question_answer

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 voltage \[{{V}_{CE}}=4V,\] the base-emitter voltage \[{{V}_{BE}}=0.6\,V\] and   the   current amplification factor \[{{\beta }_{DC}}=100.\] Calculate the value of \[{{R}_{L}}\] and \[{{R}_{B}}.\]

Answer:

\[{{R}_{L}}=1k\,\Omega ,\] \[{{R}_{B}}=185\,k\,\Omega \] From the figure, We can written as,             \[{{V}_{CE}}={{V}_{CC}}-{{I}_{C}}{{R}_{L}}\] \[\therefore \] \[{{R}_{L}}=\frac{{{V}_{CC}}-{{V}_{CE}}}{{{I}_{C}}}=\frac{8V-4V}{4\times {{10}^{-3}}A}={{10}^{3}}\Omega =1k\Omega \] as         \[{{\beta }_{DC}}=\frac{{{I}_{C}}}{{{I}_{B}}}\] \[\therefore \]      \[{{I}_{B}}=\frac{{{I}_{C}}}{{{\beta }_{DC}}}=\frac{4\times {{10}^{-3}}A}{100}=4\times {{10}^{-5}}A\] Now, base-emitter voltage, \[{{V}_{BE}}={{V}_{CC}}-{{I}_{B}}{{R}_{B}}\] \[\therefore \] \[{{R}_{B}}=\frac{{{V}_{CC}}-{{V}_{BE}}}{{{I}_{B}}}=\frac{8V-0.6V}{4\times {{10}^{-5}}A}=1.85\times {{10}^{5}}\Omega \] \[=185k\Omega \]