3

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!

winny
  • 13,064
  • 6
  • 46
  • 63
Suxsem
  • 59
  • 3
  • 7
    Rotating an electrolytic capacitor can centrifuge the electrolyte out of the capacitor – D Duck May 21 '23 at 10:28
  • @DDuck didn't thought about it!! So it's better to mount it at a specific angle? – Suxsem May 21 '23 at 10:37
  • 2
    I don't think that there is an angle that can always work - you want \$R\$ to be small so you want it on the axis of rotation. Obviously if you do your calculations and the 'g' your capacitor pulls is about 1 g then you should be ok, but depending on the geometry and \$\omega\$ you can pull 10,000 g. – D Duck May 21 '23 at 11:41
  • @dduck the apparatus will spin at 1200 rpm and the electronic will be placed at 0.2 mt from the center, so the centrifugal acceleration will be about less then 500g. Do you think that I should worry about the electrolyte? – Suxsem May 21 '23 at 13:00
  • 2
    Dang! That’s fast. Very few manufacturers rate their components under constant G-force. Film capacitors is probably a safer bet in the long-run. – winny May 21 '23 at 14:53
  • Dunno if it might work in your application, but might be worth considering moving some of the electronics off the rotating wheel. Like, maybe replace some of the lights with fiber-optic cords, then have a non-rotating light-source provide light to one end of the fiber-optic cord, instead of it coming from a rotating LED? – Nat May 21 '23 at 17:33
  • 1
    The rotation also makes me wonder about having a dynamo/alternator/etc. generating electricity instead of trying to connect it. – Nat May 21 '23 at 17:38
  • 2
    @winny I expect most manufacturers rate their components under a constant g force of 9.81 m s\$^{-2}\$ – D Duck May 21 '23 at 17:55
  • @DDuck Ah yes, that force. Fun fact: I saw a documentary about some supercap where they wanted to set a new speed record. The lap was oval and the sustained sideways force in the last corner before the max speed on the straight was so high they had to order a custom gearbox for it and the driving direction has to be set before. Spin launches into space comes to mind too. They said tall components were a problem but most SMDs could take substantial G-loads. – winny May 21 '23 at 19:40
  • I cannot find cap datasheet rating the g force, can you please provide an example? – Suxsem May 23 '23 at 06:53

2 Answers2

7

  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.

winny
  • 13,064
  • 6
  • 46
  • 63
  • Why not 5V SCAP? https://a.aliexpress.com/_EG6XfwH – Suxsem May 21 '23 at 10:41
  • I'm using a led driver power supply, from what I know they should have current limiting built in, let's hope they do :D – Suxsem May 21 '23 at 10:43
  • 1
    @Suxsem Hoping isn’t engineering. Check the datasheet and find out. Those are two capacitors in series. Hopefully with some protection against voltage imbalance but I doubt it. Only buy components from reputable suppliers With datasheets. – winny May 21 '23 at 10:57
  • You might want to use a much higher voltage and non-electrolytic capacitor and a switcher (buck) as energy stored goes as \$E = \frac{1}{2} C V ^2\$ – D Duck May 21 '23 at 11:47
  • What do you think about the LTC3350 IC? it is a high current UPS controller. I'm worried that the fast on-off frequency caused by the slip ring is not suited for this kink of IC that expects a clear loss of power instead of a bumpy input... What do you think? – Suxsem May 21 '23 at 13:06
  • Could perhaps do the job, but how much voltage can your slip-ring take in the first place? I would try to exploit that and use standard SMPS components first rather than to make do with 5 V and supercapacitors. – winny May 21 '23 at 13:24
  • I don't think voltage will be a problem https://a.aliexpress.com/_EGdRa3F :D (we couldn't use a standard slip ring because motor is not hollow so we opted for a copper ring outside of the rotor coupled with these kind of brushes). Do you think the SMPS will smooth the bumpy power supply or should I also add some caps at the input of the SMPS? – Suxsem May 21 '23 at 13:45
  • Find out how much voltage both you and your home made slip ring is happy with. Be aware that your product will fall in a different category if you have high enough voltage, even isolated, on any user touchable surface. Can you get away with 48 V DC and a decent capacitor to filter out the inherent noise? – winny May 21 '23 at 14:08
  • This is one of a kind product for on-stage events with qualified people operating it so this is not a problem. 48 V DC shouldn't be a problem for the slip ring, any recommendation about the cap? – Suxsem May 21 '23 at 14:26
  • 1
    Qualified stage-hands does not matter to the regulatory body (UL, CSA, Jet, CE) in question. Touchable = comes with stipulations. Hard to say, sweep the rotational speed at max load and log the secondary side voltage. How deep and long was the longest and deepest dip? – winny May 21 '23 at 14:44
1

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.

winny
  • 13,064
  • 6
  • 46
  • 63
shklj
  • 27
  • 4
  • 7
    If the LED is on the rotating platform and there is someone viewing it, the image of the LED is scanned across the retina of the viewer they can detect us pulses of light. – D Duck May 21 '23 at 10:27
  • 1
    And if you don't see hundreds of ms dropout from a LED, you're probably blind and couldn't see the LED in the first place. – pipe May 21 '23 at 10:50