Q.An amplifier with $Z_i$ = 1k$\Omega$, has a voltage gain A=1000. If a negative feedback of $\beta$=0.01 is applied to it, what shall be the input impedance of the feedback amplifier?

A high input impedance is desirable so that an amplifier does not load the preceding stage or the input voltage source. Voltage series negative feedback achieves exactly this.

Let $I_i$ be the input current, $V_i$ the net input to the basic amplifier, $V_s$ the source voltage and $V_o$ the output. The input impedance of the basic amplifier is $Z_i = V_i/I_i$, while the input impedance seen from the source with feedback is $Z_{if} = V_s/I_i$.

For negative feedback the feedback voltage opposes the source, so $V_i = V_s - \beta V_o = V_s - \beta A V_i$, which rearranges to

Substituting into $Z_{if} = V_s/I_i$ and using $Z_i = V_i/I_i$,

$Z_{if} = \frac{V_i(1+A\beta)}{I_i} = Z_i(1 + A\beta).$ …