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Overview

0、introduction

1、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.

feedback

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、A

Closeloop circuit、 gain with n.f.b、Af 、F(feedback factor)

1+AF (feedback depth)

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 distortion

Reduces noise

Change input and output impedance

Band spread

Reduces amplitude／frequency distortion

2、generate oscillation (oscilloscope)

Types of feedback

0、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

Negative and Positive feedback

Conclusion:

If the Input point---feedback point at the same terminal:

And same polarity==positive feedback

And different polarity==nagtive feedback

If the input point---feedback point at the different terminal:

And same polarity==negative feedback

And different polarity==positive feedback

Voltage and Current feedback

Conclusion:

If the output point---sampling point at the same terminal:

=voltage feedback

If 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 feedback

If the feedback point--input point at the different terminal:

=series feedback

a.c. negative feedback-----voltage and current\series and parallel

An n.f.b. amplifier has a fraction of its output signal 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)

EXAMPLE 4.1

A voltage amplifier has a voltage gain of 100 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),

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

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

EXAMPLE 4.5 amplifier the fed-back voltage is proportional to the current-flowing in the load.

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)

Current-Current Feedback(parallel-Current Feedback) amplifier the fed-back voltage is proportional to the current-flowing in the load.

Current-Voltage Feedback amplifier the fed-back voltage is proportional to the current-flowing in the load. （parallel-voltage Feedback)

EXAMPLE 4.6 amplifier the fed-back voltage is proportional to the current-flowing in the load.

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.)

Review amplifier the fed-back voltage is proportional to the current-flowing in the load.

advantages of negative feedback? amplifier the fed-back voltage is proportional to the current-flowing in the load.

4.3.1. stability of gain

4.3.2.input and output impedance

4.3.3.amplitude/frequency distortion

4.3.4. non-linearity distortion

Input point--feedback point--sample point--output point amplifier the fed-back voltage is proportional to the current-flowing in the load.

ConclusionEXAMPLE 4.7 amplifier the fed-back voltage is proportional to the current-flowing in the load.

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.

Therefore amplifier the fed-back voltage is proportional to the current-flowing in the load.

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.

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.

resistance-capacitance feedback network is chosen to have a equations(4.1)and(4.8)becomeloss／frequency characteristic

Input and output impedances equations(4.1)and(4.8)become

- 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 equations(4.1)and(4.8)become／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 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.

- 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.

EXAMPLE 4.8 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.

- 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

- 50=20 log 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.l0(voltage ratio)
50／20=2.5=logl0(voltage ratio)

Taking antilogl0 of sides

316.2=voltage ratio

Therefore times=32.7dB

so that

Change 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.)

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