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Capacitor behaviour

  1. Oct 11, 2009 #1
    my question is about capacitor response to ac voltage source .

    if we have circuit consistst of ac voltage source and a capacitor .so the variation of voltage across capacitor versus time will be as shown in the attached drawing.

    and from this drawing i see that during the postive half wave ,as source voltage increasing ,voltage across capacitor also increasing (charging) and when source voltage decreasing to zero ,voltage across capacitor also decreasing (discharging).is my explaination about this is right??

    my next question if this concept is right so why clamper circuit it shows that during positve half cycle when diode is short cicuit the cpacitor is charged to the peak voltage ????
    we can see that in the link below
    http://www.visionics.ee/curriculum/Experiments/Clamper/Clamper1.html [Broken]

    i hope u get my question
     

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    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 11, 2009 #2
    for clarification i mean if this explaination is right so the capacitor will be charged then discharged during the positive half cycle??so how it will only be charged to the peak voltage
     
  4. Oct 11, 2009 #3

    Averagesupernova

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    Gold Member

    Are you wondering why it is charged to peak instead of what? Peak x 2?
     
  5. Oct 12, 2009 #4

    vk6kro

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    In the attached diagrams, during the first negative half cycle (ie like the bottom diagram), the capacitor voltage will follow the input voltage until the voltage starts to drop. The capacitor cannot discharge back into the supply because the diode cannot conduct backwards.

    So, the capacitor holds the peak voltage of the supply.

    The capacitor has no way of losing this charge, so it now appears in series with the source voltage.

    See attached diagrams. The box on the left is a transformer giving 20 volts peak to peak, or 10 volts peak. Assume perfect diodes.

    In the first one, the capacitor is charged to 10 volts and the input voltage is rising to a peak of +10 volts, so the output is +20 volts.
    In the second one, the capacitor is still charged to 10 volts, but the input voltage is 10 volts negative, so the two cancel out and the output is zero volts.

    So an input that was varying from +10 volts to -10 volts is now varying from +20 volts to zero volts. This is what a clamp circuit does.
     

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  6. Oct 12, 2009 #5
    thx for ur reply but still what peak to peak means and whats difference between peak to peak vlaue and peak value
     
  7. Oct 12, 2009 #6

    berkeman

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    Staff: Mentor

    A sinusoidal voltage as a function of time is written like this:

    V(t) = A sin(wt)

    Where A is the amplitude, and omega "w" is the angular frequency in radians per second.

    The value A is called the peak amplitude. The peak-to-peak amplitude is 2A, because it is the difference between the positive peaks at A and the negative peaks at -A.

    A - (-A) = 2A

    Does that clear it up for you?
     
  8. Oct 12, 2009 #7

    vk6kro

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    See attached picture.
     

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    Last edited: Oct 12, 2009
  9. Oct 13, 2009 #8
    thank u all for your reply
     
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