Capacitor Shorting Question

Discussion in 'General Electronics Chat' started by jamers, Jan 24, 2008.

  1. jamers

    Thread Starter Member

    Jan 24, 2008
    Hello, my question is a fairly simple one yet one I am fairly at odds about. I want to build a device that involves changing current rapidy, from open, to dead short across a coil, for a brief instant in time. The simplest way I can think of of storing the energy to put across here is in a capacitor. Which leads me to my question.

    From what i have personally seen, whenever a high voltage is shorted across a capacitor, lets say 200VDC, the capacitor has always ended destroyed after, and my device would be building and then shorting this many times. From what I can imagine some energy would be dissapated building the magenetic field in the coil, but I cannot imagine that limiting the current.

    So my question is, am I just being shortsighted, would the capacitor not suffer much damage from this? Is the current not actually going to be the thing that actually damage this, perhaps the capacitor would charge in reverse and discharge again, could that be what actually destroys them?

    Sorry for this being kind of long winded, or if its a stupid question, Id just like to know if capacitors have any type of life expectancy in a situation like this, or if i should try to go an alternate route.
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    Shorting caps isn't a problem, unless you don't like notches on your screwdriver shaft.

    When you dump current into a coil, it builds a magnetic field. When the current stops the field collapses, and generates a voltage with a polarity opposite to the voltage that made the field. Since your cap is still attached to the coil, this voltage tries to charge the capacitor to the wrong polarity, and destroys it.

    Adding 2 diodes to the circuit will protect the cap. Place on in series with the coil so the reverse voltage will not flow into the capacitor, and place one in parallel with the coil so the reverse voltage will discharge through the diode. These will probably need to handle 6 - 8 amps at 250 volts.
  3. scubasteve_911

    Senior Member

    Dec 27, 2007
    Perhaps you can connect two polarized capacitors in series, negative to negative, and create a bipolar capacitor. You wouldn't need to be concerned about the negative voltage from the inductor.

  4. jamers

    Thread Starter Member

    Jan 24, 2008
    You have made my day thank you :).
  5. thingmaker3

    Retired Moderator

    May 16, 2005
    Another option would be to make a bank of non-polarized capacitors. Select enough of them to gain an equivalent value to your desired capacitance. This may or may not be practical, depending on how much capacitance is needed.
  6. SgtWookie


    Jul 17, 2007
    Keep in mind that if you decide to connect polarized electrolytic capacitors back-to-back (ie: negative to negative) in order to get non-polarized caps - you will lose a good deal of capacitance.

    The total capacitance of capacitors in series, are calculated like resistors in parallel. For two caps in series:
    Code ( (Unknown Language)):
    2.        C1 + C2
    3. Ctot = -------
    4.        C1 x C2
    If they were both rated at 2.2uF, the result would be 0.909... uF
  7. pebe

    AAC Fanatic!

    Oct 11, 2004
    Your formula is upside down. The answer is 1.1µF
  8. munna007

    Active Member

    Jul 23, 2008
    i think this formula is ok...its opposite to that of resistance buddy
  9. mik3

    Senior Member

    Feb 4, 2008
    No Pepe is right, the formula is upside down


    like resistors in parallel
  10. Ratch

    New Member

    Mar 20, 2007
  11. iamspook


    Aug 6, 2008
    Hi -
    Caps in series will require a highish value resistor across each one
    to balance the voltage correctly. This is because no two same-value
    caps are physically
    identical and you should avoid mismatched voltage drops across each one
    when they are in series.

    Vm should be V2/2 which is actually controlled by R, but not really
    controllable by C on their own.

    Desired: If Vs = 100v, then Vm = 50 v.
    If the caps are rated at 60v, then you should be ok, but without the
    R then Vm might be 70v due to poor tolerances of C and how they
    can vary so wildly with temperature.

    Your circuit should be designed to not suffer from the addition of these
    two R values.