Question

For a transistor amplifier in common emitter configuration for load impedance of `1 k Omega. (h_(fe) = 50 and h_(oe) = 25 xx 10^(-6))` the current gain is
A. `-24.8`
B. `- 15.7`
C. `-5.2`
D. `-48.78`

Answer 1

Correct Answer - D
In common emitter configuration currect gain
`A_(1) = (-hf_(e))/(1 + b_(o e) R_(L)) = (-50)/(1 + 25 xx 10^(-6) xx 1 xx 10^(3)) = -48.78`