Why do electrolytics have a different charge and discharge rate for the same current

Question

I am measuring the charge and discharge rate of a HV electrolytic capacitor (actually three 560uF, 250V capacitors in series) to calculate the capacitance. Note: The capacitors are hand picked for an effective combined AC capacitance of 155uF.

During charge I source 20mA and measure (digitize) from 35V to 695V. During discharge I sink 20mA and measure (digitize) from 695V to 35V.

The curves are really flat (the charge has a very slight curve).

Computed capacitance (20 runs) using C=\$I / {dV\over dt}\$ is:

• Charge - 166.13uF +/-0.7uF (2nd order polynomial fit)
• Discharge - 185.16uF +/-0.08uF (linear fit)

Why is the discharge capacitance consistently 11% higher?

Note: Reforming is not an issue as the test was repeated 20 times in quick succession with consistent results.

Test is performed by sourcing the current with a Keithley 2410 SMU. Voltage is measured using a calibrated 1000:1 divider into a waveform digitizer. The voltage divider is buffered and has a 10M input impedance.

Here are the charge and discharge curves.

Answer 1

Two suggestions:

The curves are really flat (the charge has a very slight curve).

The charge has a curve because your caps are unbalanced. One of them has lower capacitance, it charges faster than the others, then it hits breakdown voltage and begins to leak. When discharging the issue does not exist. Thus charge is curved, but discharge is not.

How to distinguish this from soakage/dielectric absorption: charge, then wait monitoring current, if dielectric absorption is not involved then current will be negligible.

Suggestion 2 is I'm wrong ;) in this case I'm interested in the real answer!

Answer 2

So ultimately The problem was with the curve

In both cases (charge and discharge), the overall rate is very similar. The problem is that there is a very slight curve in the case of the charge. This is pretty much expected. As the voltage increases, the current required to feed the leakage resistance increases thus stealing more and more current from the capacitor and thus reducing the \$ {dV\over dt}\$. Since I am doing a 2nd order polynomial fit and using the 1st order term for the capacitance calculation, this is equivalent of using the \$ {dV\over dt}\$ at t=0 which is where it is highest. This does make since this is where leakage has the least effect. Since C=\$I / {dV\over dt}\$, I will get a generally lower value of capacitance.

Here are the details of the curve fits

• Charge

2nd Order: y = -1.941x2 + 120.99x + 25.775 => C=165.3

1st Order: y = 109.07x + 37.977 => C=183.4

• Discharge

2nd Order: y = 0.6011x2 - 111.69x + 693.51 => C=179.1

1st Order: y = -107.97x + 689.66 => C=185.2

This explains the reason for the different calculated values; but, it does not explain why the curve does not show up on the discharge as well.

Note: On review, I realized that if I decide to do a 2nd order fit for the discharge I will need to make sure to use \$ {dV\over dt}\$ at y=0 which for the discharge is not t=0 and thus is not the 1st order term.