Clipper Circuit problem

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

  1. circuit2000

    Thread Starter Active Member

    Jul 6, 2006
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    :confused: # Could anyone please tell me whether the output waveform is right or not? In the diagram, solid lines indicate the period during which output is obtained and the dotted lines indicate the period during which no output is obtained. D1 is an ideal diode.
     
  2. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    It's not correct.
     
  3. circuit2000

    Thread Starter Active Member

    Jul 6, 2006
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    After you said that it was wrong, I solved it in a different way.
    Throughout the positive half cycle of the input voltage, D1 exists in forward biased condition. Now, current flows through D1. As a result Vo will be equal to -10V during the positive half cycle.
    During the negative half cycle, D1 exists in forward biased condition until the input voltage becomes greater than 10V. So during that period, the potential of A w.r.t B increases from -10V to -20V. When the input voltage increases from -10V to -20V, D1 exists in reverse biased condition. So no current flows through D1. Hence Vo varies from -10V to -20V. When the input voltage decreases from -20V to -10V, D1 exists in reverse biased condition and hence V0 varies from -20V to -10V. When the input voltage varies from -10V to 0V, D1 exists in reverse biased condition and hence Vo remains constantly at -10V. Am I right now? please help me in understanding the concepts of clipper circuits.
    I am herewith attaching the diagram based on the above reasoning.
     
  4. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Nice job ... I'm sure you seen your error.
     
  5. circuit2000

    Thread Starter Active Member

    Jul 6, 2006
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    Thanks for your guidance.
     
  6. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Not a problem. All you needed was that little nudge. I thought you were suspicious of your original answer and just wanted the nudge so you could re-evaluate.
     
  7. pebe

    AAC Fanatic!

    Oct 11, 2004
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    Sorry to disagree with Joe, but there is a problem with the circuit as drawn. As the input is shown connected directly to the output, then the output must be a sine wave of 20v p-p.

    Also, on the peak +ve input half cycle, the anode of D1 has +20v on it. Its cathode has –10v on it so the diode would surely pop!

    For the circuit to be meaningful, a source resistor needs to be shown between the input and point ‘A’ and a load resistor shown between ‘A’ and ‘B’. The values of both resistors need to be given.
     
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