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

Question

For a transistor amplifier in common emitter configuration for load impedance of 1 kΩ (hfe = 50 and hoe = 25 mA V-1), the current gain is (A) 5.2 (B) 15.7 (C) 24.8 (D) 48.78

Tardigrade Admin · Accepted Answer

For a transistor amplifier in common emitter configuration, current gain Ai = (hfe/1+hoeRL) where hfe and hoe are hybrid parameters of a transistor. ∴ Ai = (50/1 +25× 10 -6× 1× 103) = 48.78

Q. For a transistor amplifier in common emitter configuration for load impedance of $1 k Ω$ ( $h_{f e} = 50$ and $h_{oe} = 25 m A V^{- 1}$ ), the current gain is

5.2

15.7

24.8

48.78

For a transistor amplifier in common emitter configuration, current gain
$A_{i} = \frac{h _{f e}}{1 + h _{oe} R _{L}}$
where $h_{f e}$ and $h_{oe}$ are hybrid parameters of a transistor.
$∴ A_{i} = \frac{50}{1 + 25 \times 1 0 ^{- 6} \times 1 \times 1 0 ^{3}} = 48.78$

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

5.2

15.7

24.8

48.78

For a transistor amplifier in common emitter configuration, current gain Ai​=1+hoe​RL​hfe​​ where hfe​ and hoe​ are hybrid parameters of a transistor. ∴Ai​=1+25×10−6×1×10350​=48.78

Q. For a transistor amplifier in common emitter configuration for load impedance of $1 k Ω$ ( $h_{f e} = 50$ and $h_{oe} = 25 m A V^{- 1}$ ), the current gain is

For a transistor amplifier in common emitter configuration, current gain
$A_{i} = \frac{h _{f e}}{1 + h _{oe} R _{L}}$
where $h_{f e}$ and $h_{oe}$ are hybrid parameters of a transistor.
$∴ A_{i} = \frac{50}{1 + 25 \times 1 0 ^{- 6} \times 1 \times 1 0 ^{3}} = 48.78$