Voltage drop due to leakage current in capacitors in series

Question

When capacitors are connected in series in a DC circuit, the voltage drop across individual capacitors at and immediately after the initial charging period is inversely proportional to the individual capacitance of each capacitor. But afterwards, this begins to change due to leakage current.

1. How does this affect individual voltages across each capacitor?
2. Will this process eventually result in the capacitor with the lowest leakage current gaining a voltage drop nearly equal to the voltage drop across all of the capacitors in series?
3. Or will this situation eventually lead to voltages across each capacitor being entirely dependent upon the leakage currents (assuming each cap has a similar external load/or no load whichever is suitable for this assumption?) and whatever the capacitance of each cap does not affect the final voltages?

Answer 1

A simple model for a leaking capacitor is to consider an ideal capacitor in parallel with a leakage resistor, as bellow :

^{simulate this circuit – Schematic created using CircuitLab}

Based on this assumption, in steady state (ie DC), the ideal capacitors behave like open circuits, and we just have a voltage divider based on R1 and R2, and therefore Vout = Vin * R1/(R1+R2)

So the final voltage depends only on the leakage resistors of each capacitor.

So to answer your questions :

1. Yes, leakage current (or resistance) will afect the final voltage seen by each capacitor (U_C1 = Vin * R1/(R1+R2) ; U_C2 = Vin * R2/(R1+R2) )
2. No, you will not have one capacitor with nearly all the voltage, excepted if they have a completely different leakage resistance. Expect a relative difference in the same order of magnitude than the relative difference in leakage resistance (which might be tightly specified, loosely specified, or not specified at all)
3. Indeed, the capacitance of each capacitor will not mater for the final voltage (but will during transients). It might have some indirect influence : for a given manufacturing process, the leakage resistance will vary with the capacity, usually in inverse proportion (higher capacitance = bigger area or thinner insulation layer).

If you care about keeping the same voltage on each capacitor (for example because otherwise you exceed the maximal voltage ratings), then just add yourself resistors in parallel to the capacitors (with smaller values than the leakage), so it is your voltage divider that will fix the voltage on each capacitor in steady state (nb : it will increase the global leakage, so make sure to use low leakage capacitors, so you can use resistors with high resistivity)

Answer 2

1. Voltage over each capacitor will drift to match the leakage current. If both capacitors leak identical current at identical voltages, then 200V applied to capacitors stay at 100V per capacitor, if leakage currents at 100V are different, the voltages over the capacitors drift to a value where leakage current is equal. The voltages could be anything.

2. No, because it is relative to leakage current, only if the capacitor has no leakage then it will have all the voltage, but if you have capacitor with leakage current X and the other has 10X, it will be proportional, assuming leakage currents can be modeled as resistances (say 1Mohm and 100kohm).

3. Capacitance plays no role, the DC bias currents, i.e. the leakage resistance, will determine the final voltage in any RC system.