Just some insight: I am fifteen and have been researching capacitors for a while now.

I have come up with the idea of building a supercapacitor power supply unit solely to power some small electronics (likely 5 V at 500 mA at the absolute maximum).

I know this might be overkill, considering a few rechargeable AA batteries would do the same thing for around a few years, but I am looking at capacitors solely for the reason that they can be charged and discharged rapidly and can last for decades under the right conditions.

Now for the actual project itself. I plan on getting three supercapacitors and combining them for a total of 800 F at 8.4 V. I did 8.4 V because I want the first time constant of the capacitor to be directly be converted into 5 V, in theory lasting longer than a 5 V capacitor that would immediately drop voltage after a few minutes.

To regulate the voltage I plan on using a buck converter to output 5 V and some feedback LCDs. I am not entirely sure if it is even reasonable to wire a supercapacitor to a buck converter and for how long the capacitor might last under varying current draw. From what I have guessed the capacitor would slowly allow less and less maximum current, and I want a cutoff switch to account for the voltage/current drop. If anyone might have some insight on this project, let me know.

A few of the parts I have been looking at:

  • 3 x 800 F, 2.8 V capacitors
  • 1 x Power converter
  • 1 x cutoff switch (I don't know if the power converter already does that when voltage is below the operating levels)
  • 1 x charging circuit to output the voltage needed.

(Edit) I plan on changing the cap setup to 2.8 volts and 2400 Farad and hooking it up to a boost converter. In an ideal situation, the capacitors would now be parallel to each other, supposedly adding capacitance.

Here is my bad semantic trying to explain the circuit. Note that the capacitors are only 8 Fards because my simulator does not go any higher.

Circuit Diagram

  • 3
    \$\begingroup\$ LM2596 drops 1-1.5V at least, reducing your overhead; only the 8.4-6V range is usable, leaving much energy in the caps. It's also rather hungry, using 5-10mA all the time (perhaps more than a lot of your expected loads?). Assuming you have legit parts of course, which, it's one of many popular cloned/counterfeited parts, so beware what you find on eBay/Amazon/etc. The first problem is solved with a SEPIC type converter, and the latter with a newer more efficient regulator. \$\endgroup\$ Commented Oct 4, 2022 at 5:44
  • 3
    \$\begingroup\$ What capacitors you intend to connect and how? Three 800F caps in series is only 266F. \$\endgroup\$
    – Justme
    Commented Oct 4, 2022 at 5:49
  • \$\begingroup\$ My mistake, all the caps together are 800 farads. Unless I checked wrong I would add three 2.8 volt 800 F caps together in series to around 8.4 V. \$\endgroup\$ Commented Oct 4, 2022 at 6:16
  • 4
    \$\begingroup\$ 1. Caps in series are like resistors in parallel, you get less. en.wikipedia.org/wiki/Series_and_parallel_circuits#Capacitors 2. Caps in series will not necessarily share voltage nicely, you need additional circuits (resistors might work). \$\endgroup\$
    – Mattman944
    Commented Oct 4, 2022 at 7:01
  • 2
    \$\begingroup\$ Instead of using a Buck converter, consider a boost comverter: Buck will become useless once the cap voltage drops below 5V.. If you parallel the caps , they will share charge easily and will always be below 5 V, so you can easily use a boost converter. \$\endgroup\$
    – tobalt
    Commented Oct 4, 2022 at 8:31

1 Answer 1


$$V*C = Q$$ $$Q/I = t$$

V is the voltage, measured in Volts [V].

C is the capacitance, measured in Farads [F].

Q is the electric charge, measured in coulombs [C].

I is the current, measured in amperes [A].

t is the time period, measured in seconds [s].

A single ultracapacitor like a SkelCap SCA3200 is rated up to 2.85 volts at 3200 Farads, so it has 9120 Coulombs of charge or (2.85-1.5)*3200 = 4320 Coulombs before dropping below 1.5 volts where a boost converter may have a hard time boosting to the required current needs.

So, if the boost converter and the system load is about 200mA of current that 4320 Coulomb should last about 4320/0.200 seconds or 21600 seconds or 360 minutes or 6 hours or 1/4 of a day. Realistically with parasitic discharge (leakage current) and actual system draw I would think one is going to run about 300mA to 500mA of current and it will last about 3 hours or less on a single charge. To achieve a reasonable run time of about one week, one would need about 28 of them in parallel (costs around $1900 at time of writing).

  • 1
    \$\begingroup\$ Hm, 360 minutes (6 hours, 0.25 days) \$\endgroup\$ Commented Oct 4, 2022 at 23:53
  • 1
    \$\begingroup\$ I corrected the answer to remember minutes exists, oops. \$\endgroup\$
    – Motomotes
    Commented Oct 4, 2022 at 23:59
  • \$\begingroup\$ Fascinating, I plan to make my model modular so that I can add more caps in the future, but 3-6 hours is pretty good for what I am looking at. Thanks for the time to figure that out. This project is just so that I can run logic systems for prolonged periods of time, and I might change to a 3.3-volt system if that prolongs the time of the caps. I think that just about answers a few problems though. If anyone can recommend any particular parts that would be helpful, otherwise I think I can continue with this project and others to come. \$\endgroup\$ Commented Oct 5, 2022 at 17:04

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