For a transistor amplifier in common emitter configuration for load impedance of 1 kΩ (hfe = 50 and hoe = 25 /V) the current gain is (a) -24.8 (b) -15.7 (c) -5.2 (d) -48.78

Answer: (d). h-parameter current gain: Ai = -hfe1 + hoe RL hoe RL = 25 × 10⁻⁶ × 10³ = 0.025 Ai = -(50)/(1.025) = -48.78 So the answer is (d) -48.78.

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Junction Transistor

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

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Answer: (d)

h-parameter current gain: $A_i = -\dfrac{h_{fe}}{1 + h_{oe} R_L}$
$h_{oe} R_L = 25 \times 10^{-6} \times 10^3 = 0.025$
$A_i = -\dfrac{50}{1.025} = -48.78$
So the answer is (d) $-48.78$ .

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