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  1. In the following limiter circuit, an input voltage V1 = 10 sin 100pt applied. Assume that the diode drop is 0.7 V when it is forward biased. The zener breakdown voltage is 6.8 V

    The maximum and minimum values of the output voltage respectively are

  2. A.
    6.1 V, -0.7 V
    B.
    0.7 V, -7.5 V
    C.
    7.5 V, -0.7 V
    D.
    7.5 V, -7.5 V

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    Answer : Option C

    Explanation :

    During +ve part of Vi

    D1 will be forward biased

    Zener diode is reverse biased

    Thus net voltage = 6.8 + 0.7 = 4.5 V

    During -ve part of Vi

    D2 will be forward biased

    D1 will be reversed biased

    Thus net voltage = -0.7 V .


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