Question

Consider the amplifier circuit shown in the figure below:

$\text{The value of voltage gain}\frac{v_0}{v_{in}}\text{for the circuit is equal to \_\_\_\_\_\_V/V.}$
$\text{(Assume}{V}_T=25\text{mV and}\beta \text{to be very large)}$

• A. 150

Answer 1

The correct option is A 150
From the figure it can be seen that,
$I_C=I_E=0.5\text{mA}(\because \beta \text{is very large)}$
$\text{now,}{g}_m=\frac{I_C}{V_T}\approx \frac{I_E}{V_T}=\frac{0.5\times {10}^{-3}}{25\times {10}^{-3}}A/V=20\text{mA/V}$
Drawing the equivalent small signal circuit, we get,

Now, applying KVL, we get
$v_{in}+v_{\pi }=0$
$v_{in}=-v_{\pi }$
$\text{and}{v}_0=-g_m{v}_{\pi }\times R_C$
$v_0=g_m{R}_c{v}_{in}$
$\frac{v_0}{v_{in}}=g_n{R}_c=20\times {10}^{-3}\times 7.5\times {10}^3=150$