# 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, Voutput = V1.
Negative half cycle of input voltage
During negative half cycle, D1 exists in reverse biased condition. Hence Voutput = 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 Voutput = 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 dont know where I have gone wrong. Could anyone please help me with this problem?
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, Voutput = V1.
Negative half cycle of input voltage
During negative half cycle, D1 exists in reverse biased condition. Hence Voutput = 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 Voutput = 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 dont know where I have gone wrong. Could anyone please help me with this problem?
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