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NEET Physics Semiconducting Devices NEET PYQ-Semiconducting Devices

  • question_answer
    If \[\alpha \] and \[\beta \] are current gains in common-base and common-emitter configurations of a transistor, then \[\beta \] is equal to:

    A)  \[\frac{1}{\alpha }\]     

    B)                   \[\frac{\alpha }{1+\alpha }\]

    C)  \[\frac{\alpha }{1-\alpha }\]

    D)                   \[\alpha -\frac{1}{\alpha }\]

    Correct Answer: C

    Solution :

    Current gain in common-base configuration is.
                \[\alpha ={{\left( \frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}} \right)}_{{{V}_{CB}}}}\]
                Current gain in common-emitter configuration is,
                \[\beta ={{\left( \frac{\Delta {{i}_{C}}}{\Delta {{i}_{B}}} \right)}_{{{V}_{CE}}}}\]
                Also,     \[{{i}_{B}}={{i}_{E}}-{{i}_{C}}\]
                or         \[\Delta {{i}_{B}}=\Delta {{i}_{E}}-\Delta {{i}_{C}}\]
                \[\therefore \]      \[\beta =\frac{\Delta {{i}_{C}}}{\Delta {{i}_{B}}}=\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}}\times \frac{\Delta {{i}_{E}}}{\Delta {{i}_{B}}}\]
                or         \[\beta =\alpha \times \frac{\Delta {{i}_{E}}}{\Delta {{i}_{E}}-\Delta {{i}_{C}}}\]
                or         \[B=\alpha \times \frac{1}{1-\frac{\Delta {{i}_{C}}}{\Delta {{i}_{E}}}}\]
                or         \[\beta =\frac{\alpha }{1-\alpha }\]
                Note:    \[\beta \] is always greater than \[\alpha \]. Also \[\alpha <1\] and \[\beta >1\].


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