Semiconducting Devices Question 39
Question: For a transistor amplifier in common emitter configuration for load impedance of 1 kW (hfe = 50 and hoe = 25 mA/V) the current gain is
[AIEEE 2004]
Options:
A) - 5.2
B) - 15.7
C) - 24.8
D) - 48.78
Show Answer
Answer:
Correct Answer: D
Solution:
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.
