diode current

Discussion in 'Homework Help' started by kiemmadocco, Nov 25, 2010.

  1. kiemmadocco

    Thread Starter New Member

    Nov 24, 2010
    11
    0
    how to calculate the average diode current peak, and average valude of diode current in full wave rectifier with capcaitor filter parallel with resistor
    In my opinion, the average value of diode current= Io
    with Vo=2vm/pi
    Io=Vo/R
    how about the peak value?
    Ipeak=Irms*sqr(2)=Io*sqr(2)?
    any idea?
     
  2. Georacer

    Moderator

    Nov 25, 2009
    5,142
    1,266
    First of all take a look at the schematic of a full-wave rectifier with capacitor filter here, under the "output smoothing" section: http://en.wikipedia.org/wiki/Full_wave_rectifier

    Keep in mind that the diodes are not constantly conducting, even inside ther half-period. They conduct only to charge the capacitor when its voltage drops below the desired output voltage. That threshold depends on the load current consumption, I_{\tiny{L}}. The followin image depict quite well the current spikes that travel throught the diode: http://www.electro-tech-online.com/...d1243090102-diode-ripple-source-rectifier.jpg

    I will give you the formulas straight away, as I havent found them fully explained in bibliography anyway, and I don't think one should caclulate them in any other way other that geometrical approximation:

    I_{\tiny{DAV}}=I_{\tiny{L}}\cdot \left(1+\pi \sqrt{\frac{V_p}{2 V_r}} \right) \\<br />
I_{\tiny{DMAX}}=I_{\tiny{L}} \cdot \left(1+ 2 \pi \sqrt{\frac{V_p}{2 V_r}} \right)
    where V_p\ (not\ V_{pp})\ and \ V_r are input voltage amplitude and output ripple respectively.
     
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