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I have a 680uF electrolytic capacitor rated for 25 volts. Yesterday night I was curious as to how much current leaked out of it over time, so I charged it with a generic DC power supply, and measured the voltage across the two terminals. The voltage peaked at around 18.6V, and once the voltage began dropping slowly I quickly disconnected it. I then left it on my nightstand and went to bed.

9 hours later, I measured the new voltage - I got 16.8 volts. Curious as to how the capacitor behaved, I searched Wikipedia for modelling capacitors; I observed ESR and equivalent parallel resistance as basic approximations of capacitor behavior.

I chose to model the capacitor with ESR, using the series RC circuit differential equation solution \$V_t=V_0e^{\frac{-t}{RC}}\$

Substituting values, I got \$\large 16.8=18.6e^{\frac{-(3600*9)}{R(10^{-6}*680)}}\$. This equation yielded the solution R=49,097,000 ohms.

Is the following solution an accurate estimate of the internal resistance of the capacitor?

More importantly, is this a good model of a capacitor? (Would I be able to accurately estimate leakage for longer or shorter periods of time, such as 4.5 hours, or 18 hours?)

Thanks.

EDIT: If it would be helpful for me to mention the capacitor brand, it is a brown capacitor with a white rectangle logo and the letters "KZH" by this logo. I salvaged the capacitor out of a broken Dell AC/DC power adapter. I have found the capacitor's datasheet here.

UPDATE: As suggested, I have charged the capacitors and put them in a heated area to detect voltage drop. The capacitor I originally observed (which I had previously accidentally applied 6 volts in the wrong direction before both observations) dropped from 18.9V charged to 15.85V. The good capacitor I had went from 18.08V charged to 15.07. The temperature that these capacitors were at started at around room temperature but steadily rose to 57.5 degrees Celsius for most of the time. The significant rise in leakage is quite interesting. With the same parallel resistor model, I have calculated the leakage resistance of the original (supposedly damaged) capacitor to be around 931k ohms, and the leakage resistance of the other capacitor to be around 964k ohms.

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    \$\begingroup\$ The ESR of the capacitor is used when current is flowing out of the Capacitor. It is more accurate to model the leakage as a parallel/shunt current source, which itself might be voltage dependant. \$\endgroup\$ – placeholder Jan 20 '13 at 19:02
  • \$\begingroup\$ In other words without a datasheet there is no easy way to tell for sure how to model the leakage? \$\endgroup\$ – hedgepig Jan 20 '13 at 19:05
  • \$\begingroup\$ not in the least, modelling voltage over time will give you huge insight and is useful if you want to get down into the nitty gritty details. \$\endgroup\$ – placeholder Jan 20 '13 at 19:59
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You are confusing ESR, that stands for Equivalent Series Resistance, and the leakage. The first is modeled as a series resistor, and take account of leads resistance, leads-internal plates resistance and so on, and is ideally zero. The second is modeled as a resistor in parallel with the capacitor and takes account of small leakage currents in the dielectric, and is ideally infinity.

The formula you use is correct, but the value you come out with is NOT the ESR, is the leakage resistance. Once the capacitor is charged, if you leave it it slowly discharges trough the leakage resistor with a time constant \$R_{leak}\cdot C\$, so \$R_{leak}\$ is what you calculated, approximately \$50M\Omega\$, that is plausible.

To calculate the ESR you need to measure how long does it take the capacitor to discharge through a much smaller resistor, let's call it \$R_{dis}\$. When you discharghe the capacitor through \$R_{dis}\$ the total resistance through which it discharges is actually \$R_{dis}+R_{ESR}\$, so using the very same formula you used for the leakage resistance you can calculate the ESR.

But is it really that easy? Of course not.

The ESR is hopefully quite small, tenths of milliohms if you have a very good capacitor up to a few ohms. Since in the formula you have \$R_{dis}+R_{ESR}\$ you don't want an eccessive \$R_{dis}\$ to mask \$R_{ESR}\$. Ok then! Why don't we choose \$R_{dis}=0\Omega\$? Easy question:

  • \$0\Omega\$ resistance does not exist. But i can make it small!
  • Time. You need to be capable to measure how long does it take to the capacitor to discharge.

If you charge the capacitor to a certain voltage it will take \$\tau\ln{2}\approx0.7\cdot\tau\$ where \$\tau=RC\$. If \$R=R_{ESR}+R_{dis}=1\Omega+1\Omega=2\Omega\$ and \$C=680\mu F\$ that's less than 1ms. Without proper equipment, that is a properly set oscilloscope, you can't easily measure the ESR.

Last but not least, keep in mind that electrolytic capacitors values have a tolerance of \$\pm10\%\$, that leads to: $$ R_{ESR}=\frac{t_{dis}}{\left( C\pm C/10\right)\ln{2}} - R_{dis} $$ with the above numbers, t=1ms, C=\$680\mu F\$, \$R_{dis}=1\Omega\$, this translates to: $$ R_{ESR}\in\left[0.91,1.33\right]\Omega $$ That's 10% down and over 30% up.

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  • \$\begingroup\$ Thank you for distinguishing the difference between ESR and leakage resistance. You are saying to measure the ESR I would have to short the two terminals but also pass it through equipment capable of measuring the current time graphs, i.e. an oscilloscope? \$\endgroup\$ – hedgepig Jan 20 '13 at 21:07
  • \$\begingroup\$ Exactly. You should discharge it through a resistor of known value and plot the voltage across it, and finally compute tau from the graph. \$\endgroup\$ – Vladimir Cravero Jan 20 '13 at 22:45
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Your capacitor is very likely a Nippoon Chemi-Con KZH. The data sheet says:

I = 0.01 CV or 3 µA, whichever is greater.

Where, I: Max. leakage current (µA), C : Nominal capacitance (µF), V : Rated voltage (V)

It is common to specify the leakage characteristics of a capacitor in current and not by a parallel shunt resistance.

Using the capacitance and the leakage current, you can find out how much the voltage will decrease after a given time.

Note that the datasheet will likely give you a leakage specification way worse than you may observe with your real part. The reason is that good manufacturers want to make sure that all their parts are really within the specification, so most parts will be far better than the specification - especially true for things like leakage.

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  • \$\begingroup\$ Thanks for the clarification - essentially I would need more data with my actual capacitor to get accurate figures? \$\endgroup\$ – hedgepig Jan 20 '13 at 21:03
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Accurate leakage figures for an electrolytic are probably impossible to model. Leakage will depend on its age, and its history of charging and discharging and the operating temperature.

Quite a good topic for experiment though. Place it somewhere warm (say 50C) for an hour and see how the leakage compares.

To see something freaky, discharge it to 0V momentarily, and watch the voltage across it afterwards...

If you are happy to sacrifice it in the name of science, charge it wrong way round to 1V or so via a high resistor for just long enough to measure the leakage current.

Charge it to 18V again afterwards and watch the charge current; see if the leakage has changed.

And update the results here...

Some of the story is here

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  • \$\begingroup\$ Interesting. The capacitor I'm using is already partially damaged, from me accidentally passing 6 volts across the terminals in the wrong direction. I do however have another capacitor of the exact same model, which should have similar wear to this one. I've charged both capacitors up and have left them on my laptop's power supply, and put a pillow to cause it to heat. I'll try to update results. \$\endgroup\$ – hedgepig Jan 21 '13 at 16:29
  • \$\begingroup\$ Pillow won't warm it, try a central heating radiator \$\endgroup\$ – Brian Drummond Jan 21 '13 at 20:01
  • \$\begingroup\$ Oh, I just put a pillow on top of my 150 watt laptop power supply. It definitely heated up (hot to the touch); I measured the temperature with a candy thermometer. \$\endgroup\$ – hedgepig Jan 21 '13 at 21:36

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