microelectronic circuits ii ch9 :...

CNU EE 9.1-1

Microelectronic Circuits II

Ch 9 : Feedback

9.1 General Feedback structure9.2 Some properties of Negative feedback9.3 The Four basic Feedback topologies9.4 The Feedback Voltage Amplifier (Series-Shunt)

CNU EE 9.1-2

FeedbackTheory of negative feedback- Electronics engineer, Harold Black invented the feedback amplifier in 1928- Search for a methods for the design of amplifiers with stable gain for use in telephone repeaters- Feedback : negative (degenerative) or positive (regenerative)- Almost all op-amp circuits employ negative feedback

Properties of Negative feedback- Desensitize the gain

the value of gain is less sensitive to variations in the value of circuit components (Temperature change)- Reduce nonlinear distortion

the output is proportional to the input & gain is constant, independent of signal levels- Reduce the effect of noise

minimize the contribution to the output of unwanted electric signals generated- Control the input and output impedance

raise or lower the input and output impedance- Extend the bandwidth of the amplifier

Expense of the desirable properties of negative feedback- Reduction in gain- gain-reduction factor = amount of feedbackß circuit is desensitized, input impedance of a voltage amplifier is increased, bandwidth is extended

- The basic idea of negative feedback is to trade off gain for other desirable properties

CNU EE 9.1-3

The General Feedback Structure

loop gain

bAA

xxA

s

of +

=º1 Amount of feedback

if ,1>>bAb1

=fA

- Gain of the feedback amplifier is determined by the feedback network- Feedback network = passive components à accurate, predictable & stable gain

sf xA

Axb

b+

=1 ,1>>bA& if sf xx »

The gain of feedback amp.

Signal-flow diagram of the feedback amplifier- open-loop gain A, feedback factor b- negative feedback reduces the signal that appears at the input of the basic amplifier

io Axx =

of xx b=

fsi xxx -=

)()( osfsio xxAxxAAxx b-=-==

Subtractionmakes the feedback negative

Gain-with-feedback Af issmaller than the open-loopgain A by the amount offeedback, 1+AbAf : closed-loop gain

- Signal xi (=error signal) at the basic amplifier is reduced to almost zero

si xA

xb+

=1

1

CNU EE 9.1-4

- the feedback amplifier has a midband gain of AM/(1+ AM b) and upper 3-dB frequency wHf :

Some properties of Negative feedback§Gain Desensitivity : sensitivity reduction property

The percentage change in Af (due to variations in some circuit parameters) is smaller than the percentage change in A by the amount of feedback (b is constant)

§Bandwidth Extension - Amplifier whose high-frequency response is characterized by a single pole

( )>> 211 bb A

dAdAAAA ff +

=+

=( ) A

dAAA

dA

f

f

b+=

11

( )( )bw

bbw MH

MMff

H

M

AsAAsA

sAsAsA

sAsA

+++

=+

=+

=1/1

1/)()(1

)()(&1

)( >

)1( bww MHHf A+=

Desensitivity factor

wHf : Upper 3dB freq.AM : midband gain, wH : upper 3-dB freq.b : freq.-independent factor

Midband gain

- If the open-loop gain has a dominant low-frequency pole giving rise to a lower 3-dB frequency wL, then the feedback amplifier have a lower 3-dB frequency wLf :

Af (s) : closed-loopgain

bww

M

LLf A+

=1

à wHf is increased by the amount of feedback

à wLf is decreased by the amount of feedback

CNU EE 9.1-5

Some properties of Negative feedback§Bandwidth Extension

Amplifier bandwidth is increased by the amount of feedback by which its midband gain is decreased, maintaining the gain-bandwidth product at a constant value

Negative feedback works to minimize the change in gain magnitude, including its change with frequency

CNU EE 9.1-6

Some properties of Negative feedback§Noise reduction- reduces the noise or interferences in an amplifier = increase the ratio of signal to noise

n

s

VVIS =/

bb 21

1

21

21

11 AAAV

AAAAVV nso +

++

=

2AVV

IS

n

s=\

- Improvement in S/I ratio by the application of feedback is possible only if one can precede theinterference-prone stage by a (relatively) interference-free stage

- Application : reduction of the power supply hum in the output power-amplifier stage of an audioamplifier à hum at the output is reduced by the amount of the voltage gain of this added preamplifier

- Amplifier w/ gain A1, input signal Vs, noise or interference Vn

- Signal-to-interference Ratio (=S/N ratio)

- Amplifier w/ gain A2, that does not sufferfrom the noise problem à clean amplifier

- apply negative feedback b to keep theoverall gain constant

A2 times higher thanin the original case

CNU EE 9.1-7

Some properties of Negative feedback§Reduction in nonlinear distortion- Amplifier transfer characteristic is considerably linearized (i.e., made less nonlinear) because negative-feedback reduces the dependence of the overall closed-loop amplifier gain on the open-loop gain of the basic amplifier

- Change in open-loop gain : 1000 to 100- b = 0.01- Resulting transfer characteristic of the

closed-loop amplifier

9.9001.010001

10001 =

´+=fA

5001.01001

1002 =

´+=fA

bAAA f +

=1

- Order-of-magnitude change in slope is considerably reduced- Reduction in voltage gain à preamplifier should be added- Negative feedback can do nothing at all about amplifier saturation, since in saturation the gain is

small (almost zero) and hence the amount of feedback is also very small (almost zero)

CNU EE 9.1-8

Four basic Feedback topologies

Voltage-mixing voltage-sampling(series-shunt)

Current-mixing current-sampling(shunt-series)

Voltage-mixing current-sampling(series-series)

Current-mixing voltage sampling(shunt-shunt)

Based on the quantity to be amplified (voltage or current) and on the desired form of output (voltage or current)

CNU EE 9.1-9

Voltage Amplifiers

- Amplify an input voltage signal and provide an output voltage signal- Voltage controlled voltage source ( high input impedance & low output impedance are required)- Since output is voltage in the voltage amplifier, the feedback network should sample the output voltage- Because of the Thevenin representation of the source, feedback signal xf should be a voltage

à voltage xf is mixed with the source voltage in series- Suitable feedback topology for the voltage amplifier = voltage-mixing, voltage-sampling- Series connection at the input and parallel or shunt connection at the output à Series-shunt feedback - Stabilize the voltage gain- High input resistance caused by the series connection at the input- Low output resistance caused by the parallel connection at the output

Sample the output voltageMixed with the source voltage in series

§ Voltage Amplifiers

CNU EE 9.1-10

Voltage Amplifiers- Increased input resistance : Vf subtracts from Vs à smaller Vi at the input of basic amplifier à

smaller input current à larger resistance seen by Vs- decreased output resistance : feedback works to keep Vo as constant as possible à current change DIo lowers the change DVo in Vo à lower output resistance DVo /DIo

§Example of series – shunt feedback amplifiers- Noninverting op-amp configuration- Feedback network = voltage divider (R1, R2) develops Vfà negative input terminal of op-amp

- Negative feedback : Vf must be of the same polarity as Vsà a smaller signal at the input of the basic amplifier

- As Vs , Vo & voltage divider à Vf : change in Vf is ofthe same polarity as the change in Vs à negative feedback

- Two MOSFET amplifier stages in cascsade- Feedback network = voltage divider (R1, R2) develops Vfà source terminal of Q1

- Subtraction by applying Vs to the gate of Q1 & Vf to its source à amplifier input signal Vi = Vgs = Vs – Vf

- As Vs , drain voltage of Q1 = gate of Q2 à drain voltage Voà feedback network voltage divider à Vf : change in Vf isof the same polarity as the change in Vs à negative feedback

CNU EE 9.1-11

Current Amplifiers

Samplethe output current

Mixed with the source in shunt

- Input : Norton equivalent current source- Feedback network samples the output current- Feedback signal = current à mixed in shunt with the current source- Current-mixing current-sampling- parallel (or shunt) connection at the input and series connection at the output à Shunt-series feedback- Stabilize the current gain- Lower input resistance : Is – If à lower input voltage across the current source Is- Higher output resistance : negative feedback keeps Io as constant & if the voltage across RL is changed,

the resulting change in Io will be lower than it would have been without the feedback

CNU EE 9.1-12

Current Amplifiers§A CG stage Q1 followed by a CS stage Q2- Load current Io is fed to a load resistance RL- A small resistance RM in series with RL

à a sample of Io- The voltage developed across RM is fed via a

large resistance RF to the source node of Q1- The feedback current If that flows through RF is

subtracted from Is at the source nodeà input current Ii = Is – If

- For negative feedback, If must have the samepolarity as Is

Current samplingReference

direction of If :If subtracts from Is

§ Qualitative check for feedback polarityIs à Ii à drain voltage of Q1 = gate ofthe p-channel device Q2 à drain current ofQ2, Io à voltage across RM à If:: the same polarity assumed for the initial

change in Is à negative feedback

CNU EE 9.1-13

Transconductance Amplifiers

Samplethe output current

Mixed with the source in series

- Input signal : voltage- Output signal : current- Voltage-mixing current-sampling- Series-series feedback (series connection

at both the input and the output)- Series connection at input à increased Rin- Output current sampling à increased Ro

§Differential amplifier A1 followed by a CS stage Q2

- Output current Io is fed to RL & series resistance RF à RF develops Vf à Vf is applied to the positive input terminal of A1

- Subtraction of Vs – Vf by differential action

§Check that Vf & Vs have the same polarity - Positive change in Vs à negative change at

the gate of Q1 à Io à Positive change inVf :: same polarity assumed for the change inVs à negative feedback

CNU EE 9.1-14

Transresistance Amplifiers

Samplethe output voltage

Mixed with the source in shunt

- Input signal : current- Output signal : voltage- Current-mixing voltage-sampling- Shunt-shunts feedback (parallel (or

shunt) connection at both the input and the output)

- Shunt connection at input à reduces Rin- Shunt connection at output à reduced Ro

§Op-amp w/ a feedback resistance RF- RF senses Vo & provides a feedback current If- If is subtracted from Is at the input node

§Qualitative check for negative feedback- Is à input current Ii à voltage of the

negative input terminal à output voltageà If :: If & Is have the same polarity à negative feedback

CNU EE 9.1-15

Feedback Voltage Amplifier (Series-Shunt )§Ideal case

i

o

VVA º

bAA

VVA

s

of +

=º1

( )

( ) ii

sif

i

s

i

ii

si

RAIVR

RAV

RVI

AVV

b

bb

+=º

+==

+=

1

11

- Unilateral open-loop amplifier (A circuit)à input resistance Ri, voltage gain A,

output resistance Roà source & load resistances are included

inside the A circuit- Ideal voltage-mixing voltage-sampling

feedback network (b circuit)à does not load the A circuità does not change the value of A

Ideal structure

Equivalent circuit

§Input resistance with feedback (ideal case)

Series-mixing feedback increases the inputresistance by a factor equal to the amount offeedback, (1+Ab)Increased Rif is independent of the type ofsampling

Af : open-circuit voltage gain of the feedback amplifier

CNU EE 9.1-16

§Ideal situation : Output Resistance with feedback

x

xof I

VR =

Apply Vx between the output terminals

o

ixx R

AVVI -=

Set Vs = 0

xofxfi VVVfromVVV bbb ==-=-=

( )o

xx R

AVI b+=\

1

bARR o

of +=

1

- Shunt sampling (or voltage sampling) atthe output decreases the amplifier outputresistance by a factor equal to the amountof feedback, (1+Ab)

- The reduction of Rof does not depend on themethod of mixing

From the input loop

Feedback Voltage Amplifier (Series-Shunt )

CNU EE 9.1-17

§Practical situation- Feedback network is not an ideal voltage

controlled voltage sourceà resistive and hence, load the basic

amplifierà affect A, Ri & Ro

- Source resistance Rs & load resistance RLaffect A, Ri & Ro

- Derivation of A circuit & b circuit froma given series-shunt feedback amplifier (Practical series-shunt feedback amplifier)

Ideal structure•Practical series-shunt feedback amplifier

- Source and load resistances shouldbe lumped with the basic amplifier

- Two-port feedback network is represented in terms of h parameters


Practical series-shunt feedback amplifier Ri & Ro vs. Rin & Rout vs. Rif & Rof

CNU EE 9.1-18

§Derivation of A circuit & β circuit

use of h parameters (appendix C)

úû

ùêë

éúû

ùêë

é=ú

û

ùêë

é

2

1

2221

1211

2

1

VI

hhhh

IV

Feedback CircuitInput Impedance(w/ Output short)

Feedback CircuitOutput Admittance(w/ input open)

negligible

β

Practical series-shunt feedback amplifier w/ the feedback network represented by its h parameter

Feedback network isrepresented by a seriesnetwork at port 1 and aparallel network at port 2

Feedback Voltage Amplifier (Series-Shunt )Ri & Ro vs. Rin & Rout vs. Rif & Rof

CNU EE 9.1-19

§Derivation of A circuit and β circuit

02

112

1 =

º=IV

Vhb

amplifierbasic

networkfeedback hh 2121 <<

networkfeedback

amplifierbasic hh 1212 <<

Circuit in (b) with h21 neglected

- Current source h21I1 : forwardtransmission of the feedback network

- Passive feedback network à forwardtransmission is neglected in comparison tothe much larger forward transmission ofthe basic amplifier

à Omit the controlled source h21I1

- Includes h11 and h22 with the basic amplifier- If the basic amplifier is unilateral,

- Loading effect of feedback network on basicamplifier is represented by h11 & h22- h11 : impedance looking into port 1 of thefeedback network with port 2 short-circuited- h22 : conductance looking into port 2 of thefeedback network with port 1 open-circuited- loading effect of the feedback to basic Amp. :If connection is shunt, short-circuit the port;if connection is series, open-circuit it- b should be found w/ port 1 open-circuited


CNU EE 9.1-20

§Summary : the Rules for Finding A circuit and β circuit

- Ri & Ro : input & outputresistance of the A circuit- Rif & Rof : input & outputresistance of the feedbackamplifier, including Rs & RL- Actual input & outputresistance of the feedbackamplifier usually exclude Rs& RL à Rin & Rout

in if sR R R= -

1 11outof L

RR R

æ ö= -ç ÷ç ÷

è ø


01

111

2 =

ºVI

VR02

222

1 =

ºIV

IG

CNU EE 9.1-21

Example 9.3Op-amp connected in a noninverting configuration (open-loop gain m=104, differential input resistance Rid=100kW, output resistance ro=1kW). Use the feedback method to analyze the circuit taking both Ridand ro into account. Find A, b, closed-loop gain Vo/Vs, input resistance Rin & output resistance Rout.


01

111

2 =

ºVI

VR02

222

1 =

ºIV

IGA circuit

b circuit02

1

1 =

ºIV

Vb

Feedback network samples the output voltage Vo and provides a voltage signal (across R1) that is mixed in series with the input signal VsLoading effect of the feedback network at input/ output side : R11 & G22

RL=2kW, R1=1kW, R2=1MW, Rs=10kW

CNU EE 9.1-22

( )[ ]( )[ ] ( ) VV

RRRRR

rRRRRRR

VVA

sid

id

oL

L

i

o /6000////

//

2121

21 »++

×++

+=º m

VVA

AVVA

s

of /857

76000

1==

+=º

bVVRR

RVV

o

f /10 3

21

1 -»+

==b

( )W=´=

+=

kARR iif

77771111 b

( ) W»++= kRRRRR idsi 111// 21

W=-= kRRR sifin 739

W=+

= 3.951 bA

RR oof

( ) W»+= 667//// 21 RRRrR Loo

Loutof RRR //= W»100outR


Input resistance of A circuit

Input resistance of feedback amplifier

Output resistance of A circuit

Output resistance of feedback amplifier

microelectronic circuits ii ch9 :...

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