Clamper circuit problem

Discussion in 'Homework Help' started by circuit2000, Jul 7, 2006.

  1. circuit2000

    Thread Starter Active Member

    Jul 6, 2006
    # 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?
  2. pebe

    AAC Fanatic!

    Oct 11, 2004
    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).