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Consider an amplifier circuit in which a transistor is used in common-emitter mode. The load resistance 3kOmega. When, a signal of 30 mV is added to base emitter voltage, the base current is changed by 30muA and the collector current is changed by 3 mA. the power gain in this circuit will be |
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Answer» 10000 ` R = 2 K OMEGA = 30000 Omega ` Input voltage , ` V_(i) = 30 mV ` Change in base current , ` Delta I_(B) = 3 mu A = 3 xx 10^(-5) A ` Change in collector current , ` DeltaI_(e) = 3 mA = 3xx 10^(-3)A ` Current gain in common emitter mode , ` beta = (Delta I_(c))/(DeltaI_(B)) = (3xx 10^(-3))/(3xx 10^(-5)) = 100` Input resistance , ` R_("in") = (V_(i))/(Delta I_(B)) = (30xx 10^(-3))/(3xx10^(-5) ) = 30000 Omega` ` therefore ` Power gain ` (P_(V)) = beta^(2) = (R)/(R_("in")) = 100^(2) xx (3000)/(1000) = 30000` Hence , the power gain in this circuit will be 30000. |
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