A CE amplifier, has a power gain of 40 dB. The input resistance and output load resistance are `500Omega and 2kOmega` respectively. Find the value of common - emitter current gain.
`["take gain in dB "=10log((P_("out"))/(P_("in")))]`

Answer

`50`

Explanation

Given that
Power gain `(A_(p))=40dB`
`:.10log((P_(0))/(P_("in")))=40`
`:.log((P_(0))/(P_("in")))=4`
`:.(P_(0))/(P_("in"))=10^(4)`
Also `A_(p)=beta^(2)xx` Resistance gain
`=beta^(2)xx((R_("out"))/(R_("in")))`
`=beta^(2)xx2000/500` `=beta^(2)xx4`
`:.beta^(2)=(10^(4))/4`
`:.beta=100/2=50`