Capacitors and breakdown voltages

[manjuvenamma]

What is cause of capacitor breakdown?
When we combine different capacitors in series and parallel combination, how does the breakdown voltage of the system get affected?

 

[optrix]

Basically, when you exceed the rated voltage, as with any other rated component, the capacitor can't handle it and the dielectric will probably melt or something, and cause a short circuit between the plates. This is the only thing I can think you mean by "capacitor breakdown". The other use of the term "breakdown" in electronics is for breakdown voltages in diodes.

For capacitors in series, 1/C[total] = 1/C[1] + 1/C[2] + 1/C[3] +...

For caps in parrallel, C[total] = C[1] + C[2] + C[3] + ...

The current and v0ltage are related by i = C(dV/dt), which are just derived from the equation Q=CV.

The rated voltage for the capacitor will be on there.

If its an electrolytic capacitor, you might want to ramp up the voltage on it, for fun, and watch it explode! :)

 

[manjuvenamma]

Yes, I mean breakdown voltage is that only. The max. voltage for which the capacitor is rated. Now if that capacitor is combined with other capacitors in parallel/series or combination of these two, how is the max voltage affected?

 

[Oberst Villa]

Yes, I mean breakdown voltage is that only. The max. voltage for which the capacitor is rated. Now if that capacitor is combined with other capacitors in parallel/series or combination of these two, how is the max voltage affected?

for caps in parallel the breakdown voltage is not affected. all caps will see the same voltage which is the same as if only one cap would be present.

for caps in series, I think theoretically you should be able to apply a higher voltage than to a single cap, because the voltage should be divided equally between the caps (assuming all have the same value). but for practical (engineering) purposes I think this is not true (or at least it is not safe). because of imperfections of the caps you cannot be sure that the voltage is actually equally divided. most of the voltage might be on one of the caps and this one would break down much earlier than you would predict when simply dividing the voltage among them.

 

[Phrak]

The voltage of a capacitor is determined by the dielectric strength times the thickness of the dielectric.

Practically speaking..let's say you have two capacitors in series that are discarged. Some source of current is applied where they both become charged. One cap is 1uF and the other is 10uF. The 1uF will see ten times the voltage across it.

Care must be taken when charging banks of capacitors in series, even when they have the same ratings. With all the same ratings, the leakiest capacitor will eventually have the least voltage across it. The most charge will collect on the strongest capacitor.

If you series two 100uF +50/-10% electrolytic capacitors together, say, you can't expect them to charge to the same voltage unless they are the same series by the same manufacturer and out of the same production run.