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A common emitter amplifier has a voltage gain of `50`, an input impedence of `100 Omega` and an output impedence of `200 Omega`. The power gain of the of the amplifier isA. `500`B. `1000`C. `1250`D. `100` |
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Answer» Correct Answer - C `AC` power gain `= ("Change in output power")/("Change in input power")` `= (DeltaV_(c) xx Deltai_(c))/(DeltaV_(i) xx Deltai_(b)) = ((DeltaV_(c))/(DeltaV_(i))) xx ((Deltai_(c))/(Deltai_(b)))` `= A_(V) xx beta_(AC)` where `A_(V)` is voltage gain and `beta_(AC)` is `AC` current gain. Also, `A_(V) = beta_(AC) xx` Resistance gain `(=(R_(0))/(R_(i)))` Given, `A_(V) = 50, R_(0) = 200 Omega, R_(i) = 100 Omega` Hence, `50 = beta_(AC) xx (200)/(100)` `:. beta_(AC) = 25` Now, `AC` power gain `= A_(V) xx beta_(AC)` `= 50 xx 25 = 1250` |
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