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A p-n-p transistor is used in common- emitter mode in an amplifier circuit. A change of 40 muA in the base current brings a change of 2 mA in collector current and 0.04 V in base-emitter voltage. Find the : (i) inputresistance (R_("in")) and (ii) the base cuirent amplification factor (beta). If a load of 6kOmega is used, then also find the voltage gain of the amplifier. |
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Answer» Solution :Given `DeltaI_(B)=40muA=40xx10^(-6)A` `DeltaI_(C)=2mA=2xx10^(-3)A` `DeltaV_(BE)=0.04" volt, "R_(L)=6kOmega=6xx10^(3)Omega` (i) Input Resistance, `R_("in")=(DeltaV_(BE))/(DeltaI_(B))=(0.04)/(40xx10^(-6))=10^(3)Omega=1kOmega` (II) Current amplification FACTOR, `beta=(DeltaI_(C))/(DeltaI_(B))=(2xx10^(-3))/(40xx10^(-6))=50` (iii) Voltage gain in COMMON - emitter configuration. `A_(u)=beta(R_(L))/(R_("lmp"))=50xx(6xx10^(3))/(1xx10^(3))=300` |
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