power supply capacitor size

Discussion in 'General Electronics Chat' started by logmode, Sep 6, 2011.

  1. logmode

    Thread Starter Member

    Sep 5, 2011
    48
    0
    Hi, a power supply @ 25vac, 50Hz in and a variable 0-20vdc, 10a out. A ripple smoothing capacitor will be added parallel from + to – after the full wave rectifier. I found the formula. C = (5 X Io) / (Vs X F)
    C = farads
    Io = output amperes
    Vs = voltage between the capacitor & rectifier.
    F = Hz
    I want looking for a 10% ripple.
    Is this formula ok? Do you double the Hz? How does the 5 represent the 10% ripple? And would you work out the entire math, so I can study it?
    Thanks
     
  2. Hi-Z

    Member

    Jul 31, 2011
    157
    17
    Hi,

    The derivation of this is the equation dV/dt = I/C. In your case you will be needing a reservoir capacitor which can hold enough charge between diode conduction periods, such that ripple is limited to an amount which permits your voltage regulator to yield a satisfactory 20V.

    Assuming the conduction periods are short, and the desired ripple is small, you will have somewhat less than 10ms (at 50Hz, full-wave rectification) for the capacitor to discharge at 10A.

    Rearranging the above equation we have C = t*I/V (where t is time between diode conduction, I is load current and V is ripple voltage). Putting in 10ms (ie. 1/100 seconds) and 10A, we have C = 1/10V.

    If you decide to on 2 volts of ripple, then C will need to be 1/20 farads, or 50,000 uF.
     
  3. logmode

    Thread Starter Member

    Sep 5, 2011
    48
    0
    Beautiful explanation,
    Thank you
    What is the negative reason in this case for not just putting in a huge capacitor and trying to making a straight line (no ripple)?
     
  4. MrChips

    Moderator

    Oct 2, 2009
    12,432
    3,360
    We have gone through this before in a previous thread. When you increase the value of the reservoir capacitor the peak current that the rectifier diodes have to conduct goes way up.

    Another simple way of looking at it is the very large C becomes a short circuit to the 120Hz AC, hence the increase in current.
     
  5. Hi-Z

    Member

    Jul 31, 2011
    157
    17
    Yes, with large capacitors you can get very large currents, especially at power-up, when you may experience the highest dV/dt. Not so good for either capacitor or diodes - though often I suppose the transformer winding resistance will limit current.

    I remember messing about with power supplies and being amused by the markings on the capacitor which stated "Max ripple..." (referring to ripple current). I actually considered using "Max Ripple" as my forum name!
     
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