Q.A negative feedback amplifier has the following parameters, A=500, $\beta$=0.02 and $V_i$ = 5mV. Compute the following a) gain with feedback b) output voltage c) loop gain d) feedback voltage

When negative feedback is applied, a fraction $\beta V_o$ of the output is returned $180^\circ$ out of phase, so the net input to the basic amplifier is reduced to $V_i = V_s - V_f$. Working through the definitions of open-loop gain $A = V_o/V_i$, closed-loop gain $A_f = V_o/V_s$ and feedback factor $\beta = V_f/V_o$ gives the central result of the chapter:

The derivation follows from $V_i = V_s - \beta V_o = V_s - \beta A V_i$, which rearranges to $V_s = V_i(1+A\beta)$; dividing $V_o = A V_i$ by this gives $A_f$.

Since $(1+A\beta) > 1$, the closed-loop gain is always smaller than the open-loop gain — the gain decreases by the factor $(1+A\beta)$. This loss of raw gain is the price paid for the benefits of negative feedback.

The same relation is the workhorse for numerical problems. Rearranged, it gives the feedback fraction from a known pair of gains,

or the internal gain needed for a target closed-loop gain. Related quantities follow directly: the output voltage $V_o = A_f V_i$, the loop gain $A\beta$, and the feedback voltage $V_f = \beta V_o$.