Frequency Response Lesson #12 Small Signal Equivalent Circuits

BME 373 Electronics II –J.Schesser

22

Frequency Response

Lesson #12Small Signal Equivalent Circuits for

the BJTSection 8.4-8.8


23

Frequency Response• The gain of an amplifier is affected by the capacitance associated with its circuit. This

capacitance reduces the gain in both the low and high frequency ranges of operation.• The Bode Plot may look something like this where there is a low frequency band, a

midfrequency band and a high frequency band.

• The reduction of gain in the low frequency band is due to the coupling and bypass capacitors selected. They are essentially short circuits in the mid and high bands.

• The reduction of gain in the high frequency band is due to the internal capacitance of the amplifying device, e.g., BJT, FET, etc.. This capacitance is represented by capacitors in the small signal equivalent circuit for these devices. They are essentially open circuits in the low and mid bands.

• First, let’s continue to study the small signal equivalent circuits.

High Frequency BandLow Frequency Band

Mid-Frequency Band


24

Small Signal Equivalent Circuits and Parameters for the BJTrπ-β Model

• When the AC Portion of the input is small around the Q point (<<VT in value) then we can approximate the operation of transistor by an equivalent circuit consisting of a resistor, rπ=VT/IBQ and a current source, βib, where ib is the small signal component of the base current:

• A more thorough Equivalent Circuit may be needed to specify the performance of the BJT

rπ βibB

E

Co

o

o


25

Two-Port Devices and the Hybrid Model

• Generalized model for two-port devices

Two-port Device

+

v1

-

+

v2

-

i1 i2

11 input relationship; h11 v1

i1 v2 0

input resistance with output shorted-circuited

22 output relationship; h22 i2

v2 i1 0

output conductance with input open-circuited

21 forward transfer relationship; h21 i2

i1 v2 0

forward current transfer (or gain) with output short-circuited

(short-ciruit current gain)

12 reverse transfer relationship; h12 v1

v2 i1 0

reverse voltage transfer (or gain) with input open-circuited

(reverse-open-circuit voltage gain)

2221212

2121111

vhihivhihv


26

Hybrid-Parameter Model for the Common Emitter BJT

ceoebfec

cerebiebe

vhihivhihv

+

vbe

-

+

vce

-

ib icC

EE

Bhie

hoe

hfeib

hrevce

+

-

The parameters defined by this equivalent circuit as usually provided by transistor manufacturers to describe the performance of the BJT. For example, and hfe are typically given in BJT data sheets.


27

Hybrid-π Model for the BJT

• Another model typically specified by BJT manufacturers and is used for frequency analysis

• Includes– Resistance to model the base-emitter junction, the base

to collector junction, and the collector to emitter path– Capacitance to model base-emitter junction and the

base to collector junction– A dependent (forward) current source in the collector


28

Hybrid-π Model for the BJT (Continued)

• rx called the base spreading resistance and represents the resistance of the base-emitter junction• rπ represents the dynamic resistance for small signal analysis and depends on the Q-point of the

design - rπ=VT/IBQ

• rμ represents the feedback from the collector to the base and is related to the hybrid parameter hre= rπ /(rπ+ rμ)

• ro represents the resistance from the collector to the emitter and is related to the hybrid parameter hoe ≈ 1/ ro and is also related to the EARLY Voltage by VA/ICQ

• Cμ is the depletion capacitance of the collector-base junction• Cπ is the capacitance of the base-emitter junction and depends on the Q-point• gmvπ is the amplification factor and is equal to β ib.• Transition frequency, ft = β /[2πrπ(Cμ+Cπ)], when |Ic/Ib| is unity when the collector is grounded

for ac.

rπ

Cμ

CB

ro

EEgmvπ

rμ

Cπ

rx+

vπ-


29

Example• The Hybrid-π parameters of a

2N2222A:– Q-point => ICQ=10 mA, VCEQ= 10V– Assume VT = 26 mV– Average β => 225– hre => 4x10-4

– hoe => 25 mS ~ 200 mS– Cμ => 8 pF– ft => 300 MHz (transition frequency)– Collector-base time constant:

rxCμ=150 x 10-12

1981015010150

pF196

pF8103005852

2252

,k 5.22

k 40 ~ k 51

,M 5.1104

585

, 585385.225

S, 385.0m26m10

1212

6

AVG

4-

Cr

CfrC

rhr

hrr

gr

VI

g

x

t

o

oeo

re

m

T

CQm

2 ( )tf r C C


30

Analysis of CE at High FrequenciesThe Hybrid-π parameters :– Q-point => ICQ=10 mA, VCEQ= 10V– Assume VT = 26 mV– gm=.385 S– rx=19 Ω– rπ=585 Ω– ro=22.5k Ω– rμ=1.5 M Ω– Cπ=196pF– Cμ=8pF

Vs1V

+-

RC510

RE

1.3k

C2

1uFC1

1uF

CE100uF

Q2N2222AQ1

VCC 15V

Rs50 RB

10k

RL510

0

+

vn-

+

vout

-

0 00VEE –15V

E

RSC

rπ

CμB

RB

E

gmvπ

rμ

Cπ

rx+

vπ-

RC RL

+

vin

-

vs ro


31

Example Using the Miller Effect

C585

8pB

10k

E

.385vπ

1.5 M

196p

19

+

vπ-

510510

+

vin

-

vs 22.5k

50

+

vo

- in Miller

in Miller

r

C

out Miller

out Miller

r

C

585

B

10k

E

.385vπ

196p

19

+

vπ-

510510

+

vin

-

vs 22.5k

50 C+

vo

-


32

Example Calculation of the Miller Parameters and the Midband Gain

R’s=[10k||(19+rπ||rμINMILLER)]=550

Miler Effect Parameters

rμout MILLER=1.5M*Av /(Av-1) ≈ 1.5M

Av= vo /vπ=-gmR’L=-.385*252=-97

rμin MILLER=1.5M/(1-Av )=1.5M/(1+gmR’L ) ≈ 15.k

Calculations

rπ||rμin MILLER = 563

vπ /vin =rπ||rμin MILLER / ((rπ||rμin MILLER)+rx) =.97

vin/vs=R’s/(R’s+50)=0.92

Avs= vo/vs= (vo /vπ)(vπ /vin) (vin /vs)=-97(.92)(.97)=-86.1

R’L=22.5k||510||510||rμOUTMILLER=252

in Millerr out Millerr

585

B

10k

E

.385vμ

19

+

vπ-

510510

+

vin

-

vs 22.5k

50 C


33

Example Calculation of the Break Frequencies

R’s=[(50||10k)+19]||585||rμINMILLER=61.3

Miler Effect Parameters

rμin MILLER=15.k

rμout MILLER= 1.5M

Cμin MILLER=8p*(1-Av ) ≈ 784 pF

Cμout MILLER=8p*(Av-1) /Av ≈ 8.08 pF

Calculations

CT=196+CμinMiller= 980 pF

fb in=1/(2πR’sCT)=1/(2π*61.3*980x10-12)=2.65 MHz

fb out=1/(2πR’LCμout Miller)=1/(2π*252*8.08x10-12)=78.1 MHz

R’L=22.5k||510||510||rμOUTMILLER=252

in Miller

in Miller

r

C

out Miller

out Miller

r

C

585

B

10k

E

.385vπ

196p

19

+

vπ-

510510

+

vin

-

vs 22.5k

50 C


34

Example Alternative Method Using Circuit Analysis - Output Circuit

Output Pole Frequency

vo

v gmR 'L || ZCoutmiller

.385252 1

j81012

252 1j81012

.385 2521 j81012 252

.385 2521 j2.03109

fbout 1

22.03109 78.1MHz

C

RB’=rμINMILLER|| rπ = 15k||585

=563

B

10k

E

.385vπ

Cμoutmiller=8p

19

+

vπ-

+

vin

-

vs

50+

vo

-R’L=252CT=Cμinmiller +Cgs

=784p+196p=980pf

B’


35

Example Alternative Method Using Circuit Analysis - Input Circuit

'

'

''

''

'

'

' '

' ' '

'

' '

'' ' '

'

Input Pole Frequency

||||

1

|| 1 1

1(1 )

11

1

T

T

T

B

S B s

B C

B B C x

BT B

B CT B

BT

B

T B B

BB x T B Bx

T B

B B

B xx B T B x x BT

x B

v v vv v v

R Zvv R Z r

Rj C RR Z

j C RRj C

Rv j C R R

Rv r j C R Rrj C R

R RR rr R j C R r r R j C

r R

C

RB’=rμINMILLER|| rπ = 15k||585=563

B

10k

E

.385vπ

Cμoutmiller=8p

19

+

vπ-

+

vin

-

vs

50+

vo


=784p+196p=980pf

B’

'

'

' ' ' '''

' ' '

' ''

'

Input Pole Frequency|| ( || )

|| ( || )

(1 )( || )1 1 1

|| ( || ) ||1

T

T

T

T

B x B CB

s B x B C S

x T B B x B T B xBx B C x

T B T B T B

x BB

x B T B xB x B C B

T B

R r R Zvv R r R Z R

r j C R R r R j C R rRr R Z rj C R j C R j C R

r RRr R j C R rR r R Z R

j C R

' '

'

' '

'

''

' ' '

' ' ' ' ' '

1

1

( )(1 )( ) ( )

(1 ) ( ) ( )

T B x

T B

x B T B xB

T B

B xB x B T

B x B T B x x B

B T B x B T B x B x B T B B B x

j C R rj C R

r R j C R rRj C R

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

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


36

Example Input Circuit Cont’d

C +

vo

-

RB’=rμINMILLER|| rπ = 15k||585=563

B

10k

E.385vπ

Cμoutmiller=8p

19+

vπ-

+

vin

-

vs

50 R’L=252

CT=Cμinmiller +Cgs

=784p+196p=980pf

B’

''

'

' ' ' '

'''

'

' ' '

''

'

( )(1 )( )

|| ( || ) ( )|| ( || ) ( )(1 )

( )( )

( )(1 )( )

(

T

T

B xB x B T

x B

B x B C B x B T B B B xB

B xs B x B C SB x B T

x BS

B x B T B B B x

B xB x B T

x B

B x

R rR r R j Cr R

R r R Z R r R j C R R R rvR rv R r R Z R R r R j C

r R RR r R j C R R R r

R rR r R j Cr R

R r

''

'

' ' ' ' ' '' ' ' '

'

''

'

' '

( )(1 )( )

( ) ) ( ( )))(1 ) ( ( ))( )

( )(1 )( )

( ) ( )

B xB x B T

x B

B x B x B T B x B S B x B T B B B xB T S B x B T B B B x

x B

B xB x B T

x B

B x B S B x B

R rR r R j Cr R

R r R r R j C R r R R R r R j C R R R rR j C R R r R j C R R R rr R

R rR r R j Cr R

R r R R R r R

' ' '( ( ))T B x B S B B B xj C R r R R R R R r


37

C


=563

B

10k

E

.385vπ

Cμoutmiller=8p

19

+

vπ-

+

vin

-

vs

50+

vo


=784p+196p=980pf

B’

''

''

'' ' ' ' ' '

'

'

' ' ' '

Input Pole Frequency

( )(1 )( )1

( ) ( ) ( ( ))1

( ) ( ) ( (

B xB x B T

x BB B

B xS B s x B B x B S B x B T B x B S B B B xT

x B

B B

B x B S B x B T B x B S B B

R rR r R j Cv v r Rv R

R rv v v r R R r R R R r R j C R r R R R R R rj Cr R

R RR r R R R r R j C R r R R R R

'

' '

' ' ''

' '

8

)) 1 ( ))

1 2.652 6 10

B B

B x B B S B S x S B

B x B S B B B x SB xT

B x B B S B S x S B

bin

R RR r R R R R R r R R

R r R R R R R r RR r j CR r R R R R R r R R

f MHz

Example Input Circuit Cont’d


38

Example Input Circuit using Thevenin’s Theorem

C


=563

B

10k

E

.385vπ

Cμoutmiller=8p

19

+

vπ-

+

vin

-

vs

50+

vo

-R’L=252CT=Cμnmiller +Cgs

=784p+196p=980pf

B’

'

' '

'' ' '

' ''

Thevenin's method for Input Pole( || ) ||

( ) || ( ) ||

( )

( )

T S B x B

S B S B x S x Bx B B

S B S B

S B x S x BB

S B S B B x S B x B B

S B x S x B S B x S x B B S B BB

S B

b

R R R r RR R R R r R r Rr R R

R R R RR R r R r R R

R R R R R r R R r R RR R r R r R R R r R r R R R R RR

R R

f

1 2.652in

T T

MHzC R


39

Emitter Follower

VEE –15 V

RB10k

Rs

510

0

Q2N2222AQ1

RE

1.3kRL

50

C1

1uF C2

100uF

VCC 15 V

Vs1V

0 0

E

rπ

Cμ

B

RB gmvπ

rμ

Cπ

rx

+

vπ-

RL

vs ro

Rs

+

vo

-

C

RE


40

Emitter Follower (Continued)

vs

E

rπ

Cμ

B

RB gmvπ

rμ

Cπ

rx

+

vπ-

RL

ro

Rs

+

vo

-

C

RER’s=Rs||RB+rx R’L=RL||RE||ro

rπR’s b’ E

C

+

vo

-

=>

Cπ

rμ

Cμ gmvπ

+ vπ -v’s R’

L


41

Emitter Follower (Continued)

)'1()'1(

)'1('

Inputfor thensCalculatioEffect Miller

Millerin

'

'

Lmf

Lmf

b

ff

ffIN

∂Lmo∂b

∂Lmo

RgZZRgZ

vZv

ZV

I

vRgvvvvRgv

C

Cπ

rμ

rπ

Cμ gmvπ

+ vπ -

R’s

v’s

b’ E

+

vo

-

R’L

Cπ / (1+ gmR’L)

rμ

Cμ

R’s

v’s

b’ rπ(1+gmR’L)

RT CT


42

ExampleThe Hybrid-π parameters :– Q-point => IEQ=10.6 mA,

VCEQ= 15V– Assume VT = 26 mV– β=225– gm=.385 S– rx=19 Ω– rπ=585 Ω– ro=22.5k Ω– rμ=1.5 M Ω– Cπ=196pF– Cμ=8pF

The Circuit parameters :– RS=510 – RB=10 k– RE=1.3 k– RL=50

Calculations:

– R’s=Rs||RB+rx= 510||10k+19=504– R’L=RL||RE||ro= 50||1.3k||22.5k=48– RT=R’S||rμ||rπ(1+gmR’L)=483 – CT=Cμ+Cpπ /(1+gmR’L)= 18.1 pF– fb=1/(2πRTCT)=18.2 MHz

Recall for an emitter follower:– Av=(β+1)R’L/[rπ+(β+1)R’L]=.949– Rin=RB||[rπ+(β+1)R’L]= 5.33 kΩ– Avs=AvRin /(Rin+RS)= 8.66


43

Low Frequency Response of RC-Coupled Amplifiers

• Coupling Capacitors– To couple the various stages of a multi-stage amplifier

• For AC performance essentially a short circuit and AC current flows from one stage to the next stage

– To support the biasing of each stage individually: • For DC performance: open circuit and no biasing current flows from

one stage to another

• ByPass Capacitors– To support the addition of a resistor for biasing purposes only

• For DC performance: open circuit and current flows through the biasing resistor

– Short-circuit the biasing resistor for AC performance.• For AC performance: short circuit and no current flows through the

resistor (shorted out/bypassed)


44

Capacitors

Vs1V

+-

RC

510

RE

1.3k

C2

1uFC1

1uF

CE

100uF

Q2N2222AQ1

VCC 15V

Rs50 RB

10k

RL

510

0

+

vn-

+

vout

-

0 00VEE –15V

Coupling Capacitors Bypass Capacitors


45

Coupling Capacitors

+

-+

-

+

-

+

-VS VX

VY=Avo VX

VoVYRin

RsRo

RL

C2C1

1

1

1

1

1

11

2

2

2

22

1

1 1 ( )

( / ) ,1 ( / )1

2 ( )( / ) ,1 1 ( / )

1 ,2 ( )

,

( / )1

O OX Yvs

S S X Y

in inX

S in Sin S

in

in S

S in

O L L

Y L oL o

o L

Xvo

S

invs

in S

A

R j C Rj C R RR R

j CR j f f

R R j f f

fR R C

R R j f fR R j f fR R

j C

fR R C

A

R j f fAR R

V VV VV V V V

VV

VV

VV

2

1 2

1 2

1 2

( / )( / ) 1 ( / )

( / ) ( / )1 ( / ) 1 ( / )

Lvo

L o

vs vsmid

in Lvsmid vo

in S L o

R j f fAj f f R R j f f

j f f j f fA Aj f f j f f

R RA AR R R R

f2 f1

20log|Avsmid|


46

Bypass Capacitors

• The value of a bypass capacitor is chosen to provide a short circuit at a frequency sufficiently low than the band pass of amplifier design

• For a emitter follower, it can be shownf1=1/(2π R’E CE )

whereR’E is the equivalent resistance reflected into the

emitter circuit


60

Homework

• Hybrid-– Problems: 8.28,31

• Common Emitter– Problems: 8.40-41

• Emitter Follower– Problems: 8.56-7

• Low Frequency Response– Problems: 8.60-2

Documents

Frequency Response Lesson #12 Small Signal Equivalent Circuits