chapter 8 - feedback

Chapter 8 - Feedback

1 - Desensitize The Gain

2 - Reduce Nonlinear Distortions

3 - Reduce The Effect of Noise

4 – Control The Input And Output Impedances

5 – Extend The Bandwidth Of The Amplifier

Basic structure of a feedback amplifier. To make it general, the figure shows signal flow as opposed to voltages or currents (i.e., signals can be either current or voltage).

The open-loop amplifier has gain A xo = A*xi

Output is fed back through a feedback network which produces a sample (x f) of the output (xo) xf = xo

Where is called the feedback factor

The input to the amplifier is x i = xs – xf (the subtraction makes feedback negative) Implicit to the above analysis is that neither the feedback block nor the load affect the amplifier’s gain (A). This not generally

true and so we will later see how to deal with it. The overall gain (closed-loop gain) can be solved to be:

A is called the loop gain 1+A is called the “amount of feedback”

Source A

Loadxs

xf

xi xo

A

A

x

xA

s

of

1

The General Feedback Structure

The General Feedback Structure (Intro to Simulink)

xs xi xf xo

A

xo A xi

xi xs xf

xf xo

Af

xo

xs

A

1 A

A 1 A

feedabck factor loop gain amount of feedabck

x fA

1 A x s

Finding Loop Gain

Generally, we can find the loop gain with the following steps:

– Break the feedback loop anywhere (at the output in the ex. below) – Zero out the input signal xs

– Apply a test signal to the input of the feedback circuit – Solve for the resulting signal xo at the output

If xo is a voltage signal, xtst is a voltage and measure the open-circuit voltage

If xo is a current signal, xtst is a current and measure the short-circuit current

– The negative sign comes from the fact that we are apply negative feedback

A

xs=0

xf

xi

xoxtst

Ax

x

AxAxAxx

xx

xx

tst

o

tstfio

fi

tstf

gain loop

0

Negative Feedback Properties

Negative feedback takes a sample of the output signal and applies it to the input to get several desirable properties. In amplifiers, negative feedback can be applied to get the following properties

–Desensitized gain – gain less sensitive to circuit component variations

–Reduce nonlinear distortion – output proportional to input (constant gain independent of signal level)

–Reduce effect of noise

–Control input and output impedances – by applying appropriate feedback topologies

–Extend bandwidth of amplifier

These properties can be achieved by trading off gain

Gain Desensitivity

Feedback can be used to desensitize the closed-loop gain to variations in the basic amplifier. Let’s see how.

Assume beta is constant. Taking differentials of the closed-loop gain equation gives…

Divide by Af

This result shows the effects of variations in A on Af is mitigated by the feedback amount. 1+Abeta is also called the desensitivity amount

We will see through examples that feedback also affects the input and resistance of the amplifier (increases Ri and decreases Ro by 1+Abeta factor)

21 A

dAdAf

A

dA

AA

A

A

dA

A

dA

f

f

1

11

1 2

A

AAf

1

Bandwidth Extension

We’ve mentioned several times in the past that we can trade gain for bandwidth. Finally, we see how to do so with feedback… Consider an amplifier with a high-frequency response characterized by a single pole and the expression:

Apply negative feedback beta and the resulting closed-loop gain is:

•Notice that the midband gain reduces by (1+AMbeta) while the 3-dB roll-off frequency increases by (1+AMbeta)

H

M

s

AsA

1

MH

MMf As

AA

sA

sAsA

11

1

1

Basic Feedback Topologies

Depending on the input signal (voltage or current) to be amplified and form of the output (voltage or current), amplifiers can be classified into four categories. Depending on the amplifier category, one of four types of feedback structures should be used (series-shunt, series-series, shunt-shunt, or shunt-series)

Voltage amplifier – voltage-controlled voltage source

Requires high input impedance, low output impedance

Use series-shunt feedback (voltage-voltage feedback)

Current amplifier – current-controlled current source

Use shunt-series feedback (current-current feedback)

Transconductance amplifier – voltage-controlled current source

Use series-series feedback (current-voltage feedback)

Transimpedance amplifier – current-controlled voltage source

Use shunt-shunt feedback (voltage-current feedback)

series-shunt

shunt-series

series-series

shunt-shunt

Examples of the Four Types of Amplifiers

• Shown above are simple examples of the four types of amplifiers. Often, these amplifiers alone do not have good performance (high output impedance, low gain, etc.) and are augmented by additional amplifier stages (see below) or different configurations (e.g., cascoding).

vIN

RD

vOUT

Vb

RD

vOUT

iIN

Vb

iOUT

iIN

vIN

iOUT

VoltageAmp

TransconductanceAmp

TransimpedanceAmp

CurrentAmp

vIN

RD

vOUTVb

RD

vOUT

iIN

Vb

iIN

vIN

iOUT

RD

iOUT

RD

lower Zout lower Zout higher gain higher gain

Series-Shunt Feedback Amplifier(Voltage-Voltage Feedback)

Samples the output voltage and returns a feedback voltage signal

Ideal feedback network has infinite input impedance and zero output resistance

Find the closed-loop gain and input resistance

The output resistance can be found by applying a test voltage to the output

So, increases input resistance and reduces output resistance makes amplifier closer to ideal VCVS

ARV

AVVR

V

VR

RV

V

I

VR

A

A

V

VA

VVAV

VVV

VV

ii

iii

i

si

ii

s

i

sif

s

of

oso

fsi

of

1

1A

RR o

of

1

The Series-Shunt Feedback Amplifier

The Ideal Situation

The series-shunt feedback amplifier:

(a) ideal structure; (b) equivalent circuit.

AfVo

Vs

A

1 A

RifVs

Ii

Vs

Vi

Ri

RiVs

Vi Ri

Vi A Vi

Vi

Rif Ri 1 A

Zif s( ) Zi s( ) 1 A s( ) s( )

Z of s( )Z o s( )

1 A s( ) s( )

Series-Series Feedback Amplifier(Current-Voltage Feedback)

For a transconductance amplifier (voltage input, current output), we must apply the appropriate feedback circuit

Sense the output current and feedback a voltage signal. So, the feedback current is a transimpedance block that converts the current signal into a voltage.

To solve for the loop gain:

Break the feedback, short out the break in the current sense and applying a test current

To solve for Rif and Rof

Apply a test voltage Vtst across O and O’

Gm

RF

Iout

ZL

Itst

Vf

A

A

V

IA

s

of

1

)mG called (also i

o

V

IA

fmtst

out RGI

IA Gain Loop

AR

I

VAIR

I

IIR

I

VV

I

VR i

i

iii

i

oii

i

fi

i

sif

1

otst

otsttst

tst

oitst

tst

tstof RA

I

RIAI

I

RAVI

I

VR

1

Shunt-Shunt Feedback Amplifier(Voltage-Current Feedback

• When voltage-current FB is applied to a transimpedance amplifier, output voltage is sensed and current is subtracted from the input

– The gain stage has some resistance

– The feedback stage is a transconductor

– Input and output resistances (Rif and Rof) follow the same form as before based on values for A and beta

A

A

I

VA

VA

VIII

I

VA

s

of

oo

fis

i

o

1

ARR iif 1

A

RR o

of

1

Shunt-Series Feedback Amplifier(Current-Current Feedback)

• A current-current FB circuit is used for current amplifiers

– For the b circuit – input resistance should be low and output resistance be high

• A circuit example is shown

– RS and RF constitute the FB circuit

• RS should be small and RF large

– The same steps can be taken to solve for A, Abeta, Af, Rif, and Rof

• Remember that both A and b circuits are current controlled current sources

Iout

Iin

RS

RF

RD


MATLAB / SIMULINK

Anyone interested to put together an Introduction to MATLAB / Simulink PowerPoint Presentation plus some examples?


Exercise 8.1

Af 10 A 104

a ) R1

R1 R2

b ) AfA

1 A

1

given

AfA

1 A

Find 0.1

R1

R1 R20.1

R2

R19

Amount_Feedback 20 log 1 A c )

Amount_Feedback 60

Vs 1 Vo Af Vs Vo 10d )

Vf Vo Vf 0.999

Vi Vs Vf Vi 10 104

e ) A 0.8 104

AfA

1 A Af 9.998

10 9.99810

100 0.02


Exercise 8.1

Some Properties of Negative Feedback

Gain Desensitivity

AfA

1 A

deriving

dAfdA

1 A ( )2

dividing by AfA

1 A

dAf

Af

1

1 A ( )

dA

A

The percentage change in Af (due to variations in some circuit parameter) is smaller than the pecentage cahnge in A by the amount of feedback. For this reason the amount of feedback

1 A

is also known as the desensitivity factor.


Bandwidth Extension

High frequency response with a single pole

A s( )AM

1s

H

AM denotes the midband gain and H the upper 3-dB frequency

Af s( )A s( )

1 A s( )

Af s( )

AM

1 AM

1s

H 1 AM

Hf H 1 AM Lf

L

1 AM


Noise Reduction, Reduction of Nonlinear Distortion

Read and discuss in class

The Shunt-Series Feedback Amplifier

Example 8.1

AVo'

Vi'

RL R1 R2( )[ ]

RL R1 R2

RL R1 R2( )[ ]

RL R1 R2ro

Rid

Rid RsR1 R2

R1 R2

Rif Ri 1 A Vf

Vo''

R1

R1 R2Ri Rs Rid

R1 R2R1 R2

Rin Rif RsAf

Vo

Vf

A

1 A

RofRo

1 A

Ro

roRL R1 R2( )RL R1 R2



Example 8.1

Af 857.92AfA

1 A

Rin 7.666 105Rin Rif Rs

Rif 7.766 105Rif Ri 1 A

Ri 1.11 105Ri Rs Rid

R1R2R1 R2

9.99 104

R1

R1 R2

A 6.002 103A

RL R1 R2( )[ ]

RL R1 R2

RL R1 R2( )[ ]

RL R1 R2ro

Rid

Rid RsR1R2

R1 R2

RL 2 103 10

4ro 103Rid 10

5Rs 104R2 10

6R1 103

Ro



Rof

Ro

1 A Rof 95.228

Ro 666.223

Rout 10

given

Rof

Rout RL

Rout RL

Find Rout 99.989


Example 8.1


Exercise 8.5


Example 8.2


Exercise 8.6

The Shunt-Shunt Feedback Amplifier

Exercise 8.3


Example 8.4


Exercise 8.7

Determining The Loop Gain

Exercise 8.8-9

The Stability Problem and Margins

Closed-LoopTransfer Function

Nyquist

Root-Locus

Bode

Polar Plot

Af s( )A s( )

1 A s( ) s( )The Nyquist Plot intersects the negative real axis at 180. If this intersection occurs to the left of the point (-1, 0), we know that the magnitude of the loop gain at this frequency is greater than the unity and the system will be unstable. If the intersection occurs to the right of the point (-1,0) the system will be stable.It follows that if the Nyquist encircles the point (-1,0) the amplifier will be unstable.

The Stability Problem

Closed-LoopTransfer Function

Nyquist

Root-Locus

BodeMagnitude and Phase

Gain MarginPhase Margin - If at the frequency of unity loop-gain magnitude, the phase lag is in excess of 180 degrees, the amplifier is unstable.


T he N yquist P lo t

w 100 99.9 100 j 1 s w( ) j w f w( ) 1

G w( )50 4.6

s w( )3

9 s w( )2 30 s w( ) 40

2 1 0 1 2 3 4 5 65

0

5

Im G w( )( )

0

Re G w( )( )


Bode plot for loop gain Abeta

– Gain margin is an indication of excess gain before instability

– Phase margin is an indication of excess phase before -180° phase shift at unity gain

Stability Study Using Bode Plots

Stability Study Using Bode Plots

w .1 .11 2 K 2 j 1

G w( )K

j w j w 1( ) j w 2( )

Bode1 w( ) 20 log G w( )

0 0.5 1 1.5 220

0

20

Bode1 w( )

w

Open Loop Bode Diagram

T w( )G w( )

1 G w( )

0 0.5 1 1.5 220

0

20

Bode1 w( )

w

T w( )G w( )

1 G w( )

Bode2 w( ) 20 log T w( )

0.5 1 1.5 220

10

0

10

Bode2 w( )

w

Closed-Loop Bode Diagram


Exercise 8.10

Effect of Feedback on the Amplifier Poles

Effect of Feedback on the Amplifier Poles

Exercise 8.11

chapter 8 - feedback

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