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Capacitor Discharge

  1. Jun 11, 2009 #1
    I have been testing this equation with various different capacitors and capacitor bank values for V, R and C. I have found that when combining two capacitors in series (V doubles, C is divided in half, R is doubled) the following equation yields the same I-curve over t as a single capacitor. Wouldn't the area of the I curve be equal to the capacitor's (or bank's) total energy? In other words, shouldn't U=It?

    I = V/R * e^(-t/RC)
     
  2. jcsd
  3. Jun 11, 2009 #2

    berkeman

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    Staff: Mentor

    Just putting two caps in series does nothing to the voltage and resistance. you must mean something else in your question?
     
  4. Jun 11, 2009 #3
    Ahhh, I guess I've been gravely mistaken. I was under the belief that when you add two capacitors in:

    series, you double the ESR (because two resistors in series double the resistance,) divide capacitance by two, and double the voltage

    parallel, you double capacitance, voltage remains the same, and you divide resistance in half

    Would it be too much to ask for you to kindly provide me with the proper values of said properties when combining two capacitors in parallel and in series?
     
  5. Jun 11, 2009 #4

    berkeman

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    Ah, you didn't say that the R in your question was the ESR of the cap(s). Yes, series connecting two identical caps will halve the capacitance and double the ESR. By saying that it will double some voltage, I assume you mean if they are charged up initially and then placed in series still charged. Yes, that will double the voltage.

    Are you then shorting them out to use the equation that you posted? The only "R" is the ESR values?
     
  6. Jun 11, 2009 #5
    By doubling voltage, I meant that if you place two 450V max. caps in series, you can now charge the bank up to 900V rather than only 450V.

    Yes, they will be shorted-out and there is no additionally resistance in the circuit, other than ESR.
     
  7. Jun 11, 2009 #6

    berkeman

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    Oh, that's different. In general you want to use a single cap with the full voltage rating. If you are placing caps in series to try to get a higher overall voltage rating, you need to do something additional to ensure that the total voltage divides equally across the two lower voltage capacitors. Otherwise, one could have its voltage rating exceeded and blow, and then the other will blow.

    You would generally ensure the equal voltage division with large-value resistors that are placed around each cap. Something like 100kOhms around each cap.

    Can you not obtain a single cap with the full voltage rating?
     
  8. Jun 11, 2009 #7
    Well, I'm designing a cap bank for a rail gun so relatively high ampere currents are desired. Though not yet exacted, said current will probably require a higher voltage than I can find in a cap of sufficient capacitance. So, at this point, I'm trying to configure the best cap bank with regard to series, parallel or a combination.
     
  9. Jun 11, 2009 #8
    The problem with two caps in series is that during discharge, the voltage on one could drop faster than the other, you could overvoltage one, or even reverse voltage one.. If you do have series caps, put bleeder resistors across them, or if you have many cap-pairs in series, tie the inter-cap connections together.
     
  10. Jun 11, 2009 #9
    I see, thank you guys greatly for the tips. I will be sure to implement them.

    On a separate note, since two caps in series contain the same amount of stored energy as two caps in parallel, wouldn't the current-time curve have the same area under the function that is specified in the first post?
     
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