Question

For a transistor amplifier in common emitter configuration for load impedance of 1 k ( $h_{fe}$ = 50 and $h_{oe}$ = $25\times {10}^{-6}$) the current gain is

• A. -48.78
• B. -15.7
• C. -24.8
• D. -5.2

Answer 1

The correct option is A -48.78

Given,

Transistor amplifier in $C-E$ configuration.

Load impedance ${}^{\prime }{R}_L^{\prime }=1K\Omega$

$h_{fe}=50$

$h_{oe}=25\times {10}^{-6}$

To find the current gain $=A_i=?$

$A_i=-\frac{h_{fe}}{1+h_{oe}.R_L}=\frac{50}{1+25\times {10}^{-6}\times 1\times {10}^3}$

$=\frac{50}{1+25\times {10}^{-3}}$

$=-48.76$