Q. For a transistor amplifier in common emitter configuration for load impedance of 1 k$\Omega$ ($h_{fe} = 50\ and\ h_{oe} = 25\ \mu\ {AV}^{- 1}$), the current gain is
A) $-$5.2
B) $-$15.7
C) $-$24.8
D) $-$48.78
Explanation
(d) For a transistor amplifier in common emitter configuration, current gain
$A_{i} = \frac{h_{fe}}{1 + h_{oe}R_{L}}$
where $h_{fe}\ $and $h_{oe}\ are\ hybrid\ parameters$ of a transistor.
$\therefore\ A_{i} = \frac{50}{1 + 25 \times 10^{- 6} \times 1 \times 10^{3}} = - 48.78$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.