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InterviewSolution
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An n-p-n transistor in a common emitter mode is used as a simple voltage amplifier with a collector current of 4 mA. The terminal of a 8 V battery is connected to the collector through a load resistance R_L and to the base through a resistance R_B The collector-emitter voltage V_(CE)= 4 V, base-emitter voltage V_(BE)= 0.6 V and base current amplification factor B_(d.c) = 100. Calculate the values of R_L and R_B |
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Answer» Solution :Potentialdifferenceacross ` R_L` ` (##AKS_NEO_CAO_PHY_XII_V02_C_C06_SLV_039_S01.png" width="80%"> ` = 8 VV_(CE)` `= 8V - 4V ` `= 4 V ` Now` I_CR_L= 4V ` `R_L = (4)/( 4 x 10^(-3)) = 10^3 Omega= 11 kOmega ` Furtherfor baseemitterequation, ` V_(C C )= I_BR_B+V_(BE)` or ` I_B R_B `= potential differenceacross`R_B ` ` V_(C C ) =- V_(BE)= 8-06= 7.4V ` Again ` ,I_B= (I_C )/(beta) = ( 4 xx 10 ^(-3))/( 100 )= 4 xx 10^(-5) A` ` thereforeR_B= ( 7.4)/( 4 xx 10^(-5)) = 1.85 xx 10^5Omega= 185k Omega ` |
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