# Clamper circuit problem

#### circuit2000

Joined Jul 6, 2006
33
# Draw the output waveform for the following clamper circuit. Assume that 5y = 5RC >> T/2 where y = time constant of the RC circuit, T = time period of the input square wave voltage.

I solved it in the following way:

Positive half cycle of input voltage
During positive half of input voltage when V=V1, D1 exists in forward biased condition. Hence, V-output = V1.

Negative half cycle of input voltage
During negative half cycle, D1 exists in reverse biased condition. Hence V-output = Potential difference across resistance R = (-V) + (-V) where one V is due to the input voltage and the other V is due to the charged capacitor. Hence V-output = -2V. But the difference between my output waveform and the output waveform given in my book is that in my book they have taken 2V downwards from V1 whereas I have taken it from the origin. As per the book answer, the output voltage at a time T/2 is (2V  V1). How can that be possible when an output voltage of -2V is obtained during the negative half cycle of input voltage?
But in a clamper circuit, the output waveform has the same voltage swing as the input waveform. So, my answer is wrong but I dont know where I have gone wrong. Could anyone please help me with this problem?

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#### pebe

Joined Oct 11, 2004
626
The book answer shown against the waveform is also wrong! When the waveform is at its maximum +ve excursion, the diode conducts and only allows the RH side of the capacitor to rise to +V1 volts. The wave form is square wave and (in the perfect case) the waveform will change instantly from +V to –V, a change of 2V. So as the input falls the diode will cut off. C will not have time to change its charge and its RH side will fall by 2V from V1 to –(2V-V1).