Folks - I'm looking at using an Adafruit 12 wire slip ring: https://www.adafruit.com/product/1196

It is rated at 240V, 2 Amp. It has 28 awg wire coming out of it.

I was planning to only run 5 VDC through it. I plan to run 100 LEDs, each drawing 20 mA (total 2 Amps).

Tech support at Adafruit says this is ok.... but I thought you would have to "de-rate" the current if the voltage went down....

For example, the slip-ring is rated 240V, 2 A. According to good ole' Ohm's law, V=IR - so at 240 V and I=2A, then R= 120 ohms.

If V=5, and assuming the load is constant at 120 ohms, you wold have 5 V/120 Ohm = 41.67 mA max.

Or is it somehow a function of power?

Sorry for the dumb questions....

What am I doing wrong????

 

The slip rings know nothing about the voltage between them, only the current through them. As long as that doesn't exceed the stated maximum all will be well.

 

The slip rings know nothing about the voltage between them, only the current through them. As long as that doesn't exceed the stated maximum all will be well.

Thank you. But when I look up wire sizes, the ampacity of this gage (28 awg) wire is about 0.226 amps. How can they handle 2 amps?

 

Where did you find that number?
https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizes
https://www.powerstream.com/Wire_Size.htm

 

Where did you find that number?
https://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_AWG_wire_sizes
https://www.powerstream.com/Wire_Size.htm

I actually got it from the powerstream website - max amp for power transmission.

Regardless, both of your sites show a max of 1.4 amps.... then how can this be rated to carry 2 amps?

I'm not trying to argue - but am trying to understand.

 

Yeah, for some reason I was looking at 18AWG.
However, the datasheet for the product in your link says the wires are 26AWG which, from Powerstream is rated 2.2A, so all is well.

 

For example, the slip-ring is rated 240V, 2 A. According to good ole' Ohm's law, V=IR - so at 240 V and I=2A, then R= 120 ohms.

But that's the rated voltage from slip-ring to slip-ring (or slip-ring to ground), not the voltage across the slip-ring (which likely is only a few tenths of a volt at most) so good ole' Ohm's law doesn't apply (at least not the way you applied it).

 

Gentlemen and Ladies,

Thank you so much for your help!

Still learning about all of this and want to do it right. I Live in New Orleans and am working on a Mardi Gras float. Besides wanting everything to work - safety comes first. So I just wanted to make sure all was safe before creating a disaster....

I am 57 years old and have always loved electronics. I have a BA, BS and a MBA. But I've gone back to our community college to get an AS in EE for fun and knowledge.... So if I, going to spend the effort trying to get "educated", it never hurts to get it RIGHT!

Thanks for all of your help!

 

It all depends on how long the wire is too 800 mA for 28 awg if you going to run it long run's
Point is you may have to run bigger wire to it if the run is long.

 

Can you double-up the contacts?

Use more than 1 circuit in parallel, make it redundant and reduce the current per contact.

If you only need to supply power, use 6 circuits for Ground and 6 for Power.
Then it's really reliable.

 

Yes, using 2 slip rings in parallel will double the current capacity, or reduce the current through each brush/ring.

The 240 volt value could become important (not in your application), because above that, flashover could occur between rings.

Don't wire with anything smaller than 18 ga.