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.

Charge WaveformDischarge Waveform

  • 1
    \$\begingroup\$ Too confused by what you are actually measuring and how you are doing the math here to comment. \$\endgroup\$
    – Trevor_G
    Commented Aug 28, 2017 at 20:01
  • 1
    \$\begingroup\$ results indicate you are discharging with more energy than you put in as C increased on discharge, suggesting error somewhere. \$\endgroup\$ Commented Aug 28, 2017 at 20:50
  • 3
    \$\begingroup\$ How do you generate the constant current source and sink? Could it be possible that the two directions have 11% difference? Handling the 695 V level is not an easy feat... \$\endgroup\$ Commented Aug 28, 2017 at 21:16
  • 1
    \$\begingroup\$ Did you try your experiment with just a resistor, same during charge, and discharge? \$\endgroup\$ Commented Aug 28, 2017 at 21:21
  • 1
    \$\begingroup\$ Possibly @markshancock, unfortunately, as a question on here, it is impossible to comment with any degree of certainty on your results when the math involved, with whatever errors are introduced due to measurement cycle lag etc. is indeterminate. We cant compare apples and oranges. Either way, something is off if your results are stating you are getting 10% more charge out of them than you are putting in. If you can productize that phenomenon you will be rich indeed. \$\endgroup\$
    – Trevor_G
    Commented Aug 29, 2017 at 19:55

2 Answers 2


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!

  • \$\begingroup\$ +1 Do you know what is the spice model for overvoltaged electrolytic cap? I'd like to see this scenario in simulations.. \$\endgroup\$ Commented Aug 28, 2017 at 22:50
  • \$\begingroup\$ I don't know and I'm pretty sure if I asked the manufacturer for a model of a part used outside its specifications they would ask... what the wat? So, no idea. All I can tell is when I over-voltag-ed some caps smelly things happened... and on other times it worked fine. Wouldn't trust my life on it though.. \$\endgroup\$
    – bobflux
    Commented Aug 28, 2017 at 22:58
  • \$\begingroup\$ Would advise the OP to check the voltage on each cap during charge compared to ratings. I mean, I rode a bicycle downhill 80 km/h. Maybe more, I didn't have time to look at the speedo due due to trees and shit blurring past at insane speed. That's a bit above ratings. Really fun to do... but I wouldn't advise it for, like, everyday use, y'know. \$\endgroup\$
    – bobflux
    Commented Aug 28, 2017 at 23:06
  • \$\begingroup\$ Working voltage for each cap is 250V and surge is 300V. I am staying below both. Capacitors are hand selected to have closely matched capacitances (supposedly within 2%) - I am sure they are within 10%. \$\endgroup\$ Commented Aug 28, 2017 at 23:13
  • 1
    \$\begingroup\$ This test does not monitor each capacitor; but, I have another that does. Typically the voltages across the capacitors differ by <5V. \$\endgroup\$ Commented Aug 28, 2017 at 23:18

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.