Question

A npn transistor in a common emitter mode is used as a simple voltage amplifier with a collector current of `5mA`. The terminal of `10V` battery is connected to a collector through a load resistance `R_(L)` and to the base through a resistance `R_`. The collector emitter voltage `V_(CE)=5V`, base emitter voltage, `V_(BE)=0.5V` and base current amplification factor `beta_(d.c.)=100`. Calculate the value of `R_(L)` and `R_`.

Answer 1

Potential difference across `R_(L)` is
`=I_(C)R_(L)=10V-V_(CE)=10V-5V-5V`
or `R_(L)=5/I_c=(5V)/(5xx10^(-3)A)=10^(3)Omega=1kOmega`
Here, `I_(B)=I_C/beta_(d.c.)=(5xx10^(-3))/100=5xx10^(-5)A`
Potential difference across `R_(B)` is, `I_(B)R_(B)=10-V_(BE)=10-0.5=9.5V`
or `R_(B)=9.5/I_(B)=(9.5V)/(5xx10^(-5)A)=1.9xx10^(5)Omega=190kOmega`