Supercapacitor to prevent temporary power supply voids

Question

Hello and sorry for the improper title!

We need to transmit about 50 W (5 V 10 A) through a slip-ring-ish to drive some rotating electronics and LEDs. We are using a 5 V 25 A LED power supply. We are worried about the slip ring temporary losing contact during the high speed rotation, potentially causing the electronics to misbehave or the leds to flicker. We were thinking about adding a supercapacitor after the slip ring to handle any temporary (milliseconds I think) power voids caused by the slip ring. But I have some questions:

1. Is 1-2 F enough for my purpose?
2. Do I need a supercapacitor charger IC to prevent damages to the power supply? If so, should I switch to a higher voltage power supply to compensate for the voltage drop in the charing IC?
3. I know they capacitors have almost linear discharge rate, so in the case of a power void I would have the full 5 V of the capacitor only for a brief period of time, then the voltage would drop also if the energy in the capacitor is still there, probably causing the electronics to missbehave for the undervoltage. Should I use some kind of boost converter to get a constant 5 V from the capacitor? Do you have any IC recommendation?

And finally, it's seems to me that I'm reinventing the wheel :) do you know any IC that basically works as a UPS with a supercapacitor capable of handling 50-100 W at 5 V?

Sorry for the vague question, we are in the early stages of the development... Thank you!

Answer 1

1. Capacitors follow I=C*dV/dt. Your 1 F is good for 1 A for 1 s in you allow voltage to fall 1 V. Or in your case, 100 ms at 10 A and drop to 4 V. 4 V too low, try 50 ms at 4.5 V or 25 ms at 4.75 V. You get very diminishing returns the closer to 5 V you need to stay.

2. If the power supply has a constant current limit, no need. If not, then yes. But the elephant in the room is that there are currently no 5 V supercapacitors on the market, only 3.5 V ones so you need to place two in series and use active and/or passive means to make sure none of them go above their specified max voltage.

3. Indeed. I would use the slip ring at the highest rated voltage, place a bog standard electrolytic capacitor rated for that voltage on the output/secondary side and buck myself down to 5 V from there.

Answer 2

Not a direct answer to your question. But consider persistence of vision, if, of course, it is relevant in your application (tens, even hundreds, of ms will likely be unnoticeable). And if you have control ICs on the rotor you can just have an extra beefy bypass capacitor on the rail with a diode to prevent the discharge thought the LEDs.

1. If not so, then having a boost converter and a charge controller on the rotor is a lot of complexity. But let's assume you take 2 F super-capacitor with a boost converter that can boost to 5 V from 3 V at 50% efficiency thought out the cycle. Then you'll have 2 J of energy to harvest. A 5 ms void at 50 W is 0.25 J. So, yes, 2 F is barely enough (you may have wider voids). But the ESR of the super-capacitor is also an important consideration, it may limit current and become useless. So you need to study some datasheets.

2. You have 15 A to spare on the power supply, so I'd say you are safe if you decide not to use a charge controller.