3EJ4 Set 03 Feedback MJD

8/4/2019 3EJ4 Set 03 Feedback MJD

1/24

ECE 3EJ4Electronic Devices & Circuits II

Lecture Set 3Lecture Set 3 NegativeNegative

FeedbackFeedback

Prof. M. Jamal DeenProf. M. Jamal Deen

Professor and Senior Canada Research Chair

Dept. of Electrical and Computer Engineering

McMaster University Hamilton, ON L8S 4K1, Canada


2/24

3-2

Topics in Feedback

Text - Section 10.1 to 10.7. Unreferenced figures are form the text.

Lecture notes

Practice problems

Exercise 10.1

Exercise 10.3

Exercise 10.4

Exercise 10.5

Exercise 10.6

Exercise 10.7

Exercise 10.8

Exercise 10.9

Exercise 10.12

Exercise 10.15

Exercise 10.17

Try other problems from the text, pages 890 onwards.


3/24

3-3

Introduction to NegativeIntroduction to Negative

FeedbackFeedbackSection 10.1, pages 804Section 10.1, pages 804--808808


4/243-4

Feedback - Introduction

Feedback is present in most

physical systemsCar's cruise control maintain desired speed

System vehicle

Output vehicle speed

Control variable throttles position of engine

which determines engines torque output

Two types of feedback

Negative or degenerative feedbackNegative feedback - describe act of reversing

any discrepancy between desired and actual

output

Examples - Thermostat control, phase-locked

loop, hormonal regulation, body temperatureregulation in animals, regulating blood glucose

levels (diabetes)

Thermostat - When Temp. T in heated room

reaches a certain upper limit, room heating

switched off - T begins to fall. When T drops to

lower limit, heating switched on.

Positive or regenerative feedback

Stable system

Unstable system

http://www.physicalgeography.net/fundamentals/4f.html


5/24

3-5

Feedback - Characteristics

Stabilizes the amplifier gain less sensitive to

variations in values of components parameters,

active device variations

Reduces non-linear distortion - improves linearity

Modifies Rinput and Routput by selecting appropriate

feedback topology

Increases bandwidth

Reduce noise

Amplifier Input Resistance RIN Output Resistance ROUT

Voltage-to-Voltage RIN - Large, ROUT - Small, 0Current-to-Voltage RIN - Small, 0 ROUT - Small, Voltage-to-Current RIN - Large, ROUT - Large, 0Current-to-Current R

IN

- Small, 0 ROUT

- Large,

Reduces the amplifier gain

- reduction ~ amount of

gain stabilization,

distortion reduction etc.

Solution use morestages

Can cause stability

problems if not properly

compensated


6/24

3-6

What is Negative Feedback

Negative feedback Tends to maintain output at relatively constant level

Trades gain for improvement in systems characteristics

1

:1

;

1f

of

s

x A

A

Gain with FB A

A

x A

for

= =+

; fi

s f

o

i

ox x

x

x x

x x

A =

=

=

1

1

f s

f s

and x x for A

AFeedback Signal x x

A

=

+


7/24

3-7

Example 10.1

2

1

1

1f

R

A R= +

1

1 2

11;

1

1

ff A for

RA

R

A

R

AAA

+

=+

4

2

3

4

4

4

1

2

1

1

10 10 11

10=1+

0 0 09991 10 10

10 =10 9f

f

A ;

AA

A

R

RrA

R

.

oR

= = =

=

=+

=

+

4=1 1 10 0 1 10 001 6Fe dBedback A .+ = + = =

1

1 2

f

o

V R

V R R = =

+

40 0999 10 0 999

100 001

10f io

oVV V . .A

. V V V= = == = =

10 11 10o f ss V A V V V V V= = ==

4

4

4 0 8 109 9975

1 0 8 10 0 090

90

98 1 f

.A . V

. .. V

= =

+ =

Assume that the gain A decreases

by 20%. What is the new Vo?


8/24

3-8

Basic Properties ofBasic Properties of

Negative FeedbackNegative FeedbackSection 10.2, pages 809Section 10.2, pages 809--814814


9/24

3-9

Gain Desensitivity

( )2

d 1

d

1:

1

ff

A AGain with FB A

A A A=

+ +=

( )

1

1

d df

f

AA

A A A=

+

1 Desensitivity Factor A= +


10/24

3-10

Bandwidth Extension

Consider HF response of amplifier characterized by single pole

( )1

o M

s H

x AA s

x s

= =+

( )

( )

1:

1

M

H

M

H M

Mid band Gain A

New A

A +

= +

( )( )

( )1f

A sA s

s=

+

( )[ ]( )[ ]( )

1

11

M H

M Hf

A s

As

sA

+

++ =

( )

[ ]( )

1

1 1M

M

M

MH

f

A

As

A

A

A s

+ +

+=


11/24

3-11

Signal/Interference Ratio

- - :

s

n

Signal to Interference Ratio

VS

I V=

1 2

1 2

1

1 21

1o

n

s

A AV V A A

AV

A A+

+

= +

2

- - :

s

n

New Signal to Interfer

S

I

ence Ratio

V

V

A=

1 1o s nV V A V A= +


12/24

3-12

Non-linear Distortion

With FB, amplifier is now more linear

Penalty reduction in gain

Can restore gain by adding another stage

(a) vO vI of amplifier. Gain = 1000, 100, 0Large amount of non-linear distortion

(b) vO

vS

of amplifier with FB, =0.01

( )11000

1 1009

00

10 09f

.A .==

+

( )2 100

1 100 0 0150fA

.+ ==

1

1

1f

f

A

A

r

A

A fo A

=+


13/24

3-13

Basic FeedbackBasic Feedback

TopologiesTopologies

VoltageVoltage

CurrentCurrent

TransconductanceTransconductance

TransresistanceTransresistance


14/24

3-14

Feedback Topologies

ConfigurationFeedback

Input Output

Voltage - Vo/Vi Series Shunt

Current - Io/Ii Shunt Series

Transconductance - Io/Vi Series Series

Transresistance - Vo/Ii Shunt Shunt


15/24

3-15

1. FB Voltage Amplifier

Series-shunt connection

Voltage-mixing or voltage

sampling topology

Stabilizes voltage gain

Higher RIN Lower ROUT

Read Section 10.3.1, pages 814 - 816.


16/24

3-16

Series-Shunt FB Amplifier

( );

1 of

i s

o V AAV

VA

V =

+

( )( )1

1 oof

xx

oi x

V AV V

RI

RR= + ==

+

( )1s

if ii

AV

R RI + = =1 1

1; =s si

i i i i

VV V

AI

R RV

A=

+ +=

i

;; ;x i fi

xx

f ox

ofo

xV VV AV

I VV

IV VR

R= = = =

Read Section 10.4.2, pages 825-827

FB Voltage Amplifier


17/24

3-17

Examples FB Voltage Amplifier

( )

( )

1

1 2

=

1

f o o

i s f

o

s

V V VR

V

R

V

R

AV

A

V V=

=

+

=

+

( )

( )

1

1 2

;= ;

f o o i

s D f

f

o

sV V V V V V

Let

R

R R

V VV V Q

=

+

=

=

( )

( ) ( )

1

21

1

2

;

=

=;f i G S s s

D fG S

i

o

S

f o

S

o

V V L eV

V V

tV V V

V

R

R

V

R

V

& VVV

V

Q Q

+

=

=

=

Solve Exercise 10.6, page 816

CG

Amp.

Cascade

CS Amps.


18/24

3-18

2. FB Current Amplifier Shunt-series connection

Current-mixing or current

sampling topology

Stabilizes current gainLower RIN

Higher ROUT

CG stage Q1 followed by CS stage Q2

Io fed to RL V(RM) is fed via very large RF to S of Q1 If is subtracted from Is Ii = (Is If) For neg. FB, Ifmust have same polarity as Is

( ) ( ) ( )

( ) ( )1 1 1

2

1

1

1

2

1

:

;

s

X fgs F o f

s f

X

D D D

o

s

f

D

M

f

V Q

I Q

I I

V I

V

I

I

Node X I I I

I I Q

V I R I I R

I

R

V

=

= +

=

+

=

+

Read Section 10.3.2, pages 816-819.

I1

XY


19/24

3-19

Shunt-Series FB Amplifier

( )

1- ;: ; of

s

fo

i o

ISC I gain

I

I

I A

I I AAA = =

+

( ) ( )1 1oo' o of oV A I R A R = + = +

1

iif

RR =+ { }( )1s o

i ii i

s

of

i

I I I AV RR R

I I A A= = =

+

( )0

; ;s

i ooo'

o oo' o Iif o

o

V I AI RV

IRI

I=

= =

Read Section 10.7.2, pages 855-863

FB Current Amplifier


20/24

3-20

FB Current-Amplifier Analysis

Please try example 10.2, pages 818-

819 in text. Derive expressions.

( )2o m D i I g R I =

2

21 1

m D

Fm D

of

s

M

g R

Rg RR

IA

I

+ +

=

f M

o F M

I R

I R R

+

2

o

i m D

I

I RA g =


21/24

3-21

3. FB Transconductance Amplifier

Series-series connection

Voltage-mixing or current sampling

topology

Higher RIN

Higher ROUT

Diff. amp Q1 followed by CS stage Q2

Io fed to RL - develops feedback voltage VF

V(RF) is fed via RF to S of Q1

VF is subtracted from Vs Vi = (Vs Vf) For neg. FB, Ifmust have same polarity as Is

CS amplifier Q1 in cascade with CS stage Q2

Io fed to RL - develops feedback voltage VF

V(RF) is fed to S of Q

1 VF is subtracted from Vs Vi = (Vs Vf)

Read section 10.3.3, pages 819-820. Try exercise 10.8


22/24

3-22

Series-Series FB Amplifier

- : ;fo

om

s

ISC g gain A

V

V

I =

( ) ( )

( )1 x x o x o

f

i x

o oR A

VV I A R I RI

R

A

=

+

=

=

( ) ( )1i i ii i

ifi

I R RR

AI

IA R

+ += =

( )sif

i

i i

i

oi ii

i

AVIIVR

I

I

I

R R

I

+ =

+ ==

;xof ix

x o x

VR

IV V I I = = =

Read Section 10.5, pages 834 - 838.

Try Example 10.5, pages 838 - 842.

( );

1

o

f s

I AA

V A =

+

FB Transconductance Amplifier


23/24

3-23

4. FB Transresistance Amplifier

Shunt-shunt connection

Current-mixing or voltage

sampling topology

Lower RIN

Lower ROUT

Op-amp with RF RF senses Vo and provides feedback

current If If is subtracted from Is Ii = (Is If) For negative FB, Ifmust have same

polarity as Is

Read Section 10.3.4, pages 821-822.

Try Exercise 10.9, page 822.


24/24

3-24

FB Transresistance Amplifier Shunt-Shunt case

Read Section 10.6, pages 846 - 849.

Try Example 10.7, pages 850 - 854.

- : ;f

o

otrans

i

VOC r gain

I VA

I =

( ) ( )1 1i

ifi

iVRI A

R

A=

+=

+

i iiif

s f i o i

i

is

VV V

I V I AI I

V

I IR = = =

( );

1

o

f s

V AA

I A =

+

( )

( )

1

1

xx

oo

xx

f

xo

o

IVV R

AI

RA

A

R

V

R =

+

=

=

+

; ;xox i

xox

i xf I VAI

RR

VI

V

I

= =

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3EJ4 Set 03 Feedback MJD