I'm considering hiring an electrical engineer to design an emergency power supply for use in embedded devices. This power supply only needs to hold ~15 seconds of charge (enough time to safely shut down the devices). I would like to support 12V and up to 2A. Juice4Hault produces a Supercapacitor UPS, but thier prices are very high. Batteries aren't reliable since they degrade and corrode over time.

I found some cheap parts that I hope will keep production costs low:

2.7V 100F Supercapacitors ($1.50 each)

LM2623 ($0.50 each)

Basically I would like to use low voltage supercaps and a voltage booster to convert extra amps into higher voltage output. Where I'm lost is interpreting the LM2623 datasheet. Could it reliably boost these Supercaps into 12V 2A 15-second output?

UPDATE: I'm considering using two capacitors in series. Then I'd have 5.4V 50F at my disposal. The Samwha Green-Caps that I am considering can pull 5A continuous current. So 5A ∙ 3V(Min) ∙ 80% efficiency = 12V ∙ 1A, sustainable for 10 seconds, which I could live with. I would have to use a more expensive booster:

TPS55332-Q1 ($2.50 each)

Is there an IC that can safely support this?

  • 1
    \$\begingroup\$ Supercap backups are easy to make and they have ICs dedicated to the purpose. However, if you're going to hire an electrical engineer to design it, give them a price target and allow them to choose the components. If you want to design it yourself, look a bit more into those dedicated ICs and make a first pass at a schematic then ask some more specific questions. The dedicated supercap chargers will balance the charge on the caps and also be able to trigger alarms/interrupts when power is down. \$\endgroup\$
    – scld
    Commented Feb 22, 2016 at 23:02
  • 1
    \$\begingroup\$ 12V⋅2A⋅15s=360J \$\endgroup\$ Commented Feb 22, 2016 at 23:21
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    \$\begingroup\$ 360J in a capacitor with Vmax of 2.5V is 115F, so capacitance wise it is do-able. But assuming 80% converter efficiency you would need the caps to be rated for pulling 12A minimum continuously, and have a boost converter capable of that. But the capacitance would also have to be higher as step-up at that kind of load would be unlikely to run at less than 1V input so couldn't fully discharge the capacitors. \$\endgroup\$ Commented Feb 23, 2016 at 0:10
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    \$\begingroup\$ Make sure you know the ESR (internal resistance) of your supercap before you part with money. Some are in the tens of ohms which make great low-current long term backup supplies but would kill this app... \$\endgroup\$
    – user16324
    Commented Feb 23, 2016 at 1:12
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    \$\begingroup\$ @skibulk 12V@2A=24W out. @80% efficiency, that means 24/08=30W in. [email protected]=12A. \$\endgroup\$ Commented Feb 23, 2016 at 14:37

2 Answers 2


you want 12v x 2A = 24Watts of output power for 15 seconds if we assume efficiency of 90% and an input voltage of 5v, the average input current would be 24/0.9/5 = 5.33A this would be the average current, the peak current would be much worse!(typically 2 to 3 times more) internal mosfet resistance is 0.17ohm according to datasheet, so the dissipation would be too much for this tiny package even for 15seconds. you should use another IC which supports external mosfet or another part with much lower internal mosfet resistance.

  • \$\begingroup\$ Could we safely produce 12V 1A 10-second output? Assuming measures are taken to cut the power as soon as the capacitors drain from 5.4V down to 3V. At 3 volts, we are drawing 5A. The TPS55332-Q1 has a switch current limit of 5.7A. \$\endgroup\$
    – skibulk
    Commented Feb 23, 2016 at 21:24
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    \$\begingroup\$ in boost topology you first store the energy from your input into an inductor and then release the inductors energy to the output. so you cant draw 5A continously, what you do is that you draw burst of 10-15A of current and the current limitng would kick in. so you cant use that part either \$\endgroup\$
    – Ali80
    Commented Feb 23, 2016 at 21:30

I'd simplify and remove the regulator entirely. Maxwell makes a 16V 58F cap module. One of those will supply two amps for fifteen seconds with roughly half a volt drop. You don't specify a tolerance on the 12V so I'm guessing that's acceptable. You still need a current limited source to charge the cap, but a 100W wirewound resistor would be a good guess. Put a small schottky around the resistor to discharge back to the load. Easy.

  • \$\begingroup\$ I planned on having a standard 12V 5.5mm x 2.1mm power source, but I guess I could settle for 24V. I'd like this to be used in embedded applications, so preferably the UPS should not take up a lot of space. The Raspberry Pi Zero is the size of a credit card while The Maxwell 16V 58F Capacitor Module measures 9 inches! \$\endgroup\$
    – skibulk
    Commented Feb 24, 2016 at 3:34
  • \$\begingroup\$ Oh, well, yes, leaving 90% of the energy in the caps won't be terribly space efficient... \$\endgroup\$ Commented Feb 24, 2016 at 4:17

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