Q.The gain of an amplifier is 50 and its output resistance is 2.5kOhm. A negative feedback is applied so that the output impedance reduces to 500 Ohm. What is the value of $\beta$? If the bandwidth before feedback is 200 kHz, what is the new bandwidth?

A low output impedance is desirable so that an amplifier can deliver voltage or power to a load without much internal loss. Voltage series negative feedback, which samples the output voltage, provides this.

To find the output impedance, the input is short-circuited ($V_s = 0$) and a hypothetical voltage source $V_o$ is applied at the output, driving a current $I_o$. The amplifier is modelled as a dependent source $A V_i$ in series with its internal output impedance $Z_o$. By definition $Z_{of} = V_o/I_o$, and applying KVL around the output loop gives

With the source shorted, the net input is just the opposing feedback voltage, $V_i = -V_f = -\beta V_o$. Substituting,

so

$Z_{of} = \frac{Z_o}{1 + A\beta}.$ …