BCECE Medical BCECE Medical Solved Papers-2011

  • question_answer
    The input resistance of a common emitter transistor amplifier, if the output resistance is\[500\,\,k\Omega \], the current gain \[\alpha =0.98\] and the power gain is \[6.0625\times {{10}^{6}}\], is

    A)  \[198\,\Omega \]

    B)  \[300\,\Omega \]

    C)  \[100\,\Omega \]

    D)  \[400\,\Omega \]

    Correct Answer: A

    Solution :

    Power gain = Current gain x volume gain Volatge gain \[{{A}_{V}}=\beta \frac{{{R}_{2}}}{{{R}_{1}}}\] Also, current gain \[\beta =\frac{\alpha }{1-\alpha }=\frac{0.98}{1-0.98}=49\] \[\therefore \] \[{{A}_{V}}=(49)\left( \frac{500\times {{10}^{3}}}{{{R}_{1}}} \right)\] Power gain \[=6.0625\times {{10}^{6}}\]                 \[=49\times \left( \frac{500\times {{10}^{3}}}{{{R}_{1}}} \right)\times 49\] \[\Rightarrow \] \[{{R}_{1}}=198\,\,\Omega \]


You need to login to perform this action.
You will be redirected in 3 sec spinner