question_answer

A npn transistor in a common emitter mode is used as a simple voltage amplifier with a collector current of \[4mA\]. the terminal of a \[8V\] 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 \[{{V}_{CE}}=4V,\] base-emitter voltage \[{{V}_{BE}}=0.6\] and the base current amplification factor \[{{\beta }_{dc}}=100,\]calculate the value of \[{{R}_{B}}.\]

A)  \[{{R}_{B}}=15\,k\Omega \]

B)  \[{{R}_{B}}=200\,k\Omega \]

C)  \[{{R}_{B}}=1\,k\Omega \]

D)  \[{{R}_{B}}=185\,k\Omega \]

Correct Answer: D

Solution :

[d] By the definition \[\therefore \,\,\,\,\,\,\,\,\,\,{{i}_{B}}=\frac{{{i}_{C}}}{{{\beta }_{dc}}}=\frac{4mA}{100}=0.04mA\] From figure \[{{i}_{B}}{{R}_{B}}+{{V}_{BE}}=\,\,\,\,{{V}_{CC}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{R}_{B}}=\frac{{{V}_{CC}}-{{V}_{BE}}}{{{i}_{B}}}=\frac{(8-0.6)}{0.04\times {{10}^{-3}}}=185\,k\Omega \] Also \[{{R}_{L}}{{i}_{C}}+{{V}_{CE}}={{V}_{CC}}\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,{{R}_{L}}=\frac{{{V}_{CC}}-{{V}_{CE}}}{{{i}_{C}}}=\frac{8-4}{8\times {{10}^{-3}}}=1\,k\Omega \]