I have procured some supercaps. No datasheet was provided. The only data I have is what is printed on the caps themselves: "4.0F 5.5V cda®"

Since my DMMs don't measure capacitance in Farads, I setup the following circuit shown below. Two "10Ω 1% 2W" (20.2Ω and 20.3Ω measured) caps in series with the cap.


simulate this circuit – Schematic created using CircuitLab

Note that I did not actually use a switch, I simply connected a banana plug. I did that to avoid extra variables (switch resistance and power rating, as well as power supply startup time).

That being said, with a large expected time period of 80+ seconds, I figured that using a stopwatch and monitoring my DMM would suffice.

The RC time constant is: R*C = (20.3Ω)*(4) = 80.2 seconds

Which I take to mean that if I apply 5V as shown in the circuit and close SW1, the cap should reach 5V * 0.632 = 3.16V at 80.2 seconds. The current limit on the supply was set to 2A (more than enough).

The cap went from 0V to 3.16V in approximately 38 seconds, at which point I removed the cap from the circuit.

Solving for capacity: C = (38 seconds)/(20.3Ω) = 1.87 F, only 47% of the 4F labeled!

About a minute after being removed from the circuit, the voltage on the cap had stabilized at about 1.28V. Should I be using this value instead? That would suggest 6.43F, so I'm guessing "No".

I then tried the same test with another cap of the same specs... same result.

I next tried a discharge test, going from 4.8V to 1.64V. That should have taken 87 seconds, but instead took only 28 seconds, hinting at a capacity of only 1.27F.

However, by leaving the cap to discharge for 55 seconds showed 1.1V, suggesting a capacity of 1.82F. That's odd to me because it means that it's not following the predicted curve. And that would mean that I will end up calculating a different capacity depending on the time I record at. But that shouldn't be.

The following image is from hyperphysics.phy-astr.gsu.edu:

Capacitor charge time graph

I'm wondering what margin of error I should expect from a test like this. Assuming my multimeter is calibrated beyond practically needed, and that recorded times are valid, is it possible that the capacitance is closer to 4F?

  • 3
    \$\begingroup\$ Some supercaps have high ESR, in the tens of ohms. So model the C as a C and R in series, and find the R. For example, charging from 0V, does the voltage increase instantaneously to 1.5V then build slowly to 3.16V? That instantaneous voltage together with charging voltage, R1,R2 values will give a measure of the ESR. Then maybe the rest of it will make more sense. \$\endgroup\$ – Brian Drummond Feb 5 '17 at 15:48
  • \$\begingroup\$ Is the test method not given by the manufacturer? These are double layer caps So C is complex result of 2 caps in parallel wih the absorption capacitance. try an 1 hour time constant or a different method using frequency and current. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 5 '17 at 15:49
  • \$\begingroup\$ @TonyStewart.EEsince'75 - I have updated my question to inlude the branding on the caps. The only data I have is what is printed on the caps themselves: "4.0F 5.5V cda®". I don't even know if that's a real brand. I can't find any more info on these caps. \$\endgroup\$ – Bort Feb 5 '17 at 15:57
  • 1
    \$\begingroup\$ I ran into your exact same problem. It's called soakage. I have seen large electrolytics do the same thing. Robert Pease dealt with it. See m.electronicdesign.com/analog/whats-all-soakage-stuff-anyhow \$\endgroup\$ – Lance Beasley Feb 5 '17 at 19:23
  • \$\begingroup\$ @Bort. Try the soak test for 1hr and measure my way and record dv every few seconds to compute changes in C down to 2.5v and pls report back although it is better to use LM317 as a load with current sense and output shorted. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 7 '17 at 13:40

"Simple" Solution to get close to Maxwell 6 step method.

Use Ic = C/10 [A] or Ic=CdV/dt or dV/dt= 0.1 V/s test slew rate with 3 Ohms approx for duration after charging for one hour and discharge to 50% for the test.


simulate this circuit – Schematic created using CircuitLab Simulation of above with estimated values, not actual to show lag effects.

  • i.e. dV/dt=Ic/C = C/10/C = 1/10 V/s for Supercaps after soaking for 1 hr to < 50% drop , e.g. try 2V drop in 20 seconds , use ~3 Ohm series R after full charge for 1 hour.
  • Alternate measuring voltage drop on 3 Ohm series R and actual Cap voltage at regular time intervals and to compute slope C = dt*Ic/dV and note that C changes from initial to final voltage and effective value is the average.

Activation of the absorption layer is a key pre-requisite to this test.


Somewhat like batteries which have a useful range from Vi to Vf, "Double-layer Effect" Supercaps must be charged near full Voltage for a sustained time and full energy stored is obtained by this process. It is not a 100% efficient process, but unlike batteries can be repeated a million cycles within the C/10 current range.

Therefore to get maximum useful energy storage time ought to be in a similar manner as batteries from full to 50% initial voltage. Otherwise the C value is effectively reduced.


simulate this circuit

This is a suggestive approach to problem solving not a complete answer.

The correct method is to use the same method used in the datasheet with guaranteed values.

C2 is often called Dielectric Absorption or the "Double-layer Capacitor".

Maxwell's documented test method.

Using dV/dt=Ic/C for Ic=C/10 with I [A] and C in [F]

enter image description here

  • • Short circuit the cell at least one hour before the test
  • • Rest cells for more than 4 hours between different tests

  • "The results of the second cycle are used to calculate the capacitance and ESR with the following formulas. The first cycle data are not used because the cell has not been activated and the measured capacitance and ESR values are different compared to the second cycle and the cycles after." Maxwell

Suggested test method to follow Maxwell's approach.

Follow 6 Step Process

  • using test design I made for you below.
  • Step 1 Safely Discharge Vc to 0V

    • then short for 1 hr.
  • Below for Step 2,3,4 of 1st cycle

    enter image description here

  • Below for Step 5,6 of 1st cycle.

enter image description here

| improve this answer | |
  • \$\begingroup\$ I see a circuit, I don't see what I'm supposed to do with it though. \$\endgroup\$ – Bort Feb 5 '17 at 16:00
  • \$\begingroup\$ Do you know any methods for testing two time constants in frequency domain or time domain from your studies? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 5 '17 at 16:08
  • \$\begingroup\$ Maxwell website indicates a different test method using constant current. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 5 '17 at 16:15
  • 1
    \$\begingroup\$ I had no expectations. No disrespect but that seems irrelevant. My question is about finding the capacitance. Imagine my question as "I have a cap with nothing but a voltage rating. How do I find the capacitance?" I imagine that is something I could do, regardless of a datasheet, provided that I have suitable equipment. \$\endgroup\$ – Bort Feb 5 '17 at 16:34
  • 3
    \$\begingroup\$ Should be noted, this isn't just Maxwell's test procedure, it's the standard laid out in IEC62391-1. \$\endgroup\$ – Matt Young Feb 5 '17 at 19:19

Your approach would be valid for ideal components, but I think you are not getting expected results because of parasitics, such as ESR of the super cap, and internal resistance of the power supply. When you ran the test, did you happen to measure the actual voltage at R1 on the supply side? E.g. if you used long leads from the supply to the circuit, there may have been some voltage drop that resulted in voltage less than 5V, which would affect the charge curve.

Another factor is the dielectric absorption of the cap, which you observed seeing the voltage of the cap drop after being removed from the circuit for some time. That's going to throw off your results too.

To get more accurate measurements that are not as affected by these non-ideal factors, try using lower charge current (i.e. higher series resistance).

| improve this answer | |
  • \$\begingroup\$ Activation of the absorption layer is a key pre-requisite to this test . \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 5 '17 at 17:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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