clipper circuit help

Discussion in 'General Electronics Chat' started by jut, Sep 25, 2009.

  1. jut

    Thread Starter Senior Member

    Aug 25, 2007
    224
    2
    [​IMG]

    I understand how the above circuit works; with the diode switched out of the circuit, the output, which is green in the scope, ch2, is phased shifted by about 6 degrees to the left. I confirmed this with phasor analysis.

    [​IMG]

    But when I switch the diode into the circuit, the output is cutoff at anything more than 0.7 volts. I get this, the diode is forward biased, and it drops 0.7 volts while "shorting" out the resistor. What I don't get is why the negative voltage swing of channel 2 gets all the way down to -8 volts.
     
    Last edited: Sep 25, 2009
  2. mentaaal

    Senior Member

    Oct 17, 2005
    451
    0
    It might be very useful to observe that the voltage across the capacitor is doing. As you can imagine, if the voltage across the diode is being clipped, over 0.7 volts, the capacitor would then charge more than usual than when the diode is reverse biased. I suspect, without doing any calculations that the abnormal charging of the capacitor is causing the waveform to be shifted down like that.
     
  3. jut

    Thread Starter Senior Member

    Aug 25, 2007
    224
    2
    I think I got it now. The cap charges up on the positive half cycle while the diode is acting like a short. But on the negative half cycle, the cap now has to discharge through the resistor. But since the time constant of RC (2 ms) is greater than the period of the input wave form (1 ms), the cap doesn't have time to discharge. So it's as if the cap is stuck with the charged voltage.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Yep, you got it. ;)
     
  5. jut

    Thread Starter Senior Member

    Aug 25, 2007
    224
    2
    Thanks for the confirmation.

    This guy named "caolinlin" is leaving little snippets of spam at the end of threads. He needs to be banned.
     
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