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Ripple Voltage from Constant Power Load

Discussion in 'Electronic Design' started by D from BC, Feb 11, 2008.

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  1. D from BC

    D from BC Guest

    Here's something I'm stuck on...


    120VAC----Bridge
    60Hz ----Rectifier-----+------+ <<<What's the ripple voltage?
    | | |
    | 180uF 90W Constant
    | | real power load
    | | |
    +----------+------+

    A constant power load will draw whatever current to maintain 90Watts
    of dissipation.
    As the capacitor discharges, the current increases to maintain the
    power dissipation.

    I could spice for an answer but..
    I haven't figured out yet how to model a constant power load in
    LTSpice.. :(


    D from BC
    British Columbia
    Canada.
     
  2. legg

    legg Guest

    For a capacitive rectifier, the droop in the capacitor will be the
    same for constant power as it will be for equal average power.

    The problem is determining the exact intersect between the rectified
    sine and the capacitor droop to get the discharge period. That sounds
    like integral calculus to me, but its probably just algebra and trig.

    For rough estimation you can assume 2mSec charging in a 6mSec period,
    that leaves ~4mSec to discharge.

    C/2 x ( V1^2 - V2^2) = delta j delta j = P x t j = CV^2/2

    1.8E-4/2 x ( 160^2 - V2^2) = 90 x 4E-3

    root{ 160^2 - ( 180 x 4E-3 / 1.8E-4 ) = V2 = 147V

    ppk ripple = 166 - 147 = 20V

    Worst case ripple occurs at low line as the initial energy stored in
    the caps reduces.

    RL
     
  3. D from BC

    D from BC Guest


    I figured out a model to confirm.
    http://www.members.shaw.ca/chainsaw/SED/constantpwrload.jpg
    470kb jpeg
    Got Vripple = ~21.77Vpp
    Those pesky diodes... :p

    Good math.. Your rough estimate was accurate enough for my app. :)

    Thanks


    D from BC
    British Columbia
    Canada.
     
  4. Fred Bloggs

    Fred Bloggs Guest

    If V is the DC capacitor voltage then the DC load current is I=P/V
    making the incremental negative conductance dI/dV=-P/V^2 which is pretty
    small. Taking the average V to be say 90% of Vpeak~170V, this gives
    -4mA/V incremental or -250 ohms, which is less than 1Watt incremental,
    and not worth considering in more detail.
     
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