Question

Consider the circuit shown in the figure below.

$\text{The characteristic equation of the BJT can be given as}{I}_c=I_0\text{exp}\left(\frac{V_{BE}}{1.5V_T}\right)\text{. If the transistor is biased with}{I}_c=3\text{mA then the value of small signal voltage gain in equal to\_\_\_\_\_\_\_\_\_\_\_\_.}$
$\text{Assume no early effect and}{V}_T=25\text{mV}$

• A. -80

Answer 1

The correct option is A -80
$g_m=\frac{\partial I_c}{\partial V_{BE}}=\frac{\partial }{\partial V_{BE}}.I_0\text{exp}\left(\frac{V_{BE}}{1.5V_T}\right)$

$g_m=\frac{1}{1.5V_T}.I_c=\frac{3\times {10}^{-3}}{1.5\times 25\times {10}^{-3}}=80\text{mA/V}$

$\therefore A_B=-g_m{R}_c=-80\times {10}^{-3}\times 1\times {10}^3$
$=-80$