Question 8.14: The open-loop gain of an amplifier changes by 20 per cent du...

The amount of feedback expressed in decibels is

20 \log _{10}\left|\frac{A_{f}}{A} \right| =-20

or,                     \log _{10}\left|\frac{A_{f}}{A} \right| =-1

A_{f}=\frac{A}{1+βA}.

\frac{A_{f}}{A}=\frac{1}{1+βA}.

Therefore,

\log _{10}\frac{1}{1+βA}=-1

\log _{10}(1+βA)=1

(1+βA)=10

Again,

or, v                                  \frac{dA_{f}}{A_{f}}=2  per cent.

The change in gain with negative feedback is reduced to 2 per cent only.