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`
Text Solution
Verified by Experts
The correct Answer is:
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`
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