Question

Consider the common emitter circuit shown in the figure below:

$\text{If the value of}\beta =99\text{, then the value of the lower 3-dB frequency is \_\_\_\_\_\_\_\_Hz.}$
$\text{(Assume}{V}_T=25\text{mV})$

• A. 7.95

Answer 1

The correct option is A 7.95
The input resistance,

$r_{\pi }=\frac{V_T}{I_B}=\frac{\beta V_T}{IC}=\frac{(\beta +1)V_T}{I_E}$

$=\frac{100\times 25\times {10}^{-3}}{1\times {10}^{-3}}=2.5\text{k}\Omega$

$\therefore R_{in}=\left(20\text{k}\Omega ||20\text{k}\Omega ||2.5\text{k}\Omega \right)=2\text{k}\Omega$

$\therefore f=\frac{1}{2\pi R_{eq}{C}_{i}n}=\frac{1}{2\pi \times 10\times {10}^{-6}\times 2\times {10}^3}=7.95\text{Hz}$