# What is the “CV” unit for leakage current in a capacitor datasheet?

I've looked at a few leakage current specifications for electrolytic capacitors, and they all seem to specify the value as something like this:

I < 0.01 CV or 3 (μA) after 2 minutes, whichever is greater

Here are a few example datasheets: Panasonic, Multicomp, Nichicon, Rubycon.

Am I right in thinking that the leakage current is a product of capacitance and voltage, i.e. for a 100µF cap on a 5V supply I'd be looking at a leakage current of $I = 0.01\times100µF\times5V=5\times10^{-6}A = 5µA$.

Or is that CV unit something totally different?

Additionally, why the long time delay for this rating when a capacitor typically charges in seconds or less?

• Note that the actual units of C x V are "charge", so there is an implicit "per second" to convert this to a current. – Dave Tweed Dec 3 '16 at 16:19

The leakage spec- in this case 0.01CV (or 3$\mu$A) is the product of rated voltage and rated capacitance, not applied voltage. The 3$\mu$A, of course, means "whichever is higher" (aka "worse"). So if your cap is rated at 10V/100$\mu$F, leakage would be less than 10$\mu$A.

SP's rule #1 of data sheet interpretation is:

If a spec can be interpreted in two ways, and one is worse than the other, the worse one is the correct way.

The actual leakage of an electrolytic cap may be much less than the rated value or a bit less. Chances are a higher voltage rated capacitor will have lower leakage when operated at a much lower than rated voltage, but it is not guaranteed, nor will it necessarily last if the capacitor is continuously operated at lower than rated voltage.

The (relatively) long time is, of course, because the initial leakage may be quite a bit higher than spec and it may take some time to drop down to the guaranteed value. This is because the dielectric in an electrolytic cap is actually a very, very thin oxide layer on the etched aluminum plates and it can develop pinholes etc. that are anodized away when voltage is applied.

Here is what United Chemicon has to say about leakage:

Leakage Current (DCL)

The dielectric of a capacitor has a very high resistance which prevents the ﬂow of DC current. However, there are some areas in the dielectric which allow a small amount of current to pass, called leakage current. The areas allowing current ﬂow are due to very small foil impurity sites which are not homogeneous, and the dielectric formed over these impurities does not create a strong bond. When the capacitor is exposed to high DC voltages or high temperatures, these bonds break down and the leakage current increases. Leakage current is also determined by the following factors:

1. Capacitance value
2. Applied voltage versus rated voltage
3. Previous history

The leakage current is proportional to the capacitance and decreases as the applied voltage is reduced. If the capacitor has been at elevated temperatures without voltage applied for an extended time, some degradation of the oxide dielectric may take place which will result in a higher leakage current. Usually this damage will be repaired when voltage is reapplied

A strong 'forming' effect of this type is relatively uncommon with modern parts, and seemed to happen a lot more often in olden days when parts were sitting for some time before being used. Maybe the modern electrolyte is better controlled or more pure, or has preservative additives.

Edit: Note @Dave's comment that the units of the 0.01 parameter must be 1/s.

• Excellent answer, thanks. This seems to tally with what I've seen in datasheets that explicitly provide the leakage current in microamps for each specific model, rather than providing a coefficient. – Polynomial Dec 3 '16 at 16:49

The leakage current depends on the area of plate (so is proportional to capacitance), or inversely proportional to the plate separation (so proportional to capacitance) and on the applied voltage, so yes, the leakage current is proportional to CV.

Electrolytic capacitors have an interesting 'long time constant' which is related to both mechanical movement at the plates, and polarisation effects in the electrolyte. It's most effectively demonstrated by charging a big electrolytic capacitor, leaving it for a few minutes, discharging it quickly, then watching its voltage over the next few minutes with a high impedance DVM. The voltage rises from 0, and can get to a surprisingly large fraction of the original charge voltage. This voltage-recovery experiment is worth doing, if only to demonstrate the non-ideality of an electrolytic capacitor.

What this means is that if we are trying to measure low leakage current in a large electrolytic, it will be swamped by the effects of voltage recovery following any change in voltage. Hence the specified 2 minute delay, which the manufacturer has presumably found sufficient to remove voltage recovery as a significant source of measurement error.