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JEE Main & Advanced Physics Semiconducting Devices Question Bank Critical Thinking

  • question_answer
    On applying a potential of ? 1 volt at the grid of a triode, the following relation between plate voltage Vp (volt) and plate current \[{{I}_{p}}(\text{in }mA)\] is found \[{{I}_{p}}=0.125\,{{V}_{p}}-7.5\]If on applying ? 3 volt potential at grid and 300 V potential at plate, the plate current is found to be 5mA, then amplification factor of the triode is

    A)            100                                           

    B)            50

    C)            30  

    D)            20

    Correct Answer: A

    Solution :

                       At \[{{V}_{g}}=\,-\,3V,\,{{V}_{p}}=300\,V\] and \[{{I}_{p}}=5mA\] At \[{{V}_{g}}=-1V,\] for constant plate current i.e. \[{{I}_{p}}=5mA\] From \[{{I}_{p}}=0.125\,{{V}_{p}}-7.5\] Þ \[5=0.125{{V}_{p}}-7.5\,\] Þ \[{{V}_{p}}=100V\] \ change in plate voltage \[\Delta {{V}_{p}}=300-100=200V\] Change in grid voltage \[\Delta {{V}_{g}}=-1-(-3)=2V\] So, \[\mu =\frac{\Delta {{V}_{p}}}{\Delta {{V}_{g}}}=\frac{200}{2}=100\]


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