How to size a capacitor?

Discussion in 'Electronics & Electrics' started by Supremo, Sep 18, 2006.

  1. Supremo

    Supremo Member

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    Hi guys,

    I need help in figuring out how to size a capacitor.

    I have a 24Vdc device that runs as long as the 24Vdc is applied to its terminals. However, there is a chance that the 24Vdc *may* drop out for a few milliseconds. I need to know what size capacitor I can use to ensure that when the 24Vdc drops out, the device continues to run.

    What are the different factors I need to take into consideration when sizing the capacitor?

    Thanks.
     
  2. Goth

    Goth Grumpy Member

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    1 farad is 1 coulomb per volt, thus

    C in farads = Q / V

    C = (It)/24 (I in amps)
    t = 4ms

    C = I*4 / 24E3

    C = I / 6E5

    C in uF = I * 5/3

    I'd multiply that up a few times to be sure the voltage doesn't drop off.

    The usual rule of thumb for PS design is to use the biggest capacitor you can easily find.
     
    Last edited: Sep 19, 2006
  3. The CK

    The CK Member

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    well you can never have too big a capacitor :)

    I'm trying to remember my power electronics course here but I think the voltage ripple Vr = I/(f*C) - where I is current, f is frequency and C is capacitance...

    so say a 2 volt drop is okay, the longest the power drops out for is 10ms (so 100hz), and your device draws 1 amp. Then 2 = 1/(100*c) so C = .005F = 5000uF

    anyway I think that equation is right... so adjust it for your situation (then double it to make sure) . For 24v I would want a 35 or 50v rated capacitor.
     
  4. munka

    munka Member

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    The primary factor we need to know is the load. Sizing is based on the RC constant of the circuit. Give us R and I can work out C. R can also be worked out by finding out the current and using OHMs law. Basically for caps if you make a few assumptions then double it you'll more than likely be fine, if its a big cap and cost is a concern then let me know R or the Amps and ill have a better look.
     
  5. OP
    OP
    Supremo

    Supremo Member

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    the tech specs of this device lists its current draw as 500 mA.
    Going by the above formula I get a capacitor size of 83.33 uF.

    That seems a bit small to me.

    Since the current is 500mA, R would be 4800 Ohms.
     
  6. nux

    nux Member

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    Chuck in the biggest capacitor you can find. When power is disconnected, the voltage will fall at a speed inverse to the size of the capacitor.
     
  7. mAJORD

    mAJORD Member

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    no, it would be 48 ohms

    you're after 24 / 0.5
     
  8. dakiller

    dakiller (Oscillating & Impeding)

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    Capacitors discharge at an exponential rate defined by -

    V(t) = Vo * e^(-t/RC)

    So we need to know -
    • the starting voltage, Vo, 24v here
    • R, 48ohm
    • the lowest voltage that is needed to be maintained, V(t), I'll just work with 22v here
    • and the length of time of the 'dropout', t, another value I'll just take 10mS
    Then solve for C

    rearrange the formula we get

    C = -t/R*LOGe(Vt/Vo)

    C = 2400uF
     
  9. nudge

    nudge Member

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    After you're done with the pen and paper calculations, whip up a spice simulation to verify they were correct!
     
  10. Lauhead

    Lauhead Member

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    Or if your an uni student, you design using spice :tongue:

    Always go bigger when in doubt for caps. But if you go too big you run into other problems, such is inrush current when you turn it on.
     
  11. OP
    OP
    Supremo

    Supremo Member

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    Cool, Thanks dakiller!!

    That was the formula I was looking for...
    However, when I substitute the values and solve for C, I get a value of 18.127uF.
    C=-((10*10^(-3))/48)/(ln(22-24))

    LOGe(22/24) == ln(22/24) or Log(22/24)/Log(e) where e=2.718
    is this right?

    Ill try out the capacitor and let you know how I go. :)

    Would the formula work for AC volatge too? so if I were to apply it to a 240Vac supply Id use the same formula but ensure that the capacitor is rated for 240V?
     
    Last edited: Sep 21, 2006
  12. munka

    munka Member

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    Nup, caps don't work with AC.
     
  13. Goth

    Goth Grumpy Member

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    Well, i wouldn't say that, but i don't think you could use a capacitor to compensate for a brownout in the AC supply.
     
  14. dakiller

    dakiller (Oscillating & Impeding)

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    I would say, capacitors only 'work' with AC. When given DC they 'do' nothing at all
    ;)
     
  15. OP
    OP
    Supremo

    Supremo Member

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    :( I just edited my reply and you guys have already answered!

    :) so do caps work with AC?
     
  16. nux

    nux Member

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    No, they will not power an AC device when the AC power drops out in the same way. Putting a cap in will do bad things.
     
  17. munka

    munka Member

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    Hehe i just came back to edit my reply before i noticed how many people had jumped in. What i wrote should say, caps don't work the same way on AC. Effectively caps used this way resist change in the voltage AC is always changing so on a large cap you would only get the average (0V), or a theoretical short circuit.

    No you can't use it to do the same thing.

    How does a cap help with brownouts?
     
  18. OP
    OP
    Supremo

    Supremo Member

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    hmm I dont quite understand. :confused:

    so how would one prevent a drop out if using ac volatge?
     
  19. nux

    nux Member

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    Something like an UPS.

    AC is Alternating Current. "An alternating current (AC) is an electrical current whose magnitude and direction vary cyclically, as opposed to direct current, whose direction remains constant." From Wikipedia.

    Someone else will no doubt explain it much better, but basically a capacitor will try to resist the AC wave, and in a brownout won't supply the same AC waveform required (in Australia, 50Hz at ~230Vrms).
     
    Last edited: Sep 21, 2006
  20. OP
    OP
    Supremo

    Supremo Member

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    ok got it.

    makes sense. thanks nuxie! :)
     

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