Calculate the closed-loop gain of a negative feedback amplifier with A = 10,000 and β = 0.01. Show that the closed-loop gain does not change much even when the gain of the basic amplifier is changed by ± 50 per cent.

Question

Accepted Answer

[latex]A_{f}=\frac{A}{1+A \beta } [/latex]

[latex]A_{f}=\frac{10,000}{1+10,000 imes0.01} [/latex]

Substituting values, [latex]=\frac{10,000}{101}=99.00 [/latex]

For the changed condition of A from 10,000 to 15,000,

[latex]A_{f}=\frac{15,000}{1+15,000 imes 0.01}=\frac{15,000}{151}=99.33 [/latex]

For changed condition A from 10,000 to 5,000,

[latex]A_{f}=\frac{5,000}{1+5,000 imes 0.01}=\frac{5,000}{51}=98.03 [/latex]

This shows that when the open-loop gain, i.e. the gain of the basic amplifier, A changes by ± 50 per cent, the gain of the feedback amplifier changes by a very small percentage. Thus, the negative feedback stabilizes the amplifier voltage gain.
The condition for stabilization of negative feedback amplifier gain is

[latex]\left|A\beta ight| >> 1 [/latex]

i.e., [latex]A_{f}=\frac{A}{1+A β} ightarrow \frac{1}{\beta } [/latex] .

Again, A β>>1

or, [latex] A >>\frac{1}{β } [/latex]

This means that A >>[latex]A_{f} [/latex].
Thus, for satisfactory operation of the negative feedback amplifier (with stabilized gain), the open-loop voltage gain must be much higher than the required closed-loop gain.

Question 8.1: Calculate the closed-loop gain of a negative feedback amplif...

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