Question

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

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

Answer 1

Answer: (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.
$\therefore \quad A_{i} = \frac{50}{1 +25\times 10 ^{-6}\times 1\times 10^{3}} = 48.78$