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I am attempting to replace a wall plug power supply with a supercapacitor. From analysing the behaviour of the wall plug I can see that the current drawn is .25A with 800ms spikes of 1.5A (as determined by an arduino board).

However when I try to use the supercapacitor (15F, 7VDC), the voltage appears fine, but not enough current is being drawn. The highest current I have seen drawn on the multimeter is .11A, so the system in question does not work.

I think this may be down to the internal resistance of the supercapacitor. So the next step I will try is to increase the number of supercapacitors and place them in parallel, thereby decreasing the resistance.

However as supercapacitors are advertised to be capable of charging in a matter of seconds, surely this means they can handle a high current (at least 1.5A anyway). Or will supercapacitors draw more current during charging than they would be capable of supplying during discharge?

Any help here is greatly appreciated.

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    \$\begingroup\$ The internal resistance (ESR) comes in affect when you draw current from the capacitors and will drop voltage (V=IR) depending on the current and the ESR. When you say the voltage appears fine, how are you observing at it? \$\endgroup\$
    – R.Joshi
    Mar 29, 2018 at 11:38
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    \$\begingroup\$ What is the output voltage of this "wall plug power supply"? \$\endgroup\$
    – Olin Lathrop
    Mar 29, 2018 at 11:42
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    \$\begingroup\$ In the datasheet of the capacitor, you should find the ESR specified. For a Vishay MAL219691205E3 which has similar specs than yours, the ESR at DC is 12.5Ohm. Supercaps have high ESR, and if (as I suspect) you bought a cheap one on ebay, it is probably much worse than this (but you most likely can't tell, since there's no datasheet provided at all, is there?). So that is probably your problem, indeed. \$\endgroup\$
    – dim
    Mar 29, 2018 at 11:45
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    \$\begingroup\$ @R.Joshi The voltage across the cap and at the load where the current is needed appears as it should. The output voltage of the wall plug is 5.5VDC, so I wish for the supercapacitor to replace that. \$\endgroup\$
    – TRJ94
    Mar 29, 2018 at 12:04
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    \$\begingroup\$ @dim, you may have found my answer. The supercapacitor is a Vishay and is indeed similar to the one you have found. The ESR of the one I have is 12.5ohm, so at a low voltage this makes sense that the current is limited. \$\endgroup\$
    – TRJ94
    Mar 29, 2018 at 12:07

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This makes little sense. A power supply plugged into wall power can supply 250 mA indefinitely. A capacitor has a fixed amount of charge, and a even smaller amount of charge it can deliver usefully. In addition, the voltage on the capacitor goes down over time, since it is proportional to the charge in the cap at any one time.

You can't just "replace" a power supply with a capacitor. The two do different things, and are not equivalent.

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  • \$\begingroup\$ I understand this, however the supercapacitor should be able to provide the power for a period of time, even if just for a number of seconds, right? \$\endgroup\$
    – TRJ94
    Mar 29, 2018 at 11:37
  • \$\begingroup\$ They may not be equivalent (I fully agree), but one of these might do the job for a while... ioxus.com/files/6214/6981/0438/… \$\endgroup\$
    – Peter Smith
    Mar 29, 2018 at 14:05
  • \$\begingroup\$ "you can't just "replace"" was the point i think. \$\endgroup\$
    – R.Joshi
    Mar 29, 2018 at 14:37
  • \$\begingroup\$ @TRJ94, Re, "provide the power...for a number of seconds." That depends on what you mean by "provide the power." The wall-wart most likely is designed to supply a constant voltage. A capacitor can not do that for any length of time. The voltage of the capacitor is proportional to its charge. It will start falling immediately as you deplete the charge. The amount of time the capacitor can provide useful power to your circuit depends on the range of input voltages that will allow your circuit to work. \$\endgroup\$
    – Solomon Slow
    Mar 29, 2018 at 15:15
  • \$\begingroup\$ Look at R. Joshi's suggestion to use a DC-DC converter between the capacitor and your device. \$\endgroup\$
    – Solomon Slow
    Mar 29, 2018 at 15:17

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