# Is it possible for electrolytic capacitor to develop a short in one way only?

I've got a 1 F electrolytic capacitor and it seems to have developed a short.

I never used it before and its brand new. It reads zero resistance only when I connect positive lead of my DMM to its negative lead and negative lead of my DMM to its positive lead. In the other direction, there's no short.

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Why the need for the 1F Cap? Doing audio work? – Dean Mar 29 '11 at 8:21
@Dean I'm hoping to use it instead of a battery for a very low power device. – AndrejaKo Mar 29 '11 at 8:44
A 1 F electrolytic capacitor? o_O I don't believe supercapacitors are "electrolytic". – endolith Mar 29 '11 at 14:49
@endolith Datasheet listed them as electrolytic, so that's why I used electrolytic in the question. It also looks like any other electrolytic capacitor and has same markings as electrolytic capacitors do. I'll see if I can dig up the link to datasheet. – AndrejaKo Mar 29 '11 at 15:40
The Panasonic 0.1 to 1F Gold Capacitors (also called Electric Double Layer Capacitors) are in fact electrolytic so maybe OP has one of this kind. – jpc Mar 29 '11 at 21:47

It sounds like the capacitor is functioning correctly.

Large capacitors are normally polarised with positive/negative terminals, so they only work one way which is why you not getting a short in one direction.

The reason you appear to have a short in the other direction is because the capacitor is not charged, and your multimeter is charging the capacitor as it reads the resistance. If you held your multimeter on the capacitor for long enough it will become fully charged and your meter would change to read an open circuit. Don't bother trying this with a 1 farad though, it will likely take a long time.

So it sounds like everything is good with your capacitor.

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Does it really take that long to charge a 1F cap to such a low voltage? – Kellenjb Mar 29 '11 at 3:41
The time needed to charge to a certain level (1/e? I'm not exactly sure) is the RC time. The charging time therefore depends on the R of the system. Considering that resistance measuring is often used for continuity (0 ohm), I presume there is a hefty resistor inside the meter. A resistor of 1kOhm would mean 1000 seconds (20 minutes) to charge to 60-ish percent. – drxzcl Mar 29 '11 at 7:54
@Kellenjb For this particular capacitor, the maximum current is just 1 mA and voltage is 5.5 V, so to push it to its limits, 5.5 $k\Omega$ resistor is needed. The time to charge it to 63.2% is then 1.52 hours. – AndrejaKo Mar 29 '11 at 8:45
@AdrejaKo Wait, what? Are you measuring your cap in side of a circuit? Surely your DMM isn't measuring at 5.5v. They usually have a much smaller voltage. – Kellenjb Mar 29 '11 at 13:09
@Kellenjb No, when I detected the problem, I was checking continuity on unpowered circuit between one pin for IC and power connector. It turns out that I got continuity to both pins of the power connector. The only part which could have made a short was the capacitor, so I removed it from the circuit and investigated it separately. – AndrejaKo Mar 29 '11 at 22:09

The design of electrolytic capacitors requires them to be polarized. With an electrolytic, the end with the stripe on the body is negative.

When you connect your DMM to it in that way, the current flows correctly.

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It is not the size of the capacitor that makes it polarized, instead it is the chemistry of the capacitor. Any electrolytic is polarized. – Kellenjb Mar 29 '11 at 1:24
@kellen, though you can place two back-to-back and get an effectively non-polarized electrolytic – Nick T Mar 29 '11 at 4:02
@Nick - Not so, according to this answer – Kevin Vermeer Mar 29 '11 at 12:36

AFAIK electrolytic capacitors (like any other capacitors) will work in both polarizations, but DC voltage of the wrong direction will degrade (like in electrolysis) the oxide layer. After being degraded for a while the oxide layer will cause short and a violent explosion.

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