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# Clamper Circuit problem

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# 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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