I have a device which is operated from a 12 VDC 10 A power supply and draws 8 A in operation.

I need to run this device for at least 20 seconds when the 12 V power supply is lost upon a power failure. I thought of using capacitors or supercapacitors for this purpose.

  1. How can I calculate the needed capacitance and the voltage of the capacitors?

  2. How should I use supercapacitors for this application?

  3. Will it affect the power supply when charging if a higher capacitor is used?

Through research, I found the supercapacitor voltages are less than 2.7 V.

  • 7
    \$\begingroup\$ To calculate capacitor size, you must define what is the voltage range your device works with. Is it 11 to 13V or 11.9 and 12.1V or something else. However, it is unlikely that you actually want to use any capacitors at all to power a 12V 8A device for 20 seconds. \$\endgroup\$
    – Justme
    Apr 6 at 9:31

4 Answers 4


If your operating power is 12 V * 8 A = 96 watts, and you want to run for 20 s, you need to be able to deliver 20x96 = 1920 J, which is a huge amount of energy for capacitors. Depending on the volume you have available, supercaps and rechargable batteries are your only realistic options.

One important thing is to note that I didn't say that you needed to store 1920 J, I said you needed to deliver 1920. As the voltage of capacitors varies considerably with the stored energy, you'll need to store rather more than that figure. Swinging between max voltage and 50% of max voltage allows you to deliver 75% of your stored energy, with a reasonable voltage swing into your SMPS. This means you'd need to store 1.33 x your delivered energy, or perhaps 1.5 x taking into account the SMPS efficiency. So we'll aim to store 3 kJ.

Let's do some sums.

Take a 10 F supercap as an example. This will store \$\frac{1}{2}CV^2\$ = 0.5 x 10 x 2.7^2 = 36 J. You would need 82 of those for a total of 3 kJ.

Take a single cell 2 Ah LiPo rated at 25 C or more. That will deliver up to 50 A, 150 W when down at 3 V. In 20 s at that power (so assuming a very inefficient SMPS), that will use 20/3600 x 50 = .277 Ah, not much of its capacity. Using so little capacity, if you restrict the charging and discharging voltage to well within the min and max, it would last for a very long time.

The reason I've allowed such overkill on the amount of energy storage in the battery is that the limitation is the C rate, the current you can get from the battery to deliver your power. You can get higher C batteries (60 C are fairly obtainable, and I've seen one advertised as 135C!) but they get more expensive, and the 20-40C range is probably the cheapest for power delivery if the extra weight is not a problem.

It depends on the expected lifetime you need. If you are going to have more than tens of thousands of power fail events, then capacitors would assure you of a longer life, useful if it was an unattended situation like a remote island. However a battery would be so much smaller, cheaper and easier to use, that's the way I would go.

Voltages. In practice, you would probably want a higher voltage than a single supercap or LiPo, as running a SMPS from that low voltage is less efficient than a more 'reasonable' voltage in the 12-48 range. You would arrange your supercaps in series/parallel to hit that range, or buy a higher voltage lower Ah LiPo, perhaps a 4S 500 mAh. Using a series connection means that voltage balancing would need to be used, when charging both supercaps and LiPos.

If your load can take the voltage variation from 11 to 14 V, then an easier solution would be a lead acid 12 V battery. A battery that won't start a car may still have enough oomph for 10 A for 20 s, so you may even get it for free! Float it at 13.8 V, and you're done.

  • \$\begingroup\$ thanks for the reply. In my application I have mentioned the maximum usage mostly the power will be less than that around 40W. Is there any chance I am able to use capacitors with higher voltage ratings eg:- 100mF caps with 16V ratings. since the voltage is 12V, they will charge up to 12V, according to the equation 1/2CV^2 and if 20 caps are used the energy value is 144J. The energy required to run 4s at 30W is 120J. Will this scenario work? \$\endgroup\$
    – edupulz
    Apr 6 at 10:16
  • 1
    \$\begingroup\$ In delivering those 120 J, ie leaving 24 J of the original 144 J still in the caps, the voltage will drop to 12*sqrt(24/144) = 4.9 V. Will that till run your load? \$\endgroup\$
    – Neil_UK
    Apr 6 at 10:52
  • \$\begingroup\$ thats a problem it wont run in that voltage \$\endgroup\$
    – edupulz
    Apr 6 at 11:12
  • 3
    \$\begingroup\$ +1 In case it's not obvious, 82 10F supercaps will be expensive. At volume prices you're adding at least $60USD to the BOM with a cheap option, and for a one-off it'd be closer to $80 minimum. For decent caps you could double or triple that price. \$\endgroup\$
    – J...
    Apr 7 at 0:29
  • 4
    \$\begingroup\$ Regarding 12V lead acid battery: Depending on the power supply you might even be able to simply connect it in parallel to the supply lines without any charging controller/voltage regulator required. Just make sure it doesn’t go below 11.6V \$\endgroup\$
    – Michael
    Apr 7 at 12:14


I'm not going to touch on supercapacitors. If you can live with electrolytic capacitors, there's a trick: their low voltage energy density is underwhelming. And, to extract most energy from them, you can't just discharge them by directly connecting them to the load that expects a small range of supply voltages. You'd be only using a fraction of the energy stored in them that way. You need a switching power converter to discharge as deeply as possible until diminishing returns set in.

According to this answer, you'd want to use capacitors rated for 400-450V, since per unit volume they give you most energy stored. You'll want to charge them up to 95% of the rated operating voltage, and discharge them down to 50-100V. The lower discharge voltage depends on how good a switching converter you can put together to efficiently convert the higher voltage into the low voltage output.

Thankfully, this is a solved problem: any high-efficiency 12V-output wide input range switching power supply does a good job at discharging capacitors down from a couple hundred volts, while putting out 12V at high currents. Supplies with PFC run their DC-link capacitor quite close to 400V, so you're in ideal energy density territory. How convenient!

The Need for PFC

Without a PFC, the rectifier voltage would be a bit too low to be practical when supplied in Japan/US/Canada, i.e. from 100-120VAC. You'd have to add a voltage doubler and feed the entire supply from doubled rectifier voltage, and then you end up with a 120/240VAC switch typical of old PCs. Those are not practical and not expected in modern equipment - just leverage the PFC.

Most supply designs transition from a dumb rectifier to a PFC somewhere in the 150-250W power range. Since your load is 120W, and you'd probably want to recharge at 1/3C peak, a rating of 160W would be sufficient. With some allowance for derating a 200W supply with PFC would be an OK choice. The excess capacity is used to recharge the capacitor bank while the load consumes full power.

A Simple Design for a Capacitor Bank


simulate this circuit – Schematic created using CircuitLab

The design proposed above is not ideal by any means, but it has some positive characteristics in terms of its handling of failure modes and least interference with the parent supply's operation.

  • During normal operation, with C2 fully charged and AC present, the capacitor bank is disconnected from the DC link. D2 is open and a small self-discharge current is drawn through D1.

  • During startup, C2's charging ballast R1 is disconnected until the output converter starts up successfully. An additional delay is provided by RLY1's inherent turn-on delay. This ensures that the PFC startup conditions are conservative: as-if the capacitor bank wasn't present.

  • Should the converter fail internally, the output voltage is lost, and the capacitor bank is quickly disconnected. The converter, in a failure mode, doesn't have to cope with an unanticipatedly high DC-link capacitance.

  • When the AC power is lost, the capacitor bank provides backup power to the converter with an inherent delay needed by C1 to discharge by the drop of diode D2.

  • During backup power operation, C1 and C2 are in parallel.

  • As soon as the usable backup energy capacity is spent, the converter's output turns off, and the bank C2 is switched into discharge mode. No discharge current flows during normal operation, saving energy.

  • The C2 charge indicators X1 and X2 - presumably LEDs - are configured for minimum power loss: X1 runs from 12V and doesn't waste much power in the series resistor. X2 runs from the DC-link voltage only during C2 discharge; its series dropper dissipates power that had to be dissipated anyway.


The capacitor charging circuit is simple: a series resistor R1 to limit charge current through D1 into the capacitor bank C2. If the power-up events are rare, the energy loss on R1 is not substantial and doesn't have undue impact on the energy efficiency of the device. If dictated by the requirements, a switcher-based constant current source could replace R1.

The bank C2 doesn't need to charge super fast: a charge lasting 2-3 minutes would be reasonable given the application need of 20-30s backup time, but this can be adjusted as needed. Don't forget about the extra PFC capacity needed to charge C2, and derate the 12V supply accordingly.

In the quiescent state, when the 12V output voltage is absent, the discharge relay RLY1 connects the C2 in parallel with the safety discharge resistor R2. It should be selected to discharge the bank C2 down to a safe voltage of 48VDC in a "reasonable" time - say 10 minutes. The C2-R2 time constant would then be in the ballpark of 1-2 minutes.

Fuses F1 and F2 are important. F1 protects the capacitor bank from destructive discharge (a.k.a. a big bang) should the DC link become shorted. F2 protects RLY1 and the associated wiring, since it wouldn't be rated for the full output current of 10-15A.

The unsafe voltage on C2 is indicated in an energy-saving fashion by two indicators: X1 and X2.

X1 is active when the DC output is present. Obviously, C2 is at an unsafe voltage then, or either F1 or F2 has opened - a condition to be aware of in either case.

If the DC output is lost for any reason - be it due to discharge of C2 below converter's turn-off voltage, or due to converter failure, RLY1 switches C2 into discharge mode, and X2 takes over.

If either X1 or X2 is lit, C1||C2 is presumed at an unsafe voltage.

X2 and R1 provide redundant discharge paths for the DC-link capacitors, albeit at a vastly different time constants.

Such capacitor banks are arc flash hazards, and without proper precautions, it's too easy to get seriously injured. Have a redundant charge state indicator so that you have a last-resort visual indication that things are charged and dangerous. Please familiarize yourself with arc flash hazards in capacitor/battery bank systems!

The capacitor bank will potentially stay charged for years if the discharge load fails. Treat it as dangerous until you verify it's down to a safe voltage with an external trusted instrument.

No jewelry on your hands or above the shoulders (necklaces etc). Long hair in a bun.

  • 5
    \$\begingroup\$ Excellent idea to reuse the universal input SMPS to handle the output of the many hundred volt capacitors, and R-charge from the mains. I would have suggested high voltage electrolytics in my answer but didn't think of these savings. \$\endgroup\$
    – Neil_UK
    Apr 6 at 11:05
  • 5
    \$\begingroup\$ Also worth noting: the safety hazards here can persist after the device is powered down. Do not touch or service the capacitor bank unless you can verify that it has discharged to a safe level. \$\endgroup\$
    – bta
    Apr 6 at 23:35

The formula for the voltage on a capacitor being discharged is

$$V_{c} = V_{i}{\;\cdot\;}e^{-t/RC}$$

RC is the time constant, resistance in ohms times capacitance in farads
t is elapsed time in seconds
Vi is the initial voltage across the capacitor
Vc is the voltage across the capacitor at time t

Given a capacitance of 500F, an initial voltage of 12 V, and a resistance of 1.5 ohms (12 V / 8 A), the voltage after 20 seconds will be 11.68 V.

You can buy 500F 16 volt capacitors packaged like an automotive battery. This is an option you may want to look into further to see if it fits your needs.


As others have said, the fact that the amount of energy being stored in a capacitor is a factor of the voltage squared makes having a bank of capacitors charged up to a high voltage seem appealing, though depending on the voltage level can be difficult to design around. Safety, adequate creepage and clearance, and the power supply circuitry to charge and discharge the capacitors can present significant design challenges.

Since you asked about supercapacitors and you are saying you only need 20 seconds for what seems to be a 96 Watt load, that may be possible, though it would take a reasonably large bank of high capacitance (say in the 30F plus range) supercaps and a controller to keep the bank balanced and charge/discharge it. The advantage here is that you can keep everything at a much safer voltage level and in terms of volume/weight the solution may be smaller.

Here are some example supercaps at Digi-Key: https://www.digikey.com/short/0fhqpbfb

And here is an example of such a controller that handles the buck charging and boost discharging of the supercaps when input power is lost: https://www.analog.com/en/products/ltc3351.html

Just another option if you decided you preferred a lower voltage supercap based solution.

  • 1
    \$\begingroup\$ It seems like usable capacitance at a given cost also decreases proportional to voltage squared - no free lunch! \$\endgroup\$
    – user253751
    Apr 8 at 10:30

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