# 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?