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Question

Physics Question on Semiconductor electronics: materials, devices and simple circuits

For a transistor amplifier in common emitter configuration for load impedance of 1kΩ1\,k\Omega (hfe=50h_{fe} = 50 and hoe=25mAV1h_{oe} = 25\, mA\, V^{-1}), the current gain is

A

5.2

B

15.7

C

24.8

D

48.78

Answer

48.78

Explanation

Solution

For a transistor amplifier in common emitter configuration, current gain Ai=hfe1+hoeRLA_{i} = \frac{h_{fe}}{1+h_{oe}R_{L}} where hfeh_{fe} and hoeh_{oe} are hybrid parameters of a transistor. Ai=501+25×106×1×103=48.78\therefore\quad A_{i} = \frac{50}{1 +25\times 10 ^{-6}\times 1\times 10^{3}} = 48.78