I’m building a simple garage door circuit that have two solenoid locks. Each solenoid lock is rated at 12V/3Amps and is only supposed to be activated during 3 seconds each time the “open door” button is pressed.

I'd like to use a power supply that can feed 12V/4.5Amps, so it will not be able to provide the required current for both solenoids.

I was wondering if I could add a (big) capacitor to the Power supply output that could supply the extra energy during these 3 seconds, so I wouldn’t need to buy a new power supply. Basically this capacitor would have to provide (rough figures) 12V*1.5A=18W in 3 seconds. I could easily implement a protection delay in the button so that the solenoids are not activated before the capacitor re-charges.

Would this be possible? How could I calculate the value of the capacitor?



Thanks for all the comments. This was question was a mix of real need and a theoretical exercise. I could understand its theoretically/technically possible, but not really feasible/economical due to the high value of the capacitor, afecting its price.

From a CAPEX (investment) point of view, I could understand that investing in a 12V/10A power supply could be the cheaper option. However, from a OPEX (operating expenditure) perspective would that still be the case?

My point here is that having such a powerfull power supply turned on, standing-by, 24/7 just to activate a pair of solenoids a couple of times per week would have a small but continuous electricity cost that could change the business case on the long-run. Even though the 10A aren't being drawn, the residual consumption would be bigger than a smaller PSU, right?

What I'm evaluating now is to use the original circuit to activate a relay instead of the solenoids, and that relay will turn on a bigger power supply that is directly connected to the solenoids. This way, the PSU is turned off most of the time. This seems a good approach, but I'm concerned because the PSU will "boot" everytime with a huge load - not sure if it wouldn't hurt it.

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    \$\begingroup\$ Cant you power 1 solenoid at a time, rather than both together? \$\endgroup\$
    – HandyHowie
    Commented Jan 25, 2018 at 10:17
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    \$\begingroup\$ You might be able to cheat by observing that the maximum current will be during the short pull-in time, followed by a lower current during the hold-in. Solenoid datasheets might help us with this. \$\endgroup\$
    – pjc50
    Commented Jan 25, 2018 at 10:32
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    \$\begingroup\$ 3 Amps for 3 seconds is 9 coulombs. If you're OK with the capacitor voltage collapsing from 12V to 9V, that's 3 Coulombs per Volt, i.e. 3 Farads. That's ... quite a lot. \$\endgroup\$
    – user16324
    Commented Jan 25, 2018 at 10:51
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    \$\begingroup\$ @Trevor_G, I completely agree with you but that's only if OP's starting from scratch, which they may well not be. Could be mounted, wired in and awkward work to replace. They might even just like it and are more prepared to explore other possibilities. Could be lots of things, that's what livens all this stuff up here :-) Plus it brings us the pleasure seeing your lively and exploratory mind in full flow :-D \$\endgroup\$
    – TonyM
    Commented Jan 25, 2018 at 12:13
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    \$\begingroup\$ @TonyM yes indeed, and thanks for the compliment The practicalities matter :) He may even be better with a local PSU at the solenoids. When an OP says something like "simple garage door circuit" though it makes me think the KISS scenario is best ;D \$\endgroup\$
    – Trevor_G
    Commented Jan 25, 2018 at 12:23

1 Answer 1


There is an equation but 3 Amps at 12V for 3 seconds is a lot. You get somewhere around 4 Farads (Vo=12, Vt=10, R=4, t=3). There are supercaps (Double Layer Capacitors) but they have very low voltage. Thus you would need about 4..6 of them each costing around $10..$15.

Maybe just a small re-chargeable battery would be simpler.

Reply to comments (More readable here)
First: driving home in may car I realized my calculation is a bit off. As you stated the capacitor(s) only have to deliver the missing 1.5A not the full 3A (R=4) I used. Thus the capacitor value can be ~2F.

should I put it in series with the paralel solenoids, or should it be in series

The capacitor(s) have to operate in parallel with the power supply but you can't just put them there directly as they may look like a short when they are empty. (when you switch on). I have thought about using resistors & diodes and switches but I do not see a simple solution. Every one of them fails as soon as the capacitor voltage drops below the supply voltage when it discharges, as at that point the supply will try to keep giving 12V and starts trying to deliver the full current.

It is possible if you use the supply for one solenoid and a capacitor plus switch to drive the other one solenoid:


simulate this circuit – Schematic created using CircuitLab

But maybe somebody else has better idea

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    \$\begingroup\$ Or a better power supply/solenoids... \$\endgroup\$
    – Trevor_G
    Commented Jan 25, 2018 at 10:42
  • \$\begingroup\$ How could I connect lower voltage capacitors in order to reach the desired 12V? I've found online 4F/5.5V supercapacitors. Can I connect 3 of them in series (5.5+5.5+5.5=16V)? \$\endgroup\$ Commented Jan 25, 2018 at 14:54
  • \$\begingroup\$ You have to put them in series which reduce the capacitance. e.g. to get 12Volts from 5.5 you need 3 in series giving 3x5.5=16.5 Volts. But then to get 4F you need three capacitors each 12F. (or 9 each of 4F, 5.5V). \$\endgroup\$
    – Oldfart
    Commented Jan 25, 2018 at 14:58
  • \$\begingroup\$ Just wondering, I've search for "capacitor discharge equation" circuits and they seem to have the capacitor in series with the load. I was immagining to place it in paralell to the solenoids.... If I could get a 12V/4F capacitor, should I put it in series with the paralel solenoids, or should it be in series? \$\endgroup\$ Commented Jan 25, 2018 at 16:07

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