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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.

enter image description here

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  • \$\begingroup\$ Why does your power supply have 4 output wires? Where is the data sheet for the LED strip? What does that tell you about how to apply power? The data sheets are probably required to properly answer this. \$\endgroup\$
    – Andy aka
    Jul 31, 2020 at 17:05
  • \$\begingroup\$ @Andyaka The power supply is 40A 5V with three positive ports, and three ground ports. These are just generic WS2812B LED strips I got off amazon, but I believe they're all made the same. Adafruit has a datasheet here cdn-shop.adafruit.com/datasheets/WS2812B.pdf , but my question is more related to if the powering could be done in the way I try to show. \$\endgroup\$ Jul 31, 2020 at 17:11
  • \$\begingroup\$ Then specifically, what do the 4 wires represent? The data sheet for an individual LED is not what I wanted. \$\endgroup\$
    – Andy aka
    Jul 31, 2020 at 17:14
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    \$\begingroup\$ @Andyaka The 4 wires are just two positive wires and two ground wires from the PSU that will be powering the LED strips. They both will be powering the ends, however one will split off and also provide power in-between the two strips because voltage drop makes the middle of the two connected strips dim otherwise. \$\endgroup\$ Jul 31, 2020 at 17:23
  • \$\begingroup\$ You might as well add a 3rd ground in the middle too. \$\endgroup\$
    – Passerby
    Jul 31, 2020 at 17:42

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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.

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