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Feedback
It is a process in which part of output is sampled and fed back to the input.
Feedback Amplifiers
- The amplifiers which uses feedback principle are called 'feedback amplifiers'.
Q) List and explain two types of feedback.
- Depending upon whether the feedback signal increases or decreases the input signal, there are two basic types of feedback:
a) Positive feedback
b) Negative feedback
a) Positive feedback
If the feedback signal (voltage or current) is applied in such a way that it is in phase with input signal and thus increases it, then it is called a positive feedback.
It is also known as 'regenerative feedback' or 'Direct feedback'.
The positive feedback increases the gain of amplifier. It is used in oscillators.
b) Negative feedback
If the feedback signal (voltage or current) is applied in such a way that, it is out of phase with the input signal and thus decreases it, then it is called a negative feedback.
It is also known as 'degenerative feedback' or 'Inverse feedback'.
The negative feedback reduces the gain of amplifier. But it improves the amplifier performance in many aspects. Negative feedback is used in small signal, as well as large signal amplifier circuits.
Q) List the types of feedback connecting, explain with block diagram.
There are 4 basic types of connecting the feedback signal from an amplifier output to its input.
These types are given below:
(a) Voltage series feedback connection
(b) Voltage shunt feedback connection
(c) Current-series feedback
(d) Current-shunt feedback
It means that, voltage and current can be fed back to its input either in series or parallel.
In the feedback connection types, the term 'voltage' refer to connecting the output voltage as input to the feedback network.
The term 'current' refers to tapping some i/p current through feedback network.
The term 'series' refers to connecting the feedback signal in series with the input signal voltage and the term 'shunt' refers to connecting the feedback signal in shunt(parallel) with an input current source.
(a) Voltage series feedback connection
It is also known as shunt derived-series fed feedback connection.
A block diagram of voltage-series feedback connection is as shown in fig below:
- In this connection, a part of the output voltage is applied in series with the input voltage through feedback network.
(b) Voltage shunt feedback
It is also known as shunt derived-series shunt fed feedback connection.
A block diagram of voltage-shunt feedback connection is shown in fig below:
- In this connection, a part of o/p voltage is applied in parallel with input through feedback network.
(c) Current-series feedback
It is also called a series derived series fed feedback connection.
A block diagram of current-series feedback connection is shown in fig below:
- In this connection, a fraction of o/p current is converted into a proportional voltage by the feedback network and then applied.
(d) Current-shunt feedback
It is also called a series derived shunt fed feedback connection.
A block diagram of current-shunt feedback connection is shown in fig below:
- In this connection, a fraction of o/p current is converted into a proportional voltage by the feedback network and then applied in parallel with the i/p voltage.
Effect of feedback on input and output resistance
SL.NO | Types of feedback connection | Input resistance | Output resistance |
---|---|---|---|
1 | Voltage series | Increases | Decreases |
2 | Current series | Increases | Increases |
3 | Voltage shunt | Decreases | Decreases |
4 | Current shunt | Decreases | Increases |
Comparison of positive and negative feedback
SL.NO | Parameters | Positive feedback | Negative feedback |
---|---|---|---|
1 | Overall phase shift | 0 or 360$^{\circ}$ | 180$^{\circ}$ |
2 | Feedback signal and input signal | are in phase | are out of phase |
3 | Input voltage | Increases | Decreases |
4 | Output voltage | Increases | Decreases |
5 | Voltage gain | Increases | Decreases |
6 | Stability | Poor | Better |
7 | Application | Oscillators | Amplifiers |
8 | Noise | Increases | Decreases |
9 | Bandwidth | Decreases | Increases |
10 | Input impedance | Decreases | Increases |
11 | Output impedance | Increases | Decreases |
Q) State the effect of negative feedback on following parameters:
a) Bandwidth
b) Noise
c) Gain
d) Stability
e) Distortion
(a) Effect of negative feedback on bandwidth
Bandwidth of amplifier increases with the negative feedback.
We know that, for any amplifier, product of bandwidth and gain always remain constant.
i.e. Bandwidth $\times$ Gain = constant.
As, gain value of amplifier reduces in negative feedback, to keep value of product constant, bandwidth of amplifier increases proportionally.
- The increase in bandwidth due to the negative feedback is shown in figure below.
Also, mathematically, the gain with feedback is given by,
BW' = BW( 1 + A$\beta$)
where,
BW = bandwidth without feedback
A = gain of the amplifier without feedback
$\beta$ = feedback factor
(b) Effect of negative feedback on noise
- It is observed that,with the introduction of negative feedback, noise in the amplifier will be reduced to a considerable extent.
This noise reduction will take place for any type of negative feedback.
(c) Effect of negative feedback on gain
Gain of amplifier decreases with negative feedback. For any amplifier, product of BW and gain remains constant.
i.e. Bandwidth $\times$ Gain = constant.
As, BW of amplifier increases in negative feedback, to keep value of product constant, gain of amplifier will reduce proportionally. The decrease in gain value due to negative feedback is shown in figure above.
(d) Effect of negative feedback on stability
- It is observed that, with the introduction of negative feedback, stability of amplifier will increase.
(e) Effect of negative feedback on distortion
- Due to introduction of negative feedback, the effective to amplifier reduces which helps it to operate in the linear region. This reduces the non linear distortion.
Q) List the advantages of Negative Feedback.
Advantages of Negative Feedback are listed below:
(a) Increased stability
(b) Increased bandwidth
(c) Decreased noise
(d) Less amplitude and harmonic distortion
(e) Less frequency distortion
(f) Less phase distortion
(g) Input and output resistances can be modified as desired.
Q) List the disadvantages of negative feedback
Disadvantages of negative feedback are listed below:
(a) Reduction in gain
(b) Reduction in input resistance in case of voltage shunt and current shunt type.
(c) Increase in output resistance in case of current shunt and current series type.
(d) More no. of amplifier stages are required to be used in order to obtain required gain.
Q) List different applications of negative feedback
In almost all electronic equipments
In regulated power supply
In wide band amplifier
Gain with feedback
(a) Gain with positive feedback(A$_{F}$)
- It is given by,
A$_{F}$ = $\frac{A}{1 - A\beta}$
where,
A = Gain of amplifier without feedback(open loop gain)
$\beta$ = Gain of feedback network
(b) Gain with negative feedback(A$_{F}$)
- It is given by,
A$_{F}$ = $\frac{A}{1 + A\beta}$
where,
A = Gain of amplifier without feedback(open loop gain of amplifier)
$\beta$ = Gain of feedback network
Q1) An amplifier has a bandwidth of 100kHz and voltage gain of 50. What will be the bandwidth and gain with a 4 percent negative feedback?
BW = 100kHz, A = 50, $\beta$ = 0.04
- Gain with feedback.
A$_{f}$ = $\frac{A}{1 + A\beta}$ = $\frac{50}{1 + (0.04 \times 50)}$
$\therefore$ A$_{f}$ = 16.66
- Bandwidth with feedback
BW' = BW $\times$ (1 + A$\beta$)
= 100 $\times$ [1 + (0.04 $\times$ 50)]
BW' = 300 kHz
Q2) The amplifier has a bandwidth of 150kHz and a voltage gain of 50. What will be the bandwidth and gain if an amplifier has 5% negative feedback?
Soln:
Given: BW without feedback: BW$_{1}$ = 150kHz
Gain without feedback: A = 50
Feedback factor: $\beta$ = 0.05
- Gain with feedback: A$_{f}$ = $\frac{A}{1 + A\beta}$
= $\frac{50}{1 + (50 \times 0.05)}$
$\therefore$ A$_{f}$ = 14.3
- The gain $\times$ bandwidth product of an amplifier always remains constant. Let the bandwidth with feedback be BW$_{2}$
$\therefore$ A $\times$ BW$_{1}$ = A$_{f}$ $\times$ BW$_{2}$
BW$_{2}$ = $\frac{A \times BW_{1}}{A_{f}}$ = $\frac{50 \times 150 \times 10^3}{14.3}$
$\therefore$ BW$_{2}$ = 524.47 kHz.