Chapter 23 Transistor Amplifier Low Frequency...

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© 2009 Strangeway, Lokken, Petersen, and Gassert ch. 23 page 1

Chapter 23 Transistor Amplifier Low Frequency Analysis*

by Robert Strangeway, Richard Lokken, Owe Petersen, and John Gassert 23.1 Bipolar Junction Transistor Low Frequency Analysis...................................................2

23.1.1 The Need for Low Frequency Response Analysis in BJT CE Amplifiers ..............2

23.1.2 Highpass Effects of the Coupling Capacitors ..........................................................3

23.1.3 Highpass Effect of the Emitter-Bypass Capacitor .................................................11

23.2 Field Effect Transistor Low Frequency Analysis .........................................................17

23.2.1 The Need for Low Frequency Response Analysis in FET CS Amplifiers ............17

23.2.2 Highpass Effects of the Coupling Capacitors ........................................................18

23.2.3 Highpass Effect of the Source-Bypass Capacitor ..................................................24

23.3 Chapter Appendix: Development of the 3 dB Point for Two Closely-Spaced

Break Frequencies .......................................................................30

23.4 Class Notes ........................................................................................................................32

23.5 Homework Problems .......................................................................................................59 Note: The order of coverage of sections 23.1 and 23.2, and the corresponding sub-sections of class notes within section 23.4, may be swapped—they are independent sections. Pedagogy (for the instructor):

DC bias analysis reinforces and extends mastery of Kirchhoff’s laws

DC analysis introduces the concept of an electron device that does not have an ohm’s law type of i-v relationship

AC midband analysis introduces dependent sources in a two-port “input-output” context

AC midband analysis reinforces/extends mastery of two-port circuit analysis that was begun with first-order filters and in previous electronics courses

Frequency response analysis reinforces/extends mastery of transfer functions and break frequencies that was begun with first-order filters

* This chapter may be copied and distributed during the 2008-9 academic year to students in the Electrical/Electronics Technology and Engineering Technology programs at College of Lake County, Fox Valley Technical College, Gateway Technical College, Madison Area Technical College, Milwaukee Area Technical College, Milwaukee School of Engineering, Northeast Wisconsin Technical College, Waukesha County Technical College, and Western Technical College, all in the states of Wisconsin or Illinois. Any other reproduction or distribution is prohibited without the express written consent of the first or the second named authors.

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Outcomes: As a result of studying this chapter and working the homework problems, the student should be able to:

Explain each step in the development of the transfer functions and the break frequencies that are utilized to determine the low frequency response of a CE/CS transistor amplifier.

Select and utilize the proper results for the determination of the low frequency response of a CE/CS transistor amplifier.

23.1 BIPOLAR JUNCTION TRANSISTOR LOW FREQUENCY ANALYSIS 23.1.1 The Need for Low Frequency Response Analysis in BJT CE Amplifiers The amplifier gains predicted thus far are midband gains. The capacitors and capacitances have reactance values that do not significantly change the gain over these frequencies. However, at relatively low frequencies, the reactances of coupling capacitors and the emitter-bypass capacitor (if present) become larger and may become significant relative to the relevant resistance values. The generator-side coupling capacitor, Cgen, the load-side coupling capacitor, CL, and the emitter-bypass capacitor, CE, are shown in a CE amplifier in Figure 23.1.

Qualitatively, how would these capacitors affect the gain as the frequency is decreased? The generator and load coupling capacitors form series RC highpass filters: Cgen is in series with Rgen and the input impedance of the amplifier, and CL is in series with the output load resistance. Thus, these are highpass filters. As the frequency decreases, the impedance of CE becomes large relative to the emitter resistor; hence, the emitter resistor is not bypassed at lower frequencies and the resultant gain decreases. Overall, these three capacitors result in highpass filtering that limits the low frequency response of the amplifier.

genV ~

iV

~

oV

Figure 23.1 Capacitors in a CE BJT amplifier

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At high frequencies, the coupling and emitter bypass capacitors have reactance values that are much smaller than the resistance levels and hence, are insignificant. These frequencies are in the passband of the corresponding highpass filters. However, there are other capacitances that become significant at higher frequencies. The effects of these capacitances on the high frequency response of a transistor amplifier will be examined in the next chapter.

How does one approach the analysis of an amplifier at very low frequencies, that is, below the midband frequency range? Fortunately, you already have the tools: frequency response (transfer function) and Bode plot analysis. The overall strategy is:

1) identify the equivalent RC networks formed by the capacitances and resistances in the circuit,

2) determine the transfer function of each RC network, as is done in Bode plot analysis, and 3) modify the midband gain frequency response using the break frequencies and filtering

type (highpass in the low frequency analysis).

There is a hidden assumption in the strategy above: it is assumed that each frequency roll-off effect is independent of the others, that is, that they do not interact. This is not technically true, but usually is “true enough” to give good frequency response predictions. If more exacting predictions are required, one can simulate using the more complicated transistor models that are contained in software packages. Again, the purpose of manual circuit analysis (analysis by hand calculations) is to form starting points for designs and to make reasonable but not exact predictions. The low frequency response of a BJT CE amplifier is examined in this section. The low frequency analysis of an FET CS amplifier is examined in section 23.3. The high frequency responses will be examined in chapter 24. Review transfer function and Bode plot analysis techniques as needed in order to understand the application of the techniques to determine the low frequency response of these amplifiers. 23.1.2 Highpass Effects of the Coupling Capacitors The highpass effects of the coupling capacitors are considered first. The AC model of the BJT with the coupling capacitances is shown in Figure 23.2. The emitter resistance is assumed to be bypassed per the assumption of “non-interacting” capacitances (the effect of CE will be analyzed after the coupling capacitances analysis). The “generator” is the circuit that drives the input of the amplifier. Often, it is the previous stage in the circuit. The load resistance is often either the next stage in the circuit or the device or component that is driven by the amplifier stage. The series Rgen – Cgen – Ri circuit forms a highpass filter, as does the series Ro – CL – RL circuit. Note that Ri is the midband input impedance (resistance) of the BJT amplifier and Ro is the midband output impedance. Also note that even though Rgen is not part of the amplifier, it cannot be ignored because it will affect the value of the break frequency on the generator side.

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iVgenV

bI

c bI I

oV

Figure 23.2 BJT amplifier model with coupling capacitances

Determine the transfer function. Note how the generator is incorporated into the analysis:

geno o

i i gen

VV VH j

V V V

(23.1)

A crucial step is now performed: the generator resistance-input resistance voltage divider is defined to be at midband (Cgen is effectively a short). The reason for this step is to incorporate

the effects of the generator in the frequency response via o

gen

V

V

while maintaining the transfer

function as o

i

V

V

because these variables define the actual input and the actual output of the

amplifier stage. Thus, the generator resistance-input resistance voltage divider at midband is:

gen i gen gen ii

gen i i i

V R V R RV

R R V R

(23.2)

where i B eR R r . Substitute Eq. (23.2) into Eq. (23.1):

gen gen io o o

i i gen i gen

V R RV V VH j

V V V R V

(23.3)

Use the current divider rule to determine the load current from bI . Multiply by the

load resistance to obtain the output voltage:

1

Co L L b L

C LL

RV I R I R

R R +j C

(23.4)

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Determine bI from genV by Ohm’s law and the current divider rule:

1

genB Bb gen

B e B egen i

gen

VR RI I

R + r R + rR + R j C

(23.5)

Insert Equation (23.5) into (23.4) and this result into (23.3):

1 1

gen CBo L

B egen i C L

gen L

V RRV R

R + rR + R R R +j C j C

(23.6)

1 1gen C LB

B egen i C L

gen Lgen i

i gen

V R RRR + rR + R R R +

j C j CR RH j

R V

(23.7)

Substitute B eB e

B e

R rR r

R r

for iR in the first factor and simplify:

11 1

gen i C LB

B e B egen i C L

gen LB e

R R R RRH j

R r R + rR + R R R +j C j CR r

(23.8)

11 1

gen i C L

egen i C L

gen L

R R R RH j

r R R R R j C j C

(23.9)

Clear compound fractions:

11 1

gen i gen C L L

e gen Lgen i C L

gen L

R R j C R R j CH j

r j C j CR R R R j C j C

(23.10)

1

1 1gen i gen C L L

e gen i gen C L L

j R R C j R R CH j

r j R R C j R R C

(23.11)

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The next step is taken in order to simplify the final results of this analysis (trust us!). Factor out

the constant C L

C L

R R

R R

from the numerator of the last factor in order to make the break

frequencies of the numerator and denominator equal:

1 1

gen i gen C L LC L

gen i gen e C L C L L

j R R C j R R CR RH j

j R R C r R R j R R C

(23.12)

Recall the midband gain of the BJT CE amplifier:

C Lb

C Lo C Lv

i b e e C L

R RI

R RV R RA

V I r r R R

(23.13)

1 1gen i gen C L L

vgen i gen C L L

j R R C j R R CH j A

j R R C j R R C

(23.14)

The real parts of the complex numbers are already equal to one. Thus proceed to the next step of

b

j

form:

1 1

1 11 1

gen i gen C L Lv

C L Lgen i gen

j j

R R C R R CH j A

j j

R R CR R C

(23.15)

1 1

bg bLv

bg bL

j j

H j Aj j

(23.16)

where 1 1

and bg bLC L Lgen i gen R R CR R C

are the break frequencies of the highpass

transfer function factors. Note how the generator resistance affects the generator-side break frequency, as does the load resistance on the load-side break frequency. Thus, the entire amplifier transfer function including the effects of the generator and load coupling capacitors is

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( ) ( )og v L

i

VH j H j A H j

V

(23.17)

1 1

1 1midband1 1 loaded

voltagehighpass highpass gain

gen i gen C L L

v

C L Lgen i gen

Lg

j j

R R C R R CH j A

j j

R R CR R C

H jH j

(23.18)

Thus, the approach to analyze the highpass filtering of the low frequency response of a BJT CE amplifier due to the coupling capacitors is to determine the midband loaded amplifier gain. Then determine the transfer functions of the generator-side and load-side highpass filters:

1

, where 1

bgg bg

gen i gen

bg

j

H jR R Cj

(23.19)

1

, where 1

bLL bL

C L L

bL

j

H jR R Cj

(23.20)

The total amplifier transfer function then is:

( ) ( )og v L

i

VH j H j A H j

V

(23.21)

Thus, the amplifier low frequency response due to the coupling capacitors is obtained by modifying the midband loaded voltage gain with the highpass responses of the generator and load coupling capacitances (via the break frequencies). This realization is important because one can straightforwardly determine the highpass transfer functions of the RC sections on the input and output of the amplifier circuit and, consequently, the low frequency response of the BJT CE amplifier due to the coupling capacitors. This approach is illustrated in Example 23.1.1.

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Example 23.1.1 Determine the midband and low frequency response for the BJT amplifier shown in Figure 23.3. Consider only the coupling capacitors in the low frequency response (assume the emitter resistor is bypassed).

VCC = +15 V

Cgen

R1RC

CL = 10 μF

RER2

CE

Rgen

RL

genV ~

iV

~

oV200 Ω 10 μF

50μF

1200 Ω

3300 Ω

20 kΩ

2 kΩ120Ω

Figure 23.3 Amplifier for Examples 23.1.1 and 2

Strategy: Determine bias current IE re = 0.026/IE Build the midband amplifier model Determine Av Build the coupling capacitors low frequency amplifier model Determine the high pass break frequencies Sketch and label the midband and low frequency magnitude Bode Plot.

Solution:

Determine the bias conditions, IE , and re

2

1 2

15 20001.3636 V

20k 2000CC

Th

V RV

R R

1 2

1 2

20k 20001.818 k

20k 2000Th

R RR

R R

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KVL: 1 0

E

Th B Th BE B E

I

V I R V I R

1.3636 0.737.837 A

1 1818 131 120Th BE

BTh E

V VI

R R

1 131 37.837 A 4.9567 mAE BI I

0.026 0.026

5.24544.9567 me

E

rI

Build the midband amplifier model (Figure 23.4); 1 2|| 1818BR R R

~

oV~

iV

bI

130 bIbI

Figure 23.4 Midband amplifier model Determine Av

1200 3300167.77

5.2454 1200 3300

dB 20log 20log 167.77 44.5 dB

C Lb

C Lo C Lv

i b e e C L

v v

R RI

R RV R RA

V I r r R R

A A

Build the coupling capacitor low frequency model (Figure 23.5)

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RB

CLCgen

~

oV~

iV

Rgen

genVRLre

RC

bI

130 bI200 Ω

1200 Ω 3300 Ω1818 Ω682 Ω

10 μF10 μF

Figure 23.5 Coupling Capacitors Low Frequency Model

Determine the highpass break frequencies

The input of the amplifier is a Rgen-Cgen-Ri highpass filter with 496i e BR r R .

1 1

, 144 rad/s , 23 Hz200 496 101

bgg bg bg

gen i gen

bg

j

H j fR R Cj

The output of the amplifier is a RC-CL-RL highpass filter :

1 1

, 22 rad/s , 3.5 Hz1200 3300 101

bLL bL bL

C L L

bL

j

H j fR R Cj


( ) ( )og v L

i

VH j H j A H j

V

23 3.5, 44.5 dB, and 1 1

23 3.5

g v L

f fj j

H j A H jf f

j j

The 23 Hz highpass break frequency dominates the low frequency response (Figure 23.6)

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Figure 23.6 Low frequency response of the amplifier due to coupling capacitors ____________________________________________________________________________ 23.1.3 Highpass Effect of the Emitter-Bypass Capacitor

Now the effect of the emitter-bypass capacitor, CE, on the low frequency response of the amplifier will be considered. Qualitatively, what happens in midband and higher frequencies? The emitter bypass capacitor reactance is much less than the emitter resistor resistance; hence, the emitter resistor is bypassed as far as the AC model of the amplifier is concerned. What happens at lower frequencies for which the emitter bypass capacitor reactance is not much less than the emitter resistor resistance? The emitter bypass capacitance must be considered in the AC gain analysis. The AC model of the BJT CE amplifier in Figure 23.1 is shown with CE in Figure 23.7. The generator and load-side coupling capacitors are replaced by shorts per the assumption of “non-interacting” capacitances. The AC analysis proceeds in a similar manner as it would for a CE amplifier with RE except the impedance of the capacitor in parallel with RE must be incorporated:

|| ||

11

111

b C L b C Lo

i EE b e

E E Eb e b

EE

I R R I R RVH j

V RR I rj C j R CI r I

Rj C

(23.22)

For 1, 1 , and Eq. (23.22) simplifies to:

|| ||

1 1

C L C L

E Ee e

E E E E

R R R RH j

R Rr r

j R C j R C

(23.23)

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genViV oV

bIiI

c bI I

Figure 23.7 AC BJT model with RE and CE Recall from a previous example that the midband un-bypassed amplifier voltage gain is:

||C L

v unbypassede E

R RA

r R

(23.24)

A strategic algebraic manipulation at this point can again simplify the final results of this analysis. Factor out a constant such that the remainder is the bypassed voltage gain:

|| ||c L c Le e

v unbypassed v bypassede E e E e e E

v bypassedA

R R R Rr rA A

r R r R r r R

(23.25)

Similarly, algebraically manipulate Eq. (23.23) to show Av-unbypassed and Av-bypassed :

|| || 11

1 11

C L C L E EE E

E E E e E E Ee

E E

R R R R j R Cj R CH j

R j R C r j R C Rr

j R C

(23.26)

|| 1C L E E

e E e E E

R R j R CH j

r R j r R C

(23.27)

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11

|| 1 ||

11

E EC L E E C L

e E E e Ee E

e Ee E v unbypassed

e E E

A

j

R CR R j R C R RH j

r R C r Rjr R j

r Rr Rr R C

(23.28)

3 3

4 4

1 1

1 1

o b e bv unbypassed v bypassed

s e E

b b

H jE

j jV r

H j A AV r Rj j

(23.29)

where the break frequencies are:

3

1b

E ER C (23.30)

4

1e Eb

e Ee E EE

e E

r Rr Rr R C

Cr R

(23.31)

and the portion of the transfer function due to the emitter bypass capacitor, EH j , is indicated

in Equation (23.29). Note that ωb3 < ωb4 because the total resistance of is re||RE is less than RE. Thus, the transfer function shows the “flat/increasing/flat” highpass response due to CE , as shown in Figure 23.8. If the gain above ωb4 is equal to the gain with the bypassed emitter resistor, what is the gain below ωb3? It must be the gain without the bypassed emitter resistor! Note that ωb3 often occurs below the break frequencies in the highpass responses due to the coupling capacitors.

no thru e EI Call thru e EI C

Figure 23.8 Frequency response due to CE

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Thus, the low frequency response must include the midband amplifier gain, the highpass effects due to the coupling capacitors, and the highpass effect due to the emitter-bypass capacitor:

( ) ( )og v bypassed E L

i

VH j H j A H j H j

V (23.32)

where:

1

, where 1

bgg bg

gen i gen

bg

j

H jR R Cj

(23.33)

||c L C L

v bypassede e L L

R R R RA

r r R R

(23.34)

33 4

4

11

, where and 1

e b e EE b b

e E E E e E E

b

jr r R

H jr R R C r R Cj

(23.35)

1

, where 1

bLL bL

C L L

bL

j

H jR R Cj

(23.36)

Thus, the amplifier low frequency response is obtained by modifying the midband loaded voltage gain with the highpass responses of the generator and load coupling capacitors and the highpass response of the emitter-bypass capacitor (via the break frequencies). Again, this realization is important because it makes the determination of the low frequency response of a BJT CE amplifier straightforward. The effect with the highest break frequency dominates the low frequency response of the amplifier. The previous example is now reconsidered in Example 23.1.2 to include the effect of the emitter-bypass capacitor.

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Example 23.1.2 Determine the low frequency response for the BJT amplifier shown in Figure 23.3 of Example 23.1.1. Include the effect of the emitter-bypass capacitor.

Strategy: Determine bias current IE re = 0.026/IE Build the midband amplifier model Determine Av Build the low frequency amplifier model Determine the high pass break frequencies Sketch and label the midband and low frequency magnitude Bode Plot.

The first four steps and part of the fifth and sixth steps were performed in Example 23.1.1. Build the low frequency model with CE (Figure 23.9).

re

RCRB

RE

RL

Rgen

CE

genViV oV

bIiI

c bI I 200 Ω

50 μF

1818 Ω

120 Ω

1200 Ω

CL

10 μF

Cgen

10μF

3300 Ω

682 Ω 130 bI

Figure 23.9 Low frequency BJT amplifier model The RC circuits on the input and the output were already analyzed in Example 23.1.1. Thus, proceed to the amplifier transfer function analysis with CE (Figure 23.10). Note that the generator and load coupling capacitors have been replaced with shorts.

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iV oV

bI

130 bI

Figure 23.10 Model for HE(jω) determination

3

4

1

1

e bE

e E

b

jr

H jr R j

3 3

1 1166.7 rad/s , 26.5 Hz

120 50b bE E

fR C

4 4

5.2454 1203979.5 rad/s , 633 Hz

5.2454 120 50e E

b be E E

r Rf

r R C

Note that ωb4 dominates the low frequency response because it has the largest frequency. It is also much larger than ωbg and ωbL. Hence, ωb4 dominates the entire low frequency response of the amplifier. It is shown in Figure 23.11.

Figure 23.11. The low frequency amplifier response for Example 23.1.2

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23.2 FIELD EFFECT TRANSISTOR LOW FREQUENCY ANALYSIS 23.2.1 The Need for Low Frequency Response Analysis in FET CS Amplifiers The FET amplifier gains predicted thus far are midband gains. The capacitors and capacitances have reactance values that do not significantly change the gain over these frequencies. However, at relatively low frequencies, the reactances of coupling capacitors and the source-bypass capacitor (if present) become larger and may become significant relative to the relevant resistance values. The generator-side coupling capacitor, Cgen , the load-side coupling capacitor, CL, and the source-bypass capacitor, CS , are shown in a CS amplifier in Figure 23.12.

Qualitatively, how would these capacitors affect the gain as the frequency is decreased? The generator and load coupling capacitors form series RC highpass filters: Cgen is in series with Rgen and the input impedance of the amplifier, and CL is in series with the output load resistance. Thus, these are highpass filters. As the frequency decreases, the impedance of CS becomes large relative to the source resistor; hence, the source resistor is not bypassed at lower frequencies and the resultant gain decreases. Overall, these three capacitors result in highpass filtering that limits the lower frequency response of the amplifier.

iVoVgenV

Figure 23.12 Capacitors in a CS FET amplifier

At high frequencies, the coupling and source-bypass capacitors have reactance values that are much smaller than the resistance levels and hence, are insignificant. These frequencies are in the passband of the corresponding highpass filters. However, there are other capacitances that become significant at higher frequencies. The effects of these capacitances on the high frequency response of a transistor amplifier will be examined in the next chapter. How does one approach the analysis of an amplifier at very low frequencies, that is, below of the midband frequency range? Fortunately, you already have the tools: frequency response (transfer function) and Bode plot analysis. The overall strategy is:

_____________________________________________________________________________________________________________________


1) identify the equivalent RC networks formed by the capacitances and resistances in the circuit,

2) determine the transfer function of each RC network, as is done in Bode plot analysis, and 3) modify the midband gain frequency response using the break frequencies and filtering

type (highpass in the low frequency analysis). There is a hidden assumption in the strategy above: it is assumed that each frequency roll-off effect is independent of the others, that is, that they do not interact. This is not technically true, but usually is “true enough” to give good frequency response predictions. If more exacting predictions are required, one can simulate using the more complicated transistor models that are contained in the software packages. Again, the purpose of manual circuit analysis (analysis by hand calculations) is to form starting points for designs and to make reasonable but not exact predictions. The low frequency analysis of an FET CS amplifier is examined in this section. The low frequency response of a BJT CE amplifier is examined in section 23.2. The high frequency responses will be examined in chapter 24. Review transfer function and Bode plot analysis techniques as needed in order to understand the application of the techniques to determine the low frequency response of these amplifiers. 23.2.2 Highpass Effects of the Coupling Capacitors The highpass effects of the coupling capacitors are considered first. The source resistance is assumed to be bypassed per the assumption of “non-interacting” capacitances (the effect of CS will be analyzed after the coupling capacitances analysis). The AC model of the FET with the coupling capacitances is shown in Figure 23.13. The “generator” is the circuit that drives the input of the amplifier. Often, it is the previous stage in the circuit. The load resistance is often either the next stage in the circuit or the device or component that is driven by the amplifier stage. The series Rgen – Cgen – RG circuit forms a highpass filter, as does the series RD – CL – RL circuit. Note that even though Rgen is not part of the amplifier, it cannot be ignored because it will affect the value of the break frequency on the generator side.

RG

RD

CL

Cgen

~

oV~

iV~

gsV~

m gsg V

Rgen

genV

RL

Figure 23.13 FET amplifier model with coupling capacitances

_____________________________________________________________________________________________________________________


Determine the transfer function. Note how the generator is incorporated into the analysis:

geno o

i i gen

VV VH j

V V V

(23.37)

The generator resistance-gate resistor voltage divider is defined to be at midband (Cgen is effectively a short):

i

i i

gengeni i

gen gen i

R RVV R

V R R V R

(23.38)

where

i GR R in this FET CS amplifier.

gen gen io o o

i i gen i gen

V R RV V VH j

V V V R V

(23.39)

As with the BJT amplifier analysis, use the current divider rule to determine the load

current from m gsg V . Multiply by the load resistance to obtain the output voltage:

1

Do L L m gs L

D LL

RV I R g V R

R R +j C

(23.40)

Use the voltage divider rule on the input to relate gsV to genV :

1

gen igs

gen igen

V RV

R R +j C

(23.41)

Insert Equation (23.5) into (23.40) and this result into Equation (23.3):

1 1

gen i Do m L

gen i D Lgen L

V R RV g R

R R + R R +j C j C

(23.42)

1 1gen i D L

m

gen i D Lgen Lgen io

i i gen

V R R Rg

R R + R R +j C j CR RV

H jV R V

(23.43)

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The numerator of the generator-side voltage divider can be multiplied by the gen i

i

R R

R

factor.

Then clear compound fractions:

1 1

gen i gen D L Lm

gen Lgen i D L

gen L

R R j C R R j CH j g

j C j CR R + R R +j C j C

(23.44)

11

gen i gen D L Lm

D L Lgen i gen

j R R C j R R CH j g

j R R C j R R C

(23.45)

In a manner similar to the BJT low frequency amplifier analysis, factor out the constant

D L

D L

R R

R R

from the numerator in order to make the break frequencies of the numerator and

denominator equal:

11

gen i gen D L LD Lm

D L D L Lgen i gen

j R R C j R R CR RH j g

R R j R R C j R R C

(23.46)

11

gen i gen D L LD Lm

D L D L Lgen i gen


R R j R R C j R R C

(23.47)

Recall the midband gain of the FET CS amplifier:

D Lv m

D L

R RA g

R R

(23.48)

11

gen i gen D L Lv

D L Lgen i gen


j R R C j R R C

(23.49)

The real part is already equal to one. Thus proceed to the next step of b

j

form:

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1 1

1 11 1

gen i gen D L Lv

D L Lgen i gen

j j

R R C R R CH j A

j j

R R CR R C

(23.50)

1 1

bg bLv

bg bL

j j

H j Aj j

(23.51)

where 1 1

and bg bLD L Lgen i gen R R CR R C

are the break frequencies of the highpass

transfer functions. Thus, the entire amplifier transfer function including the effects of the generator and load coupling capacitors is:

( ) ( )og v L

i

VH j H j A H j

V

(23.52)

1 1



gen i gen D L Lv

D L Lgen i gen

Lg

j j

R R C R R CH j A

j j

R R CR R C

H jH j

(23.53)

Thus, the approach to analyze the highpass filtering at the low frequency response of a FET CS amplifier due to the coupling capacitors is to determine the midband loaded amplifier gain. Then determine the transfer functions of the generator-side and load-side highpass filters:

1

, where 1

bgg bg

gen i gen

bg

j

H jR R Cj

(23.54)

_____________________________________________________________________________________________________________________


1

, where 1

bLL bL

D L L

bL

jH j

R R Cj

(23.55)


1 1

bgo bLv

i

bg bL

j jV

H j AV j j

(23.56)

( ) ( )og v L

i

VH j H j A H j

V

(23.57)

Thus, the amplifier low frequency response due to the coupling capacitors is obtained by modifying the midband loaded voltage gain with the highpass responses of the generator and load coupling capacitors (via the break frequencies). Again, this realization is important because it makes the determination of the low frequency response due to the coupling capacitors in a FET CS amplifier straightforward. This approach is illustrated in Example 23.2.1. Example 23.2.1 For the FET common source amplifier shown in Figure 23.14, determine

the midband gain and the low frequency response due to the coupling capacitors, given IDSS = 7 mA and gm = 4 mS. (Note the source bypass capacitor is not considered in the low frequency response yet—it will be considered in the next section and used in the next example.)

iVoV

genV

Figure 23.14 FET amplifier circuit for Examples 23.2.1 and 2

_____________________________________________________________________________________________________________________


Given: The FET circuit in Figure 23.14 IDSS = 7 mA gm = 4 mS

Desired: Av Low frequency response due to the coupling capacitors

Strategy: Determine Av Determine the high pass break frequency for each coupling capacitor Sketch and label the midband and low frequency magnitude Bode plot.

Solution: CS bypasses the source resistor, so the midband voltage gain is determined

in the same manner that it was in Example 23.3.2:

3 1500 2,0004 10 5.33 14.5 dB

1500 2,000o D L

v mi D L

V R RA g

V R R

The AC low frequency model with the coupling capacitors is shown in Figure 23.13. The transfer function due to the midband amplifier gain and the coupling capacitors was determined to be

1 1

bgo bLv

i

bg bL

j jV

H j AV j j

1

where bg

gen i genR R C

and

1

bLD L LR R C

Calculate the generator and load break frequencies:

6

1 10.333 rad/s

100 300k 10 10bg

gen i genR R C

6

1 17.41 rad/s

1500 12k 10 10bL

D L LR R C

Thus the transfer function and low frequency response due to the coupling capacitors is

_____________________________________________________________________________________________________________________


0.333 7.415.331 1 1 1

0.333 7.41

bgo bLv

i

bg bL

j j j jVH j A

V j j j j

This transfer function is plotted in Figure 23.15.

Figure 23.15 Low frequency response due to the coupling capacitors in Example 23.2.1 23.2.3 Highpass Effect of the Source-Bypass Capacitor

Now the effect of the source-bypass capacitor, CS, on the low frequency response of the amplifier will be considered. Qualitatively, what happens at midband and higher frequencies? The source-bypass capacitor reactance is much less than the source resistor resistance; hence, the source resistor is bypassed as far as the AC model of the amplifier is concerned. What happens at lower frequencies for which the source-bypass capacitor reactance is not much less than the source resistor resistance? Then the source-bypass capacitance must be considered in the AC gain analysis. The AC model of the FET CS amplifier in Figure 23.12 is shown with CS in Figure 23.16. Again, the coupling capacitances are replaced by shorts due to the “non-interacting” capacitors assumption. The AC analysis proceeds in a similar manner as it would for a CS amplifier with RS except the impedance of the capacitor in parallel with RS must be incorporated.

_____________________________________________________________________________________________________________________


iV

gsV

oV

m gsg V

Figure 23.16 AC FET model with RS and CS

Determine the transfer function for the amplifier shown in Figure 23.16. The dependent source current flows through ||D LR R to give the output voltage. The input voltage is the

voltage from the gate to the source plus the voltage across the parallel combination of the source resistor and source-bypass capacitor:

1

1

D Lm gs

D Lo

iS

Sgs m gs

SS

R Rg V

R R VH j

VR

j CV g V

Rj C

(23.58)

The numerator is recognized as the midband voltage gain if RS is bypassed (the same Av that was used in the previous analysis). Clear compound fractions:

1

11

v bypassedo

iS

S Sm

SS

S

AVH j

VR

j C j Cg

j CRj C

(23.59)

1

11

1

v bypassedo S S

Si S Sm

S S

AV j R CH j

RV j R Cg

j R C

(23.60)

_____________________________________________________________________________________________________________________


1 1

1 1S S S S

v bypassed v bypassedS S m S m S S S

j R C j R CH j A A

j R C g R g R j R C

(23.61)

Make the real part equal to one:

1

11

1

v bypassed S S

S Sm S

m S

A j R CH j

R Cg Rj

g R

(23.62)

The constant 1

v bypassed

m S

A

g R

is recognized as the midband amplifier gain if RS is not bypassed.

Proceed to the next step of b

j

form:

11

11

1

v bypassed S S

m S

m S

S S

j

A R CH j

g Rj

g R

R C

(23.63)

3

4

11

1 1

o bv bypassed

i m S

b

jV

H j AV g R j

H jS

(23.64)

where the break frequencies are:

3

1b

S SR C (23.65)

4

1 m Sb

S S

g R

R C

(23.66)

and the portion of the transfer function due to the source-bypass capacitor, SH j , is indicated

in Eq. (23.64). Note that ωb4 > ωb3 because the numerator of Eq. (23.66) is greater than the numerator of Eq. (23.65), and the denominators are equal. Thus, similar to the transfer function with the emitter bypass capacitor in the BJT CE amplifier, the transfer function shows the “flat/increasing/flat” highpass response due to CS , as shown in Figure 23.17. If the gain above ωb4 is equal to the gain with the bypassed source resistor, what is the gain below ωb3? It must be

_____________________________________________________________________________________________________________________


the gain without the bypassed source resistor! Note that ωb3 often occurs below the break frequencies in the highpass responses due to the coupling capacitors.

b3 b4

no thru s SI C

|H | (dB)

Av-bypassed

all thru s SI C

Av-unbypassed

Figure 23.17 Frequency response due to CS

Thus, the low frequency response must include the midband amplifier gain, the highpass effects due to the coupling capacitors, and the highpass effect due to the source-bypass capacitor:

( ) ( )og v bypassed S L

i

VH j H j A H j H j

V (23.67)

where:

1

, where 1

bgg bg

gen i gen

bg

j

H jR R Cj

(23.68)

D Lv bypassed m

D L

R RA g

R R

(23.69)

33 4

4

111 1

, where and 1 1

b m SS b b

m S S S S S

b

jg R

H jg R R C R Cj

(23.70)

_____________________________________________________________________________________________________________________


1

, where 1

bLL bL

D L L

bL

jH j

R R Cj

(23.71)

Thus, the amplifier low frequency response is obtained by modifying the midband loaded voltage gain with the highpass responses of the generator and load coupling capacitors and the highpass response of the source-bypass capacitor (via the break frequencies). Again, this realization is important because it makes the determination of the low frequency response in a FET CS amplifier straightforward. The effect with the highest break frequency dominates the low frequency response of the amplifier. The previous example is now reconsidered including the effect of the source-bypass capacitor. Example 23.2.2 For the FET common source amplifier shown in Figure 23.14, determine

the midband gain and the low frequency response including the effects of the source bypass capacitor as well as the coupling capacitors, given IDSS = 7 mA and gm = 4 mS.

Given: The FET circuit in Figure 23.14

IDSS = 7 mA gm = 4 mS

Desired: Av Low frequency response due to the coupling and bypass capacitors

Strategy: Determine Av Determine the high pass break frequencies due to each coupling capacitor

and the source-bypass capacitor Sketch and label the midband and low frequency magnitude Bode plot.

Solution: The overall transfer function for the low frequency response is


i

VH j H j A H j H j

V

3

4

11

11 1 1

bg b bLv bypassed

m S

bg b bL

j j j

H j Ag Rj j j

Av-bypassed, ωbg, and ωbL were determined in Example 23.2.1:

_____________________________________________________________________________________________________________________


3 1500 12,000

4 10 5.33 14.54 dB1500 12,000

o D Lv bypassed m

i D L

V R RA g

V R R

6

1 10.333 rad/s

100 300k 10 10bg

gen i genR R C

6

1 17.41 rad/s

1500 12k 10 10bL

D L LR R C

The break frequencies due to the source-bypass capacitor are

3 6

1 151.3 rad/s

390 50 10b

S SR C

4 6

1 1 0.004 390131.3 rad/s

390 50 10m S

bS S

g R

R C

1 1

0.3906 8.16 dB1 1 0.004 390m Sg R

1

20log 14.54 8.16 dB 6.4 dB1v unbypassed v bypassed

m S

A Ag R

Insert these values into the transfer function and plot the response (Figure 23.18):

3

4

11

11 1 1

bg b bLv bypassed

m S

bg b bL

j j j

H j Ag Rj j j

14.5 dB

2.08 6.4 dB

10.333 51.3 7.415.33 0.3906

1 1 10.333 131.3 7.41

j j jH j

j j j

One notes that the source-bypass capacitor break frequency, 4b , is the largest

highpass frequency in the low frequency response of this FET CS amplifier. Hence, this break frequency sets the low frequency end of the midband amplifier response.

_____________________________________________________________________________________________________________________


Figure 23.18 Low frequency response of the FET CS amplifier in Example 23.2.2 23.3 CHAPTER APPENDIX: Development of the 3 dB Point for Two Closely-Spaced

Break Frequencies

Often the high-end lowpass break frequencies are within one decade of each other in transistor amplifiers. Consider two break frequencies in a lowpass filter that are “closely spaced,” that is, they are less than one decade apart. Both lowpass responses must be considered in the determination of the 3 dB frequency because they are so close together. The magnitude of the transfer function for those terms (a portion of the total transfer function magnitude in dB form) is

2 2

1 2

1 120log 20log

1 1b b

f ff f

(23.72)

The sum of two logarithms is the log of the product:

2 2

1 2

120log

1 1b b

f ff f

(23.73)

Combine the square roots: 2 2

1 2

120log

1 1b b

f ff f

(23.74)

The desired quantity is the frequency at which the previous expression equals -3 dB:

_____________________________________________________________________________________________________________________


2 2

1 2

120log 3 dB

1 1b b

f ff f

(23.75)

Solve for f: 3

20

2 2

1 2

110 0.70795

1 1b b

f ff f

(23.76)

Note that an equation solver (calculator or software) could be utilized at this point to solve for f. Instead, a convenient expression will be developed. Invert and square both side:

2 2

1 2

1 1 1.9953b b

f f

f f

(23.77)

22 2 2

1 2 1 2

1 1.9953b b b b

f f f

f f f f

(23.78)

Multiply through by the common denominator and re-arrange:

24 2 2 21 2 1 20.9953 0b b b bf f f f f f (23.79)

Use the quadratic formula to solve for f 2 :

2 22 2 2 2

1 2 1 2 1 224 0.9953

2b b b b b bf f f f f f

f

(23.80)

2 22 2 2 2 21 2 1 2 1 20.5 0.25 0.9953b b b b b bf f f f f f f (23.81)

Only the plus square root will give a positive value for f 2 (can you explain why?), and only the positive root of f 2 results in a real number. Thus, the frequency at which the lowpass response drops 3 dB when the two break frequencies are within one decade of each other is

2 22 2 2 21 2 1 2 1 20.5 0.25 0.9953b b b b b bf f f f f f f (23.82)

Actually, this equation is also valid for break frequencies spaced farther than one decade, but normal Bode plot approximations are valid in those cases.

_____________________________________________________________________________________________________________________


23.4 CLASS NOTES 23.1 Bipolar Junction Transistor Low Frequency Analysis 23.1.1 The Need for Low Frequency Response Analysis in BJT CE Amplifiers Figure 23.1: Cgen, the generator-side coupling capacitor CL, the load-side coupling capacitor CE, and the emitter-bypass capacitor

genV ~

iV

~

oV

Figure 23.1 Capacitors in a CE BJT amplifier

Qualitatively, how would these capacitors affect the gain as the frequency is decreased? Qualitatively, how would these capacitors affect the gain as the frequency is increased? (High frequency analysis of the BJT amplifier will be considered in the next chapter.)

_____________________________________________________________________________________________________________________


What general approach can be taken to analyze an amplifier at very low (or very high) frequencies? Overall strategy:

1) Identify the equivalent RC networks formed by the capacitances and resistances in the circuit,

2) Determine the transfer function of each RC network, as is done in Bode plot analysis, and 3) Modify the midband gain frequency response using the break frequencies and filtering type

(highpass in the low frequency analysis). Assumption: Each frequency roll-off effect is independent of the others (they do not interact).

Technically not true, but usually “true enough” for good frequency response predictions. For more exacting predictions, simulate using the more complicated transistor models

that are contained in software packages. The purpose of manual circuit analysis (analysis by hand calculations) is to form

starting points for designs and to make reasonable but not exact predictions. 23.1.2 Highpass Effects of the Coupling Capacitors What is the circuit in Figure 23.2?____________________________________________

iVgenV

bI

c bI I

oV

Figure 23.2

_____________________________________________________________________________________________________________________


Which capacitors are considered first?___________________________________________ Note: RE is assumed to be bypassed per the assumption of “non-interacting” capacitances. Often, what is the “generator”?_________________________________________________ Often, what is the load resistance?_______________________________________________ What type of filter does the series Rgen– Cgen–Ri circuit form?__________________________ What type of filter does the series Ro – CL – RL circuit form?___________________________ Notes: Ri is the midband input impedance (resistance) of the BJT amplifier Ro is the midband output impedance Although Rgen is technically not part of the amplifier stage, why can’t it be ignored? Explain each step in the following development:

geno o

i i gen

VV VH j

V V V

(23.1)

The generator resistance-input resistance voltage divider is defined to be at midband (Cgen is effectively a short). Why?

gen i gen gen ii

gen i i i

V R V R RV

R R V R

(23.2)

i B eR R r .

gen gen io o o

i i gen i gen

V R RV V VH j

V V V R V

(23.3)

_____________________________________________________________________________________________________________________


1C

o L L b L

C LL

RV I R I R

R R +j C

(23.4)

1genB B

b genB e B e

gen igen

VR RI I

R + r R + rR + R j C

(23.5)

1 1gen CB

o LB e

gen i C Lgen L

V RRV R

R + rR + R R R +j C j C

(23.6)

1 1gen C LB

B egen i C L

gen Lgen i

i gen

V R RRR + rR + R R R +

j C j CR RH j

R V

(23.7)

What does B eB e

B e

R rR r

R r

equal?______________________________________________

11 1

gen i C LB

B e B egen i C L

gen LB e

R R R RRH j

R r R + rR + R R R +j C j CR r

(23.8)

11 1

gen i C L

egen i C L

gen L

R R R RH j

r R R R R j C j C

(23.9)

11 1

gen i gen C L L

e gen Lgen i C L

gen L

R R j C R R j CH j

r j C j CR R R R j C j C

(23.10)

_____________________________________________________________________________________________________________________


1

1 1gen i gen C L L

e gen i gen C L L

j R R C j R R CH j

r j R R C j R R C

(23.11)

The next step simplifies the final results: Factor out the constant C L

C L

R R

R R

from the numerator

of the last factor to make the break frequencies of the numerator and denominator equal. Explain how this was accomplished in the next equation:

1 1

gen i gen C L LC L

gen i gen e C L C L L

j R R C j R R CR RH j

j R R C r R R j R R C

(23.12)

What is

C Lb

C Lo C Lv

i b e e C L

R RI

R RV R RA

V I r r R R

? (23.13)

1 1gen i gen C L L

vgen i gen C L L


j R R C j R R C

(23.14)

Why was the constant C L

C L

R R

R R

factored from the numerator?

1 1

1 11 1

gen i gen C L Lv

C L Lgen i gen

j j

R R C R R CH j A

j j

R R CR R C

(23.15)

1 1

bg bLv

bg bL

j j

H j Aj j

1 1

, bg bLC L Lgen i gen R R CR R C

(23.16)

_____________________________________________________________________________________________________________________


Does the generator resistance affect the generator-side break frequency?__________________ Does the load resistance affect the load-side break frequency?__________________________

( ) ( )og v L

i

VH j H j A H j

V

(23.17)

1 1



gen i gen C L L

v

C L Lgen i gen

Lg

j j

R R C R R CH j A

j j

R R CR R C

H jH j

(23.18)

Explain the general approach to analyze the highpass filtering of the low frequency response of a BJT CE amplifier due to the coupling capacitors: Av

1

, where 1

bgg bg

gen i gen

bg

j

H jR R Cj

(23.19)

1

, where 1

bLL bL

C L L

bL

j

H jR R Cj

(23.20)

( ) ( )og v L

i

VH j H j A H j

V

(23.21)

The amplifier low frequency response due to the coupling capacitors is obtained by

modifying the midband loaded voltage gain with the highpass responses of the generator and load coupling capacitances (via the break frequencies).

Where did the break frequencies come from? ________________________________________

_____________________________________________________________________________________________________________________


Example 23.1.1 Determine the midband and low frequency response for the BJT amplifier shown in Figure 23.3. Consider only the coupling capacitors in the low frequency response (assume the emitter resistor is bypassed).

VCC = +15 V

Cgen

R1RC

CL = 10 μF

RER2

CE

Rgen

RL

genV ~

iV

~

oV200 Ω 10 μF

50μF

1200 Ω

3300 Ω

20 kΩ

2 kΩ120Ω

Figure 23.3 Amplifier for Examples 23.1.1 and 2 Strategy: (Fill in.)

Solution: (Explain each step.)

2

1 2

15 20001.3636 V

20k 2000CC

Th

V RV

R R

1 2

1 2

20k 20001.818 k

20k 2000Th

R RR

R R

KVL: 1 0

E

Th B Th BE B E

I

V I R V I R

_____________________________________________________________________________________________________________________


1.3636 0.7

37.837 A1 1818 131 120

Th BEB

Th E

V VI

R R

1 131 37.837 A 4.9567 mAE BI I

0.026 0.026

5.24544.9567 me

E

rI

~

oV~

iV

bI

130 bIbI

Figure 23.4 _______________________amplifier model ( 1 2|| 1818BR R R )

1200 3300167.77

5.2454 1200 3300

dB 20log 20log 167.77 44.5 dB

C Lb

C Lo C Lv

i b e e C L

v v

R RI

R RV R RA

V I r r R R

A A

RB

CLCgen

~

oV~

iV

Rgen

genVRLre

RC

bI

130 bI200 Ω

1200 Ω 3300 Ω1818 Ω682 Ω

10 μF10 μF

Figure 23.5 ______________________________________________ Model

_____________________________________________________________________________________________________________________


The input of the amplifier is an Rgen-Cgen-Ri ______________ filter with 496i e BR r R .

1 1

, 144 rad/s , 23 Hz200 496 101

bgg bg bg

gen i gen

bg

j

H j fR R Cj

The output of the amplifier is an RC-CL-RL ______________________ filter.

1 1

, 22 rad/s , 3.5 Hz1200 3300 101

bLL bL bL

C L L

bL

j

H j fR R Cj

( ) ( )og v L

i

VH j H j A H j

V

23 3.5, 44.5 dB, and 1 1

23 3.5

g v L

f fj j

H j A H jf f

j j

Which highpass break frequency dominates the low frequency response (Figure 23.6)?

Figure 23.6 Low frequency response of the amplifier due to coupling capacitors Why does 40 dB/dec occur below 3.5 Hz?

_____________________________________________________________________________________________________________________


23.1.3 Highpass Effect of the Emitter-Bypass Capacitor Qualitatively, what is the effect of the emitter-bypass capacitor, CE , vs. frequency? Note: The generator and load-side coupling capacitors are replaced by shorts per the

assumption of “non-interacting” capacitances. In general, how does the AC analysis proceed?________________________________________

genViV oV

bIiI

c bI I

Figure 23.7 AC BJT model with RE and CE

Explain each step:

|| ||

11

111

b C L b C Lo

i EE b e

E E Eb e b

EE

I R R I R RVH j

V RR I rj C j R CI r I

Rj C

(23.22)

|| ||

1 1

C L C L

E Ee e

E E E E

R R R RH j

R Rr r

j R C j R C

(23.23)

||C Lv unbypassed

e E

R RA

r R

(23.24)

_____________________________________________________________________________________________________________________


What algebraic strategy is performed in the next step:

|| ||c L c Le ev unbypassed v bypassed

e E e E e e E

v bypassedA

R R R Rr rA A

r R r R r r R

(23.25)

Similarly, algebraically manipulate Eq. (23.23) to show Av-unbypassed and Av-bypassed :

|| || 11

1 11

C L C L E EE E

E E E e E E Ee

E E

R R R R j R Cj R CH j

R j R C r j R C Rr

j R C

(23.26)

|| 1C L E E

e E e E E

R R j R CH j

r R j r R C

(23.27)

11

|| 1 ||

11

E EC L E E C L

e E E e Ee E

e Ee E v unbypassed

e E E

A

j

R CR R j R C R RH j

r R C r Rjr R j

r Rr Rr R C

(23.28)

3 3

4 4

1 1

1 1

o b e bv unbypassed v bypassed

s e E

b b

H jE

j jV r

H j A AV r Rj j

(23.29)

3

1b

E ER C (23.30) , 4

1e Eb

e Ee E EE

e E

r Rr Rr R C

Cr R

(23.31)

What is EH j ?______________________________________________________________

Which is larger, ωb3 or ωb4? Why? _________________________________________________ What is the gain above ωb4? ______________________________________________________ What is the gain below ωb3? ______________________________________________________

_____________________________________________________________________________________________________________________


Comment: Break frequency ωb3 often (but not always) occurs below the break frequencies in the highpass responses due to the coupling capacitors.

b3 b4

no thru e EI C

|H | (dB)

Av-bypassed

all thru e EI C

Av-unbypassed

Figure 23.8 Frequency response due to CE

( ) ( )og v bypassed E L

i

VH j H j A H j H j

V (23.32)

1

, where 1

bgg bg

gen i gen

bg

j

H jR R Cj

(23.33)

||c L C Lv bypassed

e e C L

R R R RA

r r R R

(23.34)

33 4

4

11

, where and 1

e b e EE b b

e E E E e E E

b

jr r R

H jr R R C r R Cj

(23.35)

1

, where 1

bLL bL

C L L

bL

j

H jR R Cj

(23.36)

The amplifier low frequency response is obtained by modifying the midband loaded voltage gain with the highpass responses of the generator and load coupling capacitors and the highpass response of the emitter-bypass capacitor (via the break frequencies).

_____________________________________________________________________________________________________________________


Example 23.1.2 Determine the low frequency response for the BJT amplifier shown in Figure 23.3 of Example 23.1.1. Include the effect of the emitter-bypass capacitor.

Reminder: Strategy that was executed in Ex. 23.1.1:

Determine bias current IE re = 0.026/IE Build the midband amplifier model Determine Av Build the low frequency amplifier model Determine the high pass break frequencies Sketch/label midband and low freq. magnitude Bode plot.

What additional analysis needs is required to complete the low frequency analysis?

genViV oV

bIiI

c bI I

130 bI

Figure 23.9 Low frequency BJT amplifier model

iV oV

bI

130 bI

Figure 23.10 Model for HE(jω) determination

_____________________________________________________________________________________________________________________


From Ex. 23.1.1:

1200 3300167.77

5.2454 1200 3300

dB 44.5 dB

C Lb

C Lo C Lv

i b e e C L

v

R RI

R RV R RA

V I r r R R

A

1 1

, 144 rad/s , 23 Hz200 496 101

bgg bg bg

gen i gen

bg

j

H j fR R Cj

1 1

, 22 rad/s , 3.5 Hz1200 3300 101

bLL bL bL

C L L

bL

j

H j fR R Cj

Explain the steps in the rest of this example:

3

4

1

1

e bE

e E

b

jr

H jr R j

3 3

1 1166.7 rad/s , 26.5 Hz

120 50b bE E

fR C

4 4

5.2454 1203979.5 rad/s , 633 Hz

5.2454 120 50e E

b be E E

r Rf

r R C

Which break frequency (ωb3 , ωb4 , ωbg or ωbL) dominates the low frequency response? Why?

Figure 23.11. The low frequency amplifier response for Example 23.1.2

_____________________________________________________________________________________________________________________


23.2 Field Effect Transistor Low Frequency Analysis 23.2.1 The Need for Low Frequency Response Analysis in FET CS Amplifiers Which capacitances in Figure 23.12 affect the low frequency response of the FET amplifier? Qualitatively, how would these capacitors affect the gain as the frequency is decreased? Why? Thus, the generator and load coupling capacitors form ___________________________ filters. How does CS affect the amplifier gain as frequency decreases? Why?

iVoVgenV

Figure 23.12 Capacitors in a CS FET amplifier What is the effect of the coupling and source-bypass capacitors at high frequencies? (High frequency analysis of the FET amplifier will be considered in the next chapter.)

_____________________________________________________________________________________________________________________


What general approach can be taken to analyze an amplifier at very low (or very high) frequencies? Overall strategy:

1) Identify the equivalent RC networks formed by the capacitances and resistances in the circuit,

2) Determine the transfer function of each RC network, as is done in Bode plot analysis, and 3) Modify the midband gain frequency response using the break frequencies and filtering type

(highpass in the low frequency analysis). Assumption: Each frequency roll-off effect is independent of the others (they do not interact).

Technically not true, but usually “true enough” for good frequency response predictions. For more exacting predictions, simulate using the more complicated transistor models

that are contained in software packages. The purpose of manual circuit analysis (analysis by hand calculations) is to form

starting points for designs and to make reasonable but not exact predictions. 23.2.2 Highpass Effects of the Coupling Capacitors

RG

RD

CL

Cgen

~

oV~

iV~

gsV~

m gsg V

Rgen

genV

RL

Figure 23.13 FET amplifier model with coupling capacitances

Which capacitors are considered first?___________________________________________ Note: RS is assumed to be bypassed per the assumption of “non-interacting” capacitances.

_____________________________________________________________________________________________________________________


What type of filter does the series Rgen– Cgen–RG circuit form?__________________________ What type of filter does the series RD – CL – RL circuit form?___________________________ Notes: Ri is the midband input impedance (resistance) of the FET amplifier Ro is the midband output impedance Although Rgen is technically not part of the amplifier stage, why can’t it be ignored? Explain each step in the following development:

geno o

i i gen

VV VH j

V V V

(23.37)

i

i i

gengeni i

gen gen i

R RVV R

V R R V R

(23.38)

i GR R

gen gen io o o

i i gen i gen

V R RV V VH j

V V V R V

(23.39)

1D

o L L m gs L

D LL

RV I R g V R

R R +j C

(23.40)

1gen i

gs

gen igen

V RV

R R +j C

(23.41)

1 1gen i D

o m L

gen i D Lgen L

V R RV g R

R R + R R +j C j C

(23.42)

_____________________________________________________________________________________________________________________


1 1gen i D L

m

gen i D Lgen Lgen io

i i gen

V R R Rg

R R + R R +j C j CR RV

H jV R V

(23.43)

1 1

gen i gen D L Lm

gen Lgen i D L

gen L

R R j C R R j CH j g

j C j CR R + R R +j C j C

(23.44)

11

gen i gen D L Lm

D L Lgen i gen

j R R C j R R CH j g

j R R C j R R C

(23.45)

11

gen i gen D L LD Lm

D L D L Lgen i gen


R R j R R C j R R C

(23.46)

11

gen i gen D L LD Lm

D L D L Lgen i gen


R R j R R C j R R C

(23.47)

D Lv m

D L

R RA g

R R

(23.48)

11

gen i gen D L Lv

D L Lgen i gen


j R R C j R R C

(23.49)

_____________________________________________________________________________________________________________________


1 1

1 11 1

gen i gen D L Lv

D L Lgen i gen

j j

R R C R R CH j A

j j

R R CR R C

(23.50)

1 1

bg bLv

bg bL

j j

H j Aj j

1 1

, bg bLD L Lgen i gen R R CR R C

(23.51)

( ) ( )og v L

i

VH j H j A H j

V

(23.52)

1 1



gen i gen D L Lv

D L Lgen i gen

Lg

j j

R R C R R CH j A

j j

R R CR R C

H jH j

(23.53)

1

, where 1

bgg bg

gen i gen

bg

j

H jR R Cj

(23.54)

1

, where 1

bLL bL

D L L

bL

jH j

R R Cj

(23.55)

_____________________________________________________________________________________________________________________


1 1

bgo bLv

i

bg bL

j jV

H j AV j j

(23.56)

( ) ( )og v L

i

VH j H j A H j

V

(23.57)

The amplifier low frequency response due to the coupling capacitors is obtained by modifying the midband loaded voltage gain with the highpass responses of the generator and load coupling capacitors (via the break frequencies). Example 23.2.1 For the FET common source amplifier shown in Figure 23.14, determine the

midband gain and the low frequency response due to the coupling capacitors, given IDSS = 7 mA and gm = 4 mS. (Assume the source resistor is bypassed.)

iVoV

genV

Figure 23.14 FET amplifier circuit for Example 23.2.1 and 2

__________: IDSS = 7 mA, gm = 4 mS __________: Av, Low frequency response due to the coupling capacitors

_____________________________________________________________________________________________________________________


Strategy: (Fill in.) Solution: (Explain each step.)

3 1500 2,0004 10 5.33 14.5 dB

1500 2,000o D L

v mi D L

V R RA g

V R R

1 1

bgo bLv

i

bg bL

j jV

H j AV j j

1

where bg

gen i genR R C

and

1

bLD L LR R C

6

1 10.333 rad/s

100 300k 10 10bg

gen i genR R C

6

1 17.41 rad/s

1500 12k 10 10bL

D L LR R C

0.333 7.415.331 1 1 1

0.333 7.41

bgo bLv

i

bg bL

j j j jVH j A

V j j j j

Which highpass frequency dominates the low frequency response of this FET amplifier? Why?

_____________________________________________________________________________________________________________________


Figure 23.15 Low frequency response due to the coupling capacitors in Example 23.2.1 __________________________________________________________________________ 23.2.3 Highpass Effect of the Source-Bypass Capacitor Qualitatively, what is the effect of the source-bypass capacitor, CS , vs. frequency? Note: The generator and load-side coupling capacitors are replaced by shorts per the

assumption of “non-interacting” capacitances. In general, how does the AC analysis proceed?________________________________________ Why are the coupling capacitances replaced by shorts?

iV

gsV

oV

m gsg V

Figure 23.16 AC FET model with RS and CS

_____________________________________________________________________________________________________________________


1

1

D Lm gs

D Lo

iS

Sgs m gs

SS

R Rg V

R R VH j

VR

j CV g V

Rj C

(23.58)

1

11

v bypassedo

iS

S Sm

SS

S

AVH j

VR

j C j Cg

j CRj C

(23.59)

1

11

1

v bypassedo S S

Si S Sm

S S

AV j R CH j

RV j R Cg

j R C

(23.60)

1 1

1 1S S S S

v bypassed v bypassedS S m S m S S S

j R C j R CH j A A

j R C g R g R j R C

(23.61)

1

11

1

v bypassed S S

S Sm S

m S

A j R CH j

R Cg Rj

g R

(23.62)

The constant 1

v bypassed

m S

A

g R

is recognized as ____________________________________________

11

11

1

v bypassed S S

m S

m S

S S

j

A R CH j

g Rj

g R

R C

(23.63)

_____________________________________________________________________________________________________________________


3

4

11

1 1

o bv bypassed

i m S

b

jV

H j AV g R j

H jS

(23.64)

3

1b

S SR C (23.65) 4

1 m Sb

S S

g R

R C

(23.66)

Which is larger: ωb4 or ωb3? ______ Why? What is the gain above ωb4? ______ Why? What is the gain below ωb3? ______________________________________________ Why?

no thru s SI Call thru s SI C

Figure 23.17 Frequency response due to CS


i

VH j H j A H j H j

V (23.67)

1

, where 1

bgg bg

gen i gen

bg

j

H jR R Cj

(23.68)

_____________________________________________________________________________________________________________________


D Lv bypassed m

D L

R RA g

R R

(23.69)

33 4

4

111 1

, where and 1 1

b m SS b b

m S S S S S

b

jg R

H jg R R C R Cj

(23.70)

1

, where 1

bLL bL

D L L

bL

jH j

R R Cj

(23.71)

The amplifier low frequency response is obtained by modifying the midband loaded voltage gain with the highpass responses of the generator and load coupling capacitors and the highpass response of the source-bypass capacitor (via the break frequencies). Example 23.2.2 For the FET common source amplifier shown in Figure 23.14, determine

the midband gain and the low frequency response including the effects of the source bypass capacitor as well as the coupling capacitors, given IDSS = 7 mA and gm = 4 mS.

_____________: IDSS = 7 mA, gm = 4 mS _____________: Av, Low frequency response due to the coupling and bypass capacitors Strategy: (Fill in.)

Solution: (Explain each step.)


i

VH j H j A H j H j

V

_____________________________________________________________________________________________________________________


3

4

11

11 1 1

bg b bLv bypassed

m S

bg b bL

j j j

H j Ag Rj j j

Av-bypassed, ωbg, and ωbL were determined in Example 23.2.1:

3 1500 12,0004 10 5.33 14.54 dB

1500 12,000o D L

v bypassed mi D L

V R RA g

V R R

6

1 10.333 rad/s

100 300k 10 10bg

gen i genR R C

6

1 17.41 rad/s

1500 12k 10 10bL

D L LR R C

Calculate the remaining pertinent quantities for this example and insert into H(jω). (The results are on the next page.)

_____________________________________________________________________________________________________________________


Results: 3 51.3 rad/sb , 4 131.3 rad/sb ,

10.3906 8.16 dB

1 m Sg R

1

20log 6.4 dB1v unbypassed v bypassed

m S

A Ag R

3

4

11

11 1 1

bg b bLv bypassed

m S

bg b bL

j j j

H j Ag Rj j j

14.5 dB

2.08 6.4 dB

10.333 51.3 7.415.33 0.3906

1 1 10.333 131.3 7.41

j j jH j

j j j

Figure 23.18 Low frequency response of the FET CS amplifier in Example 23.2.2 Which break frequency dominates the low frequency response of this amplifier? _____________ Why?

_____________________________________________________________________________________________________________________


23.5 HOMEWORK PROBLEMS 1. How are coupling and bypass capacitors modeled within midband in a small signal

analysis of a transistor amplifier? 2. How are coupling and bypass capacitors modeled above midband in a small signal

analysis of a transistor amplifier? 3. How are coupling and bypass capacitors modeled below midband in a small signal

analysis of a transistor amplifier? 4. What is the criterion to determine which break frequency dominates the low frequency

response of a transistor amplifier? 5. What key assumption was made about the roll-off effect of each break frequency?

_____________________________________________________________________________________________________________________


6. For the circuit in Figure 23.6, a) determine the break frequencies and sketch/label the magnitude Bode plot showing the

dominant highpass break frequency (note: the bias solution was determined in Problem 6 of Ch. 21, and the midband analysis was performed in Problem 6 of Ch. 22),

b) simulate with a 2N3904 BJT to determine both the magnitude and phase low and

midband frequency responses, c) compare the results to parts a) and b), especially any differences (explain why), d) remove CE and re-simulate, and e) compare the results of parts b) and d) to determine the effects of CE on the frequency

response. Include a comparison of the practical bandwidth over which the phase is relatively constant.

NOTE: The phase response in MultiSIM often “jumps” between -180° and +180°. This

action is due to the method that the phase response is calculated (and is common in many programs and instrument readouts). Actually, the phase is continuing below -180° and is not to be construed as a sudden phase change.

R1 = 6.8 kΩ R2 = 1 kΩ RC = 560 Ω RE = 120 Ω RL = 820 Ω Cgen = 1 μF CL = 1 μF CE = 22 μF VCC = 9 V Rgen = 50 Ω β = 165

VCC

Cgen

R1

RC

CL

RER2CE

Rgen

RL

genV ~

iV

~

oV

Figure P23.6 BJT amplifier circuit for Problem 6

_____________________________________________________________________________________________________________________


7. For the circuit in Figure P23.7,

a) determine the break frequencies and sketch/label the magnitude Bode plot showing the dominant high pass break frequency (note: the bias solution was determined in Problem 7 of Ch. 21, and the midband analysis was performed in Problem 7 of Ch. 22),

b) simulate with a 2N3904 BJT to determine both the magnitude and phase low and

midband frequency responses, c) compare the results to parts a) and b), especially in differences (explain why), d) remove CE and re-simulate, and e) compare the results of parts b) and d) to determine the effects of CE on the frequency

response. Include a comparison of the practical bandwidth over which the phase is relatively constant.

RB = 220 kΩ RC = 330 Ω RE = 560 Ω RL = 820Ω β = 165 Cgen = 1 μF CL = 1 μF CE = 10 μF VCC = 9V Rgen = 50 Ω

genV ~

iV

~

oV

Figure P23.7 BJT amplifier circuit for Problem 7

_____________________________________________________________________________________________________________________


8. For the circuit in Figure P23.8,

a) determine the break frequencies and sketch/label the magnitude Bode plot showing the dominant high pass break frequency (note: the bias solution was determined in Problem 8 of Ch. 21, and the midband analysis was performed in Problem 8 of Ch. 22),

b) simulate with a 2N4393 JFET to determine both the magnitude and phase low and

midband frequency responses, c) compare the results to parts a) and b), especially any differences (explain why), d) remove CS and re-simulate, and e) compare the results of parts b) and d) to determine the effects of CS on the frequency

response. RG = 510 kΩ RD = 1 kΩ RS = 560 Ω RL = 1 kΩ IDSS = 20 mA

VDD = 9V Rgen = 50 Ω VGS(off) = -1.5 V Cgen = 1 μF CL = 1 μF CS = 10 μF

iVoVgenV

Figure P23.8 FET amplifier circuit for Problem 8

9. How are the low-frequency response analyses of BJT and FET amplifiers similar?

Different?

Chapter 23 Transistor Amplifier Low Frequency...

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