Electronic Devices - Diode Applications - Discussion
Discussion Forum : Diode Applications - General Questions (Q.No. 12)
12.
Determine the peak for both half cycles of the output waveform.
Discussion:
9 comments Page 1 of 1.
Vishal Mishra said:
4 years ago
Explain this clearly.
PRANAV DESHMUKH said:
6 years ago
The answer should be C.
Because during +ve cycle up to 4v diode is forward bias after that it's reverse bias up to 4v decreasing of +ve cycle and it's forward bias up to 0 that o/p for +ve cycle is 4v.
And during entire - ve cycle diode is forward bias, therefore, o/p for - ve cycle is - 16.
Because during +ve cycle up to 4v diode is forward bias after that it's reverse bias up to 4v decreasing of +ve cycle and it's forward bias up to 0 that o/p for +ve cycle is 4v.
And during entire - ve cycle diode is forward bias, therefore, o/p for - ve cycle is - 16.
Amol said:
1 decade ago
Yes it is shunt clipper circuit.
SANDIP MAURYA said:
1 decade ago
During the +ve 16v half cycle, the diode will be reverse biased so +16v goes to the output.
During the -ve 16 we cycle, a 4 volt battery connect in series with diode in opposite direction so 16-4= 12volt is short by forward biased diode so 16-12 = 4 volt in negative.
So 16 in positive and 4 volt in negative.
During the -ve 16 we cycle, a 4 volt battery connect in series with diode in opposite direction so 16-4= 12volt is short by forward biased diode so 16-12 = 4 volt in negative.
So 16 in positive and 4 volt in negative.
(2)
Titos said:
1 decade ago
During the +ve 16v half cycle, the diode will be reverse biased so +16v goes to the output.
During the -ve 16v half cycle, the diode will be forward biased then acting short and the output will be taken from the 4v battery and appears at the output as -4v.
Sameer is right, this is a shunt or parallel clipper.
During the -ve 16v half cycle, the diode will be forward biased then acting short and the output will be taken from the 4v battery and appears at the output as -4v.
Sameer is right, this is a shunt or parallel clipper.
Sameer said:
1 decade ago
Its a parallel clipper.
Samir said:
1 decade ago
Thank you umesh.
Umesh said:
1 decade ago
When input is less then -4v, the diode is on and output is -4v.
When input is greater then -4v, the diode is off and input is appear across output peak value of output=16v.
When input is greater then -4v, the diode is off and input is appear across output peak value of output=16v.
Sumanth said:
1 decade ago
How can it be -4v during negative half cycle ?
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