# Capacitor Shorting Question

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

1. ### jamers Thread Starter Member

Jan 24, 2008
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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
15,815
282
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
1,202
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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.

Steve

4. ### jamers Thread Starter Member

Jan 24, 2008
12
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You have made my day thank you .

5. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
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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 Expert

Jul 17, 2007
22,182
1,728
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)):
1.
2.        C1 + C2
3. Ctot = -------
4.        C1 x C2
If they were both rated at 2.2uF, the result would be 0.909... uF

Oct 11, 2004
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8. ### munna007 Active Member

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

9. ### mik3 Senior Member

Feb 4, 2008
4,846
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No Pepe is right, the formula is upside down

Ctot=(C1*C2)/(C1+C2)

like resistors in parallel

Mar 20, 2007
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11. ### iamspook Member

Aug 6, 2008
27
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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.

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