Electronics and Communication Engineering - Electronic Devices and Circuits - Discussion
Discussion Forum : Electronic Devices and Circuits - Section 11 (Q.No. 36)
36.
If 100 V is the peak voltage across the secondary of the transformer in a half-wave rectifier (without any filter circuit), then the maximum voltage on the reverse-biased diode is
Discussion:
8 comments Page 1 of 1.
JOPHET COLLADO SIDAYON said:
7 years ago
The output voltage across the secondary transformer is 100Vrms.
To get the average output voltage of the diode we need to use this formula(Vm=2*Vrms/)-->Vm=2*100 = 200V.
The voltage across the diode is 200V, assuming it is an ideal diode. So the PIV should also be equal to 200V. The answer should option A.
To get the average output voltage of the diode we need to use this formula(Vm=2*Vrms/)-->Vm=2*100 = 200V.
The voltage across the diode is 200V, assuming it is an ideal diode. So the PIV should also be equal to 200V. The answer should option A.
Aswathy said:
4 years ago
@Jason Garcram.
It's given that 100v is the peak value. Then how this answer option B? Explain please.
It's given that 100v is the peak value. Then how this answer option B? Explain please.
Saurabh kumar said:
5 years ago
In half-wave rectifier, PIV voltage is the maximum peak voltage.
So, the Correct option is C.
So, the Correct option is C.
Jason Gacrama said:
6 years ago
Vrms = 0.707(Vp).
Vrms = 100V,
100 = 0.707 (Vp),
Vp = 141.4 V Answer is (B).
Vrms = 100V,
100 = 0.707 (Vp),
Vp = 141.4 V Answer is (B).
(2)
Mukund said:
9 years ago
Should it not be 100V?
As PIV of half wave rectifier is Vm.
As PIV of half wave rectifier is Vm.
Sankar said:
8 years ago
Yes, C is the answer.
Kamlesh said:
8 years ago
Answer is C.
Lorlor said:
8 years ago
It's 100v.
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