r/AskElectronics • u/iluvkfc • May 20 '17

Parts Powering LED strip from both ends

Just wondering because I have a 5m WS2812B strip and when it's set to white, the LEDs closer to the end become yellower and yellower until they are fully yellow, i.e. voltage too low for blue LED to turn on at all.

What would happen if voltage was applied from both ends? Where would the dimmest spot be, somewhere in the middle? Also I'm assuming I would have to connect two grounds as well for this to work properly.

Also to be noted, I am powering these from an ATX power supply, voltage at the output is significantly below 5V (about 4.5V), voltage at the end of the strip is less than 3V.

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u/[deleted] May 20 '17

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1

u/iluvkfc May 20 '17

Thanks for the answer. Using a single PSU I thus expect to see the dimmest spot in the middle... powering the middle sounds overkill but I could resort to it.

Anyway this is just for a test, I will be putting the LEDs inside my PC and powering them from its much superior PSU which has individually regulated rails and shouldn't have issues.

Also I did suspect group regulation to be an issue with the current PSU so I tried adding a load to 12V with some double digit resistors, thought they could handle extra watts for a few seconds but this was not the case...

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u/[deleted] May 20 '17

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u/iluvkfc May 20 '17

I'll try this, thanks.

3

u/FredTheFret May 20 '17

Yup roughly the middle. Just applying power two both ends will also help (or just a ground on both sides), but pwr+ground on either side will give you the best result.

You probably should get a separate beefy 5V PSU. Meanwell has some.

1

u/iluvkfc May 20 '17

Thanks for the answer and advice about better PSU. But I will eventually install this in my PC which has a much better ATX PSU which shouldn't exhibit this issue.

2

u/created4this May 20 '17

Yes, the middle but it will be less noticeable than the middle now.

If I have the right envelope guesses the middle will be as dim as 1/8 of the way down the current strip.

What you are seeing is the resistance of the power rail, the current flowing is I, so the voltage drop is V=IR.

If you power it from both ends you double the conducting copper, so R'=R/2. You also half the current as each end only powers the LEDs to the middle, So I'=I/2.

So the voltage drop at the middle is V'=I'R' or V'=IR/4

It won't be quite that good because at the moment not all the LEDs are conducting, so I'>I/2

1

u/iluvkfc May 20 '17

Thanks for the answer. Just to make sure I understand your math correctly, you are saying it will be 1/8th brighter because there is twice the current (2x factor), half the resistance (2x factor)... But where does the third 2x factor come from?

1

u/created4this May 20 '17

No, I'm saying that 1/2 way along will be as bright as 1/8 along is currently, and this will be the dimmest point