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An n-p-n transistor in a common-emitter ...

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 terminals 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)=4V`, the base-emitter voltage `V_(BE) = 0.6 V` and the current amplification factor `beta_(dc) = 100`. Then

A

`R_L=1 kOmega, R_B = 185 kOmega`

B

`R_L=2 kOmega = R_B`

C

`R_L=2 kOmega, R_B= 15 kOmega`

D

`R_L=185 kOmega, R_B=1 kOmega`

Text Solution

Verified by Experts

The correct Answer is:
A
An n-p-n transistor in a common-emitter mode with connections as given is shown in figure. Collector emitter voltage,
`V_("CE")=V_("CC")-I_C R_L`
`therefore R_L=(V_("CC")-V_("CE"))/I_C`
`=(8V-4V)/(4xx10^(-3))=10^3 Omega=1 kOmega`
As `beta_(dc)=I_C/I_B`
`therefore I_B=I_C/beta_(dc)=(4xx10^(-3))/100A=4xx10^(-5)A`
Base-emitter voltage,
`V_(BE)=V_("CC")-I_B R_B`
`therefore R_B=(V_("CC")-V_(BE))/I_B=(8V-0.6V)/(4xx10^(-5)A)`
`=1.85xx10^5 Omega= 185 kOmega`
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