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Question

Question: For a transistor amplifier in common emitter configuration for load impedance of 1 kΩ (h<sub>fe</sub...

For a transistor amplifier in common emitter configuration for load impedance of 1 kΩ (hfe = 50 and hoe = 25) the current gain is

A

– 5.2

B

– 15.7

C

– 24.8

D

– 48.78

Answer

– 48.78

Explanation

Solution

In common emitter configuration current gain

Ai=hfe1+hoeRL=501+25×106×103A_{i} = \frac{- h_{fe}}{1 + h_{oe}R_{L}} = \frac{- 50}{1 + 25 \times 10^{- 6} \times 10^{3}}= – 48.78.