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I have an LED project with a pretty hefty power draw - 1.5 kilowatts. We're trying to figure out how to power this from a (large) 12 V battery bank.

We need to have the entire 1.5 kW output travel down a single pair of wires (slip ring).

Are there 12 V to 5 V DC converters that can do this?

I was originally thinking I could just wire together a bunch of smaller buck converters (such as these) but I'm learning that you need converters that can do "current sharing", and it's not clear to me how to identify 12 - 5 V converters that can do that.

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    \$\begingroup\$ You need 300 A at 5 V? Through a "single pair of wires"? You'd better plan on those wires being pretty hefty (on the order of 175 \$\rm mm^2\$). Whether you can get 300 A through a reasonable slip ring is beyond my knowledge. \$\endgroup\$
    – The Photon
    Jul 18, 2019 at 1:00
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    \$\begingroup\$ Maybe boost to 100 or 200 V to go through the slip ring, then buck down to 5 V on the rotating side? \$\endgroup\$
    – The Photon
    Jul 18, 2019 at 1:02
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    \$\begingroup\$ How fast will the slip ring be rotating? what if the slip ring contact has occasionally "opens"? \$\endgroup\$
    – analogsystemsrf
    Jul 18, 2019 at 1:09
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    \$\begingroup\$ Does it have to be a 12V battery bank? Could you connect batteries in series? If you can't then it's worth converting 12V --> 100V, that reduces the current to 15A which you could easily get in a carbon slipring. Then go 100VDC to whatever final supply you need. I can't imagine 5V is a good target voltage, perhaps put some of the LEDs (arrays I assume) in series and feed with a constant current supply. \$\endgroup\$
    – Jack Creasey
    Jul 18, 2019 at 1:21
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    \$\begingroup\$ @Narfanator Your buddy is wrong, losses in wires are proportional to square of current and not wattage. \$\endgroup\$
    – Rokta
    Jul 18, 2019 at 7:30

2 Answers 2

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Knowing that you need 5V for the strips, you could segment them and supply them locally with individual 5V. So maybe five groups powered by a 60A 5V supply each. Then use an inverter to make 120V from your 12V battery bank to power the slip rings.

Check this out: https://www.amazon.com/dp/B07G7S44CW/

And this: https://www.amazon.com/Power-TechOn-Inverter-Outlets-PS1002/dp/B0131L8NLM/

Just make sure you idiot-proof the slip rings. Maybe use a GFCI on them after the inverter.

And here’s a 10A slip ring: https://www.sparkfun.com/products/13063

And a 30A (!) one: https://www.amazon.com/Wires-Generator-250Rpm-Power-Collector/dp/B00MYYTWQ8

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Look for DC to DC converters that have a clock sync and/or parallelizable:

Here is a list from analog of high current DC to DC converters. None of them got to the 300A you require, but some go to from 60A to 90A, in which you could parallel three or four of them. However DC to DC converters can be difficult to size components correctly and required a lot of time reading in the datasheet and a certain amount of skill which is not for everyone (especially me, when I got the wrong mosfets and the converter doesn't work on it's first cut).

enter image description here Source: https://www.analog.com/en/products/ltc3884.html

There are also high current LED drivers like the LT3761 which can handle 100A:

enter image description here

It may be better to follow Jack Creasey's comment and buy off the shelf ones. Sometimes they can be difficult to parallel if the voltage outputs are not matched well.

It would also be better to send the 12V (or higher) across the slip ring to reduce losses from the contacts and have the 5V conversion after the slip ring to reduce I^2*R losses from the slip ring contacts.

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  • \$\begingroup\$ If each one powers its own set of leds then matching the output voltage may not be an issue. \$\endgroup\$
    – Solar Mike
    Jul 18, 2019 at 6:21
  • \$\begingroup\$ @SolarMike agree, and have suggested exactly that knowing that they're using addressable LED strips. Divide and conquer. \$\endgroup\$
    – hacktastical
    Jul 18, 2019 at 22:25

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