Question

For a transistor ampliflier in common emiter configuration for load imperdance of `1 k Omega (h_(fe) = 50 and h_(oe) = 25)` the current gain is
A. `-5.2`
B. `-15.7`
C. `-24.8`
D. `-48.78`

Answer 1

Correct Answer - D
In common emitter configuration current gain
`A_(1)=(-h_(fe))/(1+h_(oe)R_(L))=(-50)/(1+25xx10^(-6)xx10^(3))=-48.78`.