BJT Frequency Response

BJT Frequency Response 

Electronic Circuit Analysis

Agenda 

Low Frequency Predictions 

High Frequency Predictions 

Copyright 2004 Richard Lokken 2

Frequency Analysis Strategy 

Identify the equivalent RC networks formed by 

the capacitances and resistances in the circuit, 

determine the transfer function of each RC 

network, as done in Bode plot analysis, 

modify the midband gain frequency response 

using the break frequencies and filtering type 

(lowpass or highpass). 


Capacitances in a CE BJT amplifier 

I 

i 

I 

b 

R gen 

C gen 

C E 

C L 

C bc 

V 

gen 

V 

i 

R B 

βr e 

C be 

R C 

R L 

I 

c 

= βI 

b 

R B 

= R 1 

||R R E 2 

V 

o 


Typical Frequency Response 

H (dB) 

midband A v 

f (Hz) 


Low Frequency Analysis 

Consider the high pass effects of the coupling 

capacitors first. 


BJT model with Coupling C’s 

R gen 

C gen 

b 

V 

gen 

V i 

R B 

βr e 

I 

I c b 

= βI 

R C 

C L 

V 

o 

R L 

R i 


Determine the Transfer Function 

H 

V 

V 

ω = 

o 

= 

gen 

• 

( j ) 

V 

V V V 

o 

i i gen 

The generator resistance-input resistance voltage 

divider is defined to be at midband (C gen ) is effectively a 

short. 


Voltage divider at midband 

R 

V 

R + R 

i 

= 

i 

→ 

gen 

= 

gen i 

+ 

 

gen gen i i i 

V 

V 

R R V R 

R i = R B || βr e 


Transfer Function development 

H 

V V 

V R + R V 

jω = = • = • 

V V V R V 

( ) 

o gen o gen i o 

i i gen i gen 


Use CDR to determine the load I 

RC 

V 

 

= I R = −βI 

R 

1 

RC 

+ R 

L 

+ j ω C 

 

L 

o L L b L 


Determine 

I b 

from 

V gen 

I 

b 

RB 

V 

 

gen RB 

 

= I 

gen 

R 1 

B 

+ r 

= e R 

R 

B 

+ r 

 

β e 

gen 

+ + R 

β 

 

i 

jωC 

 

 

gen 


V 

o 

V 

 

gen RB 

RC 

 

= −β R 

1 

R 1 

R B 

+ r 

 

e 

gen 

+ + R β 

i 

RC + R 

L 

+ 

 

jωC 

 

gen 

jωC 

 

 

 

L 

L 

H 

V 

 

gen RB 

RC RL 

 

−β 

1 R 1 

B 

+ r 

R e 

gen 

+ + R β 

i RC R 

L 

+ 

R 

j C 

gen 

R 

+ 

i 

gen 

j C 

+ ω 

ω 

L 

ω = 

 

 

 

Ri 

 

V 

gen 

( j ) 


H 

Rgen + Ri 1 RB 

RC RL 

 

 

R 1 1 

B 

r 

 

e 

R 

R 

B 

+ r 

 

β e 

gen 

+ R 

β 

+ 

i 

RC + R 

L 

+ 

R 

j C 

B 

r 

e 

gen 

j C 

+ β ω ω 

L 

( jω ) = 

( −β) 

H 

−1 

Rgen + Ri RC RL 

 

ω = 

r 

1 

1 

e Rgen + R 

i 

+ RC + R 

L 

+ 

j C 

gen 

j C 

ω ω 

L 

( j ) 


Clear compound fractions 

H 

ω = −1 Rgen + Ri jωCgen RC RL jωCL 

 

 

r 

 

 

1 1 

e R 

j C gen 

j CL 

gen 

+ R 

i 

+ ω RC + R 

ω 

 

L 

+ 

j C 

 

gen 

j C 

 

ω ω 

L 

( j ) 

H 

( j ) 

( gen i ) gen C L L 

( ) ( ) 

−1 jω R + R C jωR R C 

ω = 

r e 

1 j Rgen Ri C 

gen 

1 j RC RL C 

 

+ ω + 

 

 

+ ω + 

L 


Simplify the final results 

H 

( j ) 

( ) −RC 

RL 

( ) ( ) 

( ) 

( ) 

jω R + R C gen i gen 

jω R + R C 

ω = C L L 


 

 

+ ω + 

gen re RC + R 

L 1+ jω RC + RL C 

L 

( ) 

( ) 

( ) 

( ) 

jω Rgen + Ri Cgen jω RC + RL CL 

 

H ( jω ) = [ Avm 

] 


 

 

+ ω + 

gen 1+ jω RC + RL C 

L 


Obtain j ω/ω b 

ω 

j 

ω 

j 

1 

1 

 

 

( Rgen Ri ) C 

 

+ 

gen 

 

( RC RL ) C 

+ 

L 

H ( jω ) = 

 

[ A 

 

v ] 

ω 

ω 

1+ j 1+ 

j 

 

1 1 

 

( Rgen + Ri ) C 

 

gen 

( RC + RL ) C 

L 

 

 

 


ω 

j 

ω 

j 

ω 

( ) 

bg ω 

bL 

H jω = A 

v 

ω 

1 j 

ω 

1 j 

 

+ 

 

+ 

 

ωbg 

ωbL 

 


Break Frequencies 

ω = 

bg 

1 

( R + ) 

R C 

gen i gen 

ω = 

bL 

1 

( + ) 

R R C 

C L L 


Example 

V CC 

= +15 V 

R gen 

200 10 F 

~ 

V 

gen 

V i 

R 

R C 

1 

C L 

= 10 F 

20 k 

C gen 

R 2 

2 k 

R E 

12 

k 

1200 

β = 130 

C E 

50 

F 

~ 

V o 

R L 

3300 


Desired 

Mid band gain 

Low Frequency Response 


Strategy 

Determine bias current I E 

r e 

= 0.026/I E 

Build the midband amplifier model 

Determine A v 

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 

V 

Th 

= V ( ) 

CCR2 

15 2000 

1.3636 V 

R + R 

= 20k + 2000 

= 

1 2 

R 

Th 

= R ( )( ) 

1R2 

20k 2000 

1.818 k 

R + R 

= 20k + 2000 

= Ω 

1 2 


KVL: V ( 1) 

Th 

− IBRTh −VBE − I 

B 

β + RE 

= 0 

 

I 

E 

I 

I 

r 

B 

E 

e 

VTh 

−VBE 

1.3636 − 0.7 

= = = 37.837 µ A 

R + β + + 

Th 

( 1) R 1818 ( 131)( 120) 

E 

= β + 1 I = 131 37.837 µ A = 4.9567 mA 

( ) ( )( ) 

B 

0.026 0.026 

= = = 5.2454 Ω 

I 4.9567 m 

E 


Midband Amplifier Model 

~ 

I 

b 

βI 

b 

130I 

b 

~ 

R B 

V o 

R C 

|| R L 

Vi 

βr 

1818 

e 880 

 

682 


Determine A v 

A 

v 

RC 

RL 

−βI 

b 

V 

( R ) 

o C 

+ RL 

−RC RL 

= = = 

V 

I 

β r r R + R 

( ) 

i b e e C L 

−1200( 3300) 

 

= 

= −167.77 

5.2454( 1200 3300) 

 

 

+ 

A 

v 

( ) ( ) ( ) 

dB = 20log A = 20log 167.77 = 44.5 dB 

v 


Coupling Capacitor 

Low Frequency Model 

V 

gen 

R gen 

200 

~ 

V i 

C gen 

C L 

I 

b 

10 F 

130I 

b 

R C 

~ 

R B βre 

V o 

10 F 

1818 

682 

1200 

R L 

3300 


Determine the High Pass 

break frequencies 

V 

gen 

R gen 

200 

~ 

V i 

C gen 

C L 

I 

b 

10 F 

130I 

b 

R 

~ 

R C 

B 

βre 

V o 

10 F 

1818 

682 

1200 

R L 

3300 


Input Coupling Capacitor 

H 

g 

( j ) 

ω = 

ω 

j 

ωbg 

ω 

1+ j ω 

bg 

1 1 

ω 

bg 

= = 

200 + 496 10µ 

( R + R ) C ( ) 

gen i gen 

= 144 rad/s , f = 23 Hz 

bg 


Output Coupling Capacitor 

H 

L 

( j ) 

ω = 

ω 

j 

ωbL 

ω 

1+ j ω 

bL 

1 1 

ω 

bL 

= = 

+ 1200 + 3300 10µ 

( R R ) C ( ) 

C L L 

= 22 rad/s , f = 3.5 Hz 

bL 


Total Amplifier Transfer Function 

H 

g 

( j ) 

ω = 

j 

1+ 

f 

23 

f 

j 

23 

A 

v 

= 

44.5 dB 

H 

L 

( j ) 

ω = 

j 

1+ 

f 

3.5 

f 

j 

3.5 


Bode Plot for Example 

H (dB) 

44.5 dB 

-20 dB/dec 

-40 dB/dec 

3.5 

23 

f (Hz) 


What about C E ? 

H 

( j ) 

V 

V 

o 

ω = = 

= 

I 

b 

i 

 

 

r 

−βI 

b ( RC || RL 

) 

 

RE 

( 1) 

 

RE 

+ 

 

I 

b ( RC || RL 

) 

RE 

( 1) 

1 

1 

j C 

E 

I 

ω 

bβ re + Ib 

β + 

1 

jωC 

 

E 

−β 

 

β 

e 

+ β + + jωREC 

 

E 

 

 


β > 100 , ( β + 1) 

≈ β 

H 

jω = 

( ) 

−β 

R 

|| 

( ) 

C 

R 

RE 

 

β re 

+ β 1 j REC 

 

+ ω 

E 

L 

= 

r 

e 

− 

R 

|| 

( ) 

C 

R 

RE 

 

+ 1 j REC 

 

+ ω 

E 

L 


AC BJT model with R E and C E 

i 

I 

b 

V 

gen 

R gen 

V 

i 

I c b 

βr e 

R 

I 

= βI 

C 

R L 

R B 

R B 

= R 1 

||R R 

2 

E 

C E 

V 

o 


Strategic Algebraic Manipulation 

A 

v−unbypassed 

−( R || ) ( || ) 

c 

RL re 

 

− Rc RL 

 

= = 

re + R 

 

E 

re R 

 

+ 

E re 

 

 

r 

e 

= 

re 

+ R 

E 

 

 

A 

 

v−bypassed 


Transfer Function 

H 

jω = 

( ) 

r 

e 

− ( R || R ) 1+ jωR C 

• R 1 + j ω R C 

+ 1 j REC 

 

+ ω 

E 

C L E E 

E E E 

= 

− 

R || R 1+ jωR C 

( )( ) 

C L E E 

r 1+ jω R C + R 

( ) 

e E E E 


H 

( j ) 

ω = 


− 

( R || R )( 1+ jωR C ) 

 

re 

+ RE 

1+ 

j 

 

( ) 

C L E E 

ωr R C 

r + R 

e 

e E E 

E 

 

 

 

ω 

1+ 

j 

1 

 

RC 

|| R 

R 

L 

EC 

 

 

E 

= − 

 

 

 

re 

R 

+ 

E ω 

1+ 

j 

re 

+ RE 

 

 

 

re REC 

 

E 


V 

ω = = 

( ) 

o 

H j Av unbypassed 

V 

− 

s 

 

1+ 

 

 

1+ 

 

j 

j 

ω 

ω 

ω 

ω 

b3 

b4 

 

 

 

 

 

ω 

1+ 

j 

re 

 

ω 

b3 

= Av −bypassed 

 

re 

R 

 

+ 

E ω 

1 j 

 

+ 

ω 

b4 

 

H ( jω) 

Copyright 2004 Richard Lokken 39 

E

Where the break frequencies are: 

ω = 

b3 

E 

1 

R C 

E 

r 

+ 

R 

ω = e E 

b4 

re R r 

EC 

= E eR 

E 

 

 

r 

e 

+ 

1 

R 

E 

C 

 

E 


Frequency Response due to C E 

H (dB) 

no 

I 

e 

A v-bypassed 

all thru 

thru 

C 

E 

I 

e 

C 

E 

A v-unbypassed 

ω 

ω b3 

ω b4 


Example 

i 

C gen 

I b 

C L 

V 

gen 

R gen 

200 

V 

i 

10 

F 

I c b 

βr e 

I 

= βI 

R L 

R 

RB 

C 

1200 

R E 

1818 

682 130I 

b 

10 F 

3300 

V 

o 

120 

C E 

50 F 


Desired 

Mid band gain 

Low Frequency Response 


Strategy 


r e 

= 0.026 / I E 


Determine A v 


Strategy 

Build the low frequency amplifier model 


Sketch and label the midband and low frequency 

magnitude Bode Plot. 


Model for H E (jω) 

I 

b 

V 

i 

βr e 

682 

R B 

R E 

C E 

130I 

b 

R C 

||R L 

880 

V 

o 

1818 

120 

50 F 



H 

E 

ω 

1+ 

j 

r 

ω 

jω = 

re 

R 

+ 

E ω 

1 j 

 

+ 

 

ωb 

4 

( ) 

e 

b3 


Break Frequencies 

1 1 

ω 

b3 

= = 

R C ( 120) 

50µ 

E 

E 

= 166.7 rad/s , f = 26.5 Hz 

b3 

re 

+ RE 

5.2454 + 120 

ω 

b4 

= = 

r R C ( 5.2454)( 120) 

50µ 

e E E 

= 3979.5 rad/s , f = 633 Hz 

b4 


Low frequency amplifier response 

44.5 dB 

H (dB) 

+20 dB/dec 

633 

f (Hz) 



AC BJT Model for 

High Frequency Response 

R gen 

I 

b 

C bc 

V 

gen 

V 

i 

R 1 

R 2 

C wi 

C be 

C wo 

βr e 

R C 

R L 

I 

c 

= βI 

b 

V 

o 


Generic Amplifier with 

Capacitance Feedback 

~ 

I f 

~ 

I inT 

C f 

~ 

I f 

~ 

I To 

~ 

V in 

~ 

I i 

A v 

~ 

I o 

~ 

V o 


Capacitances in the 

generic amplifier model 

~ 

V in 

C M 

A v 

C f 

~ 

V o 


Development of Miller Capacitance 

I = I + I 

inT i f 

I 

i 

V 

= 

R 

i 

i 


Examine Circuit 

V 

−V 

I = 

i o 

= jωC V −V 

1 

jωC 

 

f 

( ) 

f f i o 

A 

v 

V = 

V 

Copyright 2004 Richard Lokken 55 

o 

i

Insert A v 

( ) 

I = jωC V − A V 

f f i v i 

= jωC 

− 

A V 

( 1 ) 

f v i 


Insert two currents into KCL equation 

V 

I 

j C ( 1 A ) V 

R 

= 

i 

+ ω − 

inT f v i 

i 

1 

= V 

+ jωC − A 

R 

( 1 ) 

i f v 

i 

 

 

 


Solve for Total Input Admittance 

I 

1 

Y 

= 

inT 

= + jωC − A 

V 

R 

( 1 ) 

inT f v 

i i 


Solve for Total Admittance of circuit 

Y 

inT 

I 

1 

= 

inT 

= + jωC 

V 

R 

i 

i 

Mi 


By direct Comparison 

( 1 ) 

C = C − A 

Mi f v 


Perform a similar analysis at the output 

I = I − I 

oT o f 

I 

o 

V 

= 

R 

o 

o 


find current I f 

( ) 

V 

−V 

I = 

i o 

= jωC V −V 

1 

jωC 

 

f 

f f i o 


Insert the two currents 

V 

1 

I 

 

= 

o 

− jωC −1V 

R A 

oT f o 

o 

v 

1 1 

= V 

+ jωC 

1 

o 

f − 

R 

A 

o v 


Solve for Output Admittance 

Y 

oT 

I 

1 1 

= 

oT 

= + jωC 

1− 

f 

V 

 

R A 

o o v 

 

 

 


Solve for the Total Output Admittance of 

Circuit 

Y 

oT 

I 

1 

= 

oT 

= + jωC 

V 

R 

o 

o 

Mo 


By Direct Comparison 

1 

C = C 1− ≈ C 

A 

Mo f f 

v 


Determine Break Frequencies 

First develop a transfer function for the circuit. 

Determine input transfer function and break 

frequency 

Determine output transfer function and break 

frequency 


The Circuit 

R S 

~ 

I 

~ ~ 

c 

= β I 

b 

~ 

V i 

I b 

C w i 

C be 

R 1 R 2 

βr e C w o 

RC R L 

~ 

V o 

C M i C M o 

C i 

Ri 

C o 

R o 


Values from Equivalent Circuit 

C = C || C || C C = C + C 

i Mi wi be o Mo wo 

R = R || R || β r R = R || R 

i 1 2 e o c L 


Transfer Function at mid-band 

frequencies 

H 

V 

−βI 

R 

( jω ) = 

o 

= 

b o 

V 

I 

βr 

i b e 

−R 

= 

o 

= 

r 

e 

A 

v 


Incorporate the generator 

H 

V 

V 

V 

ω = 

o 

= 

gen 

• 

V V V 

( j ) 

o 

i i gen 

 

R + R V 

= 

gen i o 

• 

R V 

i gen 


Use VDR on generator side 

V 

o 

1 R 

o 

= −βI 

b 1 

j C 

R 

+ ω 

o 

o 

 

R 

o 

Ro 

 

= −βI 

b 

1+ jωRoC 

 

 

o 


Determine base current use VDR 

I 

b 

1 

V 

 

gen 

1 

+ jωCi 

 

Ri 

Ri 

V 

 

gen 

1 

1+ jωRiC 

 

i 

R 

gen 

+ 

1 

 

 

Ri 

+ jω Ci 

R 

gen 

+ 

V 

R 

i 

1 j R 

i 

iC 

 

+ ω 

i 

= = = 

 

βr βr βr 

e e e 


I 

b 

V 

 

gen Ri 

 

= 

β r 

( 1 ) 

e 

Rgen + jω RiCi + R 

i 

V 

 

gen Ri 

 

= 

β r R + R + jωR R C 

( ) 

e gen i gen i i 


Combine equations 

V 

o 

Ro 

 

= −βI 

b 

1+ jωRoC 

 

 

o 

V 

 

gen Ri Ro 

 

= −β 

 

β r ( ) 

e R 1 j R 

gen 

+ Ri + jωRgenRiC 

 

i + ω 

oC 

 

 

 

o 

 

 

 



H 

( j ) 

ω = 

 

 

 

Rgen + Ri V 

o 

• 

Ri 

V 

gen 

V 

 

gen Ri Ro 

 

− 

r 

( R ) 1 j R C 

gen 

+ Ri + jωRgenRiC 

+ ω 

 

i 

V 

R 

e o o 

gen 

+ Ri 

 

= • 

Ri 

 

gen 



H 

( j ) 

ω = 

Rgen + Ri −1 Ri Ro 

 

• 

R 

 

( ) 

i 

r 

e R 1 j R 

gen 

Ri j RgenRiC 

 

i 

oC 

 

+ + ω + ω 

 

 

o 


multiplied by the 

gen i 

factor 

 

 

R 

R 

+ 

i 

R 

 

 

H 

( ) 

( + ) 

( + ) + ω 

−1 Rgen R 

i Ro 

 

jω = 

r 

e R 1 j R 

gen 

Ri j RgenRiC 

 

i 

oC 

 

+ ω 

 

o 



H 

1 

( R ) ( ) 

gen 

+ R i 

Rgen + R 

i −R 

 

o 1 

ω = ( R ) 

1 re 1 j R 

gen 

Ri j RgenRiC 

 

 

 

+ ω 

i 

oC 

 

 

+ + ω 

o 

( R ) 

gen 

R 

 

 

+ 

i 

( j ) 

1 1 

H ( jω ) = [ A ] v 

R 

1 j R 

genR 

 

+ ω 

i 

oC 

 

 

o 

1+ j 

C 

i 

 

ω 

 

Rgen 

R 

+ 

i 


1 1 

H ( jω ) = [ A ] v 

ω ω 

1+ j 1+ 

j 

Rgen 

+ Ri 

1 

 

RgenRiC 

i 

RoC 

 

o 


1 1 

H ( jω ) = [ A ] v 

ω ω 

 

1+ j 

 

1+ 

j 

 

ωbi 

ωbo 

 

where 

Rgen 

+ Ri 

1 

ω 

bi 

= and ω 

bo 

= 

R R C R C 

gen i i o o 


Overall Transfer Function 

H ( j ) 

1 1 

[ A ] 

ω = ω v ω 

1+ j 1+ 

j 

 

ω 

ω 

bi bo 


Example 

R gen 

V 

gen 

200 

V 

i 

C i 

335.5 

pF 

R 1 

||R 2 

1818 

I 

b 

βr e 

682 

 

130I 

b 

C o 

2 pF 

R o 

880 

V 

o 

R i 

= 496 


Desired 

Frequency Response of the amplifer 


Strategy 


r e 

= 0.026 / I E 


Determine A v 


Strategy 

Build the low frequency amplifier model 


Build the high frequency amplifier model 


Strategy 

Determine the low pass break frequencies 

Sketch and label the midband, low frequency, 

and high frequency magnitude Bode Plot. 


Solution 

1 1 

H ( jω ) = [ A ] v 

ω ω 

 

1+ j 

 

1+ 

j 

 

ωbi 

ωbo 

 

( ) [ ( )] 

C = C 1− A = 2 1− − 167.77 = 335.5 pF 

Mi bc vm 

Ci = C 

Mi 

+ C 

wi 

+ Cbe 

= 335.5 + 0 + 10 = 345.5 pF 


Solution 

R = R || R || β r = 1818 682 = 496 Ω 

i 

1 2 

e 

Rgen 

+ Ri 

200 + 496 

ω 

bi 

= = 

R R C 200 496 335.5× 

10 

gen i i 

( )( )( 

−12 

) 

= 20.912 Mrad/s , f = 3.328 MHz 

bi 


Solution 

C 

Mo 

= C = 

bc 

2 pF 

C = C + C = 2 + 0 = 2 pF 

o Mo wo 

R = R || R = 880 Ω 

o C L 


Solution 

1 1 

ω = = 

bo 

R C 880 2× 

10 

o 

o 

( )( 

−12 

) 

= 568.2 Mrad/s , f = 90.4 MHz 

bo 


Total Amplifier 

Frequency Response 

|H(jω)| dB 

44.5 

dB 

633 Hz 3.3 MHz 

f(Hz)

BJT Frequency Response

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