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Q1:

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

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

ANS:C - 7.5 V, -0.7 V

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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