Q.In a negative feedback amplifier, A=100, $\beta$=0.04 and $V_i$=50mV. Find

Each part follows directly from the negative feedback relations $A_f=\dfrac{A}{1+A\beta}$, $V_o=A_fV_i$, loop gain $=A\beta$, and $V_f=\beta V_o$.

[!ANSWER]

(a) $A_f=20$ (b) $V_o=1\text{ V}$ (c) loop gain $=4$ (d) $V_f=0.04\text{ V}$

[!TLDR]

Using the four feedback relations: $A_f=20$, $V_o=1$ V, loop gain $=4$, and $V_f=0.04$ V.

Negative feedback returns a fraction $\beta$ of the output against the input, reducing the gain while stabilising it. The four required quantities each have a standard defining relation.

1. Gain with feedback:

   $A_f = \dfrac{A}{1+A\beta} = \dfrac{100}{1+100\times0.04} = \dfrac{100}{1+4} = \dfrac{100}{5} = 20$

2. Output voltage: the feedback gain multiplies the basic-amplifier input voltage,

   $V_o = A_f\,V_i = 20\times50\times10^{-3} = 1000\times10^{-3} = 1\text{ V}$

3. Loop gain is the product of open-loop gain and feedback factor,

   $A\beta = 100\times0.04 = 4$

4. Feedback voltage is the sampled fraction of the output,

   $V_f = \beta V_o = 0.04\times1 = 0.04\text{ V}$

   [!ANSWER]

(a) $A_f=20$ (b) $V_o=1$ V (c) loop gain $=4$ (d) $V_f=0.04$ V