chapter 7 amplifier frequency response

7/29/2019 Chapter 7 Amplifier Frequency Response

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Chapter 7

Amplifier Frequency Response

_____________________________________________

7.0 Introduction

The voltage gain of amplifier is a function of input signal, which is dependent

on its magnitude and frequency. In the previous chapter on the analysis of small

signal amplifiers of bipolar junction transistor BJT, junction field effect

transistor JFET, and metal oxide field effect transistor MOSFET, the effect of

capacitive reactance of coupling, bypassing capacitors, and internal capacitanceof the transistor to the gain of amplifier is assumed to be zero at mid-range

frequency. Thus, it does not cause any effect to the gain of the amplifier circuitand phase shift. However, owing to the presence of bypass and coupling

capacitors in the amplifier circuitry, the effect to the gain of amplifier can be

significant at low frequency. Similarly, the effect to the gain can be significant

at high frequency too due to presence of internal capacitance of the BJT, JFET

or MOSFET devices. Figure 7.1 shows the response of amplifier gain with the

frequency of input signal. The response range can be divided into low-

frequency range, mid-frequency range, and high-frequency range. fL is the lowcritical frequency where the effect of low frequency begins to be significant.

Likewise fH is the high critical frequency where the effect of high frequency

begins to be significant.

Figure 7.1: Amplifier gain versus frequency


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7 Amplifier Frequency Response

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7.1 Effect of Capacitance

As mentioned earlier, the effect of capacitance to the gain of amplifier is

significant at both low and high frequencies. The effects of bypass anddecoupling/coupling capacitance are significant at low frequency, whilst theeffects of internal capacitance are significant at high frequency. The study of the

effect shall be dealt with accordingly.

7.1.1 Bypass and Coupling Capacitors

The small signal amplifier circuits of common-emitter BJT and common-source

JFET transistors are shown in Fig. 7.2.

(a) Bipolar (b) JFET

Figure 7.2: The typical capacitively coupled BJT and JFET type amplifiers

The voltage gain AV at low frequency is lower because the reactance

contribution of capacitor is larger. In contrast, the opposite is true for high

frequency. Reactance of capacitor C1 is in series with resistor R1||R2 or RG andRs or Rss, which reduces the signal at base or gate. Reactance of capacitor C3 is

parallel to resistor RE and RS, which effectively increases the voltage at emitterand source, which in turn reduces the voltage gain AV and causes phase shift. As

frequency decreases, the effect is higher. Thus, capacitor C3 and C2 also act as

high-pass filter elements for the amplifier circuit.

At high frequency, internal capacitances of the device become dominant

that they reduce the gain of amplifier and cause phase shift as frequency

increases. Thus, these capacitances also act as low-pass filter element.

The bypass and coupling capacitors have values in the micro-farad range.

Thus, at high frequency, these capacitors appear to be short.


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7.1.2 Internal Capacitance of Device

To understand the effect of high frequency to the gain of an amplifier, one needs

to know the internal capacitive components of the BJT and JFET/MOSFETtransistors. The internal capacitance of the BJT and JFET/MOSFET transistor

are basically formed from the junction depletion region and gate insulation layer

of the device. The internal capacitance can be classified into two groups namely

the input and output capacitance. Cbe and Cgs are input capacitance formed

between the base and emitter, and gate and source junction, whilst Cbc and Cgd

are the output capacitance formed from the base and collector, and gate and

drain junction. Fig. 7.3 illustrates the internal capacitive components of BJT and

JFET/MOSFET transistor.

The internal capacitance has magnitude in pico to nano farad range. Thus,

they appear as open circuit at low frequency.

(a) Bipolar (b) JFET/MOSFET

Figure 7.3: Internal capacitive components of (a) BJT and (b) JFET/MOSFET transistor

Figure 7.4 illustrates the loading effect and gain effect of internal capacitance to

a specified amplifier. The effect of Cbe capacitance has reduced the input Vin

because its reactance Cbe forms the voltage-divider circuit.

The output capacitance Cbc forms a negative feedback path that reduces the

input signal because it is about 1800

out of phase.

Figure 7.4: ac equivalent circuit showing the effect of capacitance Cbe and Cbc


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In dealing with high frequency response where internal capacitance is

important, Miller's theorem can be used to simply the analysis due to the effect

of Cbc capacitance of BJT or Cgd capacitance of JFET/MOSFET transistor. The

capacitance Cbc of BJT or capacitance Cgd of JFET/MOSFET transistor can beresolved as Miller input Cin(Miller) and Miller output Cout(Miller) capacitances,

whereby Cin(Miller) = Co(1 + AV) is for the input side and Cout(Miller) = Co

+

V

V

A

1Ais

for the output side. Co is the output internal capacitance, which is either Cbc orCgd and Av is the mid-range frequency gain of the amplifier. Figure 7.5 shows

all the effective capacitances of an amplifier circuit after using Miller's theorem.

(a)Biploar junction transistor

(b) Junction field effect transsitor

Figure 7.5: ac equivalent circuit showing effective capacitances (a) for BJT and (b) for JFET

From the circuits shown in Fig. 7.5, capacitance Cbc or Cgd have greater impact

on input capacitance than its actual value since it has a (1 + Av) factor. As for

the output side, the effective output capacitance is approximately equal to Cbe orCgd since the factor (1+AV)/AV is closed to one for high voltage gain Av.


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Note that Miller theorem is applicable to the circuit that contains

component connected between the input and output of the circuit and there is

reversal of phase between the input and output signals.

7.2 Bode Plot

The voltage gain AV of the amplifier is usually expressed as the transfer

function T(s) of complex frequency s-domain. In the s-domain analysis, the

transfer function is defined as)ps)....(ps)(ps(

)zs)....(zs)(zs(K

)s(V

)s(V)s(T

n21

m21out

==

in

, where s is

equal toj. K is a constant, z1, z2, z3, ...., zm are transfer function "zeros" and p1,p2, p3, ...., pn are transfer function "poles". In the form, the resistance is R,

whilst impedance of capacitance C is replaced by 1/sC and the impedance ofinductor is replaced by sL.

The Bode plot is the plot of magnitude of voltage gain AV versus frequency

plot expressing voltage gain AV in decibel and frequency in logarithmic base 10.

Thus, the magnitude of transfer function is T|(j)|dB = 20 log10|T(j)|.

Short circuit and zero valuemethodor Open circuit and zero value methodor complex s-domain analysis can be used to analyze each RC network and to

determine its critical frequency. s-domain is a better approach because all thezeros and poles can determined. If the zero is dominant, then the -3 dB critical

frequency determined by Short circuit and zero value method becomes

incorrect. If there is zero in s-domain analysis, this shall mean the amplifier

circuit contains capacitive path directly coupling input and output. Open circuit

and zero value method is be used for the case where there are more than one

internal capacitor connected to the ground.

7.3 Gain-Bandwidth of a Bipolar Junction Transistor

As one already knows that the high frequency response of amplifier is affected

by internal capacitance C, is also called base-to-emitter capacitance Cbe, C is

also called base-to-collector capacitance Cbc, and base-to-emitter capacitance

Cce. A merit term named as unity-gainfrequency or transition frequency is used

to determine the frequency capability of the transistor. At unity gain frequency,

the product of gain and frequency is termed as gain-bandwidth. The ac

equivalent circuit of a common-emitter short circuit bipolar junction transistor

shown in Fig. 7.6 is used to determine this parameter. This is done by shortingthe output and connecting a current source to the input.


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Figure 7.6: ac equivalent circuit of a common-emitter short circuit transistor

The voltage at base-to-emitter Vbe is Vbe =

C

1||

C

1||ri

jj

i =

i)CC(r1

ri

j

++. Similarly, the current at output is io = -gmVbe. Thus, the

current gain (j) = io/ii is

(j) =)CC(

g1

)CC(r1

rg

m

L

Lm

i

o

+

+

=

++=

jji

i(7.1)

where L = gmr is the mid-range frequency current gain. By setting the

denominator of equation (7.1) equal to zero, the -3dB critical frequencyf is

f =)CC(2

g

L

m

+(7.2)

By setting |(j)| =

L

L

mg

C C1

2

+ +

( )

= 1, which is the unity-gain and

assuming 1/L is approximately zero, the transition frequencyfT or unity-gainfrequency is equal to

fT =g

C C

m

2 ( )+(7.3)

From equation (7.2) and (7.3), relationship of -3dB critical frequency and the

unity-gain frequency isf =

fT

L .


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Using the similar approach, the transition frequency fT or unity-gain

frequency of FET is equal tofT =)CC(2

g

gsgd

m

+.

7.4 Low-Frequency Analysis of A Common-Emitter

Bipolar Junction Transistor Amplifier

A typical amplifier with coupling and bypass capacitors is shown in Fig. 7.7 and

its low frequency T-model and -model ac equivalent circuits are shown in Fig.7.8 and Fig. 7.9 respectively. Without considering the effect of bypassing and

coupling capacitance, and internal capacitance, the mid-range frequency voltage

gain AV of this amplifier is AV = + +

r

R R

R R

R

R Re

L C

L C

in

in S , where Rin is Rin =

R1||R2||r, which has been shown in equation (3.44) of Chapter 3. Since1+

= ,

+= r)1(re , and

=

rg m , AV = +

+

r

R R

R R

R

R Re

L C

L C

in

in S

is also equal to

S21

21

CL

CLm

Rr||R||R

r||R||R

RR

RRg

+

+

.

Figure 7.7: A typical amplifier with coupling and bypassing capacitors


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If the bypass and coupling capacitive effects are taken into consideration, the

low frequency T-model and -model ac equivalent circuits of the amplifier shallbe as shown in Fig. 7.8 and Fig. 7.9 respectively.

Figure 7.8: Low-frequency T-model ac equivalent circuits of the common-emitter amplifier

circuit shown in Fig. 7.7

Figure 7.9: Low-frequency -model ac equivalent circuits of the common-emitter amplifiercircuit shown in Fig. 7.7

There are altogether three RC networks in the circuits. One is at input side and

two are at output side.

7.4.1 Input RC Network

After applying Short circuit method, where capacitor C3 is shorted, the low

frequency RC network circuit of the input due to capacitor C1 is shown in Fig.


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7.10. Note that input resistor r is also equal to re( + 1), where re is the acemitter resistance.

Figure 7.10: Low frequency RC network of input due to capacitor C1

The voltage seen at the base VB is equal to

in2

C

)base(in21in

1

)base(in21B V

1R

R||R||RV

RC

11R

R||R||RV

+

=

+

=

f

fj

(7.4)

where R = (R1 || R2 || Rin(base) + RS),1

CRC2

1

=f and Rin(base) = ( + 1)re = r.

When the frequencyfis equal tofC, at the point where resistance R is also

equal to reactance1C

equation (7.4) becomes

==

2R

R||R||R

V

VA

)base(in21

in

BV =

+ )RR||R||R(

R||R||R707.0

S)base(in21

)base(in21

(7.5)

This shall mean that the voltage gain is attenuated to 70.7% of the input signal

Vin or in terms of decibel, it is -3dB. The -3dB attenuation is also known as half

power point for the amplifier since dB = 20log10(AV) and dBm = [ ]PAlog10 =

[ ])mW0.1/Plog10 out . Note that dBm is referenced to 1.0mW.


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At the condition where the real impedance is equal to imaginary impedance

i.e. R = C1 =1C

1

, the frequency is called critical frequencyfc or corner or

break frequency. The input critical frequency fc(input) is determined by setting

+

RC

11

1j= 0, which is

fc(input)1RC2

1

= (7.6)

In the s-domain analysis, RC is also defined as time constant s.

Based on equation (7.4), if the voltage gain AV is one tenth of the inputsignal Vin, then the signal at output Vout is attenuated by -20dB. This also

implies that the frequencyfis equal to 0.1fC. This attenuation factor is known as

-20dB/decade or -6dB/octave. If the voltage-gain AV is one hundredth of the

input signal Vin, then the output signal is attenuated by - 40dB. This also implies

when the frequencyfis equal to 0.01fC. Figure 7.11 shows the Bode plot of the

low frequency response. Sometime -6dB/octave is used for every doubleincrease of frequency or decrease of critical frequency.

Figure 7.11: Bode plot of low frequency response

In addition to the voltage gain being reduced, the RC network also causes an

increase of phase shift through the amplifier as the frequency decreases. For

high-pass filter, the output voltage leads the input by phase , which is


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=

=

f

fc1

1

1

)input( tanRC

1tan (7.7)

Figure 7.12 illustrates the phase relationship of input and output signals.

Figure 7.12: Phase angle relationship between input and output signals

As frequency decreases, the phase increases, which is shown in Fig. 7.13 and

equation (7.7). At -3dB point, the phase is - 45o.

At -20dB point, the phase is - 84.3o. Thus, at - 5.7o phase , the frequencyf is equal to 10fC.

Figure 7.13: Bode plot of frequency and phase


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7.4.2 Output RC Network

As mentioned earlier, there are two RC networks at output, one at emitter and

one at collector. The low frequency RC network looking at the emitter is shownin Fig. 7.14.

Figure 7.14: The low frequency RC network due to capacitor C3

The deduced RC network looking at emitter is shown in Fig. 7.15.

(a) (b)

Figure 7.15: The deduced RC network at emitter

RTH is the equivalent resistance of resistor R1||R2||RS. Converting resistance RTH

from base to emitter resistance using Impedance Reflection Rule, its value

becomes RTH/(+1). This resistance value is now in series with the ac emitter

resistance re. Thus, the actual emitter resistance shall be equal to [RTH/(+1) +re]||RE. From the earlier definition, the critical frequency for emitter RC network

shall be


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fc(bypass) =1

2 1 3 (( / ( )|| )r R R Ce TH E+ +(7.8)

and the phase is defined as

(bypass) =( )

++

ETHe3

1

R||)]1/(Rr[(C

1tan (7.9)

The equivalent RC network at collector's decoupling capacitor and its

Thvenins equivalent circuit are shown in Fig. 7.16. The collector resistance

RC is shown as separated by the load resistance RL by the bypassing capacitance

C2.

(a) Nortons circuit (b) Thevnins circuit

Figure 7.16: The deduced RC network at collector due to capacitor C2

The voltage gain at output is Vout/Vbe =)]RR(C/[11

1

RR

RRg

CL2CL

CLm

++

+

j. Thus

the critical frequency for collector coupled capacitor C2 is

fc(output) =1

2 2( )R R CL C+(7.10)

and the phase is defined as

(output) =( )

+

CL2

1

RRC

1tan =

f

f )out(c1tan (7.11)

In summary, the low-frequency response function of the common-emitter BJTamplifier is equal to


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|AV(j)| LF =Sin

in

CL

CLm

RR

R

RR

RRg

+

+

+

2

)input(C1

1

f

f

2

)bypass(C1

1

+

f

f

2

)outputt(C1

1

+

f

f

(7.12)

Note that if the capacitors are treated as short then the gain of the amplifier

becomes AV =Sin

in

CL

CLm

RR

R

RR

RRg

+

+ , which is the mid-range frequency gain.

7.5 High-Frequency Analysis of A Common-Emitter BJTAmplifier

The high-frequency T-model and hybrid -model ac equivalent circuit of theBJT amplifier after applying Miller's theorem is shown in Fig. 7.17 and Fig.

7.18 respectively. Note that capacitance Cce is intentionally ignored due to its

small magnitude as compared to Miller capacitance at output Cout(Miller). Millers

theorem can be applied because C or Cbc is connected between input and output

and there is phase inversion between input and output signals.

Figure 7.17: T-model high-frequency ac equivalent circuit for a common-emitter BJT

amplifier after applying Millers theorem


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Figure 7.18: -model high-frequency ac equivalent circuit for a common-emitter BJTamplifier after applying Miller theorem

The input RC network is consist of R1||R2||RS||r and (Cin(Miller) + Cbe). The

critical frequency is

fH(input) =)CC)(r||R||R||R(2

1

be)miller(inS21 + (7.13)

The output RC network is consist of RC||RL and Cout(Miller) Thus, the critical

frequency is

fH(output) =)CC)(R||R(2

1

ce)Miller(outLC +(7.14)

As frequency increases, AV decreases. Thus, the high-frequency RC network is

also acted as low-pass filter element.

In summary, the transfer function of high-frequency response for the

common-emitter BJT amplifier is equal to

|Av(j) HF =Sin

in

CL

CLm

RR

R

RR

RRg

+

+

+

2

H(input)

1

1

f

f

2

H(output

1

1

+

)f

f

(7.15)


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Note that if the capacitors are treated as open then the gain of the amplifier

becomes AV =Sin

in

CL

CLm

RR

R

RR

RRg

+

+ , which is the mid-range frequency gain.

Example 7.1

The amplifier circuit shown in Fig. 7.7 has VCC = 12V, = 100, R1 = 60k, R2

= 20k, RS = 600, RC = 2k, RE = 1k, RL = 10k, C1 = 0.1F, C2 = 0.1F,

C3 = 4.7F, Cbe = 20pF, and Cbc = 3pF. Calculate the gain, all the criticalfrequencies of this amplifier, draw its Bode plot, and find out its bandwidth.

Solution

The base input impedance RIN(base) = ( + 1)RE = 101(1k) = 101k. This value

is very large than R2 = 20k. Thus, we can ignore this value in the calculation.

Voltage at base VB =R

R RV

k

k kV VCC

2

1 2

20

20 6012 3

+ =

+ =

. Thus, voltage at

emitter VE = VB - VBE = 3.0V - 0.7V = 2.3V. Emitter current IE shall then equal

to VE/RE = 2.3V/1k = 2.3mA.

Knowing the value of IE, ac emitter resistance is re = VT/IE = 25mV/2.3mA =

10.8.

The ac base input impedance Rin(base) = ( + 1)re = 101x10.8 = 1.09k. Thus,the ac input impedance Rin = Rin(base)|| R2||R1 = 1,007.

The ac voltage gain of the amplifier is AV = +

+

r

R R

R R

R

R Re

L C

L C

in

in S

= -95.74

From equation (7.6), the low-frequency critical frequency at input fc(input) =1

2

1

2 1607 019900

1 ( ) ( )( . ).

R R C FHz

in S+= =

.

From equation (7.8), the low-frequency critical frequency for bypass RC

network at emitter is fc(bypass) =1

2 1 3 (( / ( ))|| )r R R Ce TH E+ +. RTH = R1||R2||RS =

576.9, re + RTH/(+1) = 10.8 + 576.9/101 = 16.51 and (re +

RTH/(+1))||RE = 16.24, thenfc(bypass) = Hzk09.2F7.4)2.16(2

1=

.

From equation (7.10), the low-frequency critical frequency for RC network at

collector isfc(output) =

1

2 2( )R R CL C+ =

1

2 2 10 01 132 6 ( ) . .k k F Hz + = .


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The high-frequency Cout(Miller) = Cbc(AV + 1)/AV = 3pF(95.74+1)/95.74 = 3.03pF

and Cin(Miller) = (AV + 1)Cbc = 96.74 x 3pF = 287.22pF.

From equation (7.12), the high-frequency input critical frequency fH(input) =1

2 1 2( || || || )( )( ) ( )R R R R C CS in base in miller be+=

1

2 394 7 307 221312

( . )( . ).

pFMHz= From

equation (7.13), the high-frequency output critical frequency fH(output) =1

2( || )( )( )R R CC L out miller=

1

2 1666 7 303315

( . )( . ).

pFMHz= .

The Bode plot of this amplifier is shown in Fig. 7.16. Between critical

frequency 2.09kHz and 990Hz, the gain drop is

ln( . /2 09 20kHz / 990Hz)ln(10)

dB decade

= 6.49dB or ln( . /2 09 6kHz / 990Hz)

ln(2)dB octave

=

6.46dB. The gain drop between frequency 990.0Hz and 132.6Hz is 34.9dB in

40 dB/decade and the gain drop between frequency 31.6MHz and 1.31MHz is

27.6dB in 20dB/decade.

The bandwidth BW is 1.31MHz 2,090Hz = 1.308MHz. From the Bode plot, itmeans that the maximum voltage gain of the amplifier is -3dB for frequency

range from 1,100Hz to 1.31MHz. Beyond this range, the voltage gain is

attenuated at 20dB/decade either increase or decrease of ac frequency.

Sometime the attenuation is expressed in 6dB/octave which shall mean the

attenuation for every double increase or decrease of critical frequency.


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7.6 Low-Frequency Analysis of A Common-Source

MOSFET Amplifier

A typical common-source DE MOSFET amplifier circuit is shown in Fig. 7.19.

It has two RC networks. One is at input side and one is at output side. The mid-

range frequency voltage gain is defined as AV = +

gR R

R RmD L

D L

.

Figure 7.19: A typical common-source DE MOSFET amplifier

The input RC network due to coupling capacitor C1 of the amplifier is shown in

Fig. 7.20.

Figure 7.20: Input RC network of the common-source amplifier shown in Fig. 7.19


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From JFET theory, gate input impedance is RIN(gate) =V

I

GS

GSS

. Thus, the low-

frequency critical frequency of input RC network is

fC(MOS-gate) =1)gate(ING C)R||R(2

1

(7.16)

The output RC network circuit due to capacitor C2 is shown in Fig. 7.21.

Figure 7.21: Output RC network of amplifier shown in Fig. 7.19

The low-frequency critical frequencyfC(FET-output) of input RC network is

fC(FET-output) =1

2 2( )R R CD L+ (7.17)

7.7 High-Frequency Analysis of A Common-Source JFET

Amplifier

The high-frequency ac equivalent circuit of a typical JFET amplifier is shown inFig. 7.22.

Figure 7.22: High-frequency equivalent circuit for a JFET amplifier


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Although there is a source resistance RS, nevertheless the voltage gain is

approximately equal to AV = -gmRD since RS is small as compared to RG and

RIN(gate).

The Cgd and Cgs capacitance of the FET is usually cannot be obtained

directly. They are related with the given input capacitance Ciss, output

capacitance Coss, and reverse transfer capacitance Crss from the formulae.

Cgd = Crss (7.18)

Cgs = Ciss - Crss (7.19)

Cds = Coss - Crss (7.20)

Applying Miller Theorem to the amplifier circuit to obtain Cin(Miller) = Cgd(AV +

1) and Cout(Miller) = Cgd(AV + 1)/AV.

The high-frequency input RC network of the amplifier is shown in Fig.

7.23, whereby Rin = RG||Rin(gate).

Figure 7.23: High-frequency input RC network of JFET amplifier

The high-frequency critical frequencyfH(gate) for this RC network is

fH(gate) =1

2( || ) ( )( )R R C CS in gs in Miller +(7.21)

If RS


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The high-frequency output RC network of amplifier is shown in Fig. 7.24.

Figure 7.24: High-frequency output RC network of JFET amplifier

The high-frequency critical frequencyfH(FET-out) of this RC network is

fH(FET-output) =1

2( || ) ( )( )R R C CD L out miller ds +(7.22)

With the known low and high critical frequencies for the amplifier, the Bode

plot of this amplifier can be plotted.

7.8 Frequency Analysis of A Common-Base Bipolar

Junction Transistor Amplifier

The high frequency response of a common-base bipolar junction transistor

amplifier is analyzed by considering a common-base bipolar junction transistor

amplifier shown in Fig. 7.25 and its equivalent circuit shown in Fig. 7.26.

Figure 7.25: A common-base bipolar junction transistor amplifier


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Figure 7.26: High-frequency ac equivalent circuit of a common-base bipolar junction

transistor amplifier

From the equivalent circuit, the Miller effect due to C is zero, thus, it does not

affect the bandwidth of the amplifier. At the output loop, output voltage Vout(s)

is -gmVbeRL||RC||(C+CL) i.e.

Vout (s) =

+

+++

)CC(RR

RRs1)RR(

VRRg

L

CL

CLCL

beCLm (7.23)

Since the value of CL is large, at high frequency, it appears open. Therefore,

capacitance (C + CL) C and the output voltage shall be Vout(s)

=)R||R(Cs1

VR||Rg

CL

beCLm

+

.

At the input node E, Ie + gmVbe + )C/(1

Vbe

s + Vbe/r = 0 and voltage at emitter

Ve is also equal to Vbe. Thus, ++== Cgr/1V

I

)(Z

1m

e

e

in

ss

, which is also equal

to

++

Cr

gr1 m s or

++

Cr

1s . Therefore, the input impedance Zin(s) is equal to

+C||

1

r.


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Voltage Vbe(s) is equal to )s(VC||)1/(r||RR

C||)1/(r||Rin

ES

E

++

+. It can be shown that

the ratio of Vbe(s)/Vin(s) is equal to

[ ])1/(r||R||RsC11

)1/(r||RR)1/(r||R

)s(V)s(V

ESES

E

in

be

++

+++=

after adding the frequency

response factor. The high frequency voltage gain AV(s) shall be

Av(s)|HF = +

)R||R(Cs1

R||Rg

CL

CLm ++

+

)1/(r||RR

)1/(r||R

ES

E

[ ])1/(r||R||RCs11

ES ++ (7.24)

Equation (7.24) clearly shows that there are two poles for the amplifier at high

frequency with critical frequencies)R||R(C2

1

CLand

1

2 1 C R R rE S[ || || / ( )]+.

The low-frequency analysis of the amplifier requires a low-frequency ac

model as shown in Fig. 7.27.

Figure 7.27: Low-frequency ac equivalent circuit of a common-base bipolar junction


Short Circuit Method is used to obtain the low-frequency critical frequency.

The resistance R is1

r||RR ES

++ and capacitance C = C1. Thus, the critical

frequency due to capacitor C1 isfL1(CB) is


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++

=

1

r||RRC2

1

ES1

)CB(1Lf (7.25)

The critical frequency due to capacitor CB isfL2(CB) is

RC2

1)CB(2L

=f (7.26)

where C = CB and R = ( ) [ ]++ r)R||R)(1(||R||R ES21 .

The critical frequency due to capacitor C2 isfL3(CB) is

RC2

1)CB(3L

=f (7.26)

where C = C2 and R = ( )LC RR + .

In summary the low-frequency response of the voltage gain |Av(j)| LF(CB) of

the amplifier is

|Av(j)| LF(CB) = + CL

CL

m RR

RR

g )1/(r||RR

)1/(r||R

ES

E

++

+

+

2)CB(1L

1

1

f

f

2

)CB(2L1

1

+

f

f

2

)CB(3L1

1

+

f

f

(7.27)

7.9 Frequency Analysis of A Common-Collector Bipolar

Junction Transistor Amplifier

Consider a common-collector bipolar junction transistor amplifier shown in Fig.7.28 and its low frequency ac equivalent circuit is shown in Fig. 7.29. Its mid-

range frequency voltage gain is

++

+

LE

LE

21s

21

R||R1

r

R||R

R||RR

R||R


25/32


- 199 -

Figure 7.28: A common-collector bipolar junction transistor amplifier

Figure 7.29: Low frequency ac equivalent circuit of a common-collector bipolar junctiontransistor amplifier

The low frequency response of the amplifier shall beLFV

)(A s

=

++

+

LE

LE

21s

21

R||R1

r

R||R

R||RR

R||R

]R||R||))r)1)(R||R((R[C

11

1

21LES1 ++++

s

.

]RR||))1/()r||R||R||R[(C

11

1

LE21S2 ++

+s

, where critical frequencies are


26/32


- 200 -

respectively equal to fc1 =]R||R||))r)1)(R||R((R[C2

1

21LES1 +++and fc2 =

]RR||)}1/()r)R||R||R[{(C2

1

LE21S2 +++ .

The high frequency ac circuit of the amplifier is shown in Fig. 7.30. There

is no Miller effect since the input and output has same phase.

Figure 7.30: High frequency ac equivalent circuit of a common-collector bipolar junction


The output impedance ro and collector current gmVbe can be grouped with RE

and RL, whilst C can be grouped at input side. Thus, the rearranged high

frequency ac circuit of the amplifier is shown in Fig. 7.31.

Figure 7.31:Rearranged high frequency ac equivalent circuit of a common-collector bipolar

junction transistor amplifier


27/32


- 201 -

At the output node, the output voltage is Vout = )r||R||R()IVg( oLEbbem + .

Voltage Vbe(s) is equal to )C||r(Ib = Ib/y, where y = 1/r + sC.

The voltage at base is Vb = Vbe + Vout. This shall mean that the impedance

at the base is Zb =b

outbe

b

b

I

VV

I

V += . Knowing Vout and Vbe, the impedance Zb is

11

yg r R R r R Rm o E L o E L

[ ( || || )] ( || || )+ + , in which after replacing y, it becomes

Zb(s) =

)R||R||r(g1

sC

)]R||R||r(g1[r

1

)R||R||r(

LEomLEom

LEo

+++

(7.27)

The equivalent impedance Zb is shown in Fig. 7.32. From equation (7.27), it

shows that the effect due to capacitance C is reduced for common-collector

configuration.

Common-collector has one zero and two poles. The zero can be obtained

from Vout = Vbe(y+ gm)(RE||ro||RL) when (y+ gm) = 0. This implies that 1/r +

sC + /r = 0 and it yields the critical frequencyfc =1

21

C

r

+

. Sincer

1+ is

very small, therefore, the critical frequency is very high.

Figure 7.32: Equivalent circuit of input impedance a common-collector bipolar junction


From Fig. 7.32, the base resistance parallel to capacitance is very large as

compared to resistance RE||ro||RL. This helps to reduce the analysis by one pole.


28/32


- 202 -

The remaining pole has time constant H = RS||R1||R2||R[(RE||ro||RL) +

gm] CC

R r R gE o L m

+

+

|| ||. Therefore, the critical frequencyfH is

1

2 H. The high

frequency response of voltage gain of the amplifier shall be

HFV)s(A =

++

+

LE

LE

21s

21

R||R1

r

R||R

R||RR

R||R

2H1H

1

1

1

1

+

+

ss(7.28)

where

+

=

1

rC

11H and

H

2H

1

= .

7.10 Frequency Analysis of Multi-Stage Amplifier

When two or more stage amplifier is cascaded or cascaded to form multi-stage

amplifier for obtaining high gain and bandwidth, the overall frequency responseis determined by the frequency response of each stage depending on

relationships of the frequencies.

The low frequency response of the overall gain AVL (overall) for the multi-n-

stage amplifier shall be

+

+

+

+

=Ln

VLn

3L

3VL

2L

2VL

1L

1VL)overall(VL

1

A.....

1

A

1

A

1

A)s(A

jjjj

(7.29)

If the gain of all the n-stage are identical i.e. AVL1 = AVL2 = AVL3 =... = AVLn =

AVL, and the critical angular frequency L1 = L2 = ...= Ln = L then

AVL(overall)(s) =

( )n

n

j

+ L

VL

1

A

. Its magnitude shall be AVL(overall)|dB = 20n

logA VL

L1 + j

. The -3 dB angular frequency of this multi-stage amplifier shall be

21 L =

+

n

j anddB3L

= , which is


29/32


- 203 -

12 /1L

dB3L

=

n(7.30)

The high frequency response of overall gain AVH (overall) for the multi-n-stage

amplifier shall be

A sA A A A

VH overall

VH

H

VH

H

VH

H

VHn

Hn

( ) ( ) .....=

+

+

+

+

1

1

2

2

3

3

1 1 1 1j j j j

(7.31)

If the gain of all the n-stage are identical i.e. AVH1 = AVH2 = AVH3 =... = AVHn =

AHV, and the critical angular frequency H1 = H2 = ...=Hn =H then

AVH(overall)(s) =

( )n

n

j

+

H

VH

1

A

. Its magnitude shall be AVH(overall)|dB = 20n

log

H

VH

1

A

+ j

. The -3 dB angular frequency of this multi-stage amplifier shall be

21H

=

+

n

j anddB3H

= , which is

12

/1

HdB3H = n

(7.32)

Exercises

7.1. The output rating of an amplifier is given as 50dBm. Calculate the outputpower and power gain AP.

7.2. Explain how the RC network of the bypass and coupling capacitors of anamplifier work like high pass filter?

7.3. Explain how the internal capacitances of BJT, JFET, and MOSFET worklike low pass filter?

7.4. What will be the reactance values of internal capacitance of BJT, JFET,and MOSFET at low frequency?

7.5. What will be the reactance values of bypass and coupling capacitance ofBJT, JFET, and MOSFET at high frequency?


30/32


- 204 -

7.6. Explain the meaning and significance of -3dB point for an amplifier.7.7. Given a RC network below has R1 = 68.0k, R2 = 18.0k, = 160, C1 =

0.1F, dc emitter current IE = 1.5mA, thermal voltage VT = 25mV, andthe critical frequency equals to 580Hz.

i. Name the filtering capability of this network.ii. Calculate the value of unknown resistor RS.iii. Calculate the leaded phase of this network if the ac frequency is

200 Hz.

iv. If frequency and amplitude of Vin are 100Hz and 40mV respectively,what is the amplitude of the ac signal at point B?

7.8. The equivalent amplifier circuit shown below has RS = 1.0k, RL =4.0k,r = 2k, gm = 50mA/V and C1 = 1.0F. Using s-domain analysisto derive its gain function and show that its -3dB frequency is 53.1Hz,maximum gain asymptote is 133, and -3dB gain is 94.1.

7.9. Using the amplifier circuit shown in the figure below with known VCC =

15V, Rs = 100, R1 = 6.8k, R2 = 3.3k, RE = 3.3k, RC = 2.2k, RL =

1.8k, C1 = C2 = 0.1F, C3 = 1.0F, Cbe = 150pF, Cbc = 80pF, and =120.


31/32


- 205 -

i. Find the mid-range frequency gain of this amplifierii. Draw an ac equivalent circuit for this amplifier that covers all

frequency range.

iii. Determine all the critical frequencies of this amplifier.iv. Draw a Bode plot of this amplifier.v. Estimate the gain of this amplifier if the ac frequency of input Vin are

5kHz, 200kHz, and 1MHz?

7.10. A two-stage amplifier is made by cascading two amplifier circuits shownin question 7.9.

i. Calculate the - 3dB critical frequency.ii. Calculate the - 3dB voltage gain.

7.11. Derive the equations for critical frequencies and find the criticalfrequencies and the maximim asymptote gain of the common-collector

BJT amplifier shown in the figure. Given that C = 1.0pF, C = 15pF, gm

= 57.0mS, = 80, and r = 1.4k.


32/32


Bibliography

1. Donald A. Neamen, "Electronic Circuit Analysis and Design", secondedition, McGraw Hill Higher Education, 2001.

2. Jacob Millman and Arvin Grabel, "Microelectronics", second edition,McGraw-Hill International Editions, 1987.

3. Muhammad H. Rashid, "Microelectronic Circuits: Analysis and Design",PWS Publishing Company, 1999.4. Robert T. Paynter, "Electronic Devices and Circuits", fifth edition,

McGraw-Hill, 1997.

5. Adel S. Sedra and Kenneth C. Smith, "Microelectronic Circuits", fourthedition, Oxford University Press, 1998.

chapter 7 amplifier frequency response

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