Question

For a transistor amplifier in common emitter configuration, for a load resistance of 1 $K\Omega$ ( $h_{fe}=50$ and $h_{oe}=25\mu A/V$), the current gain is :

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

Answer 1

The correct option is D $-48.78$

Given hybrid parameters of transistor are:
$h_{fe}=50$ and $h_{oe}=25\mu A{V}^{-1}=25\times {10}^{-6}A{V}^{-1}$
$R_L=1K\Omega ={10}^3\Omega$

For a transistor amplifier in common emitter configuration, current gain is:
$A_i=-\frac{h_{fe}}{1+h_{oe}{R}_L}$

$A_i=-\frac{50}{1+(25\times {10}^{-6})\left({10}^3\right)}$

$A_i=-\frac{50}{1+0.025}=-\frac{50}{1.025}=-48.78$