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JEE Main & Advanced Physics Semiconducting Devices Question Bank Self Evaluation Test - Semiconductor Electronics: Materials, Devices and Simple Circuits

  • question_answer
    A transistor connected in common emitter con- figuration has input resistance \[{{R}_{B}}=2\,k\Omega \] and load resistance of \[5\,k\Omega \]. If \[\beta =60\] and an input signal \[12\text{ }mV\] is applied, calculate the voltage gain, the power gain and the value of output voltage

    A)  \[{{A}_{V}}=150,\,\,{{V}_{out}}=1.8\,V,\] and power gain \[=9000\]

    B)  \[{{A}_{V}}=20,\,{{V}_{out}}=1V,\]and power gain \[=2000\]

    C)  \[{{A}_{V}}=150,\,{{V}_{out}}=1.5V,\] and power gain \[=8500\]

    D)  \[{{A}_{V}}=20,\,{{V}_{out}}=1.5V,\] and power gain\[=2000\]

    Correct Answer: A

    Solution :

    [a] Voltage gain \[{{A}_{V}}=\beta =\frac{{{R}_{C}}}{{{R}_{B}}}=60\times \frac{5\times {{10}^{3}}}{2\times {{10}^{3}}}=150\] Power gain \[={{\beta }^{2}}\frac{{{R}_{C}}}{{{R}_{B}}}=60\times 60\times 2.5=9000\] Base current \[={{I}_{B}}=\frac{12\times {{10}^{-3}}}{2\times {{10}^{3}}}=6\times {{10}^{-6}}A\] \[{{I}_{C}}=\beta {{I}_{B}}=60\times 6\times {{10}^{-6}}=3.6\times {{10}^{-4}}A\] Output \[={{I}_{C}}{{R}_{L}}=3.6\times {{10}^{-4}}\times 5\times {{10}^{3}}\Omega =1.8V\]


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