make rig much greater than r,. ‘The reason for these two conditions is apparent by inspection of Eq. (10-19). When rin is much greater than r,, Eq. (10-19) reduces to Vout Us Since A is fixed because of emitter feedback, the voltage gain from source to output is fixed. We can summarize the situation as shown in Fig. 10-22. With ri. much greater than r,, almost all the source voltage reaches the amplifier input. With emitter feedback, the amplifier has a gain of rz/rz, and the output voltage is simply n Vous & — Oy Te or for Tin > Ts (10-20) This approximate formula for the source-output voltage gain depends only upon the ratio of fixed external resistances; it does not depend upon the 8 or the r’ of the transistor. For a CB circuit, the situation is reversed; that is, to get stable voltage gain from source to output, we deliberately make r, much greater than 7, the input resistance. This condition is illustrated in Fig. 10-23. The CBA-C€ Operation an ts 6 wf Mout “7,5 wh Yin [| BOE 5 Yin RS vot A'S % | | | Fin 5 ers Fig. 10-22 Stabilizing the voltage Fig. 10-23 Stabilizing the voltage gain of a CE circuit. gain of a CB circuit. amplifier has a voltage gain of rz/r{. To find the overall voltage gain we use Eq. (10-19). Your 4 Tin mT Ty % Fits 20H heen By inspection of the result, it is clear that we can swamp out the emitter junction resistance by making r, much greater than r{, Under this condition, Fest eg" for r, 4 (10-21) This same result was derived in Chap. 5 for a single-supply emitter- biased CB circuit, We have merely proved that the same result applies to any CB circuit, no matter what the particular biasing arrangement. Fig. 10-24 Stabilizing the voltage gain of a CC circuit. Finally, the voltage gain from source to output can be stabilized in a CC cireuit, as shown in Fig. 10-24. The idea is to make rin much greater than r., so that almost all the source voltage appears across the emitter- follower input. With a voltage gain of approximately unity, the emitter follower delivers an output signal almost equal to the source signal. Exampie 10-13 For the cireuit of Fig. 10-25 verify that the voltage gain from source to output is essentially fixed even though 6 changes from 20 to 200.272 Transistor Circuit Approximations SoLution First, note that the amplifier uses emitter feedback so that the volt age gain from the base to output is rz _ 10,000/|10,000 _ oe 10 The input resistance looking into the base is Tin(base) & Bz When 6 = 20, Tingoaasy = 20(500) = 10 kilohms When 6 = 200, Tinqbasey = 200(500) = 100 kilohms +20 10K IM Yout So Fig, 10-25 Example 10-13. 100 ~ 500 Inv ‘| The biasing resistor 2s appears as a resistance of Re _ 108 _ ‘ f= jo = 100 kilohms The worst case, that is, the lowest value of input resistance for the amplifier, occurs when 6 = 20 and equals Ti = Ba |[7inoasey = 100(10*)||10(10*) = 9.1 kilohms Note that 7, is only 100 ohms compared to an rin of 9.1 kilohms; therefore, almost all the source voltage reaches the amplifier input. With a I-my souree, the final output voltage will be essentially 10 mv. Thus, we have seen that the voltage gain from source to output is fixed even though the 8 changes from 20 to 200. In effect, the voltage gain is independent of the transistor characteristics.A-C Operation 273 10-8 p-n-p Operation If p-n-p transistors are used instead of n-p-n, we need only change the polarity of the supply voltages. The a-c operation is the same, so that all the results of this chapter apply to p-n-p transistors. As an example, consider the circuit of Fig, 10-26. A p-n-p transistor is used instead of an n-p-n; therefore, we must use a negative collector sup- ply, as shown, All the d-e currents and voltages in this circuit are in the opposite direction from a comparable n-p-n circuit. The magnitudes of these d-c currents and voltages are easily found by the methods of Chap. 9. For instance, the biasing arrangement is single-supply emitter bias. Fig. 10-26 Using a p-n-p transistor. The voltage divider in the base develops about 10 volts across the 10-kilohm base resistor. ‘his means that almost 10 volts appears across the 10-kilohm emitter resistor, thereby producing about 1 ma of d-c emitter current. As far as the a-c operation is concerned, we have a CB circuit. The input signal is injected into the emitter; the output signal is taken from the collector; the base is at a-c ground. To find the voltage gain from emitter to output, we note that r/ is between 25 and 50 ohms (we have already found that the d-c emitter current is about 1 ma). Therefore, the voltage gain from emitter to output is tout aw Te — 5000 _ 199 to 200 peter er The voltage gain from the source to the output is Your Th +r274 ‘Transistor Cireuit Approximations Since r, is 500 ohms and r/ is in the range of 25 to 50 ohms, the emitter diode is swamped, and we get an approximate gain of Your 7, _ 5000 _ % =r, ~ 300 ~ 1 To generalize the operation of p-n-p circuits, note: 1. The n-p-n transistors in Figs. 10-1, 10-10, 10-14, and 10-17 can be replaced by p-n-p transistors provided we use a negative Vee supply and 2 positive Veg supply. 2. All the a-c formulas shown for n-p-n circuits apply to p-n-p cireuits. 10-9 The Ground Point As pointed out in Sec. 9-9, for single-supply biasing arrangements we can ground either end of the supply. Which end of the supply is grounded is unimportant as far as the a-c operation is concerned because both ends of the supply are a-c ground points. If it is necessary to move the d-c ground point, we need only move those grounds which appear in the d-c equivalent circuit. For example, consider the cireuit of Fig. 10-27a (this is the same as Fig. 10-I/). If the opposite end of the d-e supply is grounded, we merely redraw the circuit as shown in Fig. 10-276. The a-c operation of Fig. 10-27b is identical to that of Fig. 10-270. As another example, consider the p-n-p circuit of Fig. 10-27c. If we wish to move the d-c ground point to the other end of the d-c supply, we simply draw the circuit as shown in Fig. 10-27d. The a-e operation of Fig. 10-27c and d is identical. 10-10 Maximum Signal-handling Capability In Chap. 8, we discussed the load lines of simple CE, CB, and CC circuits. Recall that the d-c load line is a graph of all the possible d-c operating points. The actual d-c operating point is somewhere along this d-c load line and can be located by the methods of Chap. 9. When an a-c signal drives a transistor, it causes changes in the tran- sistor currents and voltages. These changes take place along the a-c load line instead of the d-c load line because the a-c load seen by the transistor can be different from the d-c load. Of special importance is the maximum signal-handling capability, thatA-C Operation 215 (e} (a) Fig. 10-27. Moving the ground point. is, the largest unclipped signal that we can get from a transistor amplifier. Recall that clipping occurs at the saturation point and at the cutoff point on the a-c load line. As we saw in Chap. 8, the largest peak-to-peak un- clipped signal is 2Vcz or 2Zorz, whichever is smaller. In this present chapter, we have discussed a number of different biasing arrangements and their use in CE, CC, and CB circuits. The a-c load- line analysis of these circuits is essentially the same as the a-c load-line analysis of Chap. 8. In other words, the a-c load line passes through the d-c operating point (Vcz,/c) and has a cutoff voltage of Vex + crs.276 ‘Transistor Circuit Approximations Thus, the maximum peak-to-peak unclipped signal that we can get is Vow = 2Vex (10-22) or Veep = orn (10-23) whichever is smaller. In applying these formulas, remember that: Veg is the d-c collector-emitter voltage. Ic is the d-c collector current. rz is the a-c load resistance seen by the output terminal of the transistor. Equations (10-22) and (10-23) can be used to find the largest unclipped signal available from a transistor amplifier. The only exception worth mentioning is Fig. 10-10, where some of the emitter resistance is left un- bypassed to get a stable voltage gain. Analysis shows that the largest un- clipped signal for the circuits of Fig. 10-10 is the smaller of Vp-p = Wert (10-24) and Veep = 2Vex =2Ver forreK rz (10-25) 1 l+re/r, In most practical circuits, rz is much smaller than r;, (otherwise, the voltage gain is very low). Thus, for a first approximation, we can say that all circuits in this chapter can deliver a maximum unclipped signal of 2Zcr; or 2Vcx, whichever is smaller. Exampie 10-14 ‘The circuit of Fig. 10-7a was analyzed earlier in Example 10-1, where we found that Ig = 0.5 ma and Ves — 5 volts (this was for a 8 of 50). Find the maximum signal-handling capability of the circuit in Fig. 10-7a. Sonurion Equations (10-22) and (10-23) apply. ers, = 2(0.5)(10-*)(5)(108) = 5 volts and 2Ver = 2(5) = 10 volts Therefore, the largest unclipped peak-to-peak signal is 5 volts. Examp.e 10-15 The cireuit of Fig. 10-9 was analyzed in Example 10-5, where we found that the d-c emitter current equals 2 ma. Find the maximum signal-handling capability of this circuit.A-C€ Operation 217 So.ution Referring to Fig. 10-9, we see that a d-c emitter current of 2 ma will produce a d-c voltage from the emitter to ground of Vz = IgRz = 2(10-*)(5)(10) = 10 volts Further, the d-c collector current is approximately equal to the value of emitter current, so that the d-c collector-ground voltage is Ve = Veo — Ick, = 30 — 2(10-*)(5)(10*) = 20 volts The d-c voltage from collector to emitter is Ver = Ve — Ve = 20 — 10 — 10 volts Thus, the d-c operating point is Vex = 10 volts, Zc = 2 ma. Now we can find the maximum signal-handling capability. Using Eqs. (10-22) and (10-23), we get 2ers, = 2(0.002)(5000||50,000) = 18 volts and 2Vor = 2(10) = 20 volts The smaller value is the limitation, so that the circuit of Fig. 10-9 can deliver a maximum unclipped signal of 18 volts peak to peak. Examen 10-16 Find the maximum unclipped signal available from the circuit of Fig. 10-13a. This circuit was analyzed in Example 10-6, where we found that Ze = 1 ma for a8 of 100. SOLUTION, First, find Vex. With 1 ma of d-c collector current, the d-c voltage from collector to ground must be Vo = Voo — IcR1, = 40 — 0.001(10,000) = 30 volts and the d-c voltage from emitter to ground must be Ve = [nhs = 0.001(10,000) = 10 volts Therefore, the d-c voltage from collector to emitter is Ver = Ve — Vx = 30 — 10 = 20 volts The largest unclipped signal is the smaller of 2Zor, and 2V cz. 2er, = 2(0.001)(10,000]/30,000) = 15 volts 2Vcr = 2(20) = 40 volts278 Transistor Circuit Approximations Therefore, the largest unclipped signal that we can get from the ampli- fier of Fig. 10-13a is 15 volts peak to peak. Exampie 10-17 What is the signal-handling capability of the circuit in Fig. 10-16b? Sonurion This is an emitter follower using two-supply emitter bias. By inspec- tion, almost all the emitter supply voltage is dropped across the 20-kilohm resistor, so that the d-e collector current is Vee lo = We already know that in an emitter follower the d-c voltage from the collector to the emitter is approximately equal to the collector- ground voltage. Therefore, in Fig. 10-160, Vcw & 20 volts. Now, we calculate the maximum signal-handling capability using Eqs. (10-22) and (10-23). 2Zer, = 2(0.001)(20,000)/50) = 0.1 volt and 2Vex = 2(20) = 40 volts Therefore, the circuit of Fig. 10-16b can only deliver a peak-to-peak voltage of 0.1 volt. ExampLe 10-18 Find the largest unclipped signal available from the circuit of Fig. 10-19. So.uTion By inspection, this is a single-supply emitter-biased arrangement. The d-c voltage from base to ground is about —10 volts. Almost all this voltage appears across the 20-kilohm emitter resistor, so that the d-c emitter current is about 0.5 ma. The magnitude of Veg is clearly 15 volts, since 5 volts is dropped across the 10-kilohm collector resistor and 10 volts is dropped across the 20-kilohm emitter resistor. Thus, err = 2(0.5)(10-%)(5)(108) = 5 volts and 2Vcx = 2(15) = 30 volts The maximum unclipped signal that can be delivered to the output is therefore 5 volts peak to peak.A-C Operation 279 SUMMARY We have discussed CE, CC, and CB operation of the various bias arrange- ments of Chap. 9. A circuit arrangement is classified as CE, CC, or CB by determining which of the transistor terminals is at a-c ground. For all CE circuits discussed in this chapter, the a-c voltage gain equals r1/ri. The amplified signal at the collector is 180° out of phase with the input signal at the base. The CE circuit can be stabilized by leaving some of the emitter resist- ance unbypassed. In this case, the circuit behaves like a combination of an emitter follower and a CB amplifier. The voltage gain is rz/rg, where rp is the unbypassed part of the emitter resistance. The CC circuit normally has a voltage gain of close to unity. The out- put signal appearing at the emitter is in phase with the input signal at the base. The main advantage of an emitter follower is to increase the load resistance by a factor of 8. The CB circuit has its base at a-c ground. The output signal at the collector is in phase with the input signal at the emitter. The voltage gain of a CB circuit is r;/r', the same as that for a CE circuit. In stabilizing the source-output voltage gain of a CE circuit, the usual practice is to make the input resistance of the amplifier much larger than the source resistance. In addition, emitter feedback is used to fix the voltage gain from base to collector. To stabilize the source-output voltage gain of a CB circuit, we deliber- ately make the source resistance much greater than r’, the input resistance of the CB amplifier. Under this condition, the voltage gain becomes 71/r.. ‘The cource-output voltage gain of a CC circuit is stabilized by making the input resistance much larger than the source resistance. Under this condition, the overall voltage gain approximately equals unity. GLOSSARY a-c ground A point that is either connected directly to ground or by- passed to ground through a capacitor. biasing resistors All resistors in the d-c equivalent circuit that set the d-e collector current. degeneration Synonymous with negative feedback. In this chapter, degeneration refers to the use of an unbypassed emitter resistance. stabilizing With respect to voltage gain, this means fixing the voltage gain at some constant value, so that changes in the transistor have no effect.280 Transistor Circuit Approximations swamping With respect to the emitter diode, this means making value of rf negligible as far as voltage gain is concerned. REVIEW QUESTIONS . ACE circuit is also called a grounded-emitter circuit. Why? . Why are coupling capacitors used? 3. What is the input resistance looking directly into the base of a CE circuit whose emitter is at a-c ground? 4. Ina CE circuit in which there is a biasing resistor R, connected from collector back to base, what is the effective value of this resistance when viewed from the source? 5. A CE amplifier can be modified by leaving some of the emitter re- sistance unbypassed. What is the approximate voltage gain in this case? What should the relation between rg and r/ be to ensure that stable gain? . What is the main advantage of an emitter follower? . What condition must be satisfied if the emitter follower is to have a voltage gain of almost unity? 8. In a CB circuit, what is the approximate voltage gain from emitter to collector? If the source resistance driving the CB cireuit is much larger than r/, what is the approximate voltage gain from source to collector? 9. Toa first approximation, the voltage gains of the CE and CB circuits are the same. Why is the CE circuit more commonly used? 10. What effect does source resistance have upon the amount of signal actually reaching the input of the amplifier circuit? What relation must exist between the source resistance and the amplifier input resistance if almost all the source signal is to reach the amplifier input terminals? 11. To stabilize the voltage gain from source to output in a CB circuit, should r, be much larger or much smaller than r;,? What should the relation be in a CE circuit? n no PROBLEMS 10-1 In Fig. 10-28a, find the voltage gain and the value of vou. Use a 8 of 50 and an rf of 50 mv/Tz. 10-2 Repeat Prob. 10-1 using a 8 of 150. 10-3 Find the voltage gain in Fig. 10-28). Use an r! between 25 and 50 mv/Tg. Also find the input resistance of the entire stage. 10-4 ‘The transistor in Fig. 10-28c has an r, = 25 mv/Tz. What is the minimum and maximum voltage gain for this amplifier?A-C Operation 281 +20 30K 2h Yout Fig. 10-28 10-5 What are the minimum and maximum values of vou: in Fig. 10-28d? Use an r’ of 25 mv/Ip. 10-6 If vin = 500 nv rms in Fig. 10-29a, what is the approximate value of vour? What is the input resistance of the stage for a 8 of 100? 10-7 If the voltage gain in Fig. 10-29 is to be 15, what size should rp be? 10-8 In Fig. 10-29a, if ;, = 2 mv rms, what is the approximate value of the a-c voltage from emitter to ground?282 Transistor Circuit Approximations +30 B=50. < 100K 20K 20K Fig. 10-29 10-9 In Fig. 10-296, what is the voltage gain from base to collector for the first stage? For the second stage? The B of each transistor equals 50. 10-10 If vin = 5 wv rms in Fig. 10-296, what is the approximate value of Vout? 10-11 What is the input resistance of the amplifier shown in Fig. 10-30a? +20 10K IM Yout ¥ 30K “ine 8-50 Yn 500 Fig. 10-30A-C Operation 283 10-12 In Fig. 10-306, the value of rg is adjustable from 0 to 1 kilohm. What is the minimum possible voltage gain (use 7, = 25 mv/Tz). 10-13 The transistor in Fig. 10-30b has a 8 between 50 and 200. What is the highest possible input resistance for the entire circuit? What is the lowest possible input resistance (use r, = 25 mv/Iz)? 10-14 In Fig. 10-31a, what is the input resistance of the stage? ©+20 4 50< B< 150 ne Ww Your ‘50K 168K $820 Fig. 10-31 10-15 In Fig. 10-310, what are the minimum and maximum values of input resistance of the stage? 10-16 In Fig. 10-32a, the input voltage is 3 mv rms. Find the approxi- mate output voltage for an r, = 25/Tz. +20 —+ 130 6.8K 51K 2.2K Yout Yout 15K 68K a ca x= ira t-2 "in Join 20K OK SK 10 = 7) 1) Fig. 10-32284 Transistor Circuit Approximations 10-17 In Fig. 10-32b, the value of vin rl = 25 mv/Ig and for r, = 50 mv/Jz. 10-18 The transistor of Fig. 10-33a has a 6 between 30 and 100. What is the approximate value of vou? = 3 mv rms. Find vou for an +30 +20 60K out WK —g30K 100k 20K 21K Imv Sav rms lok T one =e => = i -20 7) (6) Fig. 10-33 10-19 In Fig. 10-33b, the transistor has a 6 between 40 and 120. What is the approximate output voltage for: (a) An r, of 600 ohms. (b) An r, of 30 kilohms. 10-20 What is the approximate output voltage in Fig. 10-34a for an input of 10 mv rms? Neglect 7. “15, 330K Fig. 10-34A-C Operation 285 10-21 In Fig, 10-348, vin is 5 mv rms. What is the approximate value of Pout? 10-22 What is the approximate output voltage in Fig. 10-35a for an input voltage of 200 uv rms? 5 100K OK out ‘50K 390, (a) (4) Fig. 10-35 10-23 If the input voltage in Fig. 10-35b is 500 pv rms, what is the approximate output voltage?11 Cascading Stages Up to now, we have analyzed a number of basic transistor circuits. We have discussed the biasing problem, that is, setting the d-c operating point. Also, we have worked out the formulas for the a-c voltage gain of typical circuits. However, we have confined our discussion to single stages. Now we want to tum our attention to the problem of cascading stages, that is, connecting the output of one transistor cireuit to the input of another. 1-1 RC Coupling The easiest and most widely used method of cascading stages is resistance- capacitance (RC) coupling. In this approach, the voltage developed across a resistance in one stage is coupled through a capacitor into the next stage. To make the discussion concrete, consider the two-stage RC-coupled amplifier shown in Fig. 11-1. A coupling capacitor is connected from the collector of the first transistor to the base of the second transistor. The purpose of this coupling capacitor is to appear as an open circuit to d-c voltage but as a short circuit to a-c voltage. As a result, there is no d-c 286Cascading Stages 287 interaction between stages; that is, the d-c voltage of the first stage does not disturb the d-c operating point of the second stage. On the other hand, any a-c voltage at the collector of the first transistor is coupled directly into the base of the second transistor. Thus, with this two-stage circuit, an a-c signal is amplified by both transistor stages. Let us find the overall voltage gain of the two-stage amplifier shown in Fig. 11-1. First, note that the d-c voltage from the base to ground +30 10K Fig. 11-1 RC-coupled amplifier. in the second stage is about 10 volts. This will produce about 1 ma of d-c emitter current in the second transistor. The rf is therefore in the range of 25 to 50 ohms. Arbitrarily, we will use 50 ohms. The voltage gain of the second stage is n Arey = v 10,0001|10,000 _ So = 100 To find the voltage gain of the first stage, we must find the a-c load resistance r, seen by the collector of the first stage. It is clear that looking out from the collector there is a 10-kilohm resistance in parallel with the input resistance of the second stage. As shown in Fig. 11-1, the ria of the second stage is approximately 5 kilohms. This is found by the methods of Chap. 10; that is, we know that the input resistance of such a stage is Tin = Rj||RallBr) = 100(10*)||50(103) ||100(50) = 5 kilohms (We are neglecting the effects of R, and Rz because they are large com- pared to the 5 kilohms looking directly into the base of the second transistor.) Here is an important point to remember. The 5 kilohms looking into288 Transistor Circuit Approximations the second stage is part of the load seen by the first stage. In other words, the first stage sees a load resistance of rr = 10(10*)|[5(10*) = 3.3 kilohms Now we can find the voltage gain of the first stage. There is also about 1 ma of d-c emitter current in the first stage, so that r/ is in the range of 25 to 50 ohms. Again, we will use the upper limit of 50 ohms. The first- stage gain is _ Tr _ 3300 _ Ai =F = iy 06 Clearly, the input signal to the first stage is amplified and appears as an input signal of Avi, to the second stage. The output of the second stage is Your = ArAavin and the amplifier has an overall gain A of Se! = A,As = 66(100) = 6600 Naturally, this result is only an approximation, since we used an r! of 50 ohms. Had we used an r/ of 25 ohms, we would have found that Ai=80 4:=200 and = A = 16,000 Thus, the two-stage amplifier of Fig. 11-1 has a voltage gain in the range of 6600 to 16,000 for a of 100 and an r! between 25 and 50 ohms. Suppose that the @ of the transistors were 50 instead of 100. How would this affect the voltage gain? Let us use an r, of 50 ohms for convenience. The voltage gain of the second stage is Ar = tb = 20,000)10,000 ae Looking directly into the base of the second stage, we see Tingbase) = Br, = 50(50) = 2500 Again, we can neglect 2, and Rz, so that the input resistance of the second stage is approximately 2500 ohms. The collector of the first stage sees 10 kilohms in parallel with the 2500-ohms input resistance of the second stage. Therefore, the voltage gain of the first stage is rr _ 10,000||2500 _ Ala eeuCascading Stages 289 The overall voltage gain is A = A1Az = 40(100) = 4000 Thus, we have seen that for r, between 25 and 50 ohms and 6 between 50 and 100, the two-stage amplifier has a voltage gain between 4000 and 16,000. ‘The 8 and r, change significantly with temperature and with transistor replacement. Therefore, a two-stage amplifier like that of Fig. 11-1 is usable only in those situations where large variations in voltage gain are tolerable. The voltage gain of the two-stage amplifier can be stabilized to some extent by using emitter feedback. For instance, suppose we add a 500-ohm resistor to each emitter, as shown in Fig. 11-2. Now the voltage gain of +30 Fig. 1-2 RC-coupled amplifier with emitter feedback. each stage is relatively free of the changes in rf since it has been swamped by the 500-ohm resistor in each emitter. The voltage gain of each stage is now given by t 1 te forrg> rt ta tr re toe Thus, the voltage gain of the second stage is The input resistance of the second stage still is a function of 8. When 8 = 50, we get Tin@oas) & Bre = 50(500) = 25 kilohms290 Transistor Circuit Approximations The input resistance of the second stage is Tin = Rill Rallrincoase = 100(10*)||50(10%)/|25(108) = 14.3 kilohms The a-c load resistance seen by the first collector is 71, = Rilirin = 10(10°)||14.3(108) = 5.9 kilohms and therefore the first-stage voltage gain is te _ 5900 Are 500 = 118 The overall voltage gain is A = AjA, = 11.8(10) = 118 Remember this result is based on a 8 of 50. If we had used a 8 of 100, we would have found that Ai = 13.3 A: =10 and A = 133 Thus, we have seen that with no feedback the overall voltage gain is in the range of 4000 to 16,000 for a 2:1 spread in 6 and r’. With 500-ohm emitter feedback resistors, the voltage gain is in the range of 118 to 133 for the same spreads in 6 and ri. EXAMPLE 11-1 Find the overall voltage gain and the input resistance of the three- stage amplifier of Fig. 11-3. Use a 8 of 100. Sonvrion ‘The voltage gain of the third stage is and the input resistance of this stage is — Bo 15, — 10° 8) = 50 ki Tin = 3 lire = 7G |100(10") = 50 kilohms ‘The a-c load resistance seen by the collector of the second stage is 1, = Ril|rin = 10(10*)||50(10") = 8.33 kilohms Therefore, the voltage gain of the second stage isCaseading Stages 291 and the input resistance of this stage is 108 8.33 Tin = Ra [Bre = [100(10*) = 54.5 kilohms 2 The a-c load resistance seen by the collector of the first stage is 7, = 10(108)|[54.5(108) = 8.45 kilohms Therefore, the voltage gain of the first stage is _ 8450 _ A= To00 = 8.45 The overall voltage gain of the amplifier is A = A,A2A; = 8.45(8.33)10 = 700 ‘The input resistance of the three-stage amplifier is equal to the input resistance of the first stage. _ Bay, _ 108 ae rin = Gy lBra = gp ||100(10") & 54 kilohms Fig. 11-3 Examples 11-1 and 11-2. EXaMpPLe 11-2 Suppose that the @ is 50 instead of 100 in the three-stage amplifier of Fig. 11-3. What is the overall voltage gain and input resistance? Sonvri0n The gain of the third stage still equals292 ‘Transistor Cireuit Approximations The input resistance, however, is lower. 10° i i ty = Fy 15010") = 33.3 kilohms The voltage gain of the second stage is now ae 10(10°) ||33.3(10*) At 1000 277 and its input resistance is : rin = 2% |[50(10*) = 36 kilohms "The voltage gain of the first stage is — 10(10*)//36(10%) _ A, = CGO 27.8 and its input resistance is 10 : ra = 22 [50(10%) 2 36 kilohms The overall gain is A = AjAsAs = 7.8(7.7)(10) = 600 and the input resistance of the amplifier equals the input resistance of the first stage: Tin = 36 kilohms Thus, we have seen that for 8 between 50 and 100, the three-stage amplifier of Fig. 11-3 hus a voltage gain between 600 and 700, and an input resistance between 36 and 54 kilohms. 11-2 Two-stage Feedback Up to this point, we have discussed single-stage feedback in the form of an unbypassed emitter resistor. It is also possible to use feedback around two stages. The most widely used two-stage feedback arrangement is shown in Fig. 11-4. The basi idea behind this amplifier is the following. The input signal vi, is amplified and inverted in the first stage. The output of the first stage is amplified and inverted again by the second stage. A portion of this second-stage output is fed back to the first stage via the voltage divider formed by rx and rz. This return signal vp is applied to the emitter of the first stage, thereby reducing the base-emitter a-c voltage of the first transistor. In other words, we have negative feedback.Cascading Stages 293 Most of us already know that negative feedback reduces the overall voltage gain. The resulting voltage gain is more stable than the gain without feedback. To understand why this is so, suppose that for some reason the 8 of the second stage becomes larger. The output signal will then try to increase; however, more signal will be fed back to the emitter of the first transistor, thereby reducing the base-emitter a-c voltage. This results in less output voltage from the first stage, which partially offsets the g increase in the second stage. Similarly, if the output voltage of the second transistor tries to become smaller, less signal is fed back to the first stage, so that the output voltage of the first stage becomes larger. This will partially offset the original change in output voltage. To find out how effective the feedback is in stabilizing the voltage gain, we must first discuss the error voltage. Note in Fig. 11-4 that the a-c Yeo Fig. 11-4 Two-stage feedback. voltage at the base of the first transistor is vj, and that the a-c voltage at the emitter is vp. The error voltage is simply the difference of these two voltages, that is, Verror = Yin — UF This is the a-c voltage actually applied to the base-emitter terminals of the first transistor.204 Transistor Circuit Approximations The error voltage is amplified by the first transistor to produce a col- lector signal of Yer = Aierror = Ar(Yin — ve) where A, is the voltage gain of the first stage, r,/r!, The voltage out of the first stage is amplified by the second stage to produce a final output voltage of Your = ArAa(vin — vp) where Ay is the voltage gain of the second stage, rz/r!. We can rearrange this equation to get Vout AyAy Note that AiA: is the product of the individual stage gains. Normally, this product is quite high, so that to a first approximation the right-hand side of the last equation is almost zero. That is, Vin — UF = Yin — vr SO or Vin = Or In other words, in a good feedback arrangement, the product of Ay and A; is large enough for the error voltage to approach zero. This is equivalent to saying that the feedback voltage vr is almost equal to the input voltage vin. To find an approximate formula for the voltage gain of Fig. 11-4, note that the output voltage is applied to a voltage divider consisting of re and rp. Thus, to a first approximation ~ te Op Spay out We showed earlier that vi, = vp. Therefore, we can rewrite the last equa- tion as tg Vin = Vout tr+ te or yislt YE TE or for >> 1 (11-1) re Equation (11-1) is quite important. It tells us that the voltage gain of the two-stage amplifier of Fig. 11-4 is essentially equal to the ratio of the feedback resistor ry to the emitter resistor rg. Since 6 and r’, do not appearCascading Stages 295 in this equation, we conclude that the amplifier gain is independent of the transistor characteristics. Thus, we have a voltage gain that is free of variations caused by temperature and transistor replacement. Admittedly, the derivation of Eq. (11-1) was only an approximation. Still, in any well-designed two-stage feedback amplifier, Eq. (11-1) is quite accurate. If we were to make a more rigorous derivation, we would obtain this expression for the voltage gain: a toe _ (Te pao fa : Ana (3 a 1) TF r/At + Oe Frode ) where A ,, is the voltage gain with feedback, vout/Vin A is the product of stage gains, that is, Ai1A2 1 ig the emitter junction resistance, 25 mv/Iz An examination of Eq. (11-2) shows that the voltage gain depends upon rl and A. But the whole point of feedback is to swamp these quantities, that is, make them unimportant in determining the voltage gain..There- fore, by deliberate design we can make retire Arn te ar <1 and «1 In Eq. (11-2), note that when these inequalities are satisfied, the equa- tion reduces to elk An=iti This is precisely what is done in any good two-stage feedback amplifier; the inequalities are deliberately satisfied to make the voltage gain essen- tially independent of the transistor characteristics. We can also show with a careful derivation that the input resistance looking into the base of the first transistor is intone) = Br [1 4+—t (3 + 4)| (11-3) retre\r, Again, we note that in a well-designed feedback amplifier the aforemen- tioned inequalities are satisfied, so that Eq. (11-3) reduces to rintoase) & A Gri incoae SG, Ore This last equation tells us that the input resistance looking into the base of the first transistor equals @r! multiplied by A/A,. For instance suppose that the voltage gain A is A = AjAs = 10,000296 Transistor Cireuit Approximations and that the voltage gain with feedback is Ay = 100 Then A/A = 10,000/100 = 100. Therefore, the input resistance is 100 times greater than fr’. Let us summarize the important results of this section. For the two- stage feedback amplifier of Fig. 11-4: 1. A = A,A2, where A, and A; are found by using the rz/r, for each stage. 2. The necessary approximation conditions are A>Z and A>Ap 3. The voltage gain is Ap=t+ie™ torre > re Te re 4, The input resistance looking into the base is A py Fingoaw 2 Br Exampte 11-3 Suppose that the amplifier of Fig. 11-4 has an rp of 10 kilohms and an rg of 100 ohms. Find the approximate voltage gain. SonuTion The voltage gain is the ratio of the feedback resistor to the emitter resistor. Exameue 11-4 Find the voltage gain and input resistance looking into the base of the first transistor in Fig. 11-5. Soxvrion The input resistance looking into the base is A Fincoase) = A BreCascading Stages 297 35K Fig. 11-5 Example 11-4, We need to find A, and As. 10(10°)|/35(108) _ |. iaanaamggiaeaaad ato). and ve wth a 10(10°)|[10/155|}5@10") = 44 re 5O Therefore, A = AjA, = 44(155) = 6820 and rinoae) = 8220 (100)(50) = 244 kilohms 11-3 Inductive Coupling Oceasionally, transistor stages will use an inductor in the place of a resistor in the collector. Figure 11-6 illustrates a two-stage amplifier of this type. The analysis of such an amplifier is straightforward. First, note that such an amplifier is inherently a high-pass amplifier; that is, it is intended to amplify frequencies that are high enough for the inductors to appear as open circuits. In this case, the a-c load seen by the collector of the first stage is simply the input resistance of the second stage (the inductor looks open). The a-c load resistance seen by the second collector is simply 10 kilohms. Let us find the voltage gain of the amplifier shown in Fig. 11-6. As already indicated, this amplifier is intended to operate at frequencies high298 Transistor Cireuit Approximations in S0KII 25K. Fig. 11-7 A-c equivalent circuit. enough for the inductors to look open. In this case, we can draw the a-c equivalent circuit as shown in Fig. 11-7. The voltage gain of the second stage is The a-c load resistance seen by the collector of the first stage is rr = Ryl|Rell8r, = 50(10*)|]25(10*)|]100(50) = 3.85 kilohms and the voltage gain of the first stage is = _ SL Atal 30 7 77 Therefore, the overall voltage gain is A = AjAz = 77(200) = 15,400 The amplifier of Fig. 11-6 has limited use since it inherently has a high- pass filter response. Note that as the frequency is reduced, the reactanceCascading Stages 299 of the inductors eventually becomes so small that the inductors appear almost as short circuits. In this case, the circuit no longer amplifies. Exampte 11-5 Find the d-c voltage from collector to ground in the stages of Fig. 11-6. The r-f chokes have 10 ohms of d-c resistance. SoLution For practical purposes, the d-c resistance of the chokes is so small that there is a negligible d-c voltage drop across them. Therefore, almost all the 30 volts from the supply appears at the collector of each transistor. Actually, there is a small d-c voltage drop across each choke. By inspection, the voltage divider in each base circuit develops about 10 volts with respect to ground. This sets up about 1 ma of d-c emitter current. In turn, the d-c collector current is about 1 ma, and this pro- duces a d-c voltage drop of about 10 mv across the 10 ohms of resistance in each coil. Therefore, the d-c collector voltage is actually 10 mv less than 30 volts. Exampte 11-6 Suppose that the inductor in the second stage of Fig. 11-6 has an L = 10 mh. Find the frequency where the inductive reactance just equals 10 kilohms. Sonvrion Xp = QxfL or 10,000 _ ,. J = 52ig = 150 kHz This means that at 159 kHz (kilocycles) the reactance of the choke is too low to neglect. The a-c load seen by the collector of the second stage is 10 kilohms of inductive reactance in parallel with the 10-kilohm load resistance. Normally, the amplifier should be operated at a much higher frequency, so that the inductive reactance is much larger than the 10-kilohm load resistor. 11-4 Transformer Coupling Sometimes, a load resistance is so small that voltage gain is impossible. For instance, consider the circuit of Fig. 11-8. The a-c load seen by the300 Transistor Cireuit Approximations 20K lok Yout Fig. 11-8 Loading down a stage. 100 10k 10K |--W \L Tf collector is 10 kilohms in parallel with 100 ohms, which is essentially 100 ohms. There is about 1 ma of d-c emitter current, so that r/ is in the range of 25 to 50 ohms. This means that the voltage gain r,/r’ will be very low, around 2 to 4. In addition, we know from Chap. 10 that the maximum peak-to-peak signal that can be obtained before clipping occurs is approxi- mately 2Vcx or 2Zcrz, whichever is smaller. In this case, 2/orz is smaller, so that the maximum peak-to-peak unclipped signal is Vo = 2orz = 2(0.001)(100) = 0.2 volts We can use a transformer in this situation like this to improve the voltage gain and the signal-handling capability. For instance, suppose we add a 10:1 transformer to the circuit, as illustrated in Fig. 11-9a. We are using an ideal transformer to simplify the analysis. Recall that for an ideal transformer whose primary-to-secondary turns ratio equals n, the impedance looking into the primary is n? times the load on the secondary. In Fig. 11-9a, the load on the secondary is 100 ohms; therefore, looking into the primary there is a resistance of tz = WR = 107(100) = 10 kilohms The a-c equivalent circuit of the transistor stage is given in Fig. 11-9b. As already indicated, r/ is in the range of 25 to 50 ohms, so that we have a voltage gain between 200 and 400. This is the base-to-collector voltage gain. To find the actual output voltage note that the voltage across the secondary is stepped down by a factor of 10. Thus, the overall voltage gain in Fig. 11-9a is between 20 and 40.Caseading Stages 301 25< 5 <50 TOK! 20K 0K Fig. 11-9 Transformer coupling. (a) Circuit; (b) transformed collector load. By using a transformer, we have not only increased the voltage gain, we have also increased the signal-handling capability. At the collector of Fig. 11-94, the maximum peak-to-peak unclipped signal is Vp-p = 2Zerz = 2(0.001)(10,000) = 20 volts (Note that 2Vcz = 40 volts, so that 2Zcr, is smaller and therefore is the limiting quantity.) With a 10:1 step-down in voltage, the secondary voltage can be as large as 2 volts peak to peak before clipping occurs. Another possible use of transformers is in a cascade of CB stages. First, consider the two-stage amplifier of Fig. 11-10. Note that there is about 1 ma of d-c emitter current in each transistor; therefore, the a-c input 10K 25 10 50 ohms i -20 Fig. 11-10 Cascading CB stages.302 ‘Transistor Circuit Approximations resistance looking into each emitter is between 25 and 50 ohms. Let us use the upper limit of 50 ohms and compute the voltage gain of the second stage. rz _ 5000 ee 50 = 100 Note that the collector of the first stage sees an a-c load 71, = 10(10*)||20(10*) [50 ~ 50 ohms In other words, the first stage is heavily loaded by the emitter of the second stage because the emitter has an input resistance of only 50 ohms. The voltage gain of the first stage is approximately unity, that is, no gain. This is one reason that CB stages are almost never RC-coupled. Some form of impedance transformation must be used between the CB stages to permit a reasonable voltage gain to take place, as well as a higher signal-handling capability. One way of transforming impedances is to use a transformer between stages, as shown in Fig. 11-11. The voltage gain of the second stage is still 100. The collector of the first stage, however, now sees an a-c load of 7, = nr, = 10*(50) = 5 kilohms Now the voltage gain from the base to the collector of the first stage is _ 5000 _ Ai = |p = 100 Of course, there is a 10:1 voltage step-down in going from the primary +20 Vy; ne Yout Fig. 11-11 Cascading CB stages with a transformer.Cascading Stages 303 to the secondary, so that the voltage gain from the base of the first stage to the base of the second stage is about 199%, or 10. The overall voltage gain of the two stages is 1000. In general, if a CB or CE stage uses a transformer to step up the imped- ance level, there is an improvement in the voltage gain. In Fig. 11-12, we see that the a-c load looking into the primary is r= WR Therefore, the voltage gain of the stage is wR The voltage is stepped down between the primary and secondary by a factor of n. Therefore, the overall voltage gain is Your _ oe (11-4) Vin ny ¥ CE or CB stage Be oe 4 tout, a8 Yin fe Fig. 11-12 Deriving the voltage gain Fig. 1-13 Example 11-7. of a transformer-coupled stage. Exampie 11-7 Find vou in the circuit of Fig. 11-13. Use an rf of 25 ohms. Sonution We get Your __ mR _ 50(10) _ % 25 and. Your = 20(1 mv) = 20 mv304 Transistor Circuit Approximations 11-5 Tuned Amplifiers Sometimes, the stages of an amplifier are designed to amplify only a narrow band of frequencies. A way of accomplishing this is shown in Fig, 11-14. The idea here is simple. At resonance, the impedance of each Ke Vee Fig. 11-14 A tuned amplifier. Bra Fig. 11-15 A-c equivalent circuit. LC tank circuit is high. Above and below the resonant frequency the impedance decreases. Therefore, the voltage gain is maximum at reso- nance because each collector load is maximum at resonance. The approximate formula for the resonant frequency is 1 foieCascading Stages 305 The 3-db bandwidth of each resonant circuit is =f BW =5 where Q is the ratio of the a-c load resistance to the inductive reactance of the coil. That is, ‘The a-c equivalent circuit is shown in Fig, 11-15. Each collector works into a parallel LC tank. At resonance, the inductive reactance cancels out the capacitive reactance, leaving a purely resistive load on cach col- lector. Above resonance, the X¢ becomes smaller than the X,; the a-c load seen by the collector is therefore reduced, and this causes the voltage gain to become lower. Similarly, the voltage gain drops off below reso- nance because the X, becomes smaller than the Xe. ~Yer Fig. 11-16 Transformer-coupled tuned stages. Another tuned amplifier is shown in Fig. 11-16. Again this is a bandpass amplifier. Transformers are used to improve the impedance match be- tween stages. The voltage gain reaches a maximum value at the resonant frequency of the LC tank circuits. EXAMPLe 11-8 In Fig. 11-14, L = 100 uh, and C = 100 pf. The a-c load resistance seen by each collector is 10 kilohms, and the r! of each transistor is306 Transistor Circuit Approximations 100 ohms. Find the following: (a) The resonant frequency. (b) The voltage gain at resonance. (c) The bandwidth of each tuned tank circuit. Souvtion (a) The resonant frequency is 1 =a, Vi000-H oO 8° Me (0) The voltage gain at resonance is A= = 10,000 10.000 10,000 (c) To find the bandwidth, we need the Q. 10,000 @ = ¥5 = seraMTOHCOOVIO=H ~ 1 Now, we can find the bandwidth. fo _ 1.59(108) _ BW = G = “jo — = 159 kHz Thus, the amplifier has a maximum voltage gain of 10,000 at a resonant frequency of 1.59 MHz. The bandwidth of each tank circuit is 159 kHz, or about 160 kHz. This means that the voltage gain of each stage is down 3 db when the frequency is 80 kHz greater or less than the resonant frequoney. The overall voltage gain will be down 6 db at these frequencies. 11-6 Direct-coupled Amplifiers There is one more type of coupling that we want to discuss. All the amplifiers discussed so far have been limited in the lowest frequency that can be amplified. In other words, as the frequency of operation is reduced, eventually we find that coupling and bypass capacitors no longer appear as short circuits. Similarly, if a transformer is used to couple between stages, there is a lower frequency limit at which its coupling properties fall off. There are many applications in which an extremely low-frequency re- sponse is needed. The use of capacitors and transformers is out of the question for extremely low frequencies because the electrical sizes of these components become prohibitively large. For instance, we may want toCascading Stages 307 amplify 0.01 Hz. In this case, the size of coupling and bypass capacitors becomes huge and impractical. One approach in obtaining a frequency response that extends all the way to zero frequency is to direct-couple from one stage to another. Many circuit arrangements are possible in direct coupling. We will examine a few of them to convey the notion of how direct coupling is accomplished. The main idea is to leave out all coupling and bypass capacitors. ‘A two-stage circuit that uses direct coupling is shown in Fig. 11-17. Since there is no coupling capacitor between stages, there will be a d-c as well as an a-c interaction between stages. In the circuit of Fig. 11-17 Fig. 11-17 Direct-coupled stages using n-p-n transistors. note that the voltage divider in the first stage develops about 3 volts from base to ground. Almost all this 3 volts appears from the emitter to ground and sets up a current of about 1 ma. This 1 ma produces a voltage drop of 24 volts across the 24-kilohm collector resistor, so that the col- lector-ground voltage is about 6 volts. This 6 volts drives the base of the second transistor. Note that the second stage loads the first stage slightly because the input resistance of the second stage is around 600 kilohms. ‘The loading effect is light, so that the base-ground voltage of the second transistor remains at about 6 volts. The emitter-ground voltage of the second transistor is approximately 6 volts, and this produces an emitter current of about 1 ma. In turn, there is a voltage drop of about 18 volts across the 18-kilohm collector resistor, so that the final output voltage is about 12 volts with respect to ground. If we apply an input voltage vin, this will change the currents and voltages throughout the two-stage amplifier of Fig. 11-17. Note that there is no lower frequency limit. The input voltage can be an extremely low308 ‘Transistor Circuit Approximations frequency. In fact, the amplifier of Fig. 11-17 can amplify a d-c change at the input. For instance, suppose that the input voltage changes from 3 to 3.1 volts. This is a change of 0.1 volt. This change is amplified by a factor of 8 in the first stage (24 kilohms divided by 3 kilohms). Therefore, the change in the collector voltage of the first stage is about 0.8 volts in the negative direction. This change is now amplified by the second stage, whose voltage gain is about 3. Thus, the change in the output voltage is around 2.4 volts. As another example of direct coupling, consider the circuit of Fig. 11-18. Note that the first stage uses an n-p-n transistor and the second stage uses a p-n-p. There is about 1 ma of current in the first transistor so that the emitter-ground voltage is about 3 volts, and the collector- ground voltage is about 24 volts. This 24 volts drives the base of the #30 Fig. 11-18 Direct-coupled stages using complementary transistors. second stage. The net voltage across the 6-kilohm emitter resistor of the second stage is about 6 volts (30 minus 24 volts). This means that the current in the second transistor is around 1 ma, and therefore the col- lector-ground voltage is about 18 volts. The voltage gain of the two-stage amplifier of Fig. 11-18 is = _ 6000 18,000 _ 4 = AAs = 3000 “Go00 = © Thus, any change in the input voltage is amplified by a factor of 6. If two power supplies are available, a direct-coupled circuit like that of Fig. 11-19 can be used. The first stage is emitter-biased; the emitter current is about 1 ma. This produces a collector voltage of about 3 volts,Caseading Stages 309 Vout 3 7 Fig. 11-19 Direct coupling with two supplies. which drives the second stage. ‘The current in the second transistor is about 1 ma, and the collector voltage is about 6 volts. Note that the voltage gain of Fig. 11-19 is 27,000 24,000 A= A\A, = =72 Probably one of the most important of the direct-coupled types of cir- cuits is the so-called difference amplifier (also known as a differential amplifier). There are many forms of the difference amplifier and we will discuss only a few of the more important ones. One way to build a differ- ence amplifier is the circuit of Fig. 11-20a. Note that there are two inputs; the output is taken between the collectors. What the circuit does is first to take the algebraic difference of the two inputs and then to amplify this difference. That is, the output of the difference amplifier of Fig. 11-204 is Your = A(v1 — v2) where A is the voltage gain of the difference amplifier. By using the ideal- transistor approximation we can show that Aw (11-5) For instance, if R, = 10 kilohms and r, = 50 ohms, we have a voltage gain of 200. Thus, large voltage gains are possible with the difference amplifier. Since it is direct-coupled, there is no lower-frequency limit. Let us obtain a qualitative understanding of how the difference ampli- fier of Fig. 10-20a works. The circuit acts like a bridge circuit. Voltages310 Transistor Circuit Approximations Nec "ee Me out # (H~ %) Your? (4-4) ve fe (a) {o) Fig. 11-20 Difference amplifiers. 2; and 22 control the currents produced by each transistor. Thus, if 01 and v, are exactly equal, the bridge will be balanced, and there will be a zero output. On the other hand, when v; and v» are unequal, the bridge is unbalanced, and there will be an output voltage equal to the voltage s the algebraic difference of v; and 2. The difference amplifier of Fig. 11-20a can be stabilized against changes in r by swamping the emitter diode, as shown in Fig. 11-208. As indicated, the output voltage becomes Yout he (m1 — v2) for rs >, (11-6) Thus, in Fig. 11-200, if 2, = 10 kilohms and rz = 500 ohms, the voltage gain will be about 20. There is no need to actually use two inputs. One of the inputs can be zero, as shown in Fig. 11-21a and b, The formulas for the voltage gain and input resistance are given in Fig. 11-21. Another important form of the difference amplifier is the circuit of Fig. 11-22a. Note that the output is referenced with respect to ground. As indicated, the output of this circuit is R. Yous = ga (V1 — 02)Cascading Stages 311 Ke aft Yout = Vin fe Tin © 2 Bre (a) Fig. 11-21 Difference amplifiers with single input. tle oc (2) Ea ° Rf, tou = Fem) ) Fig. 11-22 Difference amplifiers with single output.312 Transistor Cireuit Approximations We can swamp the emitter diodes if desired; we then have the circuit of Fig. 11-226. We have briefly examined some of the more widely used forms of the difference amplifier. One major advantage of these circuits is that they are direct-coupled circuits capable of much higher voltage gains than the previously discussed direct-coupled amplifiers. Exampie 11-9 Sketch the output waveform for the difference amplifier of Fig. 11-234. Get an approximate answer using a 6 of 100 and an ri = 50 mv/Iz. SoLution First, note the use of a small potentiometer in the collector circuit. Such an adjustment is normally added to a difference amplifier to allow us to remove any unbalance caused by differences in transistors, load resistors, and so on. We adjust this potentiometer to produce a zero output when the input is at zero. Next, note that the d-c current flowing in the 10-kilohm emitter re- sistor is about 3 ma. The reason for this is simply that with both +30 Small out lomv 266 mvt | He lo) (0) Fig. 11-23 Example 11-9.Cascading Stages 313 transistors using emitter bias, almost all the emitter supply voltage must appear across the 10-kilohm emitter resistor. Since there are two transistors sharing the 3 ma, there is about 1.5 ma in each transistor. This produces about 15 volts from each collector to ground. We can now calculate the value of r/. Using 50 mv/Ig, we get This 33 ohms is quite a bit smaller than the 250 ohms in cach emitter circuit, so that we can say the emitter diodes are reasonably well swamped out. The voltage gain from input to output is simply ~ Rr _ 10,000 =r 350 ~ 40 and the input resistance looking into the base of the left transistor is Tintbaee) & 287x = 2(100)(250) = 50 kilohms This 50 kilohms of resistance is in parallel with the 50-kilohm base return resistor, so that the input resistance of the circuit is 25 kilohms. The source has a 10-kilohm resistance; therefore the actual input voltage is oe 25,000 a . : tin = Fs = 33,999 100) = 7.15 mv pealk This input voltage is amplified by a factor of 40 to produce an output voltage of Your = 40(7.15 mv) = 286 mv peak The output wave is sketched in Fig. 11-23b. The obvious advantage of this amplifier over an RC-coupled amplifier is that it has amplified an extremely low frequency (1 Hz). SUMMARY RC coupling is the most widely used method of cascading transistor stages. The output from a resistively loaded stage is coupled through a capacitor into the input of another transistor stage. The load seen by a collector in a cascade of RC stages is its own load resistance shunted by the input resistance of the next stage. Two-stage feedback is often used to stabilize the’ overall voltage gain314 Transistor Circuit Approximations of a two-stage cascade. To a first approximation, the voltage gain equals the ratio of the feedback resistor to the emitter resistance receiving the feedback signal. Sometimes inductors are used in the place of the collector load resistors. ‘The advantage is that the only load seen by the collector is the input resistance of the next stage, which means that all the signal power is delivered to the next stage. The disadvantage is that the lower-frequency response is degraded by the inductors. We can use transformers as coupling elements between stages. They are especially useful when load is so small that it would reduce both gain and signal-handling capability. By means of a transformer, the load seen by the collector can be increased to the point where we get good voltage gain and signal-handling capability. Tuned tanks give a response that is centered on the resonant frequency of the tuned circuits. In this way, we can amplify a desired band of frequencies. When we need extremely low-frequency response, we can use direct coupling. There are a number of ways to direct-couple stages. The differ- ence amplifier is quite important in direct-coupled amplifiers because it can provide large voltage gains down to zero frequency. GLOSSARY bandwidth In a tuned amplifier, this refers to the band of frequencies between the lower and upper 3-db points. cascade An arrangement using the output of one stage as the input to another. difference amplifier A circuit that amplifies the algebraic difference of two input signals. direct coupling A circuit in which the d-c voltages of one stage are coupled into another stage. Such circuits have no capacitors. error voltage In feedback systems where part of the output is fed back to the input, the error voltage is the difference between the input signal and the feedback signal. rf chokes Inductors that look like open cireuits at radio frequency. REVIEW QUESTIONS 1. In an RC-coupled amplifier, the a-c load resistance seen by a collector consists of what resistances?Cascading Stages 315 2. In a well-designed two-stage feedback amplifier, what is the approxi- mate relation between the input signal v;, and the feedback signal vp? 3. Define the error voltage as used in this chapter. 4. What is the approximate voltage gain of the two-stage feedback ampli- fier discussed in this chapter? 5. For a cascade of inductively coupled stages operating at higher fre- quencies, what is the effective a-c load resistance seen by a collector? 6. When a load resistance is extremely small, transformer coupling is sometimes used. What are the two advantages of a transformer in this situation? 7. Why are tuned tanks sometimes used in the collector of a transistor amplifier? 8. Why are direct-coupled amplifiers necessary in some situations? 9. What is a difference amplifier? Draw the schematics of some of the difference amplifiers discussed in this chapter. PROBLEMS 11-1 Suppose that the 8 of the transistors in Fig. 11-1 is 200 instead of 100. What is the approximate voltage gain of the two-stage cascade? Use 50 mv/Tz. 11-2 In Fig. 11-2, suppose that the rz of the first stage is 300 ohms instead of 500 ohms. What is the overall voltage gain? (8 = 100.) 11-3 Work out the voltage gain of the circuit in Fig. 11-3 for transistors with a 8 of 200 and r{ of 50 mv/Tz. 11-4 The two-stage feedback amplifier of Fig. 11-4 has an rp of 10 kilohms and an rr of $2 ohms. What is the approximate voltage gain of the two-stage feedback amplifier? 11-5 Suppose that 100-mh chokes are used in Fig. 11-6. At what fre- quency will these chokes have a reactance of 10 kilohms? Should the amplifier be operated at this frequency if maximum voltage gain is desired? Why? 11-6 In Fig. 11-8, a 500-ohm load is used instead of a 100-ohm load. What is the voltage gain and maximum unclipped signal output for this 500-ohm load? 11-7 Suppose that the transistors of Fig. 11-11 have rj = 25 mv/Tx. If the input signal is 1 mv rms, what is the output signal? 11-8 In Fig. 11-14, L = 200 wh, and C = 500 pf. What is the resonant frequency of each tuned circuit? Suppose that the rz seen by each col- lector at resonance is 12 kilohms. What is the Q and the bandwidth of each tuned tank?316 ‘Transistor Cireuit Approximations 11-9 The value of Vex is —25 volts in Fig. 11-202. The value of Rg is 10 kilohms. What is the d-c emitter current in each transistor? 11-10 In Fig. 11-20a, 7 is 25 ohms and Ry is 5 kilohms. What is the approximate value of voltage gain? 11-11 In Fig. 11-206, the value of R; is 7.5 kilohms and re is 680 ohms. The emitter diode is swamped. What is the voltage gain? 11-12 The value of Rg in Fig. 11-21 is 10 kilohms. The emitter supply Vez = 20 volts. If the input signal is a 2-my-peak sine wave, what is the output voltage (use r, = 25 mv/Ip)? Use an Ry of 5000. 11-13 In Fig. 11-218, vim is a 10-my-rms signal. The emitter diodes are swamped. With an R,, of 6.8 kilohms and an rg of 390 ohms, what is the approximate value of vou.? 11-14 What is the voltage gain of the circuit in Fig. 11-22a when R, = 7.5kilohms, Rg = 10 kilohms, Veg = 25 volts, andr! = 35 mv/Ip?12 Temperature Effects We mentioned in Chap. 1 that the number of carriers in a semiconductor will increase as the temperature increases. Because of this, virtually every transistor characteristic is temperature-dependent. In some circuits, the changes caused by temperature variations aro 60 scrious that the tran- sistor cannot function normally. In this chapter, we discuss four important transistor quantities that are temperature-dependent. These are the a-c resistance r/, the 8, the base-emitter voltage Vez, and the collector leakage current. 12-1 Changes in Emitter-junction Resistance In our approximation of r} we have used 25 mv — , _ 50 my Tae ag The lower extreme in this inequality applies to abrupt p-n junctions operating at room temperature (around 25°C). The upper extreme more closely applies to diffused p-n junctions. In either case, when the tem- 317318 ‘Transistor Circuit Approximations perature of the junction changes, the value of r’ will change from its room-temperature value. By applying calculus to the general diode equation, it is possible to find out how much r’ changes with temperature. The result of such a derivation is the following: rj increases about 1 percent for each 3°C rise in ambient temperature.’ (The ambient temperature is the tempera- ture of the surrounding air.) For instance, suppose that a transistor circuit is operating in an ambient temperature of 20°C and that r! is 25 ohms. If the ambient temperature were to rise to 50°C, the ri would rise about 10 percent to a new value of 27.5 ohms. On the other hand, if the ambient temperature were to drop from 20 to —40°C (a change of 60°C), the r; would decrease about 20 percent (1 percent for each 3°C change) to a new value of 20 ohms. Recall that a CB or CE transistor stage without feedback has a voltage gain of approximately Vout TE moe By inspection of this equation, it is clear that the voltage gain will drop about 10 percent for each 30°C rise, and vice versa. If such changes are objectionable, we can swamp out the emitter diode, as discussed in earlier chapters. EXamp.e 12-1 The transistor in Fig. 12-1a has an r, of 50 ohms for an ambient temperature of 25°C. Find the approximate voltage gain at 25 and 85°C. Souvtion At 25°C, the voltage gain is approximately Vout 100 7 nz _ 10,000//10,000 _ ea 50 a When the ambient temperature rises to 85°C (a 60°C rise), the rj will increase about 20 percent to a new value of 60 ohms. The voltage gain at 85°C is approximately 83.5, Pout 7 _ 10,000{/10,000 _ Vin 7 60 1 We are assuming that the power dissipation at the emitter junction remains the same so that the junction temperature rises by the same number of degrees as the ambient temperature.Temperature Effects +30 319 +30 10K Yout vine ) 10K vined c 10K 3 Ee a Se 450 ta) Fig. 12-1 Examples 12-1 and 12-2. EXAMPLE 12-2 Suppose that we swamp the emitter diode of the transistor in the preceding example as shown in Fig. 12-15. Find the approximate volt- age gain at 25 and 85°C. Souvtron At 25°C, the circuit of Fig. 12-1) has un approximate voltage guin of _ 10,000)/10,000 _ Your Th 10 Thus, we see that by swamping the emitter diode, we have stabilized the gain. The original circuit (Fig. 12-1a) showed a change in gain of about 16 percent over the temperature range, whereas the new circuit (Fig. 12-15) changes only 2 percent over the same temperature range. Of course, we give up voltage gain when we swamp the emitter diode; however, there are many situations where it is preferable to have a stable voltage gain despite the reduction in gain.320 Transistor Circuit Approximations 12-2 Changes in § Both the d-c and the a-c 8 change with changes in temperature. No gen- eral formula can be given for 8 changes because they depend a great deal upon the techniques used in the manufacture of the transistor. Usually, as the temperature increases, both the d-c and a-c @ will increase, but there are exceptions, In some transistor structures, the a-c 8 can actually decrease as the temperature increases. About all we can say in general about the @ changes is that they can be quite large. For instance, in some transistors, the a-c 8 can change by a factor of 4 as the temperature changes —65 to 150°C. Specific information on the changes in the d-c and a-c 8 is sometimes given on the data sheet for the particular transistor. The usual approach in design is to allow for the worst-case 6 to be encountered. ‘The changes in the d-c 8 affect the d-c operating point of the transistor. In Chap. 9, we discussed the sensitivity of the d-c operating point to changes in 8. We saw that base-biased circuits were relatively sensitive to changes in 8, whereas the emitter-biased circuits showed almost no dependence upon the changes in 8. Thus, in any circuit that is to operate over a large change in temperature, emitter bias is far preferable tu base bias. The changes in the a-c primarily affect the a-c input resistance looking into the base of a CE or CC circuit. Recall the following formulas for a-c input resistance: Tacos) = Br, for a CE circuit with no feedback Fingoeae) Bre for a CE circuit with emitter feedback Tinos Br, for a CC circuit (emitter follower) All these input resistances are directly proportional to the a-c 8. Thus, the input resistance is a function of temperature. The usual design ap- proach is to use the worst-case 8 (the lowest) that can occur over the expected temperature range. The reason for using the lowest 6 as the worst case is that the input resistance will produce maximum loading on the source when the @ is at its lowest value. 12-3. Changes in Vaz In our past work, we have approximated the d-c voltage Vsz across the base-emitter diode by allowing 0.3 volt for germanium and 0.7 volt for silicon. These are room-temperature values. When the emitter-junctionTemperature Effects 321 temperature changes, the Vez drop will change. By applying calculus to the general diode equation, the following approximations can be derived: 1. For germanium transistors, Vgx decreases about 1 mv for each 1°C rise. 2. For silicon transistors, Vzz decreases about 2.5 mv for each 1°C rise. For instance, if the junction temperature rises 50°C, the Vax drop in a germanium transistor will decrease about 50 mv to a new value of 0.25 volt. Whether or not the changes in Vzz are important will depend upon the particular biasing arrangement. For instance, in Fig. 12-2a, the d-c base current is 20 = 0.7 cc anI0 We neglected the 0.7-volt drop because it is much smaller than the 20 volts driving the base circuit. Also note that even if the temperature changes, the value of Vz is still negligible. In other words, in Fig. 12-2a, the small value of Vex is swamped out by the much larger value of 20 volts, Tn = 20 pa +20 2 Fig. 12-2 Effect of Vax changes. Siticon Silicon + ‘The circuit of Fig. 12-2b is different. The supply voltage is only 2 volts; therefore, Vx is no longer negligible. At room temperature, the base current is 0.7 108 = 13 pa Since Vee is no longer negligible, changes in Vax can be important in setting the base current. For instance, if the temperature rises 60°C, the change in Vez will be AVaz = —60(2.5)(10-*) = —0.15 volt322 Transistor Circuit Approximations The base current at the elevated temperature will be Ip =P a pss = 1.45 pa This change in base current will change the d-c operating point of the transistor and can lead to clipping if a large signal is involved. It should be apparent that the shift in operating point caused by changes in Vzz can be minimized by making the supply voltages much larger than the Vyz. In other words, we must make sure that Vaz is swamped out. EXampPLe 12-3 The transistor shown in Fig. 12-3 has a Vzz drop of 0.7 volt for a junction temperature of 20°C. Find the collector-ground voltage at 20 and 70°C. SoLuTion The circuit of Fig. 12-3 is a single-supply emitter-biased arrange- ment. The voltage from base to ground is Vp = 10,000 * = 20,000 + 10,000 About 0.7 volt is dropped across the base-emitter diode at 20°C so that the voltage actually appearing across the 10-kilohm emitter resistor is 4.5 = 1.5 volts Ve =1.5 — 0.7 = 0.8 volt and the d-c emitter current is Fig. 12-3 Example 12-3.Temperature Effects 323 The d-c voltage from collector to ground is Vo = Veo — IoRy, 4.5 — 0.08(10-)(10*) = 3.7 volts When the temperature increases to 70°C, the change in Vas is AVee = —50(2.5)(10-*) = —0.125 volt The d-c collector current becomes — 15 — 0.575 To 10* = 0.0925 ma and the collector-ground voltage becomes Vo = 4.5 — 0.0925(10-4)(104) = 3.58 volts 12-4 Leakage Current in a Grounded-base Circuit In a grounded-base circuit like that shown in Fig. 12-4a, the emitter diode is forward-biased and the collector diode is back-biased. Because c (o) (6) Emitter open (ce) Fig. 12-4 Collector-base leakage current with the emitter open.324 ‘Transistor Circuit Approximations there are thermally produced minority carriers in the collector, there exists a leakage component of collector current in addition to the col- lector current produced by normal transistor action. For instance, in Fig. 12-4a, we know that as the emitter current ig is changed, the collector current ic will change. Typical characteristic curves for these changes are shown in Fig. 12-4, Note that there is some collector current even when iz = 0. In the ideal-transistor approximation, we neglected this small amount of collector current. Now we are going to take it into account. The condition ig = 0 can be represented by the equivalent circuit of Fig. 12-40, where the emitter lead has been opened. Under this condition, there can be no emitter current. However, there is still a small amount of collector current because of the reverse current in the collector diode. This reverse current is symbolized by Icno, where the subscripts CBO stand for collector to base with the emitter open. Very often, Icgo is simply written as Ico. In a grounded-base circuit the total collector current is actually the sum of two components. First, there is the component produced by normal transistor action, that is, the collector current controlled by the emitter current. Second, there is Ico, which is the result of thermally produced minority carriers. To utilize our earlier work in transistors, we ean repre- sent the effect of reverse or leakage current by using an ideal transistor in shunt with a current source of Ico, as shown in Fig. 12-5. Note that 4, (Idea!) gE Ideal fee _; fe Fig. 12-5 Equivalent cireuit for Ico leakage current. 8 the leakage component Jo is in the same direction as the ideal component of collector current. Therefore, the total collector current is Tc = Tevaesy + Ico Since Ico is the result of thermally produced carriers, we can expect Ico to increase with temperature, The amount of increase in Zco with temperature can be found by using the following approximate rules: 1. For germanium, Ico doubles for every 10°C rise in temperature. 2. For silicon, Ico doubles for every 6°C rise in temperature.Temperature Effects 325 For instance, suppose that Ico is 1 pa at 25°C. For a silicon transistor, Zco will be 2 wa at 31°C, 4 wa at 37°C, 8 wa at 43°C, ete. It is worth mentioning again that at room temperature the leakage currents in silicon transistors are generally much smaller than the leakage currents in germanium transistors. In earlier times, the germanium transistors were widely used because the silicon units were much more expensive. This is no longer the case. Silicon units are now comparably priced with germanium units. Because of this, silicon transistors are gain- ing wide acceptance. Other things being equal, the silicon transistor is far preferable to the germanium transistor because of its much lower leakage current. The leakage current Ico is undesirable. The reason for this is simply that as the temperature increases, the leakage current increases and causes a shift in the d-c operating point of the transistor. For instance, in Fig. 12-6a, the d-c emitter current is 200 10(10%) If there were no leakage current, the collector current Je would approxi- mately equal the emitter current of 1 ma. The collector-base voltage would then be Is = Ima Ve = 20 — 10-4104) = 10 volts We have shown the ideal d-c operating point in Fig. 12-6. Soturotion Actual Ideal Fig. 12-6 The effect of leakage current. Now, let us take the leakage current Ico into account. Suppose that Ico = 0.5 ma at an elevated temperature. The actual collector current is the sum of the ideal current plus the leakage current. Hence, Te = Tetaen + Ico = 1 ma + 0.5 ma = 1.5 ma326 Transistor Cireuit Approximations and the collector-base voltage is Vo = 20 — 1.5(10-*)(104) = 5 volts The actual operating point is shown in Fig. 12-6b. Thus, the operating point has shifted because of the leakage current. In fact, if the temperature were to increase, Ico would increase, and the operating point would shift upward along the d-c line. It is clear that if the temperature increases enough, the d-c operating point can shift all the way to the saturation point, which would lead to clipping of an a-c signal. The general rule for avoiding a significant shift in the d-c operating point of a grounded-base circuit is simply this: the leakage current Ico must be much smaller than the ideal collector current at the highest temperature to be encountered. Exampie 12-4 The transistor of Fig. 12-7a is made of germanium and has an Ico of 1 wa at 20°C. Calculate the value of Ves at 20 and 80°C. SorvTion The d-c emitter current is 10 Iz = 700708) ~ 0.1 ma At 20°C, the collector current is the sum of 0.1 ma and 1 ua. We can neglect the 1 ya since it is much smaller than 0.1 ma. The collector- base voltage is Ven = 10 — 0.1(10-*)(50)(10%) = 5 volts look 50K “cB Fig. 12-7 Examples 12-4 and 12-5.Temperature Effects 327 At 80°C, the temperature has risen by 60°C. Since Ico doubles for every 10°C rise, Io will double six times, so that Ico = 10-8(2*) = 64 pa = 0.064 ma The total collector current is the sum of the ideal component and the leakage component. Therefore, Ic = 0.1 + 0.064 ma = 0.164 ma and the collector voltage is Ver = 10 — 0.164(10~*) (50) (108) = 1.8 volts Note that there has been a significant shift in the d-c operating point as shown in Fig. 12-7b. If the temperature should increase further, the transistor would soon saturate. Exampte 12-5 The germanium transistor of the preceding example is replaced by a silicon transistor of similar characteristics except that Ico = 1 na (10~* amp) at 20°C. Find the value of Vcg in Fig. 12-7a for tempera- tures of 20 and 80°C. So.vuTion At 20°C, 1 na is negligible compared to the ideal collector current of 0.1 ma. Hence, Veg = 5 volts. The leakage current in silicon transistors doubles for every 6°C rise. When the temperature changes from 20 to 80°C, the leakage current will have doubled 10 times. That is, at 80°C, Teco = 10-9(2"°) = 1.02 pa Note that even at 80°C this leakage current is still negligible compared to the ideal collector current of 0.1 ma. Therefore, Veg = 5 volts as before. 12-5 Leakage Current in a Grounded-emitter Circuit Consider the grounded-emitter circuit of Fig. 12-8a. When we vary the value of base supply voltage Vx, the size of the base current will vary, which in turn changes the collector current. As we have seen before, the transistor action of this circuit can be summarized by the typical charac- teristic of Fig. 12-8). Note carefully that there is some collector current even when the base current is zero (bottom curve). The condition of zero base current is depicted by Fig. 12-Se, where we328 Transistor Cireuit Approximations have shown the base lead open. With zero base current, there is a leakage component of collector current labeled Jczo. The subscripts CEO stand for the collector to emitter with the base open. Here is an interesting phenomenon. The current Icgo is much larger than Ico. In fact, Icro = BIco. The reason for this can be seen by con- sidering Fig. 12-8d, where we have shown the real transistor as an ideal & +g tee {" ° : 0 ee Loe tog S_ (0) to} "ie | Leo Base ‘open Real c} (0) Fig. 12-8 Collector-emitter leakage current with the base open. transistor shunted by a current source Ico. Since the base lead is open, all the Zco current must flow into the base of the ideal transistor. This produces a collector current of 67co. By inspection of the circuit, Zcgo is Iczo = Blco + Ico = (8 + 1)Ico or Icxo = BIco for 6>1 (12-1) Note that for the circuit of Fig. 12-8a, Icgo is in the same direction as the normal collector current. The total collector current in the circuit is Io = Bln + IczoTemperature Effects 329 The first. component fT» is the ordinary component of collector current that we have discussed in earlier chapters; it is the component of collector current that we want and need for normal transistor action. The second component Igo is the leakage component of collector current; it is unde- sirable because it will disturb the d-c operating point of a transistor cir- cuit. Since Icgo is temperature-dependent, it can cause serious shifts in the d-c operating point as the temperature increases. This, of course, must be avoided. More is said about d-c operating-point shift in the next section. Exampe 12-6 The silicon transistor shown in Fig. 12-9 has an Ico = 1 na at 20°C. Find the value of Vcg at 20 and 80°C. (6 = 100) Sonution At 20°C, the value of Iczo is Tero = Blco = 100(10-*) = 0.1 pa The approximate value of base current is Yeo _ _20 Rn ~ 205 The total collector current at 20°C is Te = Ble + Iceo = 100(10)(10-§) + 0.1(10-*) = 1 ma Ine = 10 na and the collector-emitter voltage is Ver = 20 — 0.001(10,000) = 10 volts At 80°C, the value of Ico has doubled 10 times so that Ico = 10-*(2!°) = 1.02 pa Fig. 12-9 Example 12-6. B=100 Silicon330 Transistor Circuit Approximations Assuming that the is still 100, the value of Iczo is Toxo = BIco = 100(1.02)(10-8) = 0.1 ma The normal component of collector current BJ, is still 1 ma. To this we add 0.1 ma of leakage current to obtain a total collector current of 1.1 ma. Thus, at 80°C, the collector-emitter voltage becomes Ver = 20 — (0.0011)(10,000) = 9 volts Note that there has been a slight shift in the d-c operating point as the temperature changed from 20 to 80°C. As the temperature increases further, the shift will become more pronounced until eventually the tran- sistor saturates. 12-6 The Stability Factor Leakage current is undesirable; it causes the d-c operating point to shift as temperature increases. Ideally, we would like to have zero leakage current. More realistically, we would like to have the leakage component much smaller than the normal component of collector current. In this section, we reexamine the six biasing arrangements studied in Chap. 9. Our aim is to find out how much the d-c operating point shifts in each of these biasing arrangements. For any of the six common bias circuits, the total collector current is the sum of the ideal component and the leakage component. That is, Te = Teviaeat) + Tieskage The size of the leakage current will depend upon the bias arrangement used and the value of the resistors in the circuit. In dealing with the leakage current, it is common practice to speak of the stability factor. We will define the stability factor S by Treaxsge = Sco In other words, S is a constant of proportionality between I¢o and Theakage-' The stability factor tells us how good a particular biasing arrange- ment is as far as leakage current is concerned: the smaller the S, the better the bias arrangement. By means ot calculus, we can find the stability factor for each of the six common bias circuits discussed in Chap. 9. The results of this analysis are summarized by Figs. 12-10 and 12-11. For each circuit, the formula 1 This is approximately the same as 47¢/dIco and is more easily understood by the reader who docs not have a knowledge of calculus.‘Temperature Effects 331 +e s % $B 8 | ‘Saturation -++-____—__—_- p, BR, SE (a) +e Vip /e $ + Fg BRe. fe B BRA i ( AL soturain fy Ss BR, Re : (2) Me o ft Rs B 1s /A ii Br y Midpoint bios fe BR, = (ce) lee a Ry Fy 8 _ te (feth) B/2 A o* TR7BReR) = LE Midpoint bias i BAER) Re = a Fig. 12-10 The stability factor of base-bias circuits.332 Transistor Circuit Approximations le $ R B 1+ fy Be “16K/BRe I Pe fe fa Fe 2 ~Veg (a) $ B _ THR UR YAR | TR MAYER a 4 > fllh, () Fig. 12-11 The stability factor of emitter-bias circuits. for S is shown. Also, a sketch of S vs. Rs (or Ril|R2) is given. In all six circuits, note that the value of S lies between wnity and 8. These two extremes are the lowest and highest values of S. Since the leakage current in any circuit equals SIco, the better bias arrangements are those with low values of S. Generally speaking, the simple base-biased circuit of Fig. 12-10a is the worst way to bias a transistor since it always has an S equal to 8. This means that the leakage current in this circuit always equals BIco. The remaining base-biased circuits of Fig. 12-10 can be designed to have a value of S that is less than 8; however, very low values of S can be had only by almost saturating the transistor (this is analogous toTemperature Effects 333 obtaining low values of K by almost saturating the transistor, as dis- cussed in Chap. 9). Generally speaking, the emitter-biased circuits of Fig. 12-11 are the circuits in which we can get a very low value of S without saturating the transistor. With these circuits, we can easily set the current at the middle of the d-c load line and still have a very low value of S. For instance, in Fig. 12-12, we have a two-supply emitter-biased circuit. There is about +20 1K Fig. 12-12 Stability of a typical emitter-bias circuit. B= 100 20K 20k -20 1 ma of current in the collector, and this produces a collector-ground voltage of about 10 volts. Thus, the operating point is in the middle of the d-c load line. Since Re = Re, the stability factor of this circuit is 1+ Ro/Re | 1+1 Ly T+R,/BRs 1+ 1/8 This means that the leakage current in the collector is only 2 times the Ico of the transistor. In summary, leakage current is no longer the serious problem it was when germanium transistors were widely used almost to the exclusion of silicon transistors. However, even with the low-I¢o silicon transistors, leakage current may sometimes cause a problem. As a simple guide in determining whether leakage current is a problem, the following can be used: 1. Since S can be no greater than 8, first check to see whether Blco K Leviaeay S= at the highest junction temperature to be encountered. If this condition is satisfied, the d-c operating point is stable, and there is no need to find the S of the circuit.334 Transistor Circuit Approximations 2. If the foregoing inequality is not satisfied, it is necessary to find the S of the circuit. After S is known, check to see whether STeo K Tevaeay at the highest junction temperature to be encountered. If this condition is satisfied, the d-c operating point is stable. If not, there is a leakage- current problem, and it is necessary to reduce the S of the circuit by redesigning, or to reduce the Ico by selecting a better transistor. EXAMPLE 12-7 The germanium transistor of Fig. 12-13 has a 8 of 100 at all tempera- tures. The transistor has an Ico = 5 wa at 20°C. Determine whether or not the leakage current causes a significant shift in the d-c operating point over the temperature range of 20 to 50°C. Souution The circuit of Fig. 12-13 is a standard single-supply emitter-bias arrangement. With approximately 10 volts developed from base to ground, there is an ideal emitter current of about 2 ma. Therefore, the ideal or normal component of collector current is about 2 ma. The leakage component is to be added to the 2 ma of ideal collector current. To find the leakage current, we must first find the I¢9 at 50°C. At 50°C, Ico will have doubled three times for the germanium transis- tor. Therefore, at 50°C, Ico = 5(10-8)(2*) = 0.04 ma To determine whether leakage current is a problem, let us first make a quick check to see whether 8Zco is much smaller than 2 ma (the ideal collector current). BIco = 100(0.04)(10-*) = 4 ma This indicates that we must actually find the stability factor S for the circuit of Fig. 12-13. (Note that had 87¢o been much smaller than 2 ma, we could have concluded immediately that there was no problem at all.) The stability factor S for the circuit of Fig. 12-13 is 1+ (ill) /Re _ _ 1+ 3.33(10)/500)_ gs 1+ (Ril R)/BRe ~ TF 8.33(10%)/100(5) (05) = Therefore, the leakage current in the collector is Tieaoxe = SIco = 1.65(0.04 ma) = 0.066 ma S= Clearly, 0.066 ma is much less than the 2 ma of ideal collector current.‘Temperature Effects 335 We conclude that the leakage current is small enough at 50°C not to cause a significant shift in the d-e operating point. Note that if the same transistor had been used in a simple base-biased circuit, the stability factor would equal 100 instead of 1.65. The leakage current would then cause a large enough shift in the operating point to saturate the transistor. +30 +20 50K 4 B= 100, Germanium hy and Jpg = Sua at 20°C B=100 Silicon fog = Vn at 20°C Fig. 12-13 Example 12-7. Fig. 12-14 Example 12-8. Exampie 12-8 Determine whether or not the leakage current in the circuit of Fig. 12-14 will be a problem at 50°C. Sonution First, we need the approximate value of ideal collector current. Using the methods of Chap. 9, we get 7 Woot 20 ie °= Ri + Ra/B 50) + 109/100 In order not to be a problem, the leakage current must be much smaller than 0.333 ma at the highest temperature. At 50°C, Ico = 10-*(2°) = 32 na First, let us use the quick check of BIco & Ieviaea: BT co = 100(32 na) = 3.2 ua Since 3.2 ua is much smaller than the ideal collector current of 0.333 ma, we conclude that the leakage current is negligible. There is no point in even calculating the § value since SZco will be smaller than 3.2 na.336 ‘Transistor Cireuit Approximations SUMMARY The characteristics of a transistor change when the temperature changes. This means that the d-c and a-c operation of a transistor circuit depend to some extent upon the ambient temperature. The r, of a transistor inereases about 1 percent for each 3°C rise in ambient temperature. This will change the voltage gain of a transistor circuit unless the emitter diode is swamped. Both the d-c and a-c 6 change when the temperature changes. Usually, both of these will increase with temperature, but sometimes the a-c B shows a decrease with rising temperature. The d-c voltage drop Vas is affected by temperature change. In a germanium transistor circuit, Vax decreases about 1 mv for each 1°C rise. In a silicon transistor cireuit, Vzx decreases about 2.5 my for each 1°C rise. To minimize the effects of a changing Vsx drop, we can use a base or emitter supply voltage that is much larger than Vpz. In any biasing arrangement, the d-c collector current is the sum of the ideal collector current and a leakage component. The leakage component equals SIco, where S is the stability factor of the circuit. As a rule, we must keep the leakage current at the highest junction temperature much smaller than the ideal collector current to avoid a shift in the d-c oper- ating point. Silicon transistors are preferable to comparable germanium trans- istors because the Ico currents in silicon are much smaller than in germanium. GLOSSARY ambient temperature The surrounding temperature. The temperature of the environment in which the transistor circuit is operating. Tczo The reverse current that flows from the collector to the emitter when the base lead is open. Ico Also designated Icno. This is the reverse current that flows from the collector to the base when the emitter lead is open. leakage current This is the undesired component of collector current that flows in addition to the normal or ideal component. The leakage current in any biasing arrangement equals SIco. stability factor (S) This is the ratio of the leakage current in a given bias arrangement to the Ico of the transistor. The value of S is between 1 and for any bias circuit.‘Temperature Effects 337 REVIEW QUESTIONS 1. What is the percent change in ri for a 30°C rise in temperature? 2. When the temperature increases, what usually happens to the d-c and a-c 6? 3. In a germanium transistor circuit, how much does Vgg change for each centigrade degree rise in temperature? How much is the change for a silicon transistor circuit? 4. Why is it possible to have some collector current even when there is no emitter current? 5. What does Ieao stand for? And Ico? 6. The Ico of a germanium transistor doubles when the temperature rises how many degrees centigrade? What is the answer for a silicon transistor? 7. In general, which transistors have lower values of Ico, silicon or germanium? 8. What does Icgo stand for? 9. What is the stability factor S? 10. Which biasing arrangements usually have the lowest values of S? 11, What is the smallest possible value of S? The largest? PROBLEMS 12-1 The value of r, is 40 ohms at 25°C. What is the value of r! if the temperature rises to 65°C? 12-2 The transistor in Fig. 12-15 has an ri of 50 mv/Iz when the tem- perature is 25°C. What is the approximate voltage gain of the circuit when the temperature rises to 65°C? +20 Fig. 12-15 fe “out “om 1338 Transistor Circuit Approximations 12-3 Suppose that instead of bypassing all the 1-kilohm emitter resistor in Prob. 12-2, we bypass only 900 ohms, that is, use an rz of 100 ohms. Find the voltage gain at 25 and 65°C using Vout Th Yin e+ Te 12-4 The @ of the transistor in Fig. 12-15 equals 75 at 25°C and 125 at 65°C. Find the a-c input resistance at each temperature. Use ri = 50 mv/Tz. (Include the change in r’.) 12-5 In Fig. 12-16, suppose that the 8 of the transistor equals 50 at 25°C and 100 at 75°C. What is the value of »,,, for each of these tempera- tures? Neglect rj. 12-6 Suppose that a silicon transistor has a Vaz of 0.6 volt at 25°C. What value does Vzz have at 75°C? 12-7 Suppose a germanium transistor has a Vez drop of 0.25 volt at 25°C. What value does Vgz have at —25°C? 12-8 The silicon transistor of Fig. 12-17 has a Vag drop of 0.7 volt at 25°C. Find the collector current at 25 and 75°C (take Vx into account). 12-9 Suppose a germanium transistor is used instead of a silicon tran- sistor in Fig. 12-17. Compute the collector current at 25 and at 75°C. Use a Vag drop of 0.3 volt at 25°C. 4h \ Fig. 12-16 Fig. 12-17 12-10 Find the stability factor S$ for the circuit of Fig. 12-15. If the Ico of the transistor is 1 za, what is the leakage component of collector current? (Use a 8 of 100.)Temperature Effects 339 12-11 In Fig. 12-15, if the Ico of the transistor is 1 wa at 25°C, what would the leakage current be at 55°C if the transistor is made of ger- manium? Use a 8 of 100. 12-12 In Fig. 12-16, the transistor is made of silicon and has an Ico of 3 na at 25°C. Will leakage current be a problem in this circuit if the temperature rises to 75°C? (8 = 100.) 12-13 What is the stability factor S for the circuit of Fig. 12-16? Use a 8 of 100. 12-14 Suppose Rz = 1 megohm, R, = 10 kilohms, Rg = 1 kilohm, and 6 = 50. What is the stability factor S for the circuit of Fig. 12-100? 12-15 The circuit of Fig. 12-10c has an Ry of 5 kilohms, an Rp of 500 kilohms, and a 6 of 75. The transistor is made of silicon and has an Ico of 10 na at 25°C, Find the leakage component of collector current at 25 and 75°C. Assume the @ remains fixed with temperature change.13 Frequency Response In earlier chapters we treated all coupling and bypass capacitors as a-c shorts. Obviously, this is incorrect when the frequency of operation is so low that the capacitive reactances become comparable to the resistances in the circuit. In other words, coupling and bypass capacitors impose a lower-frequency limit on any amplifier in which they are used. We have also treated all internal transistor capacitances and stray wiring capacitances as negligibly small. When the frequency is high enough, however, this is no longer valid, because these capacitances pro- vide shunt paths for the a-c currents. In effect, the transistor and stray capacitances impose an upper-frequency limit on the operation of amplifiers. In this chapter we discuss the frequency response of single stages and cascaded stages. We are especially interested in the upper and lower cutoff frequencies of typical RC-coupled amplifiers. 18-1 Response of an RC-coupled Amplifier The RC-coupled amplifier is the most common type of amplifier. It uses resistors to develop the a-c signal and capacitors to couple and bypass 340Frequency Response 341 the a-c signal. A typical response for a fixed input voltage is shown in Fig. 13-1. For very low or very high frequencies, the output voltage drops off. Toward the middle of the frequency range, however, the output volt- age is fixed and equals a constant value K; it is in this middle range of frequencies that most RC amplifiers are normally operated. The cutoff frequencies of an amplifier are those frequencies where the output voltage equals 0.707 of its mid-frequency value. For instance, in Fig. 13-1 the output voltage equals K in the mid-frequency range; if we increase or decrease the frequency of operation, we reach a frequency where the output voltage has dropped to 0.707K. (In terms of decibels, this is equivalent to saying that the output voltage has dropped 3 db.) Yout Fig. 13-1 Typical response of an K RC-coupled amplifier. oro7k A 4 There are two cutoff frequencies f; and fy in Fig. 13-1. The lower cutoff frequency f; is caused by the coupling and bypass capacitors; the upper cutoff frequency is caused by the internal transistor capacitances and the stray wiring capacitances. The passband of an amplifier refers to the range of frequencies between frand fz, Most RC-coupled amplifiers are normally operated at frequencies well in the passband, and up to now we have analyzed the a-c operation of transistor circuits by assuming frequencies of operation well in the passhand. Throughout the remainder of this chaptor, however, we will discuss the cutoff frequencies of various transistor amplifiers. 13-2 Lower Cutoff Frequency of a Typical CB Stage In this section, we find the formulas for the lower-frequency limit of a typical CB stage. The analysis used here will be used again in the dis- cussion of CE stages. In the typical CB circuit of Fig. 13-2a, we know that the approximate voltage gain from the source to the load is Vout A % +r We derived this formula by treating the capacitors as a-c shorts. For normal amplifier operation, this is precisely how the capacitors are sup-342 Transistor Circuit Approximations posed to act; that is, they should be short circuits to a-c-current but open circuits to d-c current. In this way, the biasing of the transistor is not disturbed by the source resistance 7, or the load resistance R, and yet the a-c signal is coupled into and out of the transistor. As the frequency of the input signal becomes lower and lower, the reactances of the capacitors will increase, because 1 Xe = Oni When the reactance becomes comparable in size to the resistances in the circuit, we can expect to drop some of the a-c signal across the capacitors. In other words, the voltage gain will drop off from its normal value of rift + 1). To determine the lower-frequency limit, we draw the a-c equivalent circuit as shown in Fig. 13-2b. The input resistance ri, represents the biasing resistor Re in parallel with the input resistance looking into the emitter of the transistor. As an approximation, we can say that ri, 2 7! Examine the input side of Fig. 13-2b. When the capacitor looks like 5 & hy K om De SA i Yee tle Ag Fi "5 4 i = (a) 7) Fig. 13-2. Deriving the lower cutoff frequencies of a CB stage.Frequency Response 343 an a-c short, the emitter current is simply (13-1) However, as the frequency decreases, the capacitive reactance increases to the point where the emitter current becomes less than the value given by Eq. (13-1). Recall from basic electricity theory that in a series RC circuit, the current is down by a factor of 0.707 (3 db) when the capaci- tive reactance equals the total resistance in the circuit. In other words, in Fig. 13-2b, the emitter current is down 3 db when T+ Tin or (13-2) At this frequency the a-e emitter current is 0.707 of its value at higher frequencies. Since the emitter current is down 3 db at this frequency, the voltage gain from source to load is also down by a factor of 0.707, or 3 db. Equation (13-2) is the formula for the lower cutoff frequency produced by the input coupling capacitor. The output coupling capacitor also limits the lower-frequency response. By applying Thévenin’s theorem, we can redraw the output circuit as shown in Fig. 13-2c. The output circuit is aseries RC circuit; therefore, we again use the argument that the current is down by a factor of 0.707 when the capacitive reactance equals the total series resistance. That is, 1 aye ~ Rut R or 1 I> 350R; + RVC We can use Eqs. (13-2) and (13-3) to find the input and output cutoff frequencies. Usually, these frequencies are different. In this case, the higher of the two frequencies is more critical. For example, suppose that a circuit like that of Fig. 13-2a has an input cutoff frequency of 10 Hz and an output cutoff frequency of 90 Hz. The 90 Hz is the more critical value because the amplifier response has already dropped off at 90 Hz. If the input and output cutoff frequencies happen to be identical, the amplifier response will be down 6 db at this cutoff frequency (3 db for each capacitor). The actual cutoff frequency of the amplifier itself (where its output voltage is down 3 db) will be higher than the frequency found (13-3)344 Transistor Circuit Approximations by Eqs. (13-2) and (13-3). A useful approximation for this overall cutoff frequency is fe lof (13-4) where f is the cutoff frequency of the input and output capacitors and Ja is the cutoff frequency of the amplifier. As an example, suppose we calculate an input cutoff frequency of 50 Hz and an output cutoff fre- quency of 50 Hz. At 50 Hz, the amplifier output voltage is down 6 db. The approximate 3-db cutoff frequency of the amplifier is 1.5(50 Hz), which equals 75 Hz. Exampte 13-1 Find the lower cutoff frequency of the CB amplifier shown in Fig. 13-3. Souution First, we find the cutoff frequency produced by the input coupling capacitor. Using Eq. (13-2), fe 1 1 Qe + Fin) Cs ~ Ie (5OO F 25) (OH ~ 900 He Note that for ris we used the theoretical value of rf because the 20- kilohm biasing resistor is negligible. : Next, we find the cutoff frequency produced by the output coupling capacitor. With Eq. (13-3), we get _ 1 © 2108 104) (10=*) The input capacitor in conjunction with the source resistance and the input resistance of the stage produces a cutoff frequency of 300 Hz. On =8 Hz -20 +20 \at at Fig. 13-3 Example 13-1. %Frequency Response 345 the output side, the cutoff frequency is much lower and occurs at 8 Hz. As far as the overall amplifier response is concerned, the higher fre- quency of 300 Hz is the more critical value, and to a good approxima- tion we can say that the lower cutoff frequency of the entire stage is about 300 Hz. 13-3 Upper Cutoff Frequency of a CB Stage Consider the circuit of Fig. 13-4a. When the frequency is too low, we know that the coupling capacitors produce a lower-frequency cutoff. When we raise the frequency of operation, these coupling capacitors act like a-e shorts, and then the mid-frequency value of voltage gain is approximately Yout | Th wn +e If we continue raising the frequency of operation, we will find an upper- frequency limit where the voltage gain drops off. There are two major causes for this drop-off, and we will discuss both. Recall that the a-c a of a transistor is defined as ie ana In the ideal-transistor approximation, i. = %., so that @ = 1. In a real transistor, the a-c collector current is slightly less than the a-c emitter current. In fact, we have already indicated that most transistors have a-c a’s between 0.95 and 0.999. When the frequency of operation is high enough, however, the @ of Mee Mee Re fi, pe) é cee = g= C so ‘ Os og t G 34 - R ae | co = = Set = (a) (4) Fig. 13-4 Deriving the upper cutoff frequency of a CB stage.346 ‘Transistor Circuit Approximations the transistor begins to decrease. This drop-off in a is related to transit- time effects of the carriers as they move from the emitter to the collector. The a cutoff frequency fx is that frequency where the a has dropped to 0.707 of its low-frequency value. For instance, if a transistor has an a with a low-frequency value of 0.98 and an f. = 1 MHz, then at 1 MHz the a will equal a = 0.707(0.98) = 0.693 This means that at 1 MHz the collector current is only 0.693 times the input emitter current. Thus, the f. of a transistor represents one of the limitations on the upper-frequency response of a CR circuit. When possible, we select a transistor whose f, is much higher than the highest frequency at which we want to operate. The second major limitation on the high-frequency response of the CB circuit of Fig. 13-4a is the capacitance from the collector to ground. Even though no capacitance is shown, there is always some stray wiring capaci- tance that appears from collector to ground. In addition, the back-biased collector diode has capacitance, as discussed in Sec. 4-5. Both of these capacitive effects are represented by Cz in the a-c equivalent circuit of Fig. 13-4b. The coupling capacitors are shown as short circuits because we are now analyzing higher-frequency operation. When we increase the frequency sufficiently, the reactance of Cy eventually becomes low enough to shunt some of the a-c collector current. This means that there is less current in rz, so that the output voltage is less. The cutoff frequency occurs when the reactance of Cz equals the value of rz. That is, eee DiC, "* Solving for f gives us the cutoff frequency. 1 f= 2arpCy (13-5) There are other high-frequency effects, but the two we have discussed are the major limitations on the high-frequency response of a CB circuit like that of Fig. 13-4a. Normally, to maintain a flat frequency response up to the cutoff point, the source resistance r, is deliberately made much larger than the ri of the transistor. In other words, the emitter diode is swamped; this avoids certain inductive effects that occur in a CB circuit. The f. and the frequency given by Eq. (13-5) are usually different. In this case, the lower of the two frequencies is more critical. For instance, suppose that a CB circuit like that of Fig, 13-4a has an f, of 10 MHz and that the cutoff frequency given by Eq. (13-5) is 2 MHz. The 2 MHz isFrequeney Response 347 the more critical value because the response has already dropped off at 2 MHz. The cutoff frequency of the overall amplifier (the 3-db frequency) is approximately equal to 2 MHz. If the two frequencies happen to be exactly equal, the amplifier re- sponse will be down 6 db at this cutoff frequency. The actual cutoff fre- quency of the amplifier is lower than this cutoff frequency and is approxi- mately equal to fo = 0.65f0 As an example, suppose that the j, is 10 MHz and that the cutoff fre- quency caused by C, is also 10 MHz. The output of the amplifier will be down 6 db at 10 MHz. The approximate 3-db cutoff frequency of the amplifier will be at 0.65(10 MHz), which equals 6.5 MHz. EXaMPLe 13-2 The transistor in Fig. 13-5 has a collector capacitance C of 30 pf and anf. of 5 MHz. The stray wiring capacitance from collector to ground is 20 pf. Find the approximate value of the upper cutoff frequency. SonuTion The fz represents one of the upper-frequency limits. Since it equals 5 MHz, we know that the a-c emitter current is down 3 db at 5 MHz. Next, we determine the cutoff frequency produced by the capacitance from collector to ground. This capacitance C, is the sum of the collector capacitance and the stray wiring capacitance. Cr = Ceo + Cotray = (30 + 20)(10-") = 50 pf Using Eq. (13-5), we get ie 1 pat J = 3, (5000)(50) G0) ~ 035 KHz -20 +20 20K Fig. 13-5 Examples 13-2 and 13-3. uf 500348 ‘Transistor Cireuit Approximations The frequency limitation of 635 kHz is the more critical value. At this frequency, the output voltage is down 3 db from the mid-frequency value. At this frequency, there is no problem at all with the « cutoff of the transistor. Exampie 13-3 The transistor of Fig. 13-5 has an f, of 300 kHz and a collector capacitance C. of 2 pf. The stray capacitance from collector to ground is 3 pf. Find the approximate lower and upper cutoff frequencies of the amplifier. Sonurion In Example 13-1, we analyzed the circuit to determine the lower cutoff frequency. We found that the input coupling capacitor produces a cutoff frequeney of 300 Hz, while the output coupling capacitor caused a cutoff of 8 Hz. Thus, the lower cutoff frequency of the amplifier is approximately 300 Hz. ‘The upper cutoff frequency of the CB circuit is determined by the cutoff frequency produced by C; or by the f, of the transistor, which- ever is lower. Using Eq. (13-5), we get 1 1 ™ pe(o0o)GyCIO-R) ~ 655 MHs Since we are given that f. = 300 kHy, it is clear that in this case the fa is lower than the cutoff frequency produced by C,. Thus, the upper cutoff frequency is 300 kHz. 13-4 Lower Cutoff Frequency of a CE Amplifier In a CE amplifier like that of Fig. 13-6a, the coupling capacitors again produce a lower cutoff frequency. For the moment, let us assume that the bypass capacitor Cx is extremely large so that the emitter is held at a-c ground. In this case, we can draw the a-c equivalent circuit as in Fig. 13-6. On the input side, it is clear that the series RC circuit is similar to that analyzed for the CB amplifier. Again, we note that when the reactance of the input capacitor equals the total series resistance, the input current will be down 3 db. We can find the cutoff frequency pro- duced by the input capacitor as follows. 1 ajo. =" + tin or 1 I hip rG eeFrequency Response 349 Equation (13-6) tells us how to find the cutoff frequency produced by the input coupling capacitor. To find the cutoff frequency produced by the output coupling capacitor, we can use Thévenin’s theorem to redraw the output circuit, as shown in Fig. 13-6c. In the output cireuit, we again have a series RC circuit. This means that cutoff occurs when ee aye — Ret R 1 3a FB) By using Eqs. (13-6) and (13-7), we can find the cutoff frequencies produced by the coupling capacitors. Again, note that the higher of these (13-7) aa eS H = (a) (a) 7 BGR Vout {e) (a) Fig. 13-6 Deriving the lower cutoff frequencies of a CE stage.350 Transistor Circuit Approximations two frequencies is the more important as far as the overall low-frequency response is concerned. The bypass capacitor Cz also causes a lower-frequency cutoff to occur. The reason for this is as follows. The bypass capacitor is supposed to keep the emitter at a-c ground. At low enough frequencies, however, this capacitor no longer looks like an a-c short to the signal. Because of this, degeneration takes place at the emitter, with a resulting loss in voltage gain. To find the cutoff frequency caused by the bypass capacitor, consider Fig. 13-6d, where we have shown the essential parts of the a-c equivalent, cireuit. Looking back into the emitter of the transistor, there is a resist ance labeled roy. As far as the bypass capacitor is concerned, it sees rout in parallel with Re. To hold the emitter at a-c ground, the reactance of Cs must be much smaller than rou in parallel with Rg. But when the frequency is so low that the reactance of C'z equals rayi||Rz, the voltage gain will be down by 3 db. In other words, cutoff occurs when 1 aCe = rout|| Rx or eee (13-8) 2n(rout||Re) Ce It can be shown that the resistance rout is Pout 2 re + ol (13-9) when the emiller is bypassed to ground. When some of the emitter resistance is left unbypassed, rox: becomes roos = ta + 14 + TalRall Re B where rg is the unbypassed part of the emitter resistance. Also,when two- supply emitter bias is used, Rp takes the place of || in the last two equations. In analyzing a CE cireuit to find its lower cutoff frequency, we use Eqs. (13-6) to (13-8) to find the individual cutoff frequencies produced by the different capacitors, As usual, the largest of these frequencies is the most important because the amplifier voltage gain first begins drop- ping at this frequency. Since these cutoff frequencies are usually different, we can approximate the cutoff frequency of the overall stage by using the largest of these frequencies.Frequency Response 351 Examp.e 13-4 Find the approximate value of the lower cutoff frequency in Fig. 13-7. SoLutTion First, we find the cutoff frequency produced by the input coupling capacitor. This cutoff frequency is given by Eq. (13-6). i I" FIC The value of r, and C, are given, but we need to estimate the value of rin. Recall that for a CE stage of this type, Tin & Rall6r’, There is about 1 ma of d-c emitter current, so that ri is between 25 and 50 ohms. Arbitrarily, let us use the upper limit of 50 ohms, With a 8 of 100, ri = 5000 ohms. Thus, the input resistance of the stage is Tin & Rel|Br, = 30,000||5000 ~ 4300 Fig. 13-7 Example 13-4, OK we Slt Now, we can find the cutoff frequency produced by the input coupling capacitor. Hee 1 feet J = 3.1000 + 4300) (10-5 ~ 30 He By using Eq. (13-7), we can find the cutoff frequency produced by the output coupling capacitor. Hae 1 Haat 1 ~ J = a5; + RVC ~ B(10,000 + 10,000) 0-5 = 8 H*352 Transistor Circuit Approximations Also, we can find the cutoff frequency produced by the bypass capaci- tor. To do this, we need the value of rou. With Eq. (13-9), we calculate r4|| Bile 1000]|30,000 _ B 100 a Tou = + = 50+ 60 Now, we use Eq. (13-8) to find the cutoff frequency. oe 1 ate ie I= 2eGroul|a\Cx ~ 2¢(60]20,000) 00-5 Of the three frequencies, 30, 8, and 2650 Hz, the most important is 2650 Hz, because the amplifier response first falls off at this frequency. Therefore, the approximate lower cutoff frequency of the amplifier in Fig. 13-7 is 2650 Hz. = 2650 Hz The low-frequency response of the amplifier in Fig. 13-7 is quite poor. One way to lower this cutoff frequency is to use a larger bypass capacitor. For example, if we increase Cs from 1 to 10 uf, the cutoff frequency will change from 2650 to 265 Hz. Another approach to reducing the cutoff frequency is to leave some of the emitter resistance unbypassed. This is equivalent to increasing the size of rou, which in turn reduces the cutoff frequency. 13-5 Transistor Cutoff Frequencies In the preceding section, we discussed the a cutoff frequency of a transis- tor, which is the frequency where the a drops to 0.707 of its low-frequency value. The cutoff frequency, symbolized by fe, is another important: tran- sistor cutoff frequency. The fy is the frequency where the of the transis- tor drops to 0,707 of its low-frequency value. To bring the idea out more clearly, consider the CE connection of Fig. 13-8. A voltage source with a large source resistance drives the transistor so that the input a-c base current remains fixed at all frequencies. Here is what we find when we increase the frequency. As we approach higher frequencies, the a-c col- lector current begins to decrease. Since the a-c base current is fixed, this is equivalent to saying that the @ is decreasing. Eventually, we find a frequency at which the f has decreased to 0.707 of its low-frequency value; this frequency is called the 8 cutoff frequency fs. ‘The fr of a transistor is still another important high-frequency charac- teristic of a transistor. In Fig. 13-8, if the frequency of operation is increased above the fy of the transistor, the 8 will continue to decrease. Eventually, we find a frequency where the 6 = 1; this frequency is called the fr of the transistor.Frequency Response 353 +20 20k % Fig. 13-8 The fy and fr of a tran- i ae Ve , i pn a in sistor. ee & ‘The fr of a transistor is much higher than the fy. The relation between these two frequencies is ts -# (13-10) where 8 is the low-frequency value of 8. As an example, suppose that a data sheet gives an fr of 100 MHz and a low-frequency 6 of 50. The 8 cutoff frequency would then be _ 100 MHz _ fo= ay = 2 Ma The f, and the fr of a transistor are also related. As a rough approxi- mation, they are often treated as equal, that is, Sr =fa (13-11) Actually, the fr is less than the f,. For simple junction transistors, a better approximation is tres (13-12) The various cutoff frequencies are important in the high-frequency analysis of transistor circuits. The fs is one of the limitations of the CB amplifier. The fg and the fr are important in the analysis of a CE amplifier, which will be discussed in a later section. Example 13-5 We have two transistors. The first has an fr of 200 MHz and a low- frequency @ of 150. The second has an fr of 100 MHz and a low-fre- quency 8 of 20. Calculate the fy for each transistor. Which transistor has the higher f5?354 Transistor Circuit Approximations SoLution The first transistor has a 6 cutoff frequency of fa = -& 7 200008) = 1.33 MHz ‘The second has an “ of fo= 100009) = 5 MHz Clearly, the second transistor has the higher 8 cutoff frequency. EXAMPLE 13-6 A simple junction transistor has an f, of 50 MHz and a low-frequency 8 of 75. Estimate the fy of the transistor. SoLvtion As a rough approximation, we can treat f, and fr as equal, so that fr. fa _ 50 MHz _ a Geeg = a5 = 667 kHz We can improve this estimate by using Eq. (13-12) to get a more accurate value of fr. = Jn _ 50 MHz ir= ja = 41.6 MHz And now we can find fp. _ 41.6 MHz _ Sa= G5 = 555 ke 13-6 Base Spreading Resistance Another quantity that is important in the high-frequency analysis of a CE stage is the base spreading resistance 7. We discussed this in Chap. 5, Recall that 7, is the bulk, or ohmic, resistance of that part of the base in which base current flows. The 7 of a transistor depends upon the dimen- sions of the base as well as upon the d-c operating point of the transistor. There is no simple rule of thumb for the value of rj. Usually, the value of rf is in the range of 5 to 500 ohms, For many transistors, rf is between 50 and 150 ohms. We can improve the ideal-transistor approximation by taking rf into account, This resistance is in series with the base lead, so that we canFrequency Response 355 add it to base, as shown in the a-c equivalent circuit of Fig. 13-9. A deriva- tion similar to that given in Sec. 6-7 shows that the voltage gain is Th a eerie 13-1 m+ 7B ete Also, the input resistance is tin = 1h + Brt (13-14) In the ideal-transistor approximation, we have neglected the effect. of ron the input resistance and the voltage gain. This is usually a reasonable approximation. For instance, suppose a transistor has a A of 100. For a d-c emitter current of 1 ma, r/ is about 25 to 50 ohms. The base spreading resistance is typically 50 to 150 ohms. If we use an rf of 25 and an rj of 150, we would calculate an input resistance of Tin = 15 + Br = 150 + 100(25) = 2650 ohms By neglecting 7, as we do in the ideal-transistor approximation, we get Tm & Br, = 100(25) = 2500 Similarly, in the voltage-gain formula, Eq. (13-13), we usually find that ris much greater than rj/8, so that the effects of 7} are negligible. & Fig. 13-9 Base spreading resistance. | be 9 Occasionally, we cannot neglect rj. This is especially true at very high d-c emitter currents, where the value of r? becomes quite small. In this case, Eqs. (13-13) and (13-14) are useful in getting a more accurate pre- diction of voltage gain and input resistance. (Also, h parameters, which are discussed in the next chapter, can be used.) The base spreading resistance 7 is not too important at lower fre- quencies; however, when we increase the frequency of operation, we find that the internal transistor capacitances begin to shunt the 67! of Fig. 13-9. When this happens, the 7, does become quite important in deter- mining the upper-frequency limitations of a transistor. This is discussed in the next section.356 ‘Transistor Circuit Approximations EXAMPLE 13-7 The transistor of Fig. 13-10 has an rj of 100 ohms, a 8 of 50, and an r, of 25 mv/Tg. Find the approximate voltage gain and input resistance (including the bias resistances) in the passband of the amplifier. Soxurion In the passband, or mid-frequency range of the amplifier, all capaci- tive effects are negligible. Hence, we have straightforward calculations involving Eqs. (13-13) and (13-14). We first need the value of r/. _ 25 m Fig. 13-10 Example 13-7. 6KS “out Now, we can find the voltage gain. Vout a, Te _ 3000//6000 _ 2000 _ 5, Yin +B BF 100%) 7 The input resistance looking into the base without accounting for the biasing resistors is Tin = 7% + Brl = 100 + 50(5) = 350 The biasing resistors, 20 and 10 kilohms, appear in parallel with 350 ohms. Since the biasing resistors are large enough to neglect, the input resistance of the stage is approximately 350 ohms. 13-7 Upper-frequency Limit of a CE Stage In See. 13-3 we discussed the upper cutoff frequency of a OB stage. We saw that the high-frequency response of such a circuit is limited by the Ja or by the cutoff produced by the output capacitance.Frequency Response 387 The CE circuit also has high-frequency limitations, one of which is the cutoff produced by the capacitance from collector to ground. In Fig. 13-11 we have shown the a-c equivalent circuit of a CE stage. The capaci- tance C'; includes the stray-wiring capacitance and the collector-diode capacitance C.. That is, Cr = Cotray + Ce Fig. 13-11 Deriving the output cut- off frequency. A cutoff frequency occurs when the reactance of C;, equals the value of rz, because under this condition the current in rz is down 3 db. We can write ae Def, ™ __1 ~ Qer Cr or f (13-15) ‘The frequency given by this equation is one of the major limitations on a CE amplifier. We will call this frequency the output cutoff frequency. The decrease in 6 mentioned in a previous section is another major limitation on the high-frequency response of a CE circuit. An a-c equiva- lent circuit often used to account for this decrease in 8 is the circuit shown in Fig. 13-12a. There is an emitter capacitance C, across the emitter-base junction and a collector capacitance C, across the collector-base junction. Bo is the low-frequency value of g. Note that the current through the Bor, resistor is labeled 7. At lower frequencies, all the input base current is passes through rj and through Bur. In effect, if = a at lower frequencies. However, as the frequency increases, the reactances decrease until some of the input base current i, is diverted from the Gor, resistor. In other words, when the frequency is increased enough, the value of # begins to drop off; therefore, the Soi current source begins to decrease in value, resulting in less collector current. When a high-impedance source drives the transistor, the value of is remains fixed as the frequency increases. If we place an a-c short across the output terminals, as in Fig. 13-125, and measure i., we find that i. begins to drop off as we approach higher frequencies. The reason for this is simply that the internal capacitances shunt input base current358 Transistor Circuit Approximations Fig. 13-12 High-frequency equivalent circuits. away from the Gor, resistor. The actual current through for! then drops off, and this causes the output current oi! to drop off. The B cutoff fre- quency discussed earlier is the frequency where the collector current in Fig. 13-12b drops to 0.707 of its low-frequency value. If we connect an a-c load rz across the output terminals (Fig. 13-12c) instead of a short, we will get an output voltage. As a rough approxima- tion, we can say that the output voltage will be down 3 db from its low- frequency value when the frequency equals the fg of the transistor. The use of fy as the upper cutoff frequency for the output voltage of a CE stage is only a rough approximation, because the presence of a load resistor rz instead of short introduces a complicating effect known as the Miller capacitance. When a capacitor like C. spans from the output back to the input, as in Fig. 13-12¢, the effective capacitance seen by the source is C.(1 + A), where A equals rz/r,. This Miller capacitance is normally shown by redrawing the circuit as in Fig. 13-12d. Note that as far as the input base current is concerned, C. appears larger by a factor of | + r1/r'. This means that the shunting effect of this capacitor can be significant at frequencies that are lower than fy. To summarize our results up to this point, we can say that when a CE stage is driven by a large source impedance, the voltage gain will drop off at a frequency that is less than the fy of the transistor. When the source impedance is not large enough to hold #y fixed at all frequencies, the analysis of 2 CE stage becomes quite complicated. We will merely indicate that the following approximation can be usedFrequency Response 359 to estimate the upper cutoff frequency caused by internal transistor capacitances. 1 fe Gore fe =f (1 +o. =) Time (13-16) For convenience, we will call this the internal cutoff frequency. In estimating the cutoff frequency of a CE stage we must calculate the output cutoff frequency by using Eq. (13-15) and the internal cutoff frequency by using Eq. (13-16). The lower of these two frequencies is the more critical value; it is the frequency where the output voltage of the CE amplifier is down about 3 db from its low-frequency value. When troubleshooting or making an initial analysis of a CE stage, we often do not have the values of rj, 80, fr, etc. In this case, we can crudely estimate the internal cutoff frequency of Eq. (13-16) by the fs of the transistor. Exampte 13-8 The transistor of Fig. 13-13 has an fr of 100 MHz and a low-fre- quency 6 of 100. The stray-wiring capacitance plus the collector eapaci- tance equals 20 pf. Estimate the upper cutoff frequency of the amplifier by using the f5 of the transistor or the output cutoff frequency (Eq. 13-15), whichever is smaller. SoLution The 6 cutoff frequency is Using Eq. (13-15), we calculate an output cutoff frequency of 1 * DareOn MHz 1 fe f 2y(2000)(20)(10-*) ~ +20 Fig. 13-13 Examples 13-8 and 13-9.360 Transistor Circuit Approximations We therefore can estimate the upper cutoff frequency of the amplifier by using the lower value, which is 1 MHz. EXxaMpie 13-9 In addition to the values given in the preceding example, the tran- sistor has an 7; of 100 ohms, an r/ of 25 ohms, and a C, of 5 pf. Find the cutoff frequency of the amplifier by using the internal cutoff fre- quency or the output cutoff frequency, whichever is smaller. Sovurion We find the internal cutoff frequency by using Bq. (13-16). = 108 eS LOO(25) Fee eae Jegine (1 + 50 + 100) TF 2100) MO-*) = 2.43 MHz The stray and collector capacitances produce an output cutoff fre- quency of 4 MHz, as shown in the preceding example; therefore, as an estimate, we can say that the cutoff frequency of the amplifier is the lower value, 2.43 MHz. 13-8 Response of Cascaded States When several amplifier stages are cascaded together, the overall voltage gain is the product of the individual stage gains. In addition, the cutoff frequency of the cascade must be less than the cutoff frequency of the individual stages. For instance, if we cascade three identical stages to- gether, each of which has an upper cutoff frequency of 1 MHz, the overall amplifier has a cutoff frequency much lower than 1 MHz. There are some useful approximations for estimating the overall cutoff frequency of several identical stages in cascade. The lower cutoff fre- queney f; can be found by using free Vn (13-17) where J; is the lower cutoff frequency of the cascade f. is the lower cutoff frequency of one stage nis the number of stages For nonidentical stages, the exact formula for the cutoff frequency is complicated. We will only observe that if the cutoff frequencies of the individual stages are reasonably close together, Eq. (13-17) still applies as an approximation if we use the average of the individual cutoff fre- quencies. On the other hand, if one of the stages has a much higher cutoffFrequeney Response 361 frequency than the others, this stage predominates, and the overall cas- cade has a lower cutoff frequency approximately equal to the cutoff frequency of the dominant stage. Similarly, there is a useful approximation for the upper cutoff frequeney of ae scade of identical stages. It is fz (13-18) Livan the upper cutoff frequency of the cascade fc is the upper cutoff frequency of one stage nis the number of stages Again note that we can use this formula even when the stages are not identical. As long as the individual stages have cutoff frequencies that are close to each other, we can use the average value for f. in Eq. (13-18). Tf one of the stages has a much lower cutoff frequency than the others, it predominates in the cascade, and the overall amplifier has an upper cutoff frequency approximately equal to that of the critical stage. ExampLe 13-10 Suppose that five identical stages are cascaded. Each has a lower cutoff frequency of 100 Hz and an upper cutoff frequency of 10 MHz. Find the lower and upper cutoff frequencies of the overall amplifier. SoLution ‘The lower cutoff frequency is f= llf. Vn = 1.1(100) V5 = 250 Ha and the upper cutoff frequency is fe 10(10°) = =4 ME Livan 1175 i Thus, we sce that the passband of a single stage extends from 100 Hz to 10 MHz, whereas a cascade of five such stages has a passband of 250 Hz to 4 MHz. fie SUMMARY The cutoff frequencies of an amplifier are those frequencies where the out- put voltage drops to 0.707 of its mid-frequency value. This is equivalent to saying that the output voltage is down 3 db at the cutoff frequencies. For a simple CB amplifier, the input and output coupling capacitors362 Transistor Circuit Approximations determine the lower-frequency limit of the amplifier. Each of these capaci- tors produces a cutoff frequency, and the larger of these two frequencies is the more critical value. The f, and the output cutoff frequency are the two major limitations on the high-frequency response of a CB amplifier. The fz is the frequency where the a has dropped to 0.707 of its low-frequency value. The output cutoff frequency is the frequency where the reactance of the collector ground capacitance equals the a-c load resistance seen by the collector. The approximate cutoff frequency of a CB amplifier is the f. or the output cutoff frequency, whichever is smaller. The coupling capacitors in a CE amplifier produce lower cutoff fre- quencies. In addition, the emitter bypass capacitor (when used) produces a lower cutoff frequency. Of the three cutoff frequencies (input capacitor, output capacitor, and bypass capacitor), the largest is the most critical and can be used as an estimate of the lower cutoff frequency of a CE amplifier. The 8 cutoff frequency fy is the frequency where the @ of a transistor drops to 0.707 of its low-frequency value. The fr is the frequency where the 8 has dropped to unity. The internal cutoff frequency and the output cutoff frequency are the two major limitations on the high-frequency voltage gain of a CE ampli- fier. The internal cutoff frequency can be crudely estimated by using the fs. We can get a more accurate value for this internal cutoff frequency by using Eq. (13-16). The output cutoff frequency is that frequency where the reactance of the collector-ground capacitance equals the a-c load resistance seen by the collector. The approximate upper cutoff frequency of a CE amplifier is the internal cutoff frequency or the output cutoff frequency, whichever is smaller. GLOSSARY a cutoff frequency (fa) The frequency at which the a of a transistor equals 0.707 of its low-frequency value. base spreading resistance (rf) The resistance of that part of the base region through which base current flows. B cutoff frequency (fs) ‘The frequency at which the 6 of a transistor equals 0.707 of its low-frequency value. fr The frequency where the 8 of the transistor equals unity. The fr is sometimes called the gain-bandwidth product of the transistor. internal cutoff frequency A term used in this chapter to describe the cutoff frequency produced by the internal transistor capacitances, including the Miller effect.Frequency Response 363 output cutoff frequency The cutoff produced by the collector-ground capacitance in conjunction with the a-c load resistance rz seen by the collector. passband In an amplifier, this is the range of frequencies between the lower and upper cutoff frequencies. REVIEW QUESTIONS 1. How are the cutoff frequencies of an amplifier defined? . What is the passband of an amplifier? 3. In a CB amplificr, the input and output coupling capacitors each produce a cutoff frequency. Which of these frequencies is the more critical as far as the overall amplifier response is concerned? 4. What are the two major limitations on the high-frequency response of a CB amplifier? 5. Define the a cutoff frequency f, of a transistor. 6. Does the emitter bypass capacitor in a CE circuit produce a low or a high cutoff frequency? 7. Define the fy and the fr of a transistor. 8. Is the fr of a junction transistor much greater than, much less than, or approximately equal to the fa? 9. Define the base spreading resistance rj. What is a typical range in value for 742 10. What are the two major limitations on the high-frequency response of a CE cireuit? Which of these is the more critical cutoff frequency? 11. If four identical stages are cascaded, is the lower cutoff frequency of the cascade higher or lower than the lower cutoff frequency of one stage? What about the upper cutoff frequency of the overall amplifier? i PROBLEMS 13-1 Find the cutoff frequencies produced by the input and output coupling capacitors in Fig. 13-14a. Which of these sets the lower frequency limit on the CB amplifier? 13-2 What are the cutoff frequencies produced by the coupling capaci- tors in Fig. 13-146? To make these frequencies exactly equal, what size should the output coupling capacitor be changed to? 13-3 If the 50-ohm source resistance in Fig. 13-14 is changed to a 500- ohm resistor, what new value can be used for the input coupling capacitor to maintain the same cutoff frequency (approximately) ? 13-4 The transistor in Fig. 13-14a has an f, of 10 MHz. The collector diode capacitance is 20 pf, and the stray wiring capacitance from collector364 Transistor Circuit Approximations “10-120 12 +20 WK 25K SK 1K 10 pt Olt 50u! fe 600 15K 50 100% %s iG (0) (a) Fig. 13-14 to ground is 45 pf. Find the approximate value of the upper cutoff fre- quency of the amplifier. 13-5 The transistor in Fig. 13-14a has an f, of 300 MHz. The collector capacitance and the stray wiring capacitance total 10 pf. What is the approximate upper cutoff frequency of the amplifier? 13-6 The transistor of Fig. 13-15a has a 8 of 75 and an r} of 25 mv/Iz. Find the cutoff frequency produced by the emitter bypass capacitor. 13-7 The transistor of Fig. 13-15b has a 6 of 50 and an r! of 35 mv/Tz. What is the cutoff frequency produced by the emitter bypass capacitor? 13-8 In Fig. 13-15a, the transistor has an rf of 100 ohms and an r!, of 25 mv/TIg. The B is 100. Find the voltage gain from base to collector and the voltage gain from the source to the collector. Use Eq. (13-13).Frequency Response 365 13-9 The transistor of Fig. 13-15b has an 1 of 250 ohms and an rf of 25 mv/Ip. If the 6 is 50, what is the voltage gain from the source to the output? Use Eq. (13-13). 13-10 The fy of a transistor is 3 MHz and the @ is 75. What is the fr of the transistor? 13-11 A junction transistor has an j, of 500 MHz. What is the approxi mate value of fr? If the 8 spread is from 50 to 150, what is the spread in the fp? 13-12 The transistor in Fig. 13-15a has an fr of 250 MHz and a low- frequency @ of 75. The stray wiring capacitance plus the collector diode capacitance is 15 pf. Estimate the upper cutoff frequency by using the fo or the output cutoff frequency, whichever is lower. 13-13 In Fig. 13-16a, the transistor has the following characteristics: C. = 8 pf, fr = 300 MHz, Bo = 75, rf = 25, and rf = 50, The stray wiring capacitance from collector to ground is 7 pf. Calculate the output cutoff frequency and the internal cutoff frequency. +20 600) 50. Fig. 13-16 13-14. The transistor in Fig. 13-16b has the following parameters: 1, = 75 ohms, C. = 5 pf, 8» = 100, r} = 50 ohms, and fr = 200 MHz. ‘The stray wiring capacitance from collector to ground is 10 pf. Calcu- late the internal and output cutoff frequencies of the amplifier. 13-15 Using the data given in Prob. 13-14, caleulate the internal cutoff frequency if the source resistance is changed from 50 ohms to 5 kilohms.14 h Parameters The ideal-transistor approximation is adequate for most preliminary analysis and design, When more accurate predictions of transistor be- havior are needed, we must take into account the second-order effects that are neglected in the idcal-transistor approximation. One of the methods that takes all transistor characteristies into account is the h-parameter approach. This method of analysis is complicated and somewhat impractical; nevertheless, we must have a basic knowledge of h parameters because many transistor data sheets describe the transistor in terms of its h parameters. Further, there are times when we need as accurate a prediction as possible for the behavior of a transistor circuit. ‘The h-parameter approach, in theory at least, can give us exact answers. 14-1 Concept of the h Parameters Suppose we have a linear circuit, as shown in Fig. 14-1. (A linear circuit is one in which the resistances, inductances, and capacitances remain fixed when the voltage across them changes.) This circuit has an input voltage and current labeled v; and i}. The circuit also has an output 366h Parameters 367 voltage and current labeled v2 and iz. The upper terminals are arbitrarily shown as positive with respect to the lower terminals. Also, we have shown the currents flowing into the box. The actual voltages and currents can have different polarities and direc- tions; however, we need to agree on a convention at the outset of our dis- cussion. Therefore, we adopt the convention shown in Fig. 14-1. We assume that currents flow into the box and that voltages are positive from the upper to the lower terminals. When we analyze circuits in which the voltages are of opposite polarity or where the currents flow out of the box, we simply treat these voltages and currents as negative quantities. In Fig. 14-1, we certainly expect the voltages and currents to be related to each other. For instance, if we were to change the value of vs, we would not be surprised if ¢;, é2, and vz were to change. In other words, we expect the two currents and two voltages of Fig. 14-1 to be mathematically related. Fig. 4-1 Developing the concept of + A Linear % A parameters. . 2 circuit e One of the theorems proved in advanced circuit theory is that the voltages and currents of Fig. 14-1 can be related by the following set of equations. 1 = hurt + hasde (14-1) ig = hearts + heads (14-2) In these equations, the h’s are fixed coefficients, or constants, for any given circuit. For instance, we might have a particular circuit whose voltages and currents are related by 10% + 5v2 2% + Buz In this case, we would say that the circuit has h parameters given by Ii = 10, Ais = 5, hor = 2, and hae = 3. If we change the circuit, the set of h parameters would change. In other words, for each distinct circuit there is a set of k parameters asso- ciated with that circuit. Once these h parameters are known, we can use Eqs. (14-1) and (14-2) to find the voltages and currents in the circuit. How do we find the h parameters of a given circuit? To answer this,368 Transistor Circuit Approximations consider Fig. 14-2. A linear circuit is driven by an input voltage of v1. This input voltage sets up an input current of #1, whose size depends upon what is inside the box. Notice that we are using a zero-resistance load on the output side, that is, a short circuit. With a short on the output A Eee Fig. 14-2 Finding the forward pa- ‘i Lineor Short | Tameters, hy and ha. circuit : terminals, we can definitely say that the output voltage vp must be zero. Since v: = 0 when the output is shorted, Eqs. (14-1) and (14-2) simplify to = hats + hix(0) Aarti + he2(0) v1 dy or fair how Now we can solve for each h value to get 1 hu == forve=0 — output shorted (14-3) 7 and hy = 2 for v:=0 output shorted (14-4) 7 It is important to realize the physical meaning of these equations. Eq. (14-3) tells us that: /in: is the ratio of »; to ¢, with the output terminals shorted. The ratio of a voltage to a current is an impedance. Because of this, we interpret h1 as the input impedance with the output shorted. Equation (14-4) tells us that hs: is the ratio of iz to i; with the output terminals shorted. Since the ratio of the output current to the input cur- rent is involved, we can interpret he: as the current gain of the circuit with the output shorted. In general, if we are given the schematic of a circuit, we can find hu and hs: by calculating the input impedance and the current gain of the circuit under the condition that the output terminals are shorted together. A method for finding 1: and hs» is the following. Consider Fig. 14-3. Note that we are now driving the output terminals with a voltage v2. This sets up a current of iz. Especially note that the input terminals are open. We have deliberately left the input terminals open so that we canh Parameters 369 unequivocally state that 2: = 0. In other words, with the input terminals open, there can be no current on the input side. With i; = 0, the general h-parameter equations reduce to v1 = hy(0) + hisve tg = hey(O) + havve or V1 = hive tg = heave A Fig. 14-3 Finding the reverse pa- , rameters, hy: and hap. i Linear 4 " " a! circu 2% We can solve these equations for each h value to get lug =" fori, = 0 input open (14-5) 2 and ug = 2 for ix = 0 input open (14-6) 2 Equation (14-5) tells us that hi: is the ratio of v; to v2 with the input terminals open. We already know that the ratio of two voltages is called a voltage gain. Since the generator of Fig. 14-3 is driving the output terminals, we can say that A: is the reverse voltage gain with the input open. Equation (14-6) says that hee is the ratio of i to ve with the input ter- minals open. Current-voltage ratios represent admittances, that is, the reciprocal of impedances. Thus, hz: is the admittance looking into the output terminals when the input terminals are open. Let us summarize the physical meaning of the h parameters: hu is the input impedance with the output shorted ha is the forward current gain with the output shorted hus is the reverse voltage gain with the input open hay is the output admittance with the input open One more item. Note that hy has the dimensions of ohms because it is an impedance and ho: has the dimensions of mhos because it is an admittance. The two remaining parameters, fiz and hz, do not have dimensions; they are pure numbers.370 Transistor Circuit Approximations Exampie 14-1 Find the h parameters of the circuit in Fig. 14-4a. Sonution The circuit inside the box is a simple voltage divider consisting of a 20-ohm resistor in series with a 10-ohm resistor. To find the h param- eters, we proceed as follows. First, visualize that the output terminals are shorted, as shown in Fig, 14-40. Obviously, the input impedance of this circuit is simply 20 ohms because the 10-ohm resistor is shorted out. Thus, we have found that hi: = 20 ohms. Next, we find the forward current gain ha. With a short across the output, as in Fig. 14-46, it should be clear that any current i; flowing into the box will flow through the 20-ohm resistor and then through the shorted output as shown. (Obviously, there is no current in the 10-ohm resistor because there is zero voltage across it.) Remember that é, is the output current flowing into the box. Since the current in Fig. 14-4b is actually flowing out of the box, é2 is negative. In other words, beni As a result, iy = P= 54= -1 4 : 20 20 10° 10° (a) (0) oe b Oy pen i i, te) Fig. 14-4 Example 14-1. To find the reverse parameters, we proceed as follows. First, we visualize the circuit as in Fig. 14-4c. Any voltage v, applied to the output terminals will appear across the 10-ohm resistor. With the input terminals open, there can be no current through the 20-ohm resistor, and therefore there can be no voltage drop across this resistor. As ah Parameters 371 result, all the voltage across the 10-ohm resistor appears across the input terminals. In other words, v1 = U2 Because of this, Dy _ Va hy=2= "21 ta aaa a Finally, notice in Fig. 14-4c that the impedance looking into the output terminals with the input open is simply 10 ohms. Therefore, to find hz: we take the reciprocal to get hex = 149 = 0.1 mhos To summarize, we say that the circuit of Fig. 14-4a has the following h parameters: hi: = 20 ohms, ha: = —1, hiz = 1, and haz = 0.1 mhos. Very often, a set of h parameters is displayed in matrix form. This is nothing more than showing the h parameters in a 2 by 2 display in the same relative positions that they have in Eqs. (14-1) and (14-2). In other words, for convenience we can show a set of h parameters as follows: hu hu) __[ 20 1 hea he} [1 0.1 The position of the h’s should be memorized. In the first row, we have hy and fy». In the second row, there is ho: and he. The dimensions are usually not written, since it is understood that hi: is always in ohms, Ay2 and he; are dimensionless, and ho: is in mhos. Exampe 14-2 Find the h parameters of the cireuit in Hig. 14-5a. Sonution First, we visualize a short across the output terminals as depicted in Fig. 14-5. Under this condition, the input impedance is Iu = 2 + 2[|2 = 3 ohms Next, realize that the input current i; will divide equally at the june- tion of the 2-ohm resistors, so that the output current, will be 11/2, as shown in Fig. 14-5b. Therefore, and372 Transistor Circuit Approximations To find the reverse parameters, we visualize the circuit as in Fig. 14-5¢. Note that with the input terminals open, any voltage v2 applied to the output will be divided by a factor of 2, so that and {a} (o) 4 Open 12 tee Fig. 14-5 Example 14-2. Also, in Fig. 14-5c the impedance looking into the output terminals is 4 ohms, Therefore, hee = 14 = 0.25 mho We can summarize the set of h parameters for the original circuit of Fig. 14-5a by the following matrix: 3 0.5 —0.5 0.25 14-2 Input Impedance of a Network We know that any linear circuit with input and output terminals has a set of h parameters. What we want to do in this section is to find a formula for the input impedance in terms of the k parameters. Consider Fig. 14-6, where we have shown a linear circuit with a loadh Parameters 373 resistance of rz, across its output terminals, On the input side the voltage source v1 drives the circuit and sets up an input current of és. How can we find a formula for zin, the input impedance of the circuit, in terms of the h parameters? First, realize that the input impedance is the ratio of the input voltage to the input current. That is, ain = (14-7) oe Recall from Eq. (14-1) that the input voltage in terms of h parameters is oy = hut + hasds We can substitute this expression for v; into Eq. (14-7) to get _ hasis + hss Wy or hasv: Zin = hu + aa (14-8) 1 4 Fig. 14-6 Finding the input resist- i Linear ance in terms of A parameters. } circuit Zin This is not the final expression we want; it still contains v, and i. These quantities can be eliminated as follows. From Kq. (14-2) we know that the output current és is in = havis + hasta (14-9) Further, in Fig. 14-6 we can see that ee (14-10) TL (The minus sign is used because the actual load current is opposite to the direction of é2.) Now we can substitute Eq. (14-10) into (14-9) to get v2 A — = hart + heave "374 ‘Transistor Cireuit Approximations or —haxiy = have + 2 = vy (hs + P) fe 7 i Hae sec esiuaieas - i shee (411) Finally, we can substitute this expression into Eq. (14-8) to get Iuahas = Igy — 8 _ 5 fin = hy — jth (14-12) This result is very important. It tells us exactly how to find the input impedance of a circuit given the h parameters of the circuit and the value of load connected to the output terminals. EXampne 14-3 Find the input impedance of the circuit shown in Fig. 14-7. Sonution By inspection, we can see that the input impedance equals 20 ohms plus two 10-ohm resistances in parallel. Zin = 20 + 10||10 = 25 ohms Alternatively, we can get this same result by using Eq. (14-12). Recall that the h parameters of the circuit inside the box were found in Example 14-1. These parameters are Au hi] _[ 20 1 far hee} ~ |-1 0.1 Since the circuit is loaded by 10 ohms, we use Eq. (14-12) as follows. ee 1-1) te hu — oe Oe By using two different approaches, we have found that the input impedance in Fig, 14-7 is 25 ohms. Ordinarily, with a simple cireuit like = 20 + 5 = 25 ohms 0 10 [ $ Fig. 14-7 Example 14-3.h Parameters 375 that of Fig. 14-7, we would not use the complicated h-parameter approach because the answer can be found directly by inspection. However, there are more complicated circuits in which the input impedance cannot be found by inspection. In these circuits, Eq. (14-12) will be useful. Exampue 14-4 Find the input impedance of the circuit in Fig, 14-8 for the follow- ing values of rz: (@) rh = (b) rz © (c) r, = 30 ohms. Sonution The h parameters for the circuit inside the box are the same as those of the preceding example. We need only substitute the different values of rz into Eq. (14-12) to get the input impedance. (a) When rz = 0, -1 i Carian Oe yO Et rete 20 ohms Note that the input impedance equals hu. The second term hizhe/ (haz + 1/rz) dropped out because r, = 0. This always must happen when rz = 0, because we know that the input impedance with the output terminals shorted equals hu. (b) When rz = &, tee -1 tm = 20 — gy payee = - OF = 30 ohms 20 0 4 Fig. 14-8 Example 14-4, Zin This same answer is apparent by inspection of Fig. 14-8. It is obvious that when rz = © (an open circuit), the input impedance is simply 20 ohms plus 10 ohms.376 Transistor Circuit Approximations (c) When rz = 30 ohms, Zin = 20 = 20 + 7.5 = 27.5 ohms eect 3 ~ 01+ 30 Again, this same result can be obtained by inspection of Fig. 14-8. It is clear that when r, = 30 ohms, the input impedance must be 20 ohms plus the parallel combination of 10 ohms and 30 ohms. That is, Zin = 20 + 10/|30 = 27.5 ohms 14-3 Voltage Gain Using h Parameters Another useful formula involving the h parameters is the formula for the voltage gain of a linear circuit that is loaded by a resistance of rz. Refer again to Fig. 14-6. The voltage gain of this circuit is simply _ oe Yy The input voltage v: must equal the input current i; times the input impedance zin. That is, M1 = tizin Hence, we can express the voltage gain as (14-13) Zin In the preceding section, we found an expression for v2/i; in terms of the h parameters (Eq. 14-11). If wo substitute this cxpression for v2/it into Eq. (14-13), we get fo eee an ae Hin(Fian + T/r2) This equation is quite useful; it tells us exactly how to find the voltage gain of a circuit given its h parameters and the value of load resistance. Thus, given a circuit, we first find the input impedance zis by using Eq. (14-12). Then, we ean find the voltage gain by using Eq. (14-14). (14-14) EXAMPLE 14-5 Find the voltage gain v2/v; for the circuit of Fig. 14-9. Sonution The h parameters of the circuit inside the box were found in Example 141.h Parameters 377 hu hv] _f 20 1 fia hee -1 01 In Example 14-3 we found the input impedance of the circuit to be 25 ohms. (This should also be apparent by inspection of Fig. 14-9.) To find the voltage gain, we use Eq. (14-14). They are: The A= Fn + 17) Thus, the output voltage is one-fifth of the input voltage. ‘This should be apparent by inspection of Fig. 14-9. The two 10-ohm resistors have . 20 + Fig. 14-9 Example 14-5. 4 0 % So a net resistance of 5 ohms. Therefore, we have a voltage divider con- sisting of a 20-ohm resistor in series with a 5-ohm resistor, which implies that the output voltage will be one-fifth of the input voltage. Again, note that for a simple circuit like that of Fig. 14-9, we would not ordinarily use h parameters to find the voltage gain because we can get the answer more easily by a direct analysis of the circuit. Up to now, we have used the h-parameter approach merely to illustrate the use of Eqs. (14-12) and (14-14). In succeeding sections, however, we will use these equations to find the input impedance and voltage gain of transistor circuits. 14-4 The h Parameters of a Transistor We have seen that any linear circuit has a set of h parameters associated with it. When the linear circuit is terminated by a load of rz, we can find the input impedance and voltage gain by using Eqs. (14-12) and (14-14). In a class A transistor amplifier, the transistor is biased to some con- venient d-c operating point. We can then inject an a-c signal into the transistor to get an amplified output signal. If small a-c signals are in- volved, the transistor is a linear device because the output a-c signal is directly proportional to the input a-c signal. Under this condition, the378 Transistor Circuit Approximations a-c operation of the transistor can be described by the h parameters. In other words, for small a-c signals, each transistor has a set of h parameters associated with it. These k parameters and the a-c load r, seen by the transistor can be used to find the input impedance and the voltage gain. The set of 4 parameters will, of course, depend upon which transistor connection is used, that is, CB, CE, or CC. For instance, suppose that the CB circuit of Fig. 14-10a has an hy; = 25 ohms. If we use the same transistor in a CE circuit (Fig. 14-100), we will find that the new value (a) (6) Fig. 14-10 CB and CE connections. of has is quite different. Typically, the value of hi: might be around 2500 ohms. Similarly, we will find that the remaining h parameters (iz, ha, and hz:) also depend upon which connection is used. To distinguish the h parameters of a transistor for its three possible connections, we use the following notation. For CB connection, we can add the subscript 6 to each of the h parameters. Thus, to indicate a set of CB h parameters we write hin, hiss, haw, how. For CE connections of a transistor, we add a subscript ¢ to get Rite, Mane lates hare For CC connections of a transistor, we add a subscript ¢ to get hie Ares here, here. The foregoing is one of the notational systems used for distinguishing the three different transistor connections. Another notational system is the following. Recall the meaning of the h parameters: Au. = input impedance with output shorted ho = forward current gain with output shorted Juz = reverse voltage gain with input open he = output admittance with input open Note that the first letter on the right-hand side of each equation is italicized. These letters are i, f, r, and o. In order to simplify the notation of the h parameters the numerical subscripts can be replaced by the corresponding first letters. That is, we can replace numbers by letters