For a transistor ampliflier in common em...

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

`-5.2`

`-15.7`

`-24.8`

`-48.78`

Text Solution

Verified by Experts

The correct Answer is:

`A_(C)=-((h_(fe))/(1+h_(oe)R_(L)))=-((50)/(1+25+10^(-6)xx(10^(3))))`
`=-48.78`

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