Microelectronic Circuits SJTU Yang Hua Chapter 8 Feedback Introduction 8.1 The general feedback...

Microelectronic Circuits SJTU Yang Hua

Chapter 8 Feedback

Introduction8.1 The general feedback structure8.2 Some properties of negative feedback8.3 The four basic feedback topologies8.4 The series-shunt feedback amplifier8.5 The series-series feedback amplifier8.6 The shunt-shunt and shunt-series feedback amplifier8.10 Stability study using bode plot8.11 Frequency compensation

SJTU Yang HuaMicroelectronic Circuits

Introduction

It’s impossible to think of electronic circuits without some forms of feedback.

Negative feedback Desensitize the gain Reduce nonlinear distortion Reduce the effect of noise Control the input and output impedance Extend the bandwidth of the amplifier

The basic idea of negative feedback is to trade off between gain and other desirable properties.

Positive feedback will cause the amplifier oscillation.


Three Parts:

PartI: The basic concept and some Properties of negative feedback

PartII: The four basic feedback and analysis PartIII: The loop gain, stability problem

and frequency compensation


PartIPartI

The basic conceptThe basic concept Judgment and PropertiesJudgment and Properties

of feedbackof feedback

examplesexamples


PartI: The basic concept and some Properties of negative feedback

This is a signal-flow diagram, and the quantities x represent either voltage or current signals.

In electronic circuits, part of or all output signal is fed back to input, and affects the input signal value, which is called feedback.

8.1 The General Feedback Structure


The feedback judgment for amplifier circuits


Negative feedback and positive feedback ：According to the effecting of feedback1) positive feedback increases the signal that appears at the input of the basic amplifier2) negative feedback reduces the signal that appears at the input of the basic amplifier

DC feedback and AC feedback ：1) Feedback quantity only contains DC quantity， is called DC feedback2) Feedback quantity only contains AC quantity， is called AC feedbackUsually AC feedback and DC feedback are concomitant

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The feedback judgment ：

(a) No feedback(b) Feedback exists(c) No feedback


Instantaneous polarity method ：

The judgment of feedback parity

1) Regulate the polarity of input signal relative to ground at sometime.

2) Decide all points’ parity step by step, at last get the parity of output signal.3) According to the parity of output signal decides the parity of amount of feedback.

4) If amount of feedback increases the signal that appears at the input of the basic amplifier, the circuit inducts the positive feedback. Otherwise, it inducts the negative feedback.


To integrated operational amplifiers ， the input quantity can be UD or iN （ iP ）


To discrete components amplifiers ， the input quantity can be Ube or ib


AC feedback ， no DC feedback

DC feedback ， no AC feedback

The judgment of DC feedback and AC feedback


Example ： feedback ？ Positive or negative ？ DC or AC ？

AC and DC negative feedback


The General Feedback Equation

of

io

xx

Axx

AA

x

xA

s

of

1

Open loop gain A Feedback factor βLoop gain AβClosed loop gain Af

Amount of feedback (1+ Aβ)


The General Feedback Equation

1fA

sf

sf

xx

xA

Ax

1

If Aβ >>1, The gain of the feedback amplifier is almost entirely determined by the feedback network. If Aβ >>1, which implies that the signal Xi at the input of the basic amplifier is reduced to almost zero.


8.2 Some Properties of Negative Feedback

1. Gain desensitivity

A

dA

AA

dA

A

dAdA

f

f

f

1

1

)1( 2

the percentage change in Af (due to variations in some circuit parameter) is smaller than the percentage change in A by the amount of feedback. For this reason the amount of feedback, 1 + Aβ, is also known as the desensitivity factor.


2. Bandwidth extension

Some Properties of Negative Feedback

Note that the amplifier bandwidth is increased by the same factor by which its midband gain is decreased, maintaining the gain-bandwidth product at a constant value.



3. Noise reduction



4. Reduction in nonlinear distortion

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4. Reduction in nonlinear distortion


Homework:

May 11th, 2008

8.1; 8.7;


PartII:The four basic feedback and analysis


8.3 The Four Basic Feedback Topologies

Voltage amplifier---series-shunt feedback

voltage mixing and voltage sampling


The Four Basic Feedback Topologies

Current amplifier---shunt-series feedback

Current mixing and current sampling


Example:

Figure 8.5 A transistor amplifier with shunt–series feedback. (Biasing not shown.)


Transconductance amplifier---series-series feedback

Voltage mixing and current sampling



Example:

Figure 8.6 An example of the series–series feedback topology. (Biasing not shown.)


Transresistance amplifier---shunt-shunt feedback

Current mixing and voltage sampling



Example:

Figure 8.7 (a) The inverting op-amp configuration redrawn as (b) an example of shunt–shunt feedback.


Homework:

May 11nd, 2008

8.14; 8.15; 8.17; 8.19


8.4 The Series-Shunt Feedback Amplifier

The ideal situation The practical situation Summary


The Ideal Situation

A unilateral open-loop amplifier (A circuit).

An ideal voltage mixing voltage sampling feedback network (β circuit).

Assumption that the source and load resistance have been included inside the A circuit.


The Ideal Situation

Equivalent circuit.

Rif and Rof denote the input and output resistance with feedback.


Input and Output Resistance with Feedback

Input resistance

In this case, the negative feedback increases the input

resistance by a factor equal to the amount of feedback. Output resistance

In this case, the negative feedback reduces the output

resistance by a factor equal to the amount of feedback.

)1( ARR iif

AR

R oof

1


The Practical Situation

Block diagram of a practical series–shunt feedback amplifier.

Feedback network is not ideal and load the basic amplifier thus affect the values of gain, input resistance and output resistance.



The circuit in (a) with the feedback network represented by its h parameters.

Omit the controlled source h21I102

112

1

I

V

Vh



The circuit in (b) with h21 neglected.



The load effect of the feedback network on the basic amplifier is represented by the components h11 and h22.

The loading effect is found by looking into the appropriate port of the feedback network while the port is open-circuit or short-circuit so as to destroy the feedback.

If the connection is a shunt one, short-circuit the port. If the connection is a series one, open-circuit the port. Determine the β.

02

112

1

I

V

Vh



Summary

Ri and Ro are the input and output resistances, respectively, of the A circuit.

Rif and Rof are the input and output resistances, respectively, of the feedback amplifier, including Rs and RL.

The actual input and output resistances exclude Rs and RL.

Loutof

sinif

RRR

RRR

//


Example of Series-Shunt Feedback Amplifier



Op amplifier connected in noninverting configuration with the open-loop gain μ, Rid and ro

Find expression for A, β, the closed-loop gain Vo/Vi , the input resistance Rin and the output resistance Rout

Find numerical values





21

1

RR

R

V

V

s

f


8.5 The Series-Series Feedback Amplifier

The ideal situation The practical situation Summary


The Ideal Situation

Transconductance gain i

o

V

IA


The Ideal Situation

o

f

I

VTranresistance feedback factor


Input and Output Resistance with Feedback

Input resistance

In this case, the negative feedback increases the input resistance by a factor equal to the amount of feedback.

Output resistance

In this case, the negative feedback increases the output resistance

by a factor equal to the amount of feedback.

)1( ARR iif

)1( ARR oof



Block diagram of a practical series–series feedback amplifier.

Feedback network is not ideal and load the basic amplifier thus affect the values of gain, input resistance and output resistance.



The circuit of (a) with the feedback network represented by its z parameters.



A redrawing of the circuit in (b) with z21 neglected.



The load effect of the feedback network on the basic amplifier is represented by the components Z11 and Z22.

Z11 is the impedance looking into port 1 of the feedback network with port 2 open-circuited.

Z22 is the impedance looking into port 2 of the feedback network with port 1 open-circuited.

Determine the β.

02

112

1

I

I

Vz



Summary

Ri and Ro are the input and output resistances, respectively, of the A circuit.

Rif and Rof are the input and output resistances, respectively, of the feedback amplifier, including Rs and RL.

The actual input and output resistances exclude Rs and RL.

Loutof

sinif

RRR

RRR

'


Example of Series-Series Feedback Amplifier





112

2E

EFE

E

o

f RRRR

R

I

V



）33

2312

//)(1(

)1(

1)//(

rRrgrR

ARR

h

RrRRRR

ofomoout

oof

fe

CeEFEo


8.6 The Shunt-Shunt and Shunt-Series Feedback Amplifiers

Fig8.19. Ideal structure for the shunt-shunt feedback amplifier.

A

RR

A

RR

A

A

I

VA

oof

iif

S

of

1

1

1


The practical Shunt-Shunt feedback amplifier



Example 8.3



The shunt-series configuration

Fig 8.22 Ideal structure for the shunt–series feedback amplifier.

ARR

A

RR

A

A

I

IA

oof

iif

S

of

1

1

1


A practical shunt-series feedback amplifier



Example 8.4



summary


The method of finding A circuit

① h,z,r,g parameter method for two-port feedback network.

② Equivalent circuit method: Find the feedback network; Find the feedback network equivalent load Resistance to amplif

ier input, for output voltage sampling feedback, output short-circuit (Vo=0); for current sampling feedback, output Io open-circuit (Io=0).

Find the feedback network equivalent load Resistance to amplifier output, for input voltage mixing feedback, it should be disconnected from input to feedback network (Ii=0); for input current mixing feedback, let input connect to ground to disconnect from input signal to feedback network.


引入负反馈的一般原则

为稳定静态工作点，应引入为稳定静态工作点，应引入直流负反馈；直流负反馈；为改善电路动态性能，应引入为改善电路动态性能，应引入交流负反馈；交流负反馈；当信号源为恒压源或内阻较小的电压源时，当信号源为恒压源或内阻较小的电压源时，

应引入应引入串联负反馈；串联负反馈；当信号源为恒流源或内阻较大的电压源时，当信号源为恒流源或内阻较大的电压源时，

应引入应引入并联负反馈；并联负反馈；


当负载需要稳定的电压信号时，应引入当负载需要稳定的电压信号时，应引入电压负反馈；电压负反馈；

当负载需要稳定的电流信号时，应引入当负载需要稳定的电流信号时，应引入电流负反馈；电流负反馈；

若将电流信号转换成电压信号，应引入若将电流信号转换成电压信号，应引入电压并联负反馈电压并联负反馈 (shunt-shunt)(shunt-shunt) ；；

若将电压信号转换成电流信号，应引入若将电压信号转换成电流信号，应引入电流串联负反馈电流串联负反馈 (series-series)(series-series) 。。


Example:1 ）减小放大电路从信号源索取的电流并增强带负载的能力

2 ）将输入电流转换成与之成稳定稳定线性关系的输出电流 3 ）将输入电流转换成稳定的输出电压

答：引入 series-shunt ( 电压串联负反馈 ) ： 8 连 10 ， 3 连 9 ， 4 连 6

答：引入 shunt-series( 电流并联负反馈 ) ： 7 连 10 ， 2 连 9 ； 4 连 6

答： shunt-shunt( 电压并联负反馈 ) ： 2 连 9 ， 8 连 10 ， 5 连 6


Homework:

May 13th, 2008

8.20; 8.25; 8.30; D8.37


The Stability Problem

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when

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when

1)2n(

2n

ejjAjjAjL

jjA

jA

ssA

sAsAf

)()()()()(

11

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The Stability Problem

The condition for negative feedback to oscillate

Any right-half-plane poles results in instability. Amplifier with a single-pole is unconditionally stable. Amplifier with two-pole is also unconditionally stable. Amplifier with more than two poles has the possibility

to be unstable. Stability study using bode plot

1)()()( jjAjL


Balance condition:Magnitude condition:Phase Condition:

1|| A

12 nA

1|| A

Start up oscillation condition:

Judgment method of amplifier stability:

1) If ω180 is not exist, then the Amplifier is stable;2) If ω180 is exist and ω180< ω1, then the amplifier is not stable;3) If ω180 is exist and ω180> ω1, then the amplifier is stable;


The Definitions of the Gain and Phase margins

180lg20

AGm

Gain margin represents the amount by which the loop gain can be increased while stability is maintained.

Unstable and oscillatory

Stable and non-oscillatory

Only when the phase margin exceed 45º or gain margin exceed 6dB, can the amplifier be stable.

1

180

Am


Example:


Effect of phase margin on closed-loop response To see the relationship between phase margin and

Close-Loop gain, consider a feedback amplifier with a large low-frequency loop gain.

45 ,1

3.1

1

/1

1

180,1

1 ,1

1

1

11

1

mf

j

j

f

mj

fo

jA

e

e

jA

jAjA

ejA

AA


Stability analysis using Bode plot of |A|.

1log20log20log20 jAA

Hz6102.3

Hz5106.5


Stability Analysis Using Bode Plot of |A|

Gain margin and phase margin The horizontal line of inverse of feedback factor in dB. A rule of thumb:

The closed-loop amplifier will be stable if the 20log(1/β) line intersects the 20log|A| curve at a point on the –20dB/decade segment.

The general rule states:

At the intersection of 20log[1/ | β (jω)| ] and 20log |A(jω)| the difference of slopes should not exceed 20dB/decade.


Frequency Compensation

The purpose is to modifying the open-loop transfer function of an amplifier having three or more poles so that the closed-loop amplifier is stable for any desired value of closed-loop gain.

Theory of frequency compensation is the enlarge the –20dB/decade line.

Implementation Capacitance Cc added Miller compensation and pole splitting


Frequency Compensation

xcxH

xxH

RCCf

RCf

)(2

1

2

1

'1

1


45 ,

0log20 ,

1

2

m

H

ffthatshowsthat

Affwhen


Frequency Compensation-Miller compensation

A gain stage in a multistage amplifier with a compensating capacitor connected in the feedback path

An equivalent circuit.

Miller compensation can reduce the value of C


Homework:

May 20th, 2008

8.51; 8.54; 8.63; 8.66

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Microelectronic Circuits SJTU Yang Hua Chapter 8 Feedback Introduction 8.1 The general feedback...

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