ECE 442 Solid State Devices Circuits 16. Feedbackjsa.ece.illinois.edu/hcmut/ece342/notes/Lect_16.pdfECE 442 –Jose Schutt ... Feedback – Analysis so that ... amplifier network hh

ECE 442 – Jose Schutt‐Aine 1

ECE 442Solid‐State Devices & Circuits

16. Feedback

Jose E. Schutt-AineElectrical & Computer Engineering

University of [email protected]


• Why Use Feedback?

1. To desensitize the gain

2. To reduce nonlinear distortion

3. To reduce the effect of noise

4. To control terminal impedances

5. To increase the bandwidth

Feedback


Feedback – Basic Concepts


out

ia

vAv

=

out

in

vGv

=

out out

in ia

v vG Av v

= = =

0ia in inv v v= − =

The ideal amplifier gain is defined by

The overall gain of the feedback circuit is defined by

If input 2 is grounded

We then get

This is the open-loop gain

Feedback – Analysis


( )out in out inv A v v Gvβ= − =

1 β= =

+out

in

v AGv A

ia in outv v vβ= −

When the switch is closed, then

Feedback – Analysis

so that

from which we get

The closed-loop gain is always less than the open-loop gain for negative feedback

which is the closed-loop gain


( )1 /ω

=+

M

H

AA ss

The high-frequency response of an amplifier (single-pole) is given by:

Feedback – Bandwidth Extension

AM is the midband gain and ωH is the upper 3-dB frequency. With negative feedback, we get

( )( )1 ( )β

=+fA sA s

A s

/(1 )( )1 / (1 )

βω β

+=

+ +M M

fH M

A AA ss A

After substitution,


The feedback amplifier will have a midband gain of

Feedback – Bandwidth Extension

and an upper 3-dB frequency of

(1 )β+M

M

AA

(1 )ω β+H MA

Bandwidth is increased by factor equal to amount of feedback. It can also be shown that the lower 3dB frequency is

(1 )ω

β+L

MAThe gain-bandwidth product is constant


Series-Shunt Feedback

Voltage mixing-voltage sampling feedback


Shunt- Series Feedback

Current mixing-current sampling feedback


Series-Series Feedback

Voltage mixing-current sampling feedback


Shunt-Shunt Feedback

Current mixing-voltage sampling feedback


Transfer Function Representation

Use a two-terminal representation of system for input and output


Y-parameter Representation

1 11 1 12 2

2 21 1 22 2

I y V y VI y V y V

= += +


Y Parameter Calculations

2 2

1 211 21

1 10 0V V

I Iy yV V

= =

= =

To make V2= 0, place a short at port 2


Z Parameters

1 11 1 12 2

2 21 1 22 2

V z I z IV z I z I

= += +


Z-parameter Calculations

2 2

1 211 21

1 10 0I I

V Vz zI I

= =

= =

To make I2= 0, place an open at port 2


H Parameters

1 11 1 12 2

2 21 1 22 2

V h I h VI h I h V

= += +


H Parameter Calculations

To make V2= 0, place a short at port 2

2 2

1 211 21

1 10 0V V

V Ih hI I

= =

= =


G Parameters

1 11 1 12 2

2 21 1 22 2

I g V g IV g V g I

= += +


G-Parameter Calculations

2 2

1 211 21

1 10 0I I

I Vg gV V

= =

= =

To make I2= 0, place an open at port 2


Series-Shunt Feedback - Ideal

1o

fs

V AAV Aβ

≡ =+

Negative feedback decreases the gain

o

i

VAV

≡


Series-Shunt Feedback – Equivalent Circuit

( )1if iR R Aβ= +

Negative feedback increases the input resistance and decreases the output resistance by a factor equal to the feedback

( )/ 1of oR R Aβ= +


Series-Shunt Feedback - Actual


Series-Shunt Feedback : h-Parameters

Account for all 4 h parameters


Series-Shunt Feedback : h-Parameters

12 12basic feedbackamplifier network

h h 21 21feedback basicnetwork amplifier

h h


Series-Series Feedback - Ideal

1o

fs

I AAV Aβ

≡ =+


o

i

IAV

≡


Series-Series Feedback – Equivalent Circuit

( )1if iR R Aβ= + ( )1of oR R Aβ= +

Negative feedback increases the input resistance and increases the output resistance by a factor equal to the feedback


Series-Series Feedback - Actual


Series-Series Feedback: Z Parameters

Account for all 4 z parameters



z z21 21feedback basic

network amplifierz z

Series-Series Feedback: Z Parameters


Shunt-Shunt Feedback - Ideal

o

i

VAI

≡


Shunt-Shunt Feedback

( )/ 1if iR R Aβ= + ( )/ 1of oR R Aβ= +

Negative feedback decreases the input resistance and decreases the output resistance by a factor equal to the feedback


y y 21 21feedback basicnetwork amplifier

y y

1o

fs

V AAI Aβ

≡ =+



Shunt-Series Feedback - Ideal

o

i

IAI

≡


Shunt-Series Feedback

( )/ 1if iR R Aβ= + ( )1of oR R Aβ= +

Negative feedback decreases the input resistance and increases the output resistance by a factor equal to the feedback

1o

fs

I AAI Aβ

≡ =+



g g 21 21feedback basicnetwork amplifier

g g


Rules for Series-Shunt Feedback

1 11 1 12 2

2 21 1 22 2

V h I h VI h I h V

= += +

2 2

1 211 21

1 10 0V V

V Ih hI I

= =

= =

1 1

1 212 22

2 20 0= =

= =I I

V Ih hV V


Rules for Series-Series Feedback

1 11 1 12 2

2 21 1 22 2

V z I z IV z I z I

= += +

2 2

1 211 21

1 10 0I I

V Vz zI I

= =

= =

1 1

1 212 22

2 20 0= =

= =I I

V Vz zI I


Rules for Shunt-Shunt Feedback

2 2

1 211 21

1 10 0V V

I Iy yV V

= =

= =

1 11 1 12 2

2 21 1 22 2

I y V y VI y V y V

= += +

1 1

1 212 22

2 20 0= =

= =V V

I Iy yV V


Rules for Shunt-Series Feedback

1 11 1 12 2

2 21 1 22 2

I g V g IV g V g I

= += +

2 2

1 211 21

1 10 0I I

I Vg gV V

= =

= =

1 1

1 212 22

2 20 0= =

= =V V

I Vg gI I


Example - Feedback

Differential stage followed by an emitter follower, with series-shunt feedback supplied by the resistors R1 and R2. Perform DC analysis and find A, β, Af, Rinand Rout


1 2 0.5= =E EI I mA

2 10.7 0.5 20 0.7= − × = +cV V

30.7 0= − =o BEV V

3 5=EI mA

1 2 50= = = ΩTe e

C

Vr rI

3 5= Ωer

Example - FeedbackModel asseries-shunt


( )( )2 3

31 1

1 2

20 || 1 (2 ||10) (2 ||10) 85.7 /10 (1|| 9) (2 ||10)1 1

β

β β

⎡ ⎤+ +⎣ ⎦= = × =++ + +

+ +

eo

s ee e

rVA V VV rr r

A - Circuit


1 2 4( 1)( ) ||β= + + + +i s e e ER R r r R R

( )10 101(50 50) 1|| 9 21= + + + = ΩiR k

32

202 ||10 || 1811β

⎡ ⎤= + = Ω⎢ ⎥+⎣ ⎦

o eR r

A - Circuit – cont’


' ' 1/ 0.19 1

β = = =+f oV V V

85.7 8.96 /1 1 85.7 0.1

of

s

V AA V VV Aβ

= = = =+ + ×

( )1 1 21 9.37 201ifR R A kβ= + = × = Ω

201 10 191IN if sR R R k= − = − = Ω

( ) 181|| 18.81 9.57

oof out L

RR R RAβ

= = = = Ω+

19.1outR = Ω

β - Circuit


3-Stage Amplifier with Feedback


3-Stage Amplifier with Feedback Model asseries-series


( )( )

1 1 21

1 1 2

Cc

i e E F E

R rVV r R R R

πα−=

⎡ ⎤+ +⎣ ⎦

Gain in first stage is:

First stage parameters are: IC1 = 0.6 mA, re1=41.7 Ω, Ic2=1 mA rπ2= hfe/gm2=100/40 = 2.5 kΩ

Use α1=0.99, Rc1=9 kΩ, RE1=100 Ω, RF=640 Ω, and RE2=100 Ω

1 14.92 /c

i

V V VV

= −



( ) ( )( ){ }22 2 3 2

1

1cm C fe e E F E

c

V g R h r R R RV

⎡ ⎤= − + + +⎣ ⎦

Gain in second stage is:

Use gm2=40 mA/V, RC2= 5 kΩ, hfe=100, re3=25/4,= 6.25 Ω, RE2=100 Ω, RF = 640 Ω, and RE1 = 100 Ω, which gives

2

1

131.2 /= −c

c

V V VV



( )( )3

2 3 3 2 1

1o e

c b e E F E

I IV V r R R R

= =+ +

Gain in third stage is:

Combining the 3 stages

314.92 131.2 10.6 10 20.7 /o

i

IA A VV

−= = − × − × × =

( )2

1 10.6 /6.25 100 740

o

c

I mA VV

= =+



21

2 1

f EE

o E F E

V R RI R R R

β = = ×+ +

determining feedback

100 100 11.9100 640 100

β = × = Ω+ +



20.7 83.7 /1 1 20.7 11.9

of

s

I AA mA VV Aβ

≡ = = =+ + ×

Closed-loop gain:

3 33

o C C o Cf C

s s s

V I R I R A RV V V

− −= = = −

383.7 10 600 50.2 /o

s

V V VV

−= − × × = −



( )1if iR R Aβ= +

Input resistance

( ) ( )1 1 21 13.65i fe e E F ER h r R R R k⎡ ⎤= + + + = Ω⎣ ⎦

( )13.65 1 20.5 11.9 3.34ifR M= + × = Ω

Output resistance

( ) 22 1 3 1

Co E F E e

fe

RR R R R rh

⎡ ⎤= + + +⎣ ⎦ +Ro=143.9 Ω



( )1 143.9(1 20.7 11.9) 35.6of oR R A kβ= + = + + = Ω

output resistance

( )( )3 31out o m o ofR r g r R rπ= + +

( )( )25 1 160 25 35.6 0.625 2.5outR M= + + × = Ω



Single-Stage Amplifier with Feedback We want to determine the small-signal voltage gain Vo/Vs, the input resistance and the output resistance Rout=Rof. The transistor has β = 100

Rin

Rof-

Vo

+

Rs = 10 kΩ

RF = 47 kΩRC = 4.7 kΩ

Vs

+12 V

Model asshunt-shunt


Single-Stage – DC AnalysisFirst determine dc operating point

0.7 ( 0.07)47= + +C BV I

3.99 47C BV I= +

( )12 1 0.074.7

CB

V Iβ−= + +

Solve for IB using the following 2 equations


Single-Stage – DC AnalysisWe get

0.015BI mA 1.5CI mA 4.7CV V

1.5 60 /25

Cm

T

Ig mA VV

= = =

/ 100 / 60 1.666mr g kπ β= = = Ω

10(1.66) 1.42911.6SR r kπ = = Ω

47(1.66) 1.648.66fR r kπ = = Ω


Calculating y11 for Amplifier

111 1 0.086 /

10 1.6S f

y mA VR R rπ

= = =+ +


221

1

IyV

=

( ) ( ) ( )1 11/f fm f

S f S f

V R r V R rv g R

R R r R R rπ π

ππ π

= = −+ +


( ) ( )21 1/ f

m fS f

R ry g R

R R rπ

π

= −+

( ) ( )12 1/ f

m fS f

V R rI g R

R R rπ

π

= −+


( )211.660 0.021 8.27 /

10 1.6y mA V= − =

+




( )( )

2112

2 2 2

1 1S

S f S S

V R rvIyV R V R R r R V

ππ

π

−−= = =

+

( )( )12

1 S

S f S

R ry

R R R rπ

π

= −+


( )121.429 0.00295 /

10 47 1.429y mA V= − = −

+




( )( ) ( )

22 22

m

m S

C S f S f

g v

g V R rV VIR R r R R r R

π

π

π π

= + ++ +

( )( ) ( )

222

2

1 1m S

C S f S f

g R rIyV R R r R R r R

π

π π

= = + ++ +

( )222

2

60 1.4291 1 2.01 /4.7 1.429 47 1.429 47

Iy mA VV

= = + + =+ +


111 0.021 /

f

y mA VR

= =

211 0.021 /

f

y mA VR

= − = −

22 11y y by symmetry=

12 21y y by reciprocity=

y-parameters for Feedback Network

From Feedback Network, we get β = y12 = -0.021


0.086 0.0038.27 2.01AY

−⎡ ⎤= ⎢ ⎥

⎣ ⎦

0.021 0.0210.021 0.021FY

−⎡ ⎤= ⎢ ⎥−⎣ ⎦


y y

21 21feedback basicnetwork amplifier

y y

Feedback NetworkBasic Amplifier

Basic Amplifier vs Feedback Network


Single-Stage – Small-Signal Analysis


Single-Stage – Small-Signal AnalysisThe feedback is provided by Rf which samples the output voltage and feeds back a current to be mixed at input

( )i s fV I R R rπ π=

( )o m f CV g V R Rπ= −

( )( ) 358.7om f C s f

i

VA g R R R R r kI π= = − = − Ω

Transimpedance gain is –358.7 kΩ


Input and Output Resistances

1.4i s fR R R r kπ= = Ω

4.27o C fR R R k= = Ω


Determining β and Af

1 147

f

o f

IV R k

β = = − = −Ω

1o

fs

V AAI Aβ

≡ =+

358.7 358.7 41.61 358.7 / 47 8.63

o

s

V kI

− −= = = − Ω

+


Voltage gain is:

Single-Stage Feedback Amp

41.6 4.16 /10

o o

s s s

V V V VV I R

−= = −

The input resistance with feedback is:

1.4 162.21 8.63

iif

RRAβ

= = = Ω+

The output resistance with feedback is:4.27 495

1 8.63o

ofRRAβ

= = = Ω+


1. Most of the forward transmission occurs in the basic amplifier

2. Most of the feedback -or reverse transmission - occurs in the feedback network

3. Care should be taken in the design that these assumptions are valid

Important Remarks


1. The closed-loop transfer function is a function of frequency

2. The manner in which the loop gain varies with frequency determines the stability or instability of the feedback amplifier

3. The frequency at which the phase of the transfer function is equal to 180o will be unstable if the magnitude is greater than unity

Feedback and Frequency Dependence


1. An amplifier with a single pole response is unconditionally stable

2. An amplifier with a two-pole response is unconditionally stable

3. An amplifier with a three-pole response (or higher) can be unstable need to provide compensation

Feedback and Frequency Dependence


1. In order to insure stability, we modify the open-loop transfer function A(s) of the amplifier

2. Introduce a new pole in the function A(s) at a frequency fD such that modified open-loop gain A’(s) intersects the 20 log |1/|β|) curve with a a slope difference of 20 dB/decade

3. The closed-loop amplifier with β value (or lower) will be stable.

Feedback and Frequency Compensation


Feedback and Frequency Compensation


Miller Compensation

Common emitter amplifier with Miller compensating capacitor Cf in feedback path


Miller Compensation and Pole Splitting

R1 and C1 make up the total input impedance; R2and C2 make up the total output impedance

In the absence of the compensating capacitor, Cf, there are two poles associated with C1 and C2 fp1and fp2


Miller Compensation and Pole Splitting

1 21 1 2 2

1 12 2P Pf f

R C R Cπ π= =

( ) 1 221

f mo

i

sC g R RVI sA s B

−=

+ +

When Cf is present, the transfer function becomes

( )1 1 2 2 1 2 1 2f mA C R C R C g R R R R= + + + +

( )1 2 1 2 1 2fB C C C C C R R⎡ ⎤= + +⎣ ⎦


Pole Splitting

( )' '1 2

2 1 1 2 1 2

1 m fP P

m f f

g Cand

g R C R C C C C Cω ω

+ +

The two poles can be approximated as:

When Cf is increased, ω’P1 is reduced while ω’P2 is increased this is referred to as pole splitting

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ECE 442 Solid State Devices Circuits 16. Feedbackjsa.ece.illinois.edu/hcmut/ece342/notes/Lect_16.pdfECE 442 –Jose Schutt ... Feedback – Analysis so that ... amplifier network hh