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Electronics and Communication Engineering - Exam Questions Papers - Discussion

Discussion Forum : Exam Questions Papers - Exam Paper 3 (Q.No. 32)
Exam Questions Papers
32.
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
6.1 V, -0.7 V
0.7 V, -7.5 V
7.5 V, -0.7 V
7.5 V, -7.5 V
Answer: Option
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 .

Discussion:
1 comments Page 1 of 1.
Shashank said:   3 years ago
Net voltage in +ve has been mistyped to 4.5 instead of 7.5.

please correct it.
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