Sizing capacitors to prevent voltage dip

Discussion in 'General Electronics Chat' started by mcgyvr, Jun 28, 2012.

  1. mcgyvr

    Thread Starter AAC Fanatic!

    Oct 15, 2009
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    What is the formula or how would I go about sizing a bank of capacitors to prevent/lessen voltage dip on a power supply bus during a load fault?

    I have a DC power supply that feeds 2 different circuits. On one circuit (circuit 1) I want to include a bank of capacitors across the input to help maintain the power supply bus voltage during a load fault on this circuit.

    Basically, I don't want any load faults on circuit 1 to effect circuit 2 in any way if possible.

    To give some numbers, lets say power supply voltage is 24V and possible load current on circuit 1 is 100 Amps.
     
  2. daviddeakin

    Active Member

    Aug 6, 2009
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    Is the fault current momentary, or likely to be sustained for long periods before you realise a fault has occurred?
    If you have a 100A fault current that lasts for several minutes or even seconds then no amount of capacitance is going to help you! (Except maybe infinite capacitance, but you don't see those for sale much anymore...;)
     
  3. mcgyvr

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    The fault will last until the fuse or circuit breaker trips.
    Lets assume 1 second.
     
  4. russ_hensel

    Well-Known Member

    Jan 11, 2009
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    Calculate how much charge is used for the fault. If you do not know how to do this look it up.
     
  5. mcgyvr

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    That's why I posted :(.. I wasn't sure what forumla to use for that..
     
  6. MrChips

    Moderator

    Oct 2, 2009
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    Charge Q = C x V

    Hence if the voltage drops by dV, the drop in charge dQ is

    dQ = C x dV

    Charge Q is also I x t

    If you know the current I and the time t, the charge Q = I x t
     
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  7. ErnieM

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    Apr 24, 2011
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    The capacitor formula is:

    I(t) = C dv/dt

    It is an expression for the current as a function of time and always applies, though it's a differential (calculus) formula. However, it is still valid when I(t) is a constant and you can then use a difference equation of:

    I = C (delta V) / (delta T)

    It's a good approximation for a small change in voltage. If you solve it for C you get:

    C = I * (delta T) / (delta V)

    You can read this as when either I or (delta T) increases you need more cap, also if (delta V) decreases you also need more cap.


    To put some numbers on this say your load is 100 amps, you need to support this with no more then a 2 volt change for up to 1 second. Then:

    C = 100 amps * (1 second) / (2 volts) = 50 Farads.

    That be one big cap!
     
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  8. mcgyvr

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    Perfect..
    Thanks guys
     
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