Capacitor leakage calculation with a 10k resistor in series

[gkmaia]

I am trying to figure out how to calculate the leakage of cap using the method suggested by Xavier60 but got stuck on how to actually put that down to an equation and get the current leakage value.

"The capacitors need to be tested for leakage current at high voltage. Use any safe voltage for a start.
Measure the voltage drop across a series 10K resistor and calculate the leakage current."

I toke a 470uf 16v electrolytic that has leaked so I know it is faulty. If you look at the images before the resistor I am getting 16v. After the resistor just 13.8v.

There is a voltage drop and the capacitor is working as a voltage divider and current is flowing through it. If I were to build a divider replacing the cap with a resistor it would be at the order of 60kOhms and allowing 220uA fo flow in order to provide a voltage drop of 2.2volts.

Am I looking at it completely wrong? What is the right way to calculate the leakage?
Does that mean there is 2.2 volts leaking through the capacitor direct to ground?

[StillTrying]

Yes at 13.8V the cap leakage is 220uA, equivalent to a 62.7k resistance.

"Does that mean there is 2.2 volts leaking through the capacitor direct to ground?"

No? The cap's acting like a resistor, the missing 2.2v is across the 10k because of the 220uA flowing through it.

[bsfeechannel]

First off, when we say leakage we are referring to the leakage current that flows through the capacitor dielectric.

Electrolytic capacitors have a liquid (actually a gel) as a dielectric, the electrolyte, that can leak and produce among other nasty effects a leakage current.

However paper capacitors do not leak their dielectrics. Its paper. But paper attracts moisture and with time the dielectric that should be a good insulator becomes a resistor.

The leakage of currents should not be confounded with the leakage of electrolytes.



So, what you'll measure is the leakage current as indicated in the picture below.



Other dielectrics also suffer from degradation with time if submitted to high voltages, high temperatures, etc. and they'll all become an undesirable path to DC, instead of blocking it.

[gkmaia]

You guys rock! Thanks!

I did test a brand new Aliexpress cap and voltage drop went from 16v to 12v after the resistor... when compared to my leaky Elna that dropped from 16v to 13.8v. And when compared to a standard Nichicon the drop went from 16v to 15.98v. Good to see how bad these Aliexpress caps are.

I imagine now I need to compare the amount of current that is flowing through the cap to the allowed Leakage Current in the datasheet. If it is above that then the cap may have an issue.

For example the Nichicon datasheet says no more than 4uA. Comparing tests.

Leaky Elna - 200uA
Aliexpress cap - 411uA
Good nichicon - 7.96uA

Will the same logic apply to ceramic and film caps?

[bsfeechannel]

You guys rock! Thanks!

No wuckers.

For example the Nichicon datasheet says no more than 4uA. Comparing tests.

In my understanding, the datasheet says that the leakage current is no more than 0.03CV or 4 (µA), whichever is greater for capacitors rated between 6.3 and 100V.

So, if you have, say a 470µF x 16V cap, the leakage current will be no more than 0.03 x 470 x 10⁻⁶ x 16 A = 225.6 µA. It will be equal or less.

But if you have instead a 4.7µF x 16V cap, using the formula will give you 2.246µA, which is less than 4µA. This means that in this case the leakage current for the rated voltage can be higher than 2.246µA but no more the 4µA.

Remember that you have to apply the rated voltage across the capacitor, not across the capacitor in series with the 10k resistor, and wait for at least one minute. The rated temperature also must be taken into consideration.

Will the same logic apply to ceramic and film caps?

Ceramic and film caps have other specs.

Kemet commercial grade high voltage ceramic caps have a leakage resistance of 1000/C (Ω) or 100 GΩ, whichever lower. So, for a 100nF, you should have 1000/(100 x 10⁻⁹) Ω = 10 GΩ (which is lower than 100 GΩ).

To measure it, you have to apply 500V across the capacitor for at least two minutes @ 25°C. If you use the 10k resistor technique you should read a voltage no more than 10kΩ x 500V/10GΩ = 500µV across the resistor (which corresponds to 50nA).