Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Semiconductors and Electronic Devices

Question:

For a transistor amplifier in common emitter configuration for load impedance of 1kΩ ($h_{fe} = 50$ and $h_{oe}= 25mAV^{-1}$), the current gain is:

Options:

5.2

15.7

24.8

48.78

Correct Answer:

48.78

Explanation:

For a transistor amplifier in common emitter configuration, current gain

$A_i=\frac{h_{fe}}{1+h_{oe}R_L}$

Where $h_{fe}$ and $h_{oe}$ are hybrid parameters of a transistor.

$∴A_i=\frac{50}{1+25×10^{-6}×1×10^3}=48.78$