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Calculating the filter capacitor for a power supply

  1. Feb 15, 2017 #1
    Trying to build a simple 9V power supply using a 7809 voltage regulator and full wave rectifier. The load current from the regulator is 1A. So I = C*dv/dt, therefore,

    C = I*t/(delta V)

    uitiQ.gif

    To my understanding t is the time between the two peaks or the time the capacitor supplies the load. So if t is reduced, shouldn't the ripple increase ?, so isn't ripple (delta V) dependent on t ?, if so then why is delta V just arbitrarily chosen without considering t ?
     
  2. jcsd
  3. Feb 15, 2017 #2

    Merlin3189

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    1, t is fixed if you are using mains power - at roughly 0.01sec.
    2, ripple would increase with t, if you were using a variable frequency supply.
    This should be clear from the graph. If the peaks are further apart, the output drops more between peaks.
    3, I don't think delta V is arbitrarily chosen. It comes from the permissible ripple. You know delta t, from mains frequency, and I from your spec. Then you calculate C.
    Permissible ripple may come from the difference between, the lowest operating voltage of your regulator and the peak value your capacitor charges to.
     
  4. Feb 15, 2017 #3

    jim hardy

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    2016 Award

    look at your graph
    solve for dv
    dv = I/C X dt , therefore
    increasing Δt increases Δv

    you knew that, just you were in a hurry. Think in small patient steps.
     
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