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A npn transistor in a common emitter mode is used as a simple voltage amplifier with a collector current of 5 m A. The terminal of 10V battery is connected to a collector through a load resistance R_(L) and to the base through a resistance R_(B). The collector emitter voltage V_(CE)=5V, base emitter voltage, V_(BE)=0.5Vand base correct amplification factor beta_(d.c)=100. Calculate the values of R_(L) and R_(B) |
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Answer» Solution :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=9.5` or `R_(B)=(9.5)/(I_(B))=(9.5V)/(5xx10^(-5)A)=1.9xx10^(5)Omega=190kOmega` |
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