I've got a fairly large capacitor (hundred mF's) that has a low voltage rating (10V). And, when I charge it to that value, as soon as I remove power supply (same voltage) the capacitor drops to around 6V (60%).

Is this possibly a capacitor electrolyte leak? I've subjected the capacitor to fairly harsh conditions.

EDIT: I forgot to mention, when I reconnect the power supply, the 6V suddenly jumps back to 10V, and when I disconnect, again, down to 6V... Perhaps, the capacitor never actually reaches 10V?? But it's weird as I observed the charging from nearly 0V to 10V, in a logarithmic way (or is it right shifted upside down exponentiation??).

  • \$\begingroup\$ All this time, I had the voltmeter connected. Why am I saying this? Because for at least the voltmeter in my multimeter, it seems to let nanoamps of current through that it can discharge a capacitor, which one can tell as the voltage is dropping in whatever capacitor I measure. The real issue to me is that this large capacitor I'm doing experiments with just drops instantaneously. \$\endgroup\$ – Dehbop Mar 31 '15 at 5:06
  • \$\begingroup\$ When you remove the power supply, is the capacitor completely unconnected from everything else? Or is it part of a circuit? \$\endgroup\$ – Dan Laks Mar 31 '15 at 5:21
  • \$\begingroup\$ it's just the capacitor and the DMM. \$\endgroup\$ – Dehbop Mar 31 '15 at 7:25
  • \$\begingroup\$ Is “mF” milli farad (10⁻³ F)? Ī̲ never saw the capacitance marked in such units; usually µF are used. \$\endgroup\$ – Incnis Mrsi Aug 24 '16 at 15:55

What you have described is either a broken capacitor, a broken multimeter, an unintentionally reversed capacitor or you have left something connected.

  • \$\begingroup\$ EDIT: I forgot to mention, when I reconnect the power supply, the 6V suddenly jumps back to 10V \$\endgroup\$ – Dehbop Mar 31 '15 at 8:38

Leak would result in a slow decay of the voltage so I don't think that's it.

Does the capacitor perhaps have a high ESR? If so, leaving it connected to the supply for longer may charge it to a higher voltage.

There is also dielectric relaxation, but (if this is a normal electrolytic) I wouldn't expect quite such a big effect. Again, charging for longer should improve the final voltage.

You can also spot dielectric relaxation by discharging the capacitor (through a resistor) until it reads 0, then disconnecting the resistor, and watch the voltage build up again, maybe to a couple of volts ... spooky.


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