Powering 10 m of WS2812B RGBs

Question

I wanted to make sure that my solution would work before wiring it physically and I felt that all the great minds here would be able to do that.

So I have two WS2812B 5 m LED strips, each containing 60 LED/m, or 300 LED per strip. The manufacturer states 0.3 W per LED, and these are 5 V LEDs. That would mean each strip is 300 × 0.3 = 90 W, and requires 90/5 = 18 A at a full white bright.

If I were to just solder one of the 5 m to the end of the other 5 m, I would now have 600 LEDs in parallel correct? Which would mean a current of 36 A would be going through the top line before the respective currents go to the ground after the LEDs. I don't believe the LED strips could handle that much current.

After some testing it turns out that the voltage loss is significant enough that I can't even power one LED from one side, so I'd have to power it from both ends. In order to have two strips powered, I'd think I can power it then from three places, the beginning, the connection of the two strips, and the end of the second strip.

I drew up this quick wiring diagram, would something like this work? I would have three power sources with two grounds, and the LED strips would be fully connected.

Answer 1

The chip uses internal regulators so all you need is 3.5V Min at the end. If you get this it works for 5m but not 10m. Thus you can distribute AWG16 or suitable wire to either 2 starts or 1 start 1 end or all 3 Nodes. Testing the end supply voltage with 1 Start will tell you when all RGB are on what your drop is on both V+ and rise on Gnd which may be equal.

If you can get more than 250mW or more on each chip it may burn your finger but still work. But all you need is 3.5V not 5V. More voltage just makes it hotter. So supply enough and no more for expected performance with least temperature rise.

Thus 1 feed at each end as you have shown may be ideal. If middle is say ~4V. But if all 3 nodes are 5V and middle of 5m is say 4.5V then your are just adding heat and not brightness.