Question

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

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

Answer 1

Answer: (D)

– 48.78

Answer 2

In common emitter configuration current gain

$A_{i} = \frac{- h_{fe}}{1 + h_{oe}R_{L}} = \frac{- 50}{1 + 25 \times 10^{- 6} \times 10^{3}}$= – 48.78.