Electronics chapter 4 negative feedback Guangdong Institute of Education ---BTEC electronic Review and Overview 0 、 introduction 1 、 what is feedback?

Electronics chapter 4 negative feedback

Guangdong Institute of Education ---Guangdong Institute of Education ---BTEC electronicBTEC electronic

Review and Overview0 、 introduction1 、 what is feedback?2 、 why we need feedback in our circuit?3 、 how many types of feedback?4 、 how many types of negative feedback?5 、 how to judge all kinds of negative feedback?6 、 advantages of negative feedback? 7 、 how to analyse the negative feedback circuit?8 、 conclusion----- assignment



introduction An audio-frequency amplifier can be designed to

have a certain current, voltage or power gain together with particular values of input and output impedance.

The amplifier will add noise and distortion to the signals. The components employed in the amplifier, both passive (resistors, capacitors, etc.) and active (transistors and f.e.t.s), will vary in value with both time and change in temperature and will have manufacturing tolerances.



introduction The gain of the amplifier may therefore vary with

time, with change in ambient temperature, and when

a component has to be replaced by another of the

Same type. (manufacturing tolerances)

Any fluctuations in the power supply may also cause

the gain of the amplifier to alter.


Guangdong Institute of Education ---Guangdong Institute of Education ---BTEC electronicBTEC electronic feedback

R110kΩ

R210kΩ

R3240Ω

R41kΩ

RL

C1

4.7uF

C2100uF

C3

47uF

VCC

12V

IO1IO1

IO2

IO3

IO4

IO3

0

Q1

2N914

10

9

1

R5

470kΩKey=A

50% VCC

2

Ui

Uo

Ii

Io

1 、 what is feedback?

2 、 For many applications a more or less constant gain iS necessary ， and this can be obtained if negative feedback (n.f.b.) iS applied to the amplifier, at the expense, however,of a reduction in gain.



Openloop circuit 、 gain without n.f.b 、 ACloseloop circuit 、 gain with n.f.b 、 Af 、 F(feedback factor)

1+AF (feedback depth)

R110kΩ

R210kΩ

R3240Ω

R41kΩ

RL

C1

4.7uF

C2100uF

C3

47uF

VCC

12V

IO1IO1

IO2

IO3

IO4

IO3

0

Q1

2N914

10

9

1

R5

470kΩKey=A

50% VCC

2

Ui

Uo

Ii

Io



why we need feedback in our circuit?1 、 feedback make our circuit stabilization:Quiescent operation point;Stability of gain: current gain, voltage gain;Decrease non-linearity distortionReduces noiseChange input and output impedanceBand spreadReduces amplitude ／ frequency distortion 2 、 generate oscillation (oscilloscope)



Types of feedback0 、 d.c feedback

1 、 a.c. feedback

2 、 positive feedback

3 、 negative feedback

4 、 voltage feedback

5 、 current feedback

6 、 Series feedback

7 、 parallel feedback



D.C. and A.C. feedback



Negative and Positive feedback





Negative and Positive feedback

Conclusion:If the Input point---feedback point at the same terminal:And same polarity==positive feedbackAnd different polarity==nagtive feedbackIf the input point---feedback point at the different terminal:And same polarity==negative feedbackAnd different polarity==positive feedback



Voltage and Current feedback

Conclusion:If the output point---sampling point at the same terminal:=voltage feedbackIf the output point---sampling point at the different terminal:=current feedback



Series and parallel feedback

Conclusion:If the feedback point--input point at the same terminal:=parallel feedbackIf the feedback point--input point at the different terminal:=series feedback



Conclusion

Input point--feedback point--sample point--output point

R110kΩ

R210kΩ

R3240Ω

R41kΩ

RL

C1

4.7uF

C2100uF

C3

47uF

VCC

12V

IO1IO1

IO2

IO3

IO4

IO3

0

Q1

2N914

10

9

1

R5

470kΩKey=A

50% VCC

2

Ui

Uo

Ii

Io


Guangdong Institute of Education ---Guangdong Institute of Education ---BTEC electronicBTEC electronic a.c. negative feedback

-----voltage and current\series and parallel An n.f.b. amplifier has a fraction of its output sign

al feed back into its input terminals in antiphase with the input signal.

four types ：voltage-voltage feedback, (series-voltage feedback)

voltage-current feedback, (series-current feedback)

current-current feedback. (parallel-current feedback)

current-voltage feedback. (parallel-voltage feedback)



Voltage-Voltage Feedback (Series-voltage Feedback)



S IN OUTV V V

OUT v S v IN OUT v IN v OUTV AV A V V AV AV (1 )OUT v v INV A AV

(1 )v IN

OUTv

AVV

A

( ) (1 )OUT v

v FIN v

V AA

V A

Voltage gain with n.f.b.



EXAMPLE 4.1 A voltage amplifier has a voltage gain of 10

0 before n.f.b.is applied.Calculate its voltage gain if 3/100 of the output voltage is fed back to the input in antiphase with the input signal

Solution:

From equation(4.1),

( )

100 10025

(1 3/100 100) 4OUT

v FIN

VA

V



The voltage gain of a transistor amplifier without n.f.b.iS

The effective load resistance RL(eff) seen by the transistor in an emitter follower circuit iS the resultant of the emitter resistor R3 in parallel with the external load resistance RL.The voltage gain of an emitter follower is

i Lv

IN

ARA

R

( ) ( )( )

( ) ( )

/

1

i L eff IN i L effv F

i L eff IN i L eff

IN

AR R ARA

AR R ARR


Guangdong Institute of Education ---Guangdong Institute of Education ---BTEC electronicBTEC electronicVoltage-Current Feedback(Series-current Feedback)



When voltage-current feedback is applied to an amplifier the fed-back voltage is proportional to the current-flowing in the load.

OUT OUT F

OUT OUT L

V I R

V I R

F

L

R

R 3i

vIN

ARA

R

3

3( )

34 4

3

1 .

i

IN iv F

i IN i

IN

AR

R ARA

ARR R ARR R


Guangdong Institute of Education ---Guangdong Institute of Education ---BTEC electronicBTEC electronic EXAMPLE 4.5

An amplifier of the type shown in Fig.4.8b has the following data:R2=4.7 kΩ,R3=1 kΩand gm=5 mS.Calculate the voltage gain of the circuit.

Solution

From equation(4.6) 3 3

( ) 3 3

5 10 4.7 103.92

1 5 10 10v FA


Guangdong Institute of Education ---Guangdong Institute of Education ---BTEC electronicBTEC electronicCurrent-Current Feedback(parallel-Current Feedback)



1

1 2

OUT FOUT

F F

I RI

R R

1

1 2

F

F F

R

R R

S IN OUTI I I

( )OUT i S i IN OUTI A I A I I

(1 )OUT i i INI A A I (1 )

i INOUT

i

A II

A

( ) (1 )OUT i

i FIN i

I AA

I A


Guangdong Institute of Education ---Guangdong Institute of Education ---BTEC electronicBTEC electronicCurrent-Voltage Feedback （ parallel-voltage Feedbac

k)



/OUTOUT FI V R

OUT

OUT F

OUTOUT

L

VI R

VIR

L

F

R

R



EXAMPLE 4.6 Calculate the current gain of the circuit

shown in Fig.4.12a if the current gain of the transistor is 120.

Solution

The feedback factor is (Ans.)

Therefore

Current gain with feedback

(Ans.)

2

1

3 1

120 40

R

R

( )

120 12030

1 41 12040

i FA





Review



advantages of negative feedback?

4.3.1. stability of gain4.3.2.input and output impedance4.3.3.amplitude/frequency distortion4.3.4. non-linearity distortion



Conclusion

Input point--feedback point--sample point--output point

R110kΩ

R210kΩ

R3240Ω

R41kΩ

RL

C1

4.7uF

C2100uF

C3

47uF

VCC

12V

IO1IO1

IO2

IO3

IO4

IO3

0

Q1

2N914

10

9

1

R5

470kΩKey=A

50% VCC

2

Ui

Uo

Ii

Io


Guangdong Institute of Education ---Guangdong Institute of Education ---BTEC electronicBTEC electronic EXAMPLE 4.7

An amplifier has a voltage gain of 1000.If 3 ／ 100 of the output voltage IS applied as negative feedback,calculate the change in overall gain if the gain before feedback falls by 50 ％ .

• Solution Gain with feedback,inherent gain being 1000.

New gain with feedback,inherent gain having been reduced to 500.

( )

1000 100032.26

1 (1000 3/100) 31v FA

( )

500 50031.25

1 (500 3/100) 16v FA



Therefore

Change in gain

(Ans.)

Thus a 50 ％ fall in the inherent gain of the amplifier results in only a 3.13 ％ change in the overall gain.

32.26 31.25100 3.13%

32.26



If the loop gain is much larger than unity then equations(4.1)and(4.8)become

It is evident that, if , the gain of the amplifier iS merely a function of the feedback circuit and iS quite independent of the characteristics of the amplifier itself.Any changes in the performance of the amplifier will not now affect the overall gain.

( )

1vv F

v

AA

A ( )

1ii F

i

AA

A

1A


Guangdong Institute of Education ---Guangdong Institute of Education ---BTEC electronicBTEC electronicresistance-capacitance feedback network is chosen to have a

loss ／ frequency characteristic



Input and output impedances

• The application of negative feedback to an amplifier alters its input and output impedances.Each impedance may be either increased or decreased depending on the type of feedback used ； see Table 4.1.



Amplitude ／ Frequency Distortion

• for example,the gain of an R-C coupled amplifier falls off at both high and low frequentcles because of capacitive effects.The application of negative feedback to an amplifier reduces change in its gain and thereby reduces amplitude ／ frequency distortion.Fig.4.14



• Clearly the feed-back has made the amplifier gain much‘‘flatter” over most of the frequency band.At high and low frequencies, however,the loop gain is not much greater than unity(because the gain A has fallen)and the effect of the feedback is reduced.

• this means that the 3 dB bandwidth of the amplifier has been increased by the applied negative feedback.An increase in the upper 3 dB frequency can be obtained by the use of frequency-sensitive feedback.



Non-linearity Distortion

• For example,suppose the input signal contains frequencies,f1 and f2.If the output signal has components at 2f1,2f2,3f1,3f2,etc.,harmonic distortion has occurred ； if components at 2f1 士 f2, f1 士 2f2, etc.,exist inter-modulation distortion iS present.Often both harmonic and intermodulation distortion occur at the same time.

distortion without n.f.b.Distortion with n.f.b. applied=

1+ A



EXAMPLE 4.8

• An amplifier has a voltage gain of 50 dB.Find the change in the gain if 1/50 of the output voltage is fed back into the input in opposition to the input signal.What is then the reduction in harmonic distortion at the output of the amplifier? Phase shift in the amplifier and the feedback path may be neglected.

Solution Equation(4.1)is ( ) 1

vv F

v

AA

A



• 50=20 logl0(voltage ratio) 50 ／ 20=2.5=logl0(voltage ratio)Taking antilogl0 of sides 316.2=voltage ratioTherefore times=32.7dBso thatChange in gain=50-32.7=17.3 dB (Ans.)From equation(4.12),the reduction in harmonic distortion is

1/(1+(316.2/50))=1/7.32=0.137times (Ans.)

( )

316.243.2

1 (316.2 / 50)v

v FA

distortion without n.f.b.Distortion with n.f.b. applied=

1+ A



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Electronics chapter 4 negative feedback Guangdong Institute of Education ---BTEC electronic Review and Overview 0 、 introduction 1 、 what is feedback?