Hallo Mahmood,

the question about the internal resistance of an electrolytic capacitor is the question about the leakage current. Leakage current depends on many different influences like temperature, applied voltage, quality of fabrication, etc. Nevertheless, there's a way to make a helpful estimation, as we will see later.

First, why is leakage current so bad?
Have a look at the following equivalent schematic of a series circuit of four 'identical' capacitors:



As you can see, leakage current is represented by a resistance connected in parallel to C. Because leakage current of identical capacitors (which have identical capacitance) can be very different, in the schematic four different resistances are distinguished, Ris1 to Ris4. Even without calculation it's clear, that just the good capacitor, presenting very low leakage current is the one which has the highest risk of undergoing a voltage overrating. Because for this capacitor, Ris is highest and highest voltage drop will occur.

For preventing damage we must guarantee, that all capacitors used in this series circuit must show equal leakage current. Then voltage drops will also distribute equally. Unfortunately, leakage current of electrolytic capacitors cannot be predicted as good as needed, that for all situations and future times leakage current of those four capacitors is nearly identical. It must be forced, by the additional parallel connection of symmetrizing resistors, namely Rsym. If we keep them so small, that current through them is about ten times higher than worst leakage current, distribution of voltage drops can be made equally enough.

How Rsym is calculated?
According to DIN41240 for a 'long life grade' electrolytic capacitor operational leakage current (which is the leakage current observed after about 1hour of permanent operation) is less than

Il = 0.005A x C x U / F / V

For a 'general purpose' type about 4 times higher leakage currents are observed. Given data is only valid for 20°C! At 40°C leakage current is twice, at 60°C leakage current is 5 times higher and at 85°C leakage current is about 10 times higher. U is the rated voltage of capacitor, F is Farad and V is Volt.

If we demand the current Ir through Rsym to be ten times greater than maximum leakage current Il, we get:

Ir = U / Rsym = 10 x Il = 10 x 0.005A x C x U / F / V

This yields:

Rsym = (F / 10 / 0.005 / C) Ohm

An example: 10µF/450V 'LL' grade capacitor needs Rsym <= 2MOhm at 20°C. At 60°C Rsym must be smaller than 400kOhm.

In the above estimation it was assumed, that voltage dropping across electrolytic capacitor is about rated voltage, means in our example about 450V. If we increase the number of capacitors connected in series, we can highly reduce leakage current: If voltage drop is only 50% of rated voltage, then maximum operational leakage current is only about 10% of Il! So, if we connect 8 capacitors in series, all 22µF/450V 'LL' grade types, Rsym becomes (F / 10 / 0.0005 / C) Ohm = 9.1MOhm. This is valid under the assumption, that temperature is 20°C and maximum voltage drop across any 22µF/450V capacitor is less than 225V.

We can now compare both versions: Four 10µF/450V 'LL' grade capacitors in series make total capacitance of 2.5µF, paralleled by 4 x 2MOhm = 8MOhm. Eight 22µF/450V 'LL' grade capacitors in series make total capacitance of 2.75µF, paralleled by 8 x 9.1MOhm = 73MOhm. That's a nice improvement, isn't it?

Nevertheless, I highly recommend to choose enough safety margin! So, I would use some lower Rsym values than calculated. And I would try to get the best fabricated 'LL' grade capacitors available...


Good luck,
Kai