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A p-n-p transistor is used in common-emitter mode in an amplifier circuit. A change of 40 mu A in the base current brings a change of 2 mA in collector current and 0.04 V in base-emitter voltage. Find the : (i) input resistance (R_("in")) and (ii) the base current amplification factor (beta ). If a load of 6k Omegais used, then also find the voltage gain of the amplifier. |
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Answer» Solution :Given `DeltaI_B = 40mu A= 40 xx 10^(-6) A ` `DeltaI_C= 2mA= 2 xx 10^(-3) A ` ` Delta V_(BE )= 0.04` volt `R_L= 6 k OMEGA= 6 xx 10^2 Omega ` (i) Input Resistance, `R_("in") = (Delta V_(BE ))/( Delta I_(B )) = ( 0.04 ) /( 40 xx 10^(-6)) = 10^3 Omega1 k Omega ` (ii) Current AMPLIFICATION factor, ` beta= (DeltaI_C )/(Delta I_B) = ( 2 xx 10^(-3))/( 40 xx 10^(-6)) = 50` (iii) VOLTAGE gain in common-emitter CONFIGURATION. ` A_u = beta(R_L)/(R_("imp"))=50 xx(6 xx 10^3)/( 1 xx 10^3 ) = 300` |
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