BJT Frequency Responseecampus.matc.edu/lokkenr/pdfs/bjt frequency response.pdf · Frequency Analysis Strategy Identify the equivalent RC networks formed by the capacitances and resistances

BJT Frequency Response

Electronic Circuit Analysis

Copyright 2004 Richard Lokken 2

Agenda

Low Frequency Predictions

High Frequency Predictions


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

β re RC

RB

RB = R1||R2RE

RL

Rgen

Cgen

CE

CLCbc

Cbe

genViV oV

bIiI

c bI I= β


Typical Frequency Response

f (Hz)

H (dB)

midband Av


Low Frequency Analysis

Consider the high pass effects of the coupling

capacitors first.


BJT model with Coupling C’s

RB

CL

Cgen

iV

Rgen

genV

RLβre

RC

bI

c bI I= β

Ri

oV


Determine the Transfer Function

( ) geno o

i i gen

VV VH j

V V Vω = = •

The generator resistance-input resistance voltage

divider is defined to be at midband (Cgen) is effectively a

short.


Voltage divider at midband

gen gen ii i

gen gen i i i

V R RV R

V R R V R

+= → =

+

Ri = RB || βre


Transfer Function development

( ) gen gen io o o

i i gen i gen

V R RV V VH j

V V V R V

+ ω = = • = •


Use CDR to determine the load I

1C

o L L b L

C LL

RV I R I R

R R +j C

= = −β + ω


Determine from bI

genV

1genB B

b genB e B e

gen igen

VR RI I

R + r R + rR + R j C

= = β β + ω


1 1gen CB

o LB e

gen i C Lgen L

V RRV R

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

= −β β + + ω ω

( )

1 1gen C LB

B egen i C L

gen Lgen i

i gen

V R RR

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

H jR V

−β β + + ω ω+ ω =


( ) ( ) 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

+ ω = −β β β + + ω ω+ β

( ) 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

+ − ω = + + + + ω ω


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

+ ω ω− ω = ω ω + + + + ω ω

( ) ( )( ) ( )

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

ω + ω− ω = + ω + + ω +


Simplify the final results

( ) ( )( ) ( )

( )( )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

ω + ω + − ω = + ω + + + ω +

( ) ( )( ) [ ] ( )

( )1 1gen i gen C L L

vmgen i gen C L L

j R R C j R R CH j A

j R R C j R R C

ω + ω + ω = + ω + + ω +


Obtain j ω/ωb

( )( )

( )

[ ] ( )

( )

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

ω ω + + ω = ω ω + +

++


( )1 1

bg bLv

bg bL

j j

H j Aj j

ω ω ω ω ω = ω ω + + ω ω


Break Frequencies

( )

( )

1

1

bg

gen i gen

bLC L L

R R C

R R C

ω =+

ω =+


Example

VCC = +15 V

Cgen

R1RC

CL = 10 F

RER2

CE

Rgen

RL

genV ~

iV

~

oV200 10 F

50F

1200

3300

20 k

2 k12k

β = 130


Desired

Mid band gain

Low Frequency Response


Strategy

Determine bias current IEre = 0.026/IEBuild 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

( )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= = = Ω

+ +


( )KVL: 1 0

E

Th B Th BE B E

I

V I R V I R− − − β + =

( ) ( )( )1.3636 0.7

37.837 A1 1818 131 120

Th BEB

Th E

V VI

R R

− −= = = µ+ β + +

( ) ( )( )1 131 37.837 A 4.9567 mAE BI I= β + = µ =

0.026 0.0265.2454

4.9567 meE

rI

= = = Ω


Midband Amplifier Model

RB~

oV~

iV βre

RC || RL

bI

130 bI

880 1818 682

bIβ


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

−β+ −= = =

β +

− = = − +

= = =


Coupling Capacitor

Low Frequency Model

RB

CLCgen

~

oV~

iV

Rgen

genVRLβre

RC

bI

130 bI

200

1200 3300 1818 682

10 F10 F


Determine the High Pass

break frequencies

RB

CLCgen

~

oV~

iV

Rgen

genVRLβre

RC

bI

130 bI

200

1200 3300 1818 682

10 F10 F


Input Coupling Capacitor

( )

( ) ( )

1

1 1

200 496 10

144 rad/s , 23 Hz

bgg

bg

bg

gen i gen

bg

j

H jj

R R C

f

ωω

ω = ω+ω

ω = =+ µ+

= =


Output Coupling Capacitor

( )

( ) ( )

1

1 1

1200 3300 10

22 rad/s , 3.5 Hz

bLL

bL

bLC L L

bL

j

H jj

R R C

f

ωωω = ω+

ω

ω = =+ + µ

= =


Total Amplifier Transfer Function

( )

( )

23

123

44.5 dB

3.5

13.5

g

v

L

fj

H jf

j

A

fj

H jf

j

ω =+

=

ω =+


Bode Plot for Example

f (Hz)

H (dB)

44.5 dB

233.5

-20 dB/dec

-40 dB/dec


What about CE?

( ) ( )

( )

( )( )

||

1

11

||

11

b C Lo

iE

Eb e b

EE

b C L

Eb e

E E

I R RVH j

VR

j CI r I

Rj C

I R R

RI r

j R C

−βω = =

ω β + β + + ω

−β=

β + β + + ω


( )100 , 1β > β + ≈ β

( ) ( )

( )

||

1

||

1

C L

Ee

E E

C L

Ee

E E

R RH j

Rr

j R C

R R

Rr

j R C

−βω =

β + β + ω

−=

+ + ω


AC BJT model with RE and CE

βre RC

RB

RB = R1||R2RE

RL

Rgen

CE

genViV oV

bIiI

c bI I= β


Strategic Algebraic Manipulation

( ) ( )|| ||c L c Lev unbypassed

e E e E e

ev bypassed

e E

R R R RrA

r R r R r

rA

r R

−

−

− − = = + +

= +


Transfer Function

( ) ( )

( )( )( )

|| 1

11

|| 1

1

C L E E

E E Ee

E E

C L E E

e E E E

R R j R CH j

R j R Cr

j R C

R R j R C

r j R C R

− + ωω = •+ ω + + ω

− + ω=

+ ω +


Obtain j ω/ωb

( ) ( )( )( )

|| 1

1

11

||

1

C L E E

e E Ee E

e E

E EC L

e E

e E

e E E

R R j R CH j

r R Cr R j

r R

j

R CR R

r Rj

r R

r R C

− + ωω =

ω + + +

ω +

= − ω+ + +


( )

( )

3

4

3

4

1

1

1

1

o bv unbypassed

s

b

e bv bypassed

e E

b

H jE

jV

H j AV j

jr

Ar R j

−

−

ω

ω + ω ω = = ω + ω

ω + ω = ω+ + ω


Where the break frequencies are:

3

1b

E ER Cω =

4

1e Eb

e Ee E EE

e E

r Rr Rr R C

Cr R

+ω = = +


Frequency Response due to CE

ωb3 ωb4

no thru e EI C

H (dB)

Av-bypassed

all thru e EI C

Av-unbypassed

ω


Example

βre

RCRB

RE

RL

Rgen

CE

genViV oV

bIiI

c bI I= β 200

50 F

1818

120

1200

CL

10 F

Cgen

10F

3300

682 130 bI


Desired

Mid band gain

Low Frequency Response


Strategy

Determine bias current IE

re = 0.026 / IE

Build the midband amplifier model

Determine Av


Strategy

Build the low frequency amplifier model


Sketch and label the midband and low frequency

magnitude Bode Plot.


Model for HE(jω)

βre RC||RL

RB

RE CE

iV oV

bI

130 bI

50 F

1818

120

880 682


Transfer Function

( ) 3

4

1

1

e bE

e E

b

jr

H jr R j

ω + ω ω = ω+ + ω


Break Frequencies

( )3

3

1 1

120 50

166.7 rad/s , 26.5 Hz

bE E

b

R C

f

ω = =µ

= =

( )( )4

4

5.2454 120

5.2454 120 50

3979.5 rad/s , 633 Hz

e Eb

e E E

b

r R

r R C

f

+ +ω = =µ

= =


Low frequency amplifier response

f (Hz)

H (dB)

44.5 dB

633

+20 dB/dec



AC BJT Model for

High Frequency Response

βre RC RL

Cbc

R1

Cbe

CwiR2 Cwo

Rgen

genV

bI

c bI I= β iV oV


Generic Amplifier with

Capacitance Feedback

Av

~

inV

~

inTI

~

iI

~

fI

~

oI

~

ToI

~

oV

Cf

~

fI


Capacitances in the

generic amplifier model

Av

~

inV~

oVCM Cf


Development of Miller Capacitance

inT i fI I I= +

ii

i

VI

R=


Examine Circuit

( )1

i of f i o

f

V VI j C V V

j C

−= = ω − ω

ov

i

VA

V=


Insert Av

( )( )1

f f i v i

f v i

I j C V A V

j C A V

= ω −

= ω −


Insert two currents into KCL equation

( )

( )

1

11

iinT f v i

i

i f vi

VI j C A V

R

V j C AR

= + ω −

= + ω −


Solve for Total Input Admittance

( )11inT

inT f vi i

IY j C A

V R= = + ω −


Solve for Total Admittance of circuit

1inTinT Mi

i i

IY j C

V R= = + ω


By direct Comparison

( )1Mi f vC C A= −


Perform a similar analysis at the output

oT o fI I I= −

oo

o

VI

R=


find current If

( )1

i of f i o

f

V VI j C V V

j C

−= = ω − ω


Insert the two currents

11

1 11

ooT f o

o v

o fo v

VI j C V

R A

V j CR A

= − ω −

= + ω −


Solve for Output Admittance

1 11oT

oT fo o v

IY j C

V R A = = + ω −


Solve for the Total Output Admittance of

Circuit

1oToT Mo

o o

IY j C

V R= = + ω


By Direct Comparison

11Mo f f

v

C C CA

= − ≈


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

βre RC RL

R1CbeCw i R2 Cw o

RS

~

iV~

oV

~

bI

~ ~

c bI Iβ=

RiCi

CM i CM o

RoCo


Values from Equivalent Circuit

|| ||i Mi wi be o Mo woC C C C C C C= = +

1 2|| || ||i e o c LR R R r R R R= β =


Transfer Function at mid-band

frequencies

( ) o b o

i b e

ov

e

V I RH j

V I r

RA

r

−βω = =β

−= =


Incorporate the generator

( ) geno o

i i gen

gen i o

i gen

VV VH j

V V V

R R V

R V

ω = = •

+ = •


Use VDR on generator side

11

1

oo b

oo

o

ob

o o

RV I

Rj CR

RI

j R C

= −β + ω

= −β + ω


Determine base current use VDR

11

111

1

gen

iii

geni i

geni

i geni i ii

be e e

Vj C

RR Vj R C

R R

j C R R j R CV

Ir r r

+ ω + ω + + ω + + ω = = =

β β β


( )

( )

1gen i

be gen i i i

gen i

e gen i gen i i

V RI

r R j R C R

V R

r R R j R R C

= β + ω +

= β + + ω


Combine equations

( )

1

1

oo b

o o

gen i o

e o ogen i gen i i

RV I

j R C

V R R

r j R CR R j R R C

= −β + ω

= −β β + ω+ + ω


Transfer Function

( )

( ) 1

gen i o

i gen

gen i o

e o ogen i gen i igen i

i gen

R R VH j

R V

V R R

r j R CR R j R R CR R

R V

+ ω = •

− + ω+ + ω + = •


Transfer Function

( ) ( )1

1gen i i o

i e o ogen i gen i i

R R R RH j

R r j R CR R j R R C

+ − ω = • + ω+ + ω


( ) ( )( )

1

1gen i o

e o ogen i gen i i

R R RH j

r j R CR R j R R C

+− ω = + ω+ + ω

multiplied by the factorgen i

i

R R

R

+


Obtain j ω/ωb

( ) ( )( )

( )

( )

1

11 1

gen i gen i o

e o ogen i gen i i

gen i

R R R R RH j

r j R CR R j R R CR R

+ + − ω = + ω+ + ω +

( ) [ ]1 1

11

vo ogen i

igen i

H j Aj R CR R

j CR R

ω = + ω + ω +


( ) [ ]1 1

1 11

v

gen i

gen i i o o

H j Aj j

R R

R R C R C

ω = ω ω+ + +


[ ]1 1( )

1 1

1where and

v

bi bo

gen ibi bo

gen i i o o

H j Aj j

R R

R R C R C

ω = ω ω+ + ω ω

+ω = ω =


Overall Transfer Function

[ ]1 1( )

1 1v

bi bo

H j Aj j

ω = ω ω+ + ω ω


Example

βreR1||R2

Rgen

iVbI

130 bI

Ri = 496

Ci RoCogenV oV

200

2 pF335.5pF 880 682

1818


Desired

Frequency Response of the amplifer


Strategy

Determine bias current IE

re = 0.026 / IE

Build the midband amplifier model

Determine Av


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( )

1 1v

bi bo

H j Aj j

ω = ω ω+ + ω ω

( ) ( )[ ]1 2 1 167.77 335.5 pFMi bc vmC C A= − = − − =

335.5 0 10 345.5 pFi Mi wi beC C C C= + + = + + =


Solution

1 2|| || 1818 682 496i eR R R r= β = = Ω

( )( )( )12

200 496

200 496 335.5 10

20.912 Mrad/s , 3.328 MHz

gen ibi

gen i i

bi

R R

R R C

f

−

+ +ω = =×

= =


Solution

2 pFMo bcC C= =

2 0 2 pFo Mo woC C C= + = + =

|| 880o C LR R R= = Ω


Solution

( )( )12

1 1

880 2 10

568.2 Mrad/s , 90.4 MHz

boo o

bo

R C

f

−ω = =

×

= =


Total Amplifier

Frequency Response

44.5dB

633 Hz 3.3 MHzf(Hz)

|H(jω)|dB

Documents

BJT Frequency Responseecampus.matc.edu/lokkenr/pdfs/bjt frequency response.pdf · Frequency Analysis Strategy Identify the equivalent RC networks formed by the capacitances and resistances