Diode and capacitor

Discussion in 'Homework Help' started by gbox, May 23, 2016.

  1. gbox

    Thread Starter Member

    Dec 29, 2015
    42
    0
    צילום מסך 2016‏.05‏.23 ב‏.18.30.18.png

    I need to graph Vout.

    when Vin=V C1 and C2 charge to half the voltage of Vin, both diodes are off so Vout=Vin-C1=C2.
    But when Vin=-V the rightmost diode is on and Vout=-Vin+C1?
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,795
    What if C1 is a thousand time larger (or smaller) than C2?
     
  3. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,795
    Assuming zero leakage, that the two capacitors are equal, and that they start out initially uncharged, then they will charge to the same voltage as Vin increases from zero to some positive voltage because the right diode will be reverse biased and all of the current must flow through both capacitors. In steady state, there will be zero volts across the diode (since even a small forward voltage means some current is flowing and so both capacitors will still be charging, with the voltage across them increasing, and hence the voltage across the diode decreasing.

    But as the voltage is decreased, C1 will discharge but C2 will not.

    This is a pretty classic flying capacitor voltage multiplier circuit.
     
    Last edited: May 23, 2016
    gbox likes this.
  4. Bordodynov

    Active Member

    May 20, 2015
    637
    188
    See
    C2=100C1
    Diode_and_Capacitor.png
     
    Last edited: May 24, 2016
  5. johnmariow

    New Member

    May 4, 2016
    20
    1
    Thanks for the correction. I didn't recognize the circuit as a multiplier.
     
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