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I have recently received my order of 2 2.7V 500F super capacitors from eBay.

After much searching, my yet unanswered question is: Can I connect both of these capacitors in series and charge with 5V running absolutely no risk of explosion or damaging the caps?

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    \$\begingroup\$ Please don't put your contact information on here... \$\endgroup\$
    – user103380
    Commented Nov 5, 2019 at 4:09
  • \$\begingroup\$ Search for capacitor grading resistors? \$\endgroup\$
    – winny
    Commented Nov 5, 2019 at 8:36

3 Answers 3

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Unfortunately you cannot just connect them in series because of two effects. The first is the tolerance in the value of the capacitance. A +-20% variance is normal in capacitors (it could be bigger or smaller depending on the specific model). If one of your capacitors is 500*1.2=600F, and the other is 500*0.8=400F, then the voltage across the first will be 2V and the voltage across the second will be 3V, which will damage it and or make it explode.

The other effect is the leakage current. All capacitors have a leakage current, and on supercapacitors it can be quite large. If one capacitor leaks more than the other, which is pretty much guaranteed to happen, then the voltage on the one that leaks less can go up, possibly going above 2.7V and damaging the capacitor.

You could use resistors to deal with these issues, but the current required may be quite large and unacceptable (usually it should be at least 10X the maximum expected leakage current). You could also implement some active circuit, and there is more info here

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  • \$\begingroup\$ The problem is that I don't have a bench DC power supply. Only 5V from USB and Arduino and 3V3 from Arduino \$\endgroup\$
    – user235520
    Commented Nov 5, 2019 at 12:10
  • \$\begingroup\$ If you're talking about doing this manually, you can just use a resistor divider to generate 2.5V. It will be very inefficient, but that doesn't really matter if all you're trying to do is charge these up manually \$\endgroup\$
    – BeB00
    Commented Nov 7, 2019 at 4:34
  • \$\begingroup\$ Something like this? images.app.goo.gl/bxgRzp9adzTocc6v6 \$\endgroup\$
    – user235520
    Commented Nov 7, 2019 at 14:06
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If you connect them in series you have to ensure the voltage is divided evenly at all times so it never goes over 2.7V for each one. Normally this is true just by regular series connection, but if one shorts out or something alike then you will end up applying 5V the other capacitor.

schematic

simulate this circuit – Schematic created using CircuitLab

edit: I added the schematic of the circuit, now the values of the R are not really well thought off, but if you can switch them out once the voltage of your capacitors is close to 2.2V then their value does not matter much as long as they are equal, it is better if they are large so it doesn't load your source

I don't know what is your final design, but if you can connect a large resistor in parallel to each capacitor, they would help to ensure the voltage stays somewhat even in case anything happens. You only need them during charging so you could add a switch to each resistor to cut them out of the circuit so they dont discharge your capacitors after you turn off your voltage supply.

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  • \$\begingroup\$ Could you explain how to do that? Or add a schematic diagram? \$\endgroup\$
    – user235520
    Commented Nov 5, 2019 at 12:15
  • \$\begingroup\$ Theoretically, can I connect these super caps to 20V (only 2 caps) in parallel, and disconnect before either cap reaches 20V? \$\endgroup\$
    – user235520
    Commented Nov 5, 2019 at 21:56
  • \$\begingroup\$ I will edit in a schematic of what I mean in my post, I would not connect these caps directly to 20V under any circumstance. You will have a much easier time making a circuit that yields 2Vs rather than finding something that disconnects the capacitors before they reach 2V while connected to 20V \$\endgroup\$
    – Juan
    Commented Nov 7, 2019 at 13:05
  • \$\begingroup\$ And if I do it manually? \$\endgroup\$
    – user235520
    Commented Nov 7, 2019 at 17:05
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    \$\begingroup\$ This assumes that (1) the capacitance values of C1 and C2 are very similar (if not, the initial voltage will not divide equally), and (2) the leakage currents are similar (if not, over time the voltages will drift until either the leakage currents equalize or something goes 'bang'). \$\endgroup\$
    – Technophile
    Commented May 27, 2020 at 3:41
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Depending on charging rate, overall power consumption (is this AC line powered or e.g. solar charged battery), active clamps might be appropriate. Either two independent circuits that measure the voltage across a supercap and turn on an appropriate resistive load across that supercap if it gets close to 2.7V, or a single circuit that measures the difference in supercap voltages and can turn on a load across one or the other.

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