Chap9 Sample

513

CHAPTER 9Amplifier Frequency

Response

9.1 High-Frequency Small-Signal Models for Design

9.2 Stages with Voltage andCurrent Gain

9.3 Voltage Buffers

9.4 Current Buffers

9.5 Comparison of Single-Stage Amplifiers

9.6 Multistage Amplifiers

9.7 Differential Amplifiers

Solutions to Exercises

Chapter Summary

References

Problems

INTRODUCTION

Preceding chapters have discussed DC biasing and the small-signal midband ACperformance of amplifiers. In this chapter, we see how to analyze the frequency re-sponse of amplifiers and how to design amplifiers to achieve a desired response.We assume that the desired response is flat within some range of frequencies (themidband, as already noted in Chapter 8). We seek methods for determining thatrange of frequencies and design techniques to extend the range. Many of theanalysis techniques discussed here apply to other linear circuits as well, but ampli-fiers will be the main topic. Filters, which are used to shape the frequency responseof a system to achieve various design goals, are discussed in Chapter 11.

As in the previous two chapters, we begin with a discussion of the models need-ed for passive and active devices. We then proceed to a discussion of techniquesfor determining the midband range of a given circuit, called the bandwidth, andfinally give some specific circuit examples. We again use the generic transistor tointroduce discussions of the different circuit topologies and then present specificresults for each type of transistor.

Frequency response is important in many electronic systems and can also be observed in na-ture.The sky is blue because high-frequency blue light is scattered more than lower-frequencylight.The sunset is red because the higher-frequency light is scattered so much while taking thislonger path through the atmosphere that it gets absorbed (i.e., the energy drops to nearly zero).

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514 Chapter 9 Amplifier Frequency Response

As always in this text, a major goal of the chapter is to develop the techniquesnecessary to analyze circuits in an intuitive and reasonable manner so that we canuse the results for design. The analysis of the complete frequency response (i.e., aknowledge of the frequencies of all poles and zeros) of even a simple amplifier isusually too complex to do by hand, and even if a result is obtained, it is often toocomplex to be any use in the design process. Therefore, we will concentrate onfinding the cutoff frequencies,1 which determine the bandwidth. If a knowledge ofthe frequencies of other poles or zeros is desired (e.g., for compensation of a feed-back loop, as discussed in Chapter 10), they can sometimes be found by handanalysis; but more often, computer simulation would be used.

In this chapter, we assume that you are familiar with complex impedances andtransfer functions. We also assume you are familiar with single-pole transfer func-tions as covered in Aside A1.6, and AC- and DC-coupled transfer functions, aspresented in Sections 6.5.2 and 8.1.2.

9.1 HIGH-FREQUENCY SMALL-SIGNAL MODELSFOR DESIGN

Section 8.1 covered the small-signal midband AC models we use for active andpassive devices. In this section, we extend those models for use at higher frequen-cies. But, not all parasitic elements are explicitly presented here. For example, alldiodes and transistors have series inductance in the wires connecting them withthe rest of the circuit.These additional parasitic elements are extremely importantat very high frequencies. Nevertheless, this section focuses on models suitable foruse at moderate frequencies (up to a few hundred MHz at most). Many of the ad-ditional parasitic elements can be included in a straightforward way, based on thematerial presented here.The key issue is deciding what level of modeling is neces-sary for a given application.

9.1.1 Independent SourcesThe models for independent sources have been discussed for DC analysis inChapter 7 and for AC analysis in Chapter 8. It makes no difference whether thefrequencies are high or low for these models; therefore, the models used inChapter 8 apply equally well here.

9.1.2 Linear Passive Elements (Rs, Ls, & Cs)In analyzing the frequency response of electronic circuits, it is occasionally neces-sary to reevaluate the models used for passive components. For example, a wire-wound precision resistor also has significant inductance, which must be taken intoaccount at higher frequencies. In fact, all passive components have parasitic ele-ments in series or parallel with the desired element (e.g., the inductance of thewire-wound resistor), as shown in Figure 9-1.

At very high frequencies, it may not be possible to use lumped models for allelements.2 A lumped model is one that uses a single ideal element to model each

1 These are often called the - 3 dB frequencies. They are simply the frequencies at which the specifiedtransfer function has dropped by 3 dB (i.e., ) from its midband value.

2 The boundary is usually taken to be when the dimensions of the circuit exceed one-tenth of thewavelength of the voltage wave. Voltage waves in circuits travel at roughly 1>2 the speed of light, sothe wavelength in meters is approximately l L 150> (in MHz). For a device with a physical length of1 inch, the boundary between lumped and distributed analysis occurs at about 5.9 GHz. In lumpedanalysis, we use Kirchhoffs laws, which are only an approximation to Maxwells equations [9.1].

1>12

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R SpencerChapters 7 (DC biasing), 8 (low-frequency small-signal ac analysis and amplifiers), 9 (amplifier frequency response), 12 (low-frequency large-signal ac analysis), and 15 (transistor-level digital circuits) all begin with a discussion of the models used for hand analysis.

9.1 High-Frequency Small-Signal Models for Design 515

iRiR

Cp

Ls

R

R(Real)

vR

+

vR

+

(a)

iLiL

Cp

RS

L

L(Real)

vL

+

vL

+

(b)

iCiC

C(Real)

vC

+

vC

+

(c)

RS

RpC

LS

Figure 91 One possible model for (a) a real resistor, (b) a real inductor, and (c) a real capacitor.The elements used in the models are ideal resistance, capacitance, and inductance. Rs is the seriesparasitic resistance (caused by the leads), Rp is the parallel parasitic resistance, and Ls and Cp are theparasitic series inductance and parallel capacitance, respectively.

characteristic (e.g., resistance, inductance) of a device. (The models in Figure 9-1are lumped.) Lumped models can be quite complex when they are used to ap-proximate a distributed system. For example, a wire-wound resistor could be mod-eled by a string of Rs and Ls in series (each RL combination modeling a singleturn), with a capacitor in parallel with each series RL combination to model thecapacitance from one turn to the next. Models this complicated are not typicallyemployed, however, especially in hand analysis.

When lumped models fail, we can use distributed models instead. A distributedmodel is a model that allows the voltages and currents to be functions of physicalposition as well as time. Other techniques exist for modeling distributed systemsin a more elegant way, but they are beyond the scope of this text. Nevertheless,there are parasitic effects that are commonly encountered even at much lowerfrequencies. In this subsection, we will discuss a few of these effects and presentmodels that can be used for passive devices if the effects are significant. We willdiscuss discrete and integrated passive elements separately.

Discrete Passive Elements All real resistors have some series inductance andsome parallel capacitance, as shown in Figure 9-1(a). With the exception of wire-wound resistors, which are typically avoided except at lower frequencies, theseparasitic elements are not usually important at frequencies where lumped modelsare acceptable.

All real inductors have some series resistance and parallel capacitance, asshown in Figure 9-1(b). The parallel capacitance can usually be ignored, but thereare many cases where the series resistance needs to be accounted for. This resis-tance is usually specified by the manufacturer, often by specifying the quality fac-tor, or Q, of the inductor. The Q is a measure of how much energy is stored in theinductor relative to how much is dissipated. In the context of Figure 9-1(b), if weconsider low frequencies where Cp can be ignored, the Q is given by

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. (9.1)

This equation allows us to find Rs given the Q, the test frequency, and the induc-tance. As an example, one commonly available molded inductor of 10 H has a Qof 50 at a test frequency of 7.9 MHz. Using (9.1), we find that Rs = 10 .

All real capacitors have some series resistance and inductance and some paral-lel resistance, as shown in part (c) of Figure 9-1. The parallel resistance is a resultof leakage current through the dielectric and can be significant in some applica-tions. Electrolytic capacitors can have especially large leakage currents and areusually used only as power supply filter capacitors. The leakage currents are oftenspecified on the data sheets.

In addition to the leakage current, the series inductance of capacitors can causeproblems at surprisingly low frequencies. Because of this inductance, a capacitorwill exhibit series resonance at some frequency. For frequencies above this self-resonant frequency, the capacitor appears mostly inductive, and its impedanceincreases with increasing frequency. The self-resonant frequencies of many popu-lar capacitors are in the 10 kHz to 100 MHz range, with aluminum electrolytic ca-pacitors at the bottom of the range and silvermica capacitors at the high end. It iscommon for capacitors used as bypass capacitors to look inductive at the frequen-cies of interest. However, as long as the overall impedance is low enough, they stillserve the purpose of bypassing the elements they are in parallel with.

The impedance of a capacitor at frequencies below the self-resonant frequencycan always be modeled as just a resistor in series with a capacitor. (Both elementvalues will, in general, be functions of frequency.) If this two-element model isused, the series resistance is called the effective series resistance, or ESR.The ESRis not the same as Rs in Figure 9-1, but it yields a useful simplified model. For ex-ample, the power losses in a capacitor can be readily calculated using the ESR.Since the ESR is modeling several different effects, it is a function of frequency, asalready noted.

Frequently, the ESR is not given directly, but it is specified by giving thedissipation factor (DF) of the capacitor. The DF can be defined as

, (9.2)

where Q is the quality factor. The DF is sometimes specified as a dimensionlessnumber (6 1) and sometimes as a percentage. Using this definition, and assumingthat the capacitor is modeled by a series resistance Rs and a capacitance C, we canderive (see Problem P9.4)

, (9.3)

The DF is also sometimes given as

. (9.4)

where is defined to be the difference between the phase of the overall capacitorimpedance and -90. For an ideal capacitor, both and the DF are zero.The DF isalso equal to the ratio of the magnitudes of the voltages across the resistance andthe capacitance in the two-element model (see Problem P9.5):

DF = tan d = vRsC

DF = vRsC

DF =1

2p

energy dissipated in one cycle

peak energy stored=

1Q

Q =vL vR

=vLRs

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. (9.5)

Remembering from (9.2) that the DF is 1>Q, we see that (9.5) agrees with (9.1) forthe Q of an inductor.

DF =vR vC

= vRsC

Example 9.1

Problem:Find the ESR of a 0.001-F polyester-film capacitor with a dissipation factor of1% at 1 kHz.

Solution:Using (9.3) with DF = 0.01 (i.e., 1%), we find that Rs = 1.6 k. This value mayseem unbelievably large, but you must compare it with the reactance of the capac-itor at the same frequency: = 159 k. Note thatXC is 100 times larger than the ESR, which is exactly what (9.5) implies for a DFof 1%.

XC 11 kHz2 = 1>12p # 1k # 0.001m2

Exercise 9.1

Given a 1-H inductor with a Q of 44 measured at 25 MHz, what is RS? Ignore Cpduring your calculation.

Exercise 9.2

Measurements on a 0.01-F capacitor show that the impedance reaches a minimumvalue of 1.5 at 5 MHz. (a) Ignoring Cp , find values for RS and LS. (b) What is thedissipation factor for this capacitor at 1 kHz?

Integrated Passive Elements Resistors, capacitors, and inductors, as fabricatedin integrated circuits, have different high-frequency parasitics than do their dis-crete counterparts. We will briefly point out the differences in this section forthose readers familiar with IC technology.

As noted in Section 8.1, inductors are rarely used in ICs at this time. Neverthe-less, they are used, especially for high-frequency circuits. Integrated inductors canbe modeled in much the same way as discrete inductors.

Diffused resistors have parasitic pn junctions, as discussed in Section 8.1. Athigh-frequencies, the capacitance of a pn junction can be a significant parasiticand should be included in the model used for design. Usually, a simple lumpedmodel is adequate, although there are occasions when a more complicatedlumped approximation to a distributed model is required. Thin-film resistors alsohave parasitic capacitance to the underlying substrate, but it is typically muchsmaller than for diffused resistors.

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ASIDE A9.1 BROADBAND DECOUPLING

In designing electronic circuits, it is frequently necessary to bypass some element or a power supply over awide range of frequencies. Since any real capacitor exhibits self resonance, this task can be more difficultthan you might at first imagine. For example, if a large capacitor on the order of 10 F is needed to providean adequate bypass at low frequencies, this capacitor will usually have a self-resonant frequency between100 kHz and 10 MHz, depending on the type of capacitor chosen. If the self-resonant frequency is 500 kHz,ignoring the parallel capacitance, we determine the series inductance to be , which is 10 nH.The impedance of this capacitor will then be equal to 6.4 at 100 MHz. Depending on the application, thisimpedance may not be small enough.

One technique that is frequently employed in this situation is to put a smaller capacitor with a higherself-resonant frequency in parallel with the large capacitor. The logic is that the small capacitor will takeover and provide an acceptable bypass for frequencies above the self-resonant frequency of the larger ca-pacitor. But is this logic correct?

Consider a specific example where the large capacitor is the previously considered 10-F electrolytic witha self-resonant frequency of 500 kHz and the smaller capacitor is a 0.01-F polyester-film capacitor with aself-resonant frequency of 20 MHz. We calculate the inductance of the smaller capacitor to be 6.3 nH. Thetwo capacitors in parallel can be modeled as shown in Figure A9-1 if we neglect the series resistance.

Using this model, we find that the reactance of the large capacitor is ; similarly, the re-actance of the smaller capacitor is . The total impedance of the parallel combination isgiven by

.

This impedance is plotted, along with the reactances of each capacitor, in Figure A9-2. The figure also in-dicates the resonant frequencies of the capacitors, and we see that their reactances go through zero atthose frequencies, as expected. What may be unexpected is the peak in the total impedance, which isforced to occur between the self-resonant frequencies of the two capacitors.

Z1jv2 = -1vL1 - 1>vC121vL2 - 1>vC22j1vL1 - 1>vC12 + j1vL2 - 1>vC22

X2 = vL2 - 1/vC2X1 = vL1 - 1>vC1

Ls = 1>1vo2C2

L2

C2

L1

C1

ZFigure A9-1 A model for twobypass capacitors in parallel.

Capacitors in integrated circuits are usually made by using parallel-platestructures, where the plates are either conductors (metal or heavily doped poly-silicon) or conductive regions in the semiconductor substrate. When a conduc-tive region in the substrate is used as one of the plates of the capacitor, there isa significant parasitic capacitance from this plate to the surrounding substrateregions. These extra parasitic capacitances are often important and must be in-cluded in the models used.

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106105104 107 108

5

2.5

0

2.5

5

7.5

10

Reactance of thelarge capacitor

Total impedance

Z(in )

f (Hz)f1 f2

Reactance of thesmall capacitor

Figure A9-2 The total impedanceand individual reactances for the modelin Figure A9-1.

The peak in the total impedance can be explained in the following way: For frequencies above the self-reso-nant frequency of the large capacitor,v1, it appears inductive; whereas, for frequencies below the self-resonantfrequency of the small capacitor, v 2, it appears capacitive. Therefore, for frequencies between v1 and v 2, wehave, in effect, a parallel LC circuit, and the total impedance peaks when this parallel LC circuit becomes reso-nant (i.e., the net inductance from the large capacitor cancels the net capacitance of the smaller capacitor).

Putting these two capacitors in parallel does not therefore accomplish the desired goal. It is certainly true,as can be seen by comparing the total impedance with the reactance of the large capacitor, that the parallelcombination has a lower impedance for high frequencies; however, we have introduced a pole into the im-pedance at a frequency between the self-resonant frequencies of the two capacitors. In other words, there is anarrow band of frequencies for which this combination will not be an effective bypass at all. This can have aserious detrimental effect on the performance of an amplifier.

We should not conclude from this example that capacitors should never be put in parallelonly that we needto be careful, and as always, we need to be using the appropriate level of modeling to understand the situation.

9.1.3 DiodesBoth pn-junction diodes and Schottky diodes exhibit capacitance because ofcharge storage in the depletion region, as discussed in Section 2.4.5.The small-sig-nal capacitance caused by the change in depletion-region charge is called thejunction capacitance and is given by

, (9.6)

where Cj0 is the zero-bias value of the junction capacitance, V0 is the barrier po-tential, mj is an exponent that is a function of the doping profile in the device (typ-ically between 0.2 and 0.5), and the value given for Cj in forward bias is only anapproximation.

In addition to the junction capacitance, pn-junction diodes exhibit minority-car-rier storage in forward bias, as discussed in Section 2.4.5. This charge storage termleads to the small-signal diffusion capacitance, given by

, (9.7)Cd KdQFdVD

= tF IDVT

Cj KdQDRdVD

= L Cj011 - VD>V02mj in reverse bias2Cj0 in forward bias

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Crd

Figure 92 A small-signalmodel for a diode that is validfor high frequencies. rd is presentonly in forward bias.The value ofC depends on the type of diodeand whether it is forward or re-verse biased; see text for details.


Ccm

Coc OC M

M

rcmgmvcm

vcm

+

Ccm Coc

O

C

(a) (b)

rm aic im'

im'

Figure 93 (a) The high-frequency hybrid- model and (b) the high-frequency T model for thegeneric transistor.

where QF is the minority-carrier charge stored in forward bias and tF is the asso-ciated time constant, often called the forward-transit time. This equation is validonly in forward bias.

The small-signal model for a diode presented in Chapter 8 can be modified forhigher frequencies by adding a capacitor in parallel with the small-signal resis-tance, as shown in Figure 9-2. In forward bias, the capacitor is given by (9.6) forSchottky diodes and by (9.6) plus (9.7) for pn-junction diodes. The small-signalresistance shown is not present in reverse bias; only the capacitor is present, andits value is given by (9.6) alone. The small-signal resistance was found earlier tobe rd = nVT >ID.Exercise 9.3

A pn-junction diode has V0 = 0.6 V, Cj 0 = 2 pF, IS = 10-14 A, n = 1, mj = 0.33, andtF = 500 ps. Derive the high-frequency small-signal AC model when (a) VD = 0.65 Vand (b) VD = -3 V.

Integrated circuits usually use diode-connected transistors in place of actualdiodes, as noted in Section 8.1. Nevertheless, junction diodes do show up as para-sitic elements in transistors, as discussed in Section 8.1. When parasitic diodes arepresent, the models presented in the preceding section are used.

9.1.4 The Generic TransistorAll of the transistors presented in this book have at least two parasitic capacitorsthat are important to include in the high-frequency model. Figure 9-3 shows thosetwo capacitors added to our models for the generic transistor.The first capacitor iscalled Ccm, and it appears between the control and merge terminals. The other ca-pacitor is called Coc and appears between the output and control terminals. Notethat in the current-controlled version of the high-frequency T model the control-ling current, , is not the external merge current; it is the current through rmalone. (The external merge current flows through rm in parallel with Ccm.)

9.1.5 Bipolar Junction TransistorsBipolar junction transistors have capacitances associated both with pn junctionsand with the storage of excess minority carriers (mostly in the base) when exter-nal currents flow.The details are presented in Section 2.5.5.We are interested onlyin operation in the forward-active region in this chapter, so the collector-basejunction will be reverse biased, and will exhibit a junction capacitance given by(the same as (9.6) in reverse bias except for the subscripts)

im

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R SpencerThe "generic transistor" is used in the text to present material that is common to all types of transistors. This avoids repetition without requiring an instructor to cover a particular device first, and also fosters a modern device-independent view of circuits.


. (9.8)

C is also sometimes called Cjc . (It is called C because it is in parallel with r.)Equation (9.8) is valid for npn transistors. For pnp transistors, VBC needs to be re-placed by VCB (i.e., the voltage should be negative when the junction is reverse bi-ased). The coefficient mjc is typically between about 0.2 and 0.5, and the built-inpotential V0 is typically between 0.5 V and 1 V.

The forward-biased emitter-base junction exhibits both junction capacitanceand diffusion capacitance. Because this capacitance appears in parallel with rp inour hybrid- model, we call it Cp . This small-signal capacitance is given by

, (9.9)

where the junction capacitance in forward bias is an approximation and the timeconstant tF is frequently called the forward-transit time, or base-transit time. tF isnot exactly equal to the base-transit time, but we will ignore that detail here.

The high-frequency hybrid- model for a BJT is shown in Figure 9-4. The twocapacitors given by (9.8) and (9.9) have been included in the model, but we havestill ignored the extrinsic emitter and collector resistors and r.The base resistanceand output resistance can frequently be ignored in this model, but rb affects thehigh-frequency performance more often than it does the midband performanceand must be included in some circumstances. The nonreactive model elementswere found in Section 8.1 and are , , and .

In specifying the high-frequency model of a BJT, it is common to give the fre-quency at which the common-emitter current gain (b) goes to unity, called thetransition frequency. Aside A9.2 explains the terminology used and relates thetransition frequency to the model parameters.

Finally, we note that we can also add capacitors to the T model of the BJT, asshown in Figure 9-5. We have left rb out of this model, because it rarely affects theperformance of common-base stages. It would be included in series with the baseconnection if necessary. The other elements in the model were previously found tobe and . As noted in Section 9.1.4, thecontrolling current, is not the external emitter current; it is only the current in re .(The external emitter current flows through re in parallel with Cp .)

iere = rp 1>gm L 1>gm = VT>IEa = b>1b+12

ro = VA>ICrp = b0>gmgm = IC>VT

Cp = Cje + Cd L 2Cje0 + gmtF

Cm KCm011 - VBC>V02mjc

B

C

C

r ro

C

E

rb

gmvv

+

Figure 94 The high-frequencyhybrid- model for a BJT.

C C

E C

B

ie'

ie'

reFigure 95 The high-frequency T model of a BJT withrb omitted.

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ASIDE A9.2 THE TRANSITION FREQUENCY OF A BJT

We often desire to compare the performance of one bipolar process with another. In addition, we would liketo have a parameter that describes the high-frequency capability of a bipolar device in a more intuitive anddirect way than is possible by specifying the parameters of the high-frequency hybrid- model.The most com-mon parameter used for these purposes is the transition frequency, which is denoted fT or vT. The transitionfrequency is defined to be the frequency at which the short-circuit common-emitter current gain goes tounity. Consider the circuit shown in Figure A9-3, wherein we drive the base of the transistor with a currentsource and measure the short-circuit output current. The high-frequency hybrid- model has been used. Theoutput resistance of the transistor has been neglected, since it has a short circuit in parallel with it and willnot, therefore, have any current in it.

Notice that C and Cp are in parallel in this circuit, and we can write

. (A9.1)

Now, assuming that the current from the controlled source is much larger than the current through C (a rea-sonable assumption for most frequencies of practical interest; see Problem P9.12), we can write

. (A9.2)

Finally, we can solve for the current gain:

. (A9.3)

This can also be expressed as

, (A9.4)

where

(A9.5)

as before, and we have implicitly defined the - 3 dB frequency to be

. (A9.6)

Using (A9.4), we can solve for the frequency at which |b(jv)| = 1. This frequency is, in a sense, the largest fre-quency at which the transistor is useful, and is defined to be the transition frequency. Since the transition oc-curs at a frequency well above vb , we obtain

, b1jvT2 = 1 = b01 + j

vT

vb

L b0vbvT

vb =1

rp1Cp + Cm2

b0 = gmrp

b1jv2 = bo1 + jv>vb

b1jv2 = Ic1jv2Ib1jv2 = gmrp1 + jvrp1Cp + Cm2

Ic1jv2 L gmVp1jv2

Vp1jv2 = Ib1jv2 Crp 1Cp + Cm2 D = Ib1jv2 rp1 + jvrp1Cp + Cm2

icrb

ib Cgmv

C

v

+

r

Figure A9-3 A circuit for determiningthe short-circuit common-emitter currentgain.

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which yields

. (A9.7)

Substituting from (A9.5) and (A9.6), we find that

. (A9.8)

We can rewrite (A9.8) in a useful way by remembering that Cp = gmtF + Cje :

. (A9.9)

Using (A9.9), we can see how fT varies with bias current. (Remember that gm = IC>VT.) For large bias currents,the junction capacitance terms are negligible and vT L 1>tF. For small bias currents, the junction capacitanceterms dominate and vT L gm>(Cje + C). In this region, vT is approximately proportional to IC. If the bias cur-rent is made too large, the value of tF increases so that vT drops at very high currents as well. (The increase intF at high currents is briefly discussed in Section 2.5.5.) The net result is that vT is a maximum for some cur-rent, which depends on the individual transistor. However, it decreases for much smaller or larger bias cur-rents, as shown in Figure A9-4.

The data sheets for bipolar transistors usually specify fT at some current, as well as the junction capaci-tances Cje and Cjc (i.e., C), but they dont typically specify tF, even though it is the parameter used by SPICEand other simulation programs to model the diffusion capacitance. Therefore, (A9.9) has much practical sig-nificance if a SPICE model is needed for a device.

vT =gm

gmtF + Cje + Cm

vT =gm

Cp + Cm

vT = 2pfT = bovb

High-levelinjection

IC

T

12F

Figure A9-4 The variation of vT withcollector current.

B

r C

C

ro

C

E

rb

0 i

i

Figure 96 The current-controlledversion of the high-frequency hybrid-pBJT model.

This same notational problem is encountered in the high-frequency hybrid-model if we use the current-controlled version of the source. At midband, thesource is equal to b0ib , as shown in Chapter 8, but when capacitors are included inthe model, we cannot use this same formula, since it would lead to a constant cur-rent from the controlled source if ib were held constant. As shown in Aside A9.2,this is not what happens in the device. If, however, we let the controlled source begiven by , where is defined to be the current in rp , as shown in Figure 9-6,we obtain the proper results (see Problem P9.17).

ipb0ip

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Exercise 9.4

Consider a BJT with IS = 10-14A, Cjc0 = 2 pF, Cje0 = 1 pF, V0 = 0.6 V, mjC = 0.33, tF =200 ps, rb = 50 , VA = 60 V, and b0 = 100. Find the complete small-signal high-frequency AC hybrid- and T models when VBE = 0.67 V and VCE = 4 V.

In addition to the parasitic elements just presented, other elements may bepresent in integrated bipolar transistors. Section 8.1 discussed the different para-sitic junction diodes present in a typical integrated BJT. For high-frequencyanalysis, each of the parasitic diodes has a capacitance associated with it, which ismodeled exactly as shown for a diode in Section 9.1.3.

9.1.6 MOS Field-Effect TransistorsCharge storage in MOSFETs was discussed in Section 2.6.4.The main capacitive el-ement in a MOSFET is the gate-to-channel capacitor formed by the MOS struc-ture.This structure is a parallel-plate capacitor, and the total capacitance is given by

, (9.10)

where Pox is the dielectric constant of the oxide, tox is the thickness of the oxide,and W and L are the width and length of the channel, respectively. The gate-to-channel capacitance leads to capacitance from the gate to the source, Cgs, andfrom the gate to the drain, Cgd. When the device is operating in the linear region,Cg-ch splits equally between Cgs and Cgd. Therefore, we have

. (9.11)

When the transistor is forward active, however, the channel is pinched off at thedrain end, and the gate-to-channel capacitance appears from the gate to thesource. In addition, the value is reduced because of the linear reduction in channelcharge from the source to the point at which the channel pinches off. A detailedanalysis shows that in forward-active operation we get (see Section 2.6.4 for de-tails)

(9.12)

and

(9.13)

In addition to the gate-to-channel capacitance just described, there is a small ca-pacitance due to the gate conductor overlapping the diffused source and drain re-gions in the device. This capacitance is called the overlap capacitance, and it addsto each of the capacitances already given. The overlap capacitance is proportionalto the width of the device, but is not a function of the length. Including the overlapcapacitance, Col, leads to the following two equations for the device in the for-ward-active region of operation:

Cgd L 0.

Cgs =23

PoxWLtox

Cgs = Cgd =PoxWL

2tox

Cg - ch =PoxWL

tox

SPENC09.201361833_513-658v13 6/28/02 1:37 PM Page 524


G

S

(a) (b) (c)

DCgd

Cgs rds

G

S

DCgd

Cgs rovgs

+

gmvgs

S

1gm

G

D

Cgs Cgdis

is

Figure 97 The high-frequency small-signal models for a MOSFET: (a) in the linear region, (b) the hybrid- model forforward-active operation (i.e., saturation), and (c) the T model for forward-active operation.

(9.14)

and

. (9.15)

The high-frequency MOSFET models are shown in Figure 9-7. Part (a) of the fig-ure shows the small-signal model in the linear region, where the drain-to-sourceresistance is given by (see (7.18))

(9.16)

and the capacitors are given by (9.11) with the overlap capacitance added

. (9.17)

Part (b) of the figure shows the high-frequency hybrid- model for forward-activeoperation.The capacitors are given by (9.14) and (9.15), and the transconductanceand output resistance are given by the equations derived in Chapter 8:

, (9.18)

and

. (9.19)

Part (c) of the figure shows the high-frequency T model, which is also valid for for-ward-active operation, and the same equations apply as for the hybrid- model.As noted in Section 9.1.4, the controlling current, , is not the external source cur-rent; it is only the current in . (The external source current flows through in parallel with Cgs.) We left ro out of the T model, but it can be included.

1>gm1>gm i s

ro =1

gds=

VAID

=1lID

gm = 2K1VGS - Vth2 = 22KIDCgs = Cgd =

PoxWL2tox

+ Col

Rds =1

2K1VGS - Vth2

Cgd = Col

Cgs =23

PoxWLtox

+ Col

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ioii Cgs

gmvgs

CgdG D

S

vgs

+

Figure 98 Finding fT for a FET.

A figure of merit that is often used to specify the high-frequency capabilities ofa FET is the transition frequency, vT, or fT. The transition frequency is defined asthe frequency at which the short-circuit common-source current gain goes tounity.This frequency is, in some sense, the maximum usable frequency of the tran-sistor. Figure 9-8 shows a common-source FET driven by a current source andwith the output shorted.

Whether or not we include ro in this circuit is irrelevant, since it has a short cir-cuit in parallel with it and therefore will not have any current in it. The short-circuit current gain is found from this circuit to be

, (9.20)

where the final term represents the constant fraction of the input current thatflows through Cgd. If we ignore this usually small term and set the magnitude of(9.20) equal to unity, we find that

, (9.21)

or

. (9.22)

Note from (9.20) that the current gain of a FET goes to infinity as the frequencyapproaches zero. This is to be expected, since the DC gate current is zero if we ig-nore the extremely small leakage current present.

Exercise 9.5

Consider an n-channel MOSFET with a 250--thick gate oxide, W = 5 m, L = 1 m,Col = 2 fF, Vth = 0.7 V, and = 0.035 V1. Assume that the electron mobility in thechannel is 675 cm2>Vs. Find the complete small-signal high-frequency AC modelwhen VGS = 1.5 V and VDS = 3 V. What is the corresponding value of T?

Integrated MOS Field-Effect Transistors The high-frequency models presentedup to this point have ignored the bulkor bodyconnection of the transistors.Real MOSFETs are four-terminal devices, however, and there is capacitance fromeach of the other three terminals to the bulk. A simple high-frequency MOSFETmodel that includes the bulk connection is shown in Figure 9-9.

fT =gm

2p1Cgs + Cgd2

vT =gm

Cgs + Cgd

Ai1jv2 = Io1jv2Ii1jv2 = gm - jvCgdjv1Cgs + Cgd2 = gmjv1Cgs + Cgd2 - CgdCgs + Cgd

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vgs

+

gmvgs

Cgs

Csb

Cgd

Cgb Cdb

vbs

+

gmbvbsro

G

B

S

D

Figure 99 A high-frequency MOSFET model, including the bulk connection.

The source-to-bulk and drain-to-bulk capacitances (Csb and Cdb ,respectively) area result of the pn junctions formed by the source and drain diffusions.These capaci-tances are voltage dependent, as described for a pn-junction diode in Section 9.1.3,but for hand analysis, they are almost always modeled as constants.The gate-to-bulkcapacitance is caused by the gate conductor forming a parallel-plate capacitor withthe substrate.The gate and substrate act as the plates and the oxide is the dielectric.This capacitor is outside the active area of the transistor; otherwise, it is called thegate-to-channel capacitance. Consequently, it is no different from the parasitic ca-pacitance to the substrate found for every conductor on top of the oxide.

All three of these capacitors are parasitic elements that are not part of the in-trinsic device. Since there are other parasitics that must also be included for ouranalyses to be accurate, we will not model these elements separately and will showonly the intrinsic capacitors (Cgs and Cgd) in our models.

9.1.7 Junction Field-Effect TransistorsJFETs are modeled exactly like depletion-mode MOSFETs, except for the equa-tions used for the device capacitances and the different notation that is sometimesused for the drain current. The notational differences have been dealt with inSection 8.1. The different capacitance equations are presented here.

Because there is a reverse-biased pn junction connecting the gate to the chan-nel, the gate-to-channel capacitance is that of a reverse-biased pn-junction diode.As noted in Section 2.7.3, the total gate-to-channel capacitance has the form

, (9.23)

where the signs are correct for a p-channel JFET. For an n-channel JFET, VGCHwill be negative, and the sign in the denominator sum needs to be changed. Whena JFET is in the linear region, the gate-to-channel capacitance splits equally be-tween Cgs and Cgd , and (9.23) can be used with VGCH, taken to be the average ofVGS and VGD. In saturation, Cgs is much larger than Cgd, and we typically use

(9.24)Cgs =Cgs0

a1 + VGSV0

bmj

Cgch =Cgch0

a1 + VGCHV0

bmj

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cL cH

Amb

log

(a)

A

cH

A0

log

(b)

A

Figure 910 The log-magnitude plots of (a) an AC-coupled amplifier and (b) a DC-coupled amplifier.

and

. (9.25)

The high-frequency small-signal AC models of integrated JFETs are identical tothose presented for discrete JFETs, except for the inclusion of parasitic junctiondiodes. Precisely which parasitic diodes need to be included and how importantthey are depends on the specific structure used and will not be covered here.

9.2 STAGES WITH VOLTAGE AND CURRENT GAIN

Section 6.5.2 briefly discussed the frequency response of both DC-coupled andAC-coupled amplifiers. The log-magnitude plots for both types of amplifier arerepeated in Figure 9-10 for convenience. We are interested in being able to ana-lyze amplifier circuits to determine both the low and high cutoff frequencies, vcLand vcH. The cutoff frequencies are the frequencies where the gain drops by 3 dBfrom its midband value and so are sometimes called the -3-dB frequencies. Theyare also sometimes called the half-power frequencies, since the power deliveredto a load is one-half of the midband value. They are not necessarily equal to anypole frequency, although they may be equal or close to one.

An important characteristic of the amplifiers performance is its small-signalbandwidth.The small-signal bandwidth of an amplifier is typically defined as BW =vcH - vcL for an AC-coupled circuit and as BW = vcH for a DC-coupled circuit. Wecall it the small-signal bandwidth, since we are assuming linear circuits in drawingthe responses shown in Figure 9-10.

Small-signal midband AC analysis has already been covered in Chapter 8, so wecan use that as a starting point in finding the bandwidth of an amplifier.While per-forming the small-signal midband AC analysis, we approximated all coupling andbypass capacitors as short circuits and all parasitic capacitors in our devices asopen circuits. In other words, we assumed that for frequencies much below vcH,the capacitors that contribute to the high-frequency roll-off were small enough invalue and located in the circuit in such a way that we could open-circuit themwithout making a significant error.To determine the value of the upper cutoff fre-quency, however, we need to keep these capacitors in our circuit.

Similarly, when doing midband analysis, we assumed that for frequencies wellabove vcL, the capacitors that contribute to the low-frequency roll-off are largeenough in value and located in a way that allowed us to approximate them asshort circuits without making significant errors. To determine what the lower cut-off frequency is, however, we need to include these capacitors in the analysis.

Aside A9.3 presents a theorem that will be useful in subsequent analyses.

Cgd =Cgd0

a1 + VGDV0

bmj

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9.2 Stages with Voltage and Current Gain 529

ASIDE A9.3 MILLERS THEOREM

Consider the circuit shown in Figure A9-5, where an ideal voltage amplifier has an impedance connectedfrom its output back to its input.This feedback element prevents the overall circuit from appearing to be uni-lateral and therefore change makes the analysis more difficult and less intuitive.

However, we can calculate equivalent impedances from each side to ground and derive an equivalent circuit,as shown in Figure A9-6.

For these two circuits to be equivalent, it must be true that all the voltages and currents remain the sameunder all conditions. In the circuit in Figure A9-5 we have

(A9.10)

and

. (A9.11)

For the circuit in Figure A9-6, we have

(A9.12)

and

. (A9.13)

Equating (A9.10) and (A9.12) allows us to find the Miller input impedance

. (A9.14)

Similarly, equating (A9.11) and (A9.13) allows us to find the Miller output impedance

. (A9.15)

If we use the Miller input and output impedances from (A9.14) and (A9.15) in Figure A9-6, the circuit is com-pletely equivalent to the one in Figure A9-5, but is obviously unilateral and therefore easier to analyze.This tech-nique is useful only if we assume that vo = Avi independent of Zf . If the gain does depend on Zf, using thistheorem will usually not simplify the problem. In addition, if the gain is a function of frequency, using the theoremusually does not result in any significant simplification. Even when the algebra required to find an answer is notgreatly simplified, however, this theorem still presents a powerful way of thinking about some circuits.

ZMout =Zf

1 - 1>A

ZMin =Zf

1 - A

i2 =vo

ZMout

i1 =vi

ZMin

i2 =vo - vi

Zf=

vo11 - 1>A2Zf

i1 =vi - vo

Zf=

vi11 - A2Zf

i1

vi vo

i2

Zf

A

Figure A9-5 An ideal voltage amplifierwith a feedback impedance.

vi

ZMin ZMout

vo

i1 i2

A

Figure A9-6 An equivalent circuit.

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+

vS R2

R1

VOO

RO

RS CIN

vI

vO vL

RL

COUT

Figure 911 A common-mergeamplifier.

9.2.1 A Generic Implementation:The Common-Merge AmplifierSection 8.2 analyzed the midband performance of a common-merge amplifier.The circuit is repeated in Figure 9-11 for convenience. This section examines howto find both the low and high cutoff frequencies of that circuit.

The first step in finding either cutoff frequency is to draw the appropriateequivalent circuit. There are three small-signal equivalent circuits that we may beinterested in, depending on what we want to find. All three of them are shown inFigure 9-12 for the common-merge amplifier. Part (a) of the figure shows thesmall-signal low-frequency AC equivalent circuit, part (b) shows the small-signalmidband AC equivalent circuit, and part (c) shows the small-signal high-frequen-cy AC equivalent circuit. Part (d) of the figure shows the frequency response ofthe amplifier and indicates the frequency range in which each equivalent circuit isvalid. Because these are small-signal equivalent circuits, they all use a linearmodel for the transistor. Both the low-frequency and midband equivalent circuitsuse the low-frequency small-signal model, since they are valid for frequenciesbelow those where the small parasitic capacitances in the transistor will affect theresponse. The difference between these two equivalents is that the low-frequencyequivalent includes the coupling capacitors (and bypass capacitors, if any are pre-sent), whereas the midband equivalent is valid only for frequencies that are highenough that these capacitors are accurately modeled as short circuits.

The difference between the midband and high-frequency equivalent circuits isthat the high-frequency small-signal model for the transistor is used in the high-frequency equivalent circuit. Because these are all AC equivalent circuits, the DCpower supplies appear as grounds as they did in Chapter 8.

Once we have these equivalent circuits, we can proceed with our analyses. Anintroductory treatment is given first, followed by a more advanced method.

Introductory Treatment Using the low-frequency equivalent circuit ofFigure 9-12(a), we can derive the low cutoff frequency. We start by writing theoverall gain as a product of simpler transfer functions:

. (9.26)

We can now write each of these transfer functions by inspection using the meth-ods of Aside A1.6. First, we recognize that vcm>vs is a single-pole high-pass transferfunction. Then we note that the zero is at DC, since vcm only goes to zero at DCwhere CIN is an open circuit. The gain at high frequencies is given by a resistor di-vider, and the pole is found by shorting vs, removing CIN , and looking back intothe terminals to find the resistance seen by it. The resulting transfer function is

vlvs

=vl

vcm

vcmvs

SPENC09.201361833_513-658v13 6/28/02 1:37 PM Page 530

R SpencerAll of the sections in this chapter are divided into an "introductory treatment" that uses the Miller approximation and an "advanced treatment" that uses open-circuit time constants. Similar partitioning is used in other chapters to facilitate choosing the desired level of coverage.

R SpencerIn this chapter and elsewhere, the generic transistor is used to introduce the important concepts that are not device dependent. This treatment is then followed by sections that cover the BJT and FET specifically. An instructor may cover either first and is not at all required to cover both to have a complete treatment.


gm vcmRO

vl

RL

COUT

vcm

+

+

vs RCC rcm

RS CIN

(a)

vcm

+

+

vs RCC Ccm rcm

RS Coc

(c)

(d)

vcm

+

+

vs RCC rcmgm vcm

gm vcm

RS vl

vl

RORL

RORL

(b)

fcL

vlvs

fcH f

HFequivalent OK

Midbandequivalent OK

LF equivalent OK

Figure 912 The small-signal equivalent circuits for the common-merge amplifier: (a) low-fre-quency AC, (b) midband AC, and (c) high-frequency AC. (d) The frequency response of the ampli-fier showing the range of validity for each equivalent circuit.

. (9.27)

Moving to the output side, we again have a single-pole high-pass circuit with a DCzero. The transfer function is given by

. (9.28)

We can combine these equations and write the overall transfer function for thelow-frequency equivalent circuit. The result has two poles. To find the low cutofffrequency, we need to set the magnitude of the transfer function equal to 1> 12

Vl 1jv2Vcm 1jv2 = - jv[gm1RO RL2]jv+ 11RO + RL2COUT

Vcm1jv2Vs1jv2 =

jv a RCC rcmRCC rcm + RS

bjv+

11RS + RCC rcm2CIN

SPENC09.201361833_513-658v13 6/28/02 1:37 PM Page 531


gm vcm

vo

RL'vcm

+

+

vs RCC Cin rcm

RS

CMout

Figure 913 The equivalent circuit from Figure 9-12(c) after application of Millers theorem.

and solve for vcL. In this particular example, the contribution of each capacitor tothe low cutoff is readily seen, since the input and output circuits are separate sin-gle-pole circuits. As a practical aside, note that anytime you can calculate a timeconstant for a capacitor without having to decide what to do with any other ca-pacitor, that capacitor is independent of all other capacitors, and there is a pole atthe reciprocal of its time constant. This statement is true even if you cant see howto split the circuit up into individual units and even if the circuit is not unilateral.

This brute force method will not always yield results useful for design. Withoutmore sophisticated analysis tools, we would probably want to use computer sim-ulation if we had a more complex low-frequency equivalent circuit. For example,even a three-capacitor circuit would require solving three equations in three un-knowns (assuming that it cant be separated into single-pole circuits) and wouldyield a result too complicated to be of much use in design. In general, finding vcLin the manner just described is not very intuitive, except for the case of simplecircuits. It would also involve a lot of algebra if we had a more complicated am-plifier. A far more reasonable approximate approach will be presented in thenext part of this section.

Lets now find the high cutoff frequency for this circuit. Examination of the cir-cuit in Figure 9-12(c) reveals a difficulty: Because of the presence of Coc, the cir-cuit is not unilateral. Although we could write two nodal equations (at vcm and vl)and solve them for the transfer function, it will be easier and far more intuitive tofirst make the circuit unilateral by using Millers theorem (presented inAside A9.3). To apply the theorem, we need to find the voltage gain from vcm to vl(i.e., the gain across Coc). Unfortunately, this gain does depend on Coc to some ex-tent, so the theorem does not simplify the analysis if we apply it exactly. Never-theless, if the current through Coc is small compared with gmvcm, the gain from vcmto vl is approximately independent of Coc , and we can expect the theorem to pro-vide results that are approximately correct while simplifying the analysis signifi-cantly. Removing Coc and defining , we can write

, (9.29)

which is the midband gain. Using this gain along with the formulas in (A9.14) and(A9.15), we can redraw the circuit as shown in Figure 9-13.The overall input capac-itance is denoted Cin and is equal to Ccm in parallel with the Miller input capacitancegiven by

, (9.30)

so that

CMin = 11 + gmRL2Coc

vlvcm

= -gmR L

RL = RO RL

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3 The relative contribution of each capacitor to the cutoff frequency depends on the equivalent resis-tance seen by each capacitor, too, but the difference in the capacitors is so large that the statementmade here is usually true.

4 Actually, strictly speaking, we ignore all capacitors other than CMin when making the Miller approxi-mation. In this example, Ccm is directly in parallel with CMin , though, so it would be silly not to includeit.Also, when more capacitors are ignored, we need to figure out some way to check whether or not theapproximation is still valid. We will deal with this complication later as the need arises.

. (9.31)

This input capacitance is usually dominated by CMin if the gain of the stage is rea-sonably large. The Miller output capacitance is then

. (9.32)

We can check to see whether or not the current through Coc is small comparedwith gmvcm by examining what fraction of the output current flows through CMout,which, owing to the Miller theorem, should be about the same as the currentthrough Coc . (It would be exactly the same if we used the frequency-dependentgain instead of the midband gain to find CMout.) At DC, none of the current flowsthrough this capacitor, and at v = 1>CMout the current through CMout is one-halfof the output current. (To see this, set the capacitive reactance equal to ).Therefore, for frequencies much below v = 1>CMout , using Millers theoremwith the midband gain will provide results that are reasonable for this circuit.

We can now find the high cutoff frequency of the circuit in Figure 9-13.We needto be careful when interpreting the results we obtain with this equivalent circuit,however. The circuit is unilateral, and each part is a single-pole circuit. Conse-quently, we are tempted to write the overall transfer function as a product of twoseparate single-pole transfer functions. Because CMin and CMout both model thesame element, however (Coc), the two single-pole circuits are not independent,and we cannot treat them as such. We will see in the next section how to estimatethe high cutoff frequency with both Cin and CMout included, but for now we notethat because Cin is much larger than CMout, it tends to dominate the cutoff fre-quency (i.e., it leads to a lower cutoff frequency).3 Therefore, the high cutoff fre-quency can be estimated by calculating the pole frequency caused by Cin alone inthe circuit of Figure 9-13. The result is

. (9.33)

When we use Millers theorem with the midband gain and also ignore CMout whilefinding vcH, we say we are making the Miller approximation.

4 We still use CMoutas just outlined to determine whether or not the approximation is valid. If 1> CMout is much higher than the cutoff frequency calculated using (9.33), then(9.33) is a reasonable approximation (i.e., the midband gain is a reasonable ap-proximation for frequencies near vcH). Equivalently, we can say that if

, (9.34)1vCH

W tMout

RL

vcH L11RS RCC rcm2Cin

RLRL

RL

CMout = 11 + 1>gmRL2Coc L Coc

Cin = Ccm + 11 + gmRL2Coc

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then the approximation is acceptable. (We have deliberately avoided plugging inthe formulas for our specific example, since we want an equation that is true ingeneral.) In practice, if 1>vcH is greater than three times tMout, the approximationis usually reasonable.

Note that the frequency at which half of the current goes through CMout is thepole frequency of the single-pole circuit on the output side of the equivalentcircuit in Figure 9-13. The word pole is in quotes because, as noted, there isnt apole at this frequency, since this capacitor is not independent of Cin. In fact, ifwe find the two poles of the original circuit and compare them with (9.33) and1> CMout, we see that the dominant pole given by (9.33) is reasonably accu-rate, but the second pole may be very far away from 1> CMout. Therefore, toavoid confusion, we prefer to state the test for the validity of the Miller ap-proximation in the form given by (9.34).

RLRL

Example 9.2

Problem:Use the Miller approximation to find fcH for the circuit in Figure 9-12(c), given thatRS = 1 k, RCC = 20 k, Ccm = 5 pF, rcm = 5 k, Coc = 1 pF, gm = 10 mA>V, and =5 k. Then use exact analysis to find both pole frequencies. Compare the dominantpole with fcH and the highest pole with 1> CMout.Solution:Using (9.31), (9.32), and the given data, we find that Cin = 56 pF and CMout = 1.02 pF.Using (9.33), we then find fcH L 3.6 MHz. Finally, we calculate

Using the pole-zero analysis option in HSPICE5 reveals that the two pole fre-quencies for this circuit are 3.2 MHz and 393 MHz. The lower pole agrees quitewell with the dominant pole found using the Miller approximation, but note thatthe higher pole is more than 10 times which clearlydemonstrates that you cannot treat the circuit of Figure 9-13 as two independentsingle-pole circuits.

1>12pRLCMout2 = 31.2 MHz,

31.2 MHz.1>12pRLCMout2 =

RL

RL

5 HSPICE is a product of Cadence Corporation.

As previously mentioned, Cin is dominated by Coc if the gain is reasonably large.Therefore, Coc determines the bandwidth when the gain is large. We can see intu-itively why this should happen if we look at the original circuit (Figure 9-11). SinceCoc appears from input to output, when the input voltage increases by an amountvI, the output voltage decreases by an amount |Av|vI, where Av is the voltagegain from input to output. (Av is negative.) The current through Coc is then largerthan would be seen for a capacitor of the same size connected to ground. In fact,the total change in voltage across Coc is vCoc = vI( 1 - Av), so the current in-creases by this same factor, and the effective capacitance seen to ground is also in-creased by this same factor.

Knowing that Coc is dominant in determining the high cutoff frequency for thiscircuit when the gain is large is important from a design point of view. If we wishto extend the bandwidth of the stage, we can either reduce the voltage gain, or re-duce Coc, or reduce the resistance seen by CMin.We will also discover in Section 9.6

SPENC09.201361833_513-658v13 6/28/02 1:37 PM Page 534


that we can significantly improve the bandwidth with the addition of anothertransistor. The sections on bipolar and field-effect implementations of this circuitshow that we can extend the bandwidth of this stage in other ways as well.

Exercise 9.6

Use the Miller approximation to find fcH for the circuit in Figure 9-12(c), giventhat RS = 50 , RCC = 20 k, Ccm = 5 pF, rcm = 5 k, Coc = 0.5 pF, gm = 0.01 A>V,and = 5 k. Is the Miller approximation valid in this case?

An important figure of merit for amplifiers is the gain-bandwidth product, orGBW. The GBW is a useful means for comparing amplifiers. The GBW is definedto be the product of the magnitude of the midband gain and the bandwidth of theamplifier. For a DC-coupled amplifier, the GBW is then

. (9.35)

The overall voltage gain has been used in (9.35), since the bandwidth is always cal-culated including the source resistance RS and is therefore the bandwidth from vSto vO. For AC-coupled amplifiers, it is almost always true that vcL V vcH , and(9.35) can still be used without significant error.

For the common-merge amplifier example, we can combine (9.29), (9.33), andthe input attenuation factor to find the GBW. The attenuation factor is given by

, (9.36)

where Ri is the small-signal midband AC input resistance. After combining theseequations, the result is

(9.37)

Making use of (9.31) and simplifying, we find that

, (9.38)

which is a particularly simple result that relates the GBW product to basic quanti-ties. Equations such as this are useful for design, because they clearly show us thelimitations of a given topology. The trade-off between gain and bandwidth is im-portant and is discussed further in Aside A9.4.

GBW LgmR L

RSgmR LCoc=

1RSCoc

GBW L a RiRi + RS

b

gmR L1RS RCC rcm2Cin L a RiRi + RS

b gmR L1RS Ri2Cin = gmR LRSCin .

ai =RCC rcm

RCC rcm + RS=

RiRi + RS

GBW = aiAv vcH

RL

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75 1 Fvi

+

+

vsCin

10 pFRL

1 krin

2.5 k

RS CC

1 pF

Cfbvo

gmvi

gm = 0.04 A >V

6 The results will always be conservative for the circuit we analyze, but if we have already used otherapproximations to simplify the original circuit prior to applying the open- or short-circuit time con-stants, the resulting bandwidth is not guaranteed to be conservative.

Exercise 9.7

Consider the amplifier whose small-signal AC equivalent circuit is shown inFigure 9-14. CC is a large coupling capacitor that contributes a DC zero and alow-frequency pole. Cin , on the other hand, contributes a zero at infinite frequen-cy and a high-frequency pole. Cfb is, as the notation indicates, a feedback capaci-tor. Since there is inverting voltage gain across Cfb , it will be multiplied by theMiller effect and will contribute to the high-frequency roll-off. (It also contributesa high-frequency zero). (a) Determine approximate values for fcL and fcH.(b) What is the GBW for this amplifier?

Advanced Method: Open- and Short-Circuit Time Constants We just foundthat estimating the low cutoff frequency with our existing methods is a cumber-some task for all but the simplest amplifier. In addition, the method used up tothis point does not often lend insight into the operation of the circuit, nor is it par-ticularly well suited for design. The method used for finding the high cutoff fre-quency, on the other hand, is quite intuitive and leads to reasonable results andequations suitable for use in the design process. However, not all amplifier stagescan be analyzed by applying the Miller approximation. Therefore, we now intro-duce a new set of tools for finding approximations to the cutoff frequencies: theopen- and short-circuit time constants.

The approximations we arrive at based on these time constants have three dis-tinct advantages:

1 The time constants are relatively straightforward to find; we can often do itby inspection.

2 The results are immediately useful for design, since they clearly show uswhich part of the circuitif anydominates the bandwidth.

3 The resulting estimates of the cutoff frequencies are conservative (i.e.,vcLis always lower than we predict and vcH is always higher than we predict).

6

On the downside, the techniques only work for circuits without inductors, al-though this is not a major restriction when dealing with amplifiers. Also, the ap-proximations depend on one pole being dominant for each cutoff frequency,otherwise there may be a relatively large error (around 20% is not uncommon).

We start by examining how to find a reasonable estimate of the low cutoff frequen-cy of a circuit with a bandpass transfer function (e.g., the amplifier in Figure 9-11).When we have finished with vcL , we will consider the high cutoff frequency,vcH.

Figure 914 Small-signal AC model of an amplifier.

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ASIDE A9.4 THE RELATIONSHIP BETWEEN GAIN AND BANDWIDTH IN AMPLIFIERS

Consider a DC-coupled amplifier with a single-pole transfer function given by

. (A9.16)

The gain-bandwidth product of this amplifier is defined to be the product of the low-frequency gain and thepole frequency, namely,

, (A9.17)

and is an important measure of the performance of the amplifier. A Bode magnitude plot1 of the transferfunction is shown in Figure A9-7 and consists of two asymptotes.

For frequencies well below vp, the gain is approximately constant and equal to A(j0). For frequencies wellabove vp, the gain is approximately given by

, (A9.18)

so that the magnitude drops in proportion to 1>v.On the high-frequency asymptote described by (A9.18), the magnitude of the product of the gain and fre-

quency at any given point is

. (A9.19)

If we can change the amplifier in a way that affects either the gain or the bandwidth, but that keeps the GBWconstant, or at least approximately so, we can see from (A9.19) that we can trade gain for bandwidth. For ex-ample, if the gain is reduced by a factor of two while keeping GBW constant, then the bandwidth is doubled.If the amplifier has a steeper roll-off of gain versus frequency (i.e., it has more than one pole in the frequen-cy range of interest), the tradeoff wont be one for one, but we can still trade gain for bandwidth. Chapter 10discusses this issue further.

1 We can also draw an asymptotic approximation to the phase. In that case, we say that the phase is approximately zero for frequenciesless than 1>10 of of the pole frequency and is approximately -90 for frequencies greater than 10 times the pole frequency. We approxi-mate the phase as linear between these two points, so that it passes through -45 right at the pole frequency. (For a review of Bode plots,see Aside AD.2 in Appendix D on the CD.)

A1jv2 v = A1j02 vp = GBW

A1jv2 L vpA1j02jv

GBW = A1j02vp

A1jv2 = A1j021 +

jvvp

p

A( j0)A( j)

Figure A9-7 The Bode magnitude plot of a sin-gle-pole transfer function.

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To find an approximation to vcL , we first assume that it is always possible todraw a low-frequency equivalent circuit, as in Figure 9-12(a), so that we are dealingwith a high-pass transfer characteristic. If we further assume that one of the polesin the circuit dominates the high-pass cutoff frequency, the problem reduces tofinding an approximation for the frequency of this dominant pole in the circuit.Since we are dealing with a high-pass characteristic, the dominant pole is thehighest frequency pole in the circuit. (Dont allow the preceding statement to con-fuse you:We are still looking for the low cutoff frequency of the original circuit, butbecause we are examining only the low-frequency equivalent,vcL is approximatelythe highest frequency pole in this equivalent.) We will examine the accuracy of theapproximation when there isnt a dominant pole later.We also must assume that allof the zeros in the transfer function are at frequencies well below the dominantpole, but this assumption usually is true in amplifier circuits.

Given that we have a low-frequency equivalent circuit, we find the short-cir-cuit time constant associated with each capacitor in the circuit and then sum thereciprocals of these time constants to find the approximate cutoff frequency. Inother words,

, (9.39)

where there are n capacitors in the equivalent circuit. The short-circuit time con-stant associated with capacitor Ci is

, (9.40)

and the short-circuit driving-point resistance seen by capacitor Ci is denoted Ris.This resistance is seen when you set all of the independent sources in the circuit tozero, remove Ci, short all of the other capacitors, and look back into the terminalswhere Ci was connected. A detailed justification for this approximation is given inthe companion section on the CD, but for now we can see that the result is rea-sonable by examining (9.39). If one of the capacitors does cause a dominant pole(i.e., it is substantially higher than all the other poles in the circuit), then for fre-quencies near this pole, all of the other capacitors will have negligible impedancesand can be approximated as short circuits. Making this approximation, we can findthe pole frequency of the resulting single-pole circuit by using the method present-ed in Aside A1.6. The resulting pole frequency will be the reciprocal of the short-circuit time constant for the dominant capacitor. In this circumstance, all of theother short-circuit time constants will be larger, and the sum in (9.39) will be dom-inated by this one time constant.

Be very careful here: There is a strong tendency to associate a pole with the re-ciprocal of each time constant, and doing so is completely wrong! For example, ifwe look at a capacitor that does not cause the dominant pole, for frequencies nearthe pole caused by this capacitor, the dominant capacitor will appear as an opencircuit, and the pole cannot possibly be given by the reciprocal of the short-circuittime constant. It is possible to find time constants by using combinations of open-and short-circuiting the remaining capacitors and, by so doing, find the nondomi-nant poles as well [9.3], but this method is not often used now that computer pro-grams can find the poles and zeros of a circuit.

An individual pole frequency is equal to the reciprocal of a time constant onlyfor a single-pole network. If the time constant associated with a capacitor can befound without needing to know what to do with other capacitors (i.e., whether to

tis = RisCi

vcL L an

i = 1

1tis

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RS

RINs

RCC rcm

Figure 916 Finding theshort-circuit driving-point resis-tance for CIN.


short them out or open-circuit them), the circuit containing the capacitor is a sin-gle-pole circuit (although it may be part of a much larger circuit), and there is apole directly associated with the time constant. Also, note that the short-circuittime-constant method finds an approximation to the highest frequency pole inwhatever circuit it is applied to; the designer must make sure that it is applied tothe low-frequency equivalent circuit if an estimate of vcL is desired.

The use of short-circuit time constants is best demonstrated by example.

Example 9.3

Problem:Find an estimate of the low cutoff frequency for the common-merge amplifierwith the low-frequency equivalent circuit shown in Figure 9-15.

100 2 Fvcm

+

+

vsRCC5 k

rcm2.5 k

RS CIN

RM1 k

gmvcm

CM100 F

RO2 k

RL1 k

1 F

COUT

Figure 915 The low-frequency equivalent circuit of a common-merge amplifier.

Solution:This circuit has three capacitors, and they cannot be combined in series or paral-lel, so we need to calculate three short-circuit time constants and use (9.39) to es-timate the low cutoff frequency.

To find the time constant for CIN, we set the independent source vs to zero, shortCM and COUT, and find the resistance seen looking back into the terminals whereCIN was connected. (What we do with COUT doesnt matter for this particular cir-cuit, but CM does.) The procedure is illustrated in Figure 9-16.

The result is

1,767 ,

which leads to a short-circuit time constant of

3.5 ms.

Finding the short-circuit time constant for COUT is similar. With CIN and CM short-ed and vs set to zero, we see that vcm = 0, so the controlled current source is anopen circuit, and

3 ms.tOUTs = ROUTsCOUT = 1RO + RL2COUT =

tINs = RINsCIN =

RINs = RS + RCC rcm =

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R SpencerThe rest of this chapter has been removed from this sample.


rcm

ic

Rx

RM1 k

aimic

RORL667

RSRCC98

RMs

(a)

rcm

ic

aimicRORLRSRCC

Rx

(b)

vx+

ix

Figure 917 Finding the short-circuit driving-point resistance for CM .

The short-circuit time constant for CM is found by using the circuit shown inFigure 9-17. We have used the current-controlled version of the transistor modelin this case because it simplifies the analysis.

From part (a) of the figure, we see that

.

We can find Rx as shown in part (b) of the figure. First note that by KCL,

.

Then write KVL to obtain

.

Finally, combining these two equations results in

.

Assuming that aim = 99 for this example yields Rx = 26 . Finally, we find that tM =RMsCM = 2.5 ms.

Now using these results along with (9.39) yields

1 krad>s.vcL L 1tINs + 1tOUTs + 1tMs =

Rx =vxix

=RS RCC + rcm

1 + aim

vx = - ic1RS RCC + rcm2ic11 + aim2 = - ix

RMs = RM Rx

Lets compare the approximate result found in Example 9.3 with what is ob-tained by computer simulation. A simulation of the circuit in Figure 9-15 revealsthat vcL = 754 rad>s, which is 25% lower than the estimate (Exam3.sch).While thisis a very large error, the result is still useful for three reasons. First, the approxi-mation was conservative, so the design will still work (i.e., we usually dont mind ifthe bandwidth of an amplifier is a bit larger than we designed it to be). Second, theanalysis still shows us how to increase or decrease vcL and how much a change inany given capacitor or driving-point resistance will affect vcL. Third, we do notoften require an extremely accurate estimate of bandwidth.

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im'rm

aicim'vi vo

ZM

ZO

Figure 918 The small-signalmodel of a common-mergestage.


7 The results are conservative for the circuit we analyze. Of course, we are usually analyzing an approx-imation to a real circuit, so the results are not guaranteed to be conservative in practice, although theyusually are if our approximations are acceptable.

Even though our result is useful, the error is larger than we would like. Howev-er, the particular example given here has a large error because there isnt a domi-nant pole. In fact, exact analysis reveals that the poles of this circuit are at 66 rad>s,333 rad>s, and 612 rad>s.The highest frequency pole is only about twice the secondpole frequency; therefore, we expect a rather large error with this technique.When the short-circuit time constants are used to estimate vcL, the error is typi-cally between 10 and 25%, but the resulting estimate is conservative.7 The errorcan be very small if one pole is truly dominant, but that is not usually the case forvcL.

As a final step, we should check that the zeros in the low-frequency equivalentcircuit are all low enough to be safely ignored. We already have one confirmationof this fact, since our analytical result agrees quite well with the simulated result.In practice, this agreement would probably be sufficient. Nevertheless, for com-pleteness, we take a moment to examine the zeros in the circuit.

Both CIN and COUT contribute zeros at DC, because they are in series with thesignal and, as the frequency goes to zero, they become open circuits and cause thegain to go to zero. The emitter bypass capacitor is a bit more troubling, but canstill be handled in a very intuitive way. Consider the small-signal model of a com-mon-merge stage shown in Figure 9-18. We have used the midband T model andshown the external components in series with the output and merge terminals asgeneral impedances. The implicit assumption is that we are examining frequen-cies low enough that the transistors internal capacitances can be ignored. Thegain of this circuit is

. (9.41)

If we consider the effect of CM alone in our circuit, the output impedance is pure-ly real and equal to RO, whereas the merge impedance is RM in parallel with CM.At DC, the capacitor is an open circuit and the gain is

. (9.42)

As the frequency goes to infinity, the capacitor becomes a short and the gain in-creases to

. (9.43)Vo 1jq2Vi 1jq2 = -aicROrm = -gmRO

Vo 1j02Vi 1j02 = -aicROrm + RM

Vo 1jv2Vi 1jv2 = -aicZO 1jv2rm + ZM 1jv2

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z p

gmRO

ROrm + RM

VoVi

Figure 919 A plot of the approximatemagnitude versus frequency for the circuitin Figure 9-18 when ZO = RO and ZM =RM || CM.

100 1 Fvcm

+

+

vsRCC

20 krcm

2.5 k

RS CIN

RMB820

RMA270

gmvcm

CM47 F

RO4.7 k

RL10 k

1 F

COUT

Figure 920 The small-signal low-frequency AC equivalent circuit for a common-merge am-plifier.

Therefore, we intuitively see that we have a zero followed by a pole. We can findthe separation between the pole and zero frequencies by using a plot of the trans-fer function magnitude versus frequency, as shown in Figure 9-19. Note that thisplot uses linear axes and an asymptotic approximation for the magnitude (i.e., thestraight-line asymptotes).

The slope on the plot in between the pole and zero is unity, so the ratio of thefrequencies is equal to the ratio of the gains:

. (9.44)

If RM W rm, the pole and zero are widely separated, and our approximationshould be fine. Note that we can also derive (9.44) directly from (9.41) by pluggingin RO and for the impedances. After a bit of algebra, the result is

, (9.45)

which leads to (9.44) as well.

Exercise 9.8

Consider the small-signal low-frequency AC equivalent circuit shown in Figure 9-20.This circuit represents a common-merge amplifier and is similar to the one analyzedin Example 9.3. Assume that aim = 99, and find an estimate of fcL by using short-cir-cuit time constants.

Vo 1jv2Vi 1jv2 = -aic 1RO>rm21jv+1>RMCM2jv+1>1RM rm2CM

RM CM

vp

vz=

gmRORO>1rm + RM2 = gm1rm + RM2 L rm + RMrm

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Now we can consider how to find an approximation to the high cutoff frequen-cy, vcH. In this case, assume that it is always possible to draw a high-frequencyequivalent circuit as in Figure 9-12(c). The high-frequency equivalent will alwayshave a low-pass transfer characteristic, and if there is a pole that is significantlybelow all the other poles, then it is dominant and vcH is approximately equal tothat pole frequency. Since we are dealing with a low-pass characteristic, the domi-nant pole is the lowest frequency pole in the circuit. (We are looking for the highcutoff frequency of the original circuit, but since we are examining the high-fre-quency equivalent, vcH is approximately the lowest pole frequency in this equiva-lent.) Later we will examine the accuracy of our approximation when there isnt adominant pole. For the approximation to be reasonable, we must assume that allof the zeros in the transfer function are well above vcH, which is usually true foramplifiers.

Given a high-frequency equivalent circuit we find the open-circuit time constantassociated with each capacitor in the circuit and then take the reciprocal of the sumof these time constants to find the approximate cutoff frequency. In other words,

, (9.46)

where there are n capacitors in the equivalent circuit, the open-circuit time con-stant associated with capacitor Ci is

, (9.47)

and the open-circuit driving-point resistance seen by capacitor Ci is denoted Rio.This resistance is the resistance seen when you set all of the independent sourcesin the circuit to zero, remove Ci, open all of the other capacitors, and look backinto the terminals where Ci was connected. A detailed justification for this ap-proximation is given in the companion section on the CD, but we can see that theresult is reasonable by examining (9.46). If one of the capacitors does cause adominant pole (i.e., it is substantially lower than all the other poles in the circuit),then for frequencies near this pole, all of the other capacitors will have very largeimpedances and can be reasonably approximated by open circuits. Making thisapproximation, we can find the pole frequency of the resulting single-pole circuitby using the method presented in Aside A1.6.The resulting pole frequency will bethe reciprocal of the open-circuit time constant for the dominant capacitor. Sinceit is the largest time constant by far, it will dominate the sum in (9.46), and thepole will be approximately the same as vcH given by (9.46).

This result is extremely important. Using (9.46), we can estimate the high cutofffrequency of a circuit far more readily than we can derive the entire transfer func-tion. In addition, the resulting equations are much easier to deal with and allow usto clearly see the contributions of the individual capacitors. We must again resistthe temptation to associate a pole with the reciprocal of each time constant. Thismethod is useful only for approximating the cutoff frequency; it will not provideinformation about nondominant pole frequencies. Finally, we reiterate that theopen-circuit time constant method finds an approximation to the lowest frequen-cy pole in whatever circuit it is applied to; the designer must make sure that it isapplied to the high-frequency equivalent circuit if an estimate of vcH is desired.

This section concludes with an example of the usage of open-circuit time constants.

tio = RioCi

vcH L1

an

i = 1tio

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vcm

+

+

vsRCC5 k

Ccm10 pF

rcm2.5 k

RS100 Coc

2 pF

gm vcm

vo

RORL667

Figure 921 A common-merge amplifierhigh-frequency equivalent circuit.

Rcmo9.43

vcm

+

ix

gm vcm

RORL667

vx

io

+

Figure 922 Finding the open-circuitdriving-point resistance for Coc.

Example 9.4

Problem:Find an estimate of the high cutoff frequency for the common-merge amplifierwith the high-frequency equivalent circuit shown in Figure 9-21.This is a repeat ofFigure 9-12(c), but with component values included.

Solution:This circuit has two independent capacitors, so we need to calculate two open-cir-cuit time constants and then use (9.46) to estimate the high cutoff frequency.

To find the time constant for Ccm, we set the independent source vs to zero, openCoc , and find the resistance seen looking back into the terminals where Ccm wasconnected. The result is

94.3 ,

which leads to an open-circuit time constant of

943 ps.

The open-circuit time constant for Coc is found by using the circuit shown inFigure 9-22.

tcmo = RcmoCcm =

Rcmo = RS RCC rcm =

Writing KCL at the output node, we find that

.

We next write KVL to obtain

,

and solve to find

Assuming that gm = 0.1 A>V for this example yields Roco = 7 k. Finally, we findthat toco = RocoCoc = 14 ns.

Now using these results along with (9.46) yields

67 Mrad>s.vcH L 1tcmo + toco

=

Roco =vxix

= CRcmo + 11 + gmRcmo21RO RL2 Dvx = ixRcmo + io1RO RL2 = ix CRcmo + 11 + gmRcmo21RO RL2 D

io = ix + gmvcm = ix11 + gmRcmo2

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+

vi rcm RLgmvcm

voCoc

vcm

+

Figure 923 A simplified circuit for seeing thezero due to Coc.

In this last example, the time constant for Coc is dominant because of the Millereffect. In fact, if you reexamine the equation for Roco , you will note that it can berewritten as

, (9.48)

where Avmb is the midband gain across Coc. Multiplying the final term in (9.48) byCoc to get the time constant, we see that the result is equivalent to multiplying Cocby -Avmb to get the Miller input capacitance and then multiplying that by theequivalent resistance seen by CMin. Therefore, the results achieved by this approx-imation are very nearly the same as those achieved by the Miller approximation ifCoc is dominant. If Coc is not dominant, the open-circuit time constant method willusually yield more accurate results.

If we solve for the cutoff frequency in the high-frequency equivalent of Figure 9-21exactly or use SPICE, we find that vcH = 67 Mrad>s, as estimated (Exam4.sch). Thereason that the estimate is so accurate this time is that there is, in fact, a dominantpole in the circuit. In general, the open-circuit time-constant method yields estimatesthat are between 0 and 25% lower than the true cutoff frequency. The estimate isagain conservative, so even the larger errors are often not a serious problem.

Since our results agree well with SPICE, we could stop here and assume that thezeros must be enough higher than the dominant pole that the approximation isreasonable in this case.We will look at the zeros for completeness, however, ratherthan relying solely on our agreement with the simulator. Examining Figure 9-21,we note that Ccm produces a zero at infinity, since it forces the gain to zero as itshorts out rcm.To see what Coc does to the response, we examine the simplified cir-cuit shown in Figure 9-23.

At DC, Coc is an open circuit and the gain of this circuit is -gmRL. As the fre-quency goes to infinity, the gain becomes unity; that is, vo = vi. The resulting mag-nitude plot is shown in Figure 9-24, where we have used an asymptoticapproximation (i.e., the straight-line asymptotes) and linear scales on both axes.The ratio of the pole and zero frequencies can be found by recognizing that theslope of the middle asymptote on the plot is -1. The result is

. (9.49)vz

vp= gmRL

Roco = CRcmo + 11 + gmRcmo21RO RL2 D L gm1RO RL2Rcmo = -AvmbRcmo

p z

1

gmRL

VoVi

Figure 924 The asymptotic magnitude plotfor Figure 9-23.

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+

vS

vI

R210 k

R140 k

RL1 k

RC2.2 k

RE650

RS

VE = 1.3 V

VC = 5.6 V

VB = 2 V

CE

COUT vO

CINIC = 2 mA

VCC = 10 V

100

Figure 925 A common-emitter amplifier. (This is the same circuit as in Figure 8-33.)

+Vs( j) RBB

r

RS CIN

Ib

CERE

RLRC

Vo( j)Ib( j)Vb( j)COUT

Figure 926 The small-signal low-frequency AC equivalent circuit for the common-emitteramplifier of Figure 9-25.

Therefore, so long as the midband gain from control to output is much larger thanunity (in practice, four or more is usually good enough), this zero should be wellabove the pole caused by Coc and should not affect our results.

9.2.2 A Bipolar Implementation:The Common-Emitter AmplifierWe analyze the frequency response of the common-emitter amplifier in two ways.The introductory treatment uses the Miller approximation and standard analysistechniques. The advanced treatment requires an understanding of open-circuitand short-circuit time constants.

Introductory Treatment Consider the common-emitter amplifier shown inFigure 9-25. We already analyzed the DC biasing and small-signal midband ACperformance of this circuit in Section 8.2 (see Figure 8-33), where we determinedthe bias voltages and currents shown in the figure.

We now want to find both the low and high cutoff frequencies for this circuit.Tofind the low cutoff frequency, we draw the small-signal low-frequency AC equiva-lent circuit as shown in Figure 9-26.

Writing KVL around the base-emitter loop, we obtain

, (9.50)

where

. (9.51)ZE1jv2 = RE1 + jvRECE

Vb1jv2 - Ib1jv2rp - 1b+12Ib1jv2ZE1jv2 = 0

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+Vs ( j)

Vi (j)

RBB r

RS CIN

Ib = gmV

CERE ( + 1) + 1

RLRC

Vo( j)Ib(j)Vb( j)COUT

V( j)+

Figure 927 The circuit of Figure 9-26 with the emitter impedance reflected into the base.

8 The b used here is the midband value, b 0, but we ignore the subscript for notational simp

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