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I'm currently working on a LED project with WS2812B LED strips with 30LEDs/m. After a couple meters at full brightness the color gets red because of the voltage drops.

Is it possible to connect 5v to the start of the LED strip and Ground to the opposite end of the LED strip to save wires (as in the sketch below)? When testing it that way it worked great, I just want to make sure it's safe and doesn't have any drawbacks. Whenever I Google this topic people suggest to connect ground and 5v to both ends of the strip which is why I'm unsure.

Thanks in advance! enter image description here

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    \$\begingroup\$ Yes it is safe and will somewhat reduce voltage drop. Connecting both wires at both ends would result in less drop though. \$\endgroup\$
    – user1850479
    May 13 at 15:01
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    \$\begingroup\$ I expect that to give most uniform voltage at each "pixel". A compromise between additional conductor material and voltage uniformity would be achieved feeding one at ¼ of the length, the other at ¾. \$\endgroup\$
    – greybeard
    May 13 at 15:12

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Connecting 5 V at one end and GND at the other will dramatically reduce the differential between different LEDs, and improve the voltage at the lowest voltage LED ...

...assuming all the loads are drawing maximum current.

The resistance of the supply remains quite high though, and as the LED current drops, the voltage will rise. You will therefore still get a significant change in voltage and hence colour as you change the programmed brightness of the LEDs.

If you do have access to both ends, it's far better to supply 5 V and GND to both ends. This reduces the supply impedance and so the voltage swing with brightness significantly.

Here is a simple simulation of a power supply along a string of 4 loads drawing, in this example, 100 mA each. The supply has a 1Ω resistance wires between each load. The schematic shows the circuit in the 'alternate ends' configuration, drawing maximum current. With a very low programmed brightness, the voltage on all the loads will rise to 5 V.

enter image description here

Load


 Single end

R1 5 V, R6 GND		 Alternate ends

R5 5 V, R6 GND		 Both ends

R1+5 5 V, R6+10 GND		 5 V one end
Ground both ends
R1 5 V, R6+10 GND
I1 4.2 V 3.6 V 4.6 V 4.4 V
I2 3.6 V 3.4 V 4.4 V 4.0 V
I3 3.2 V 3.4 V 4.4 V 3.8 V
I4 3.0 V 3.6 V 4.6 V 3.8 V

Re-reading your question, it appears your diagram shows supply at one end, but ground connected to both ends. I've added this to the simulation. It's rather better than one end, and a bit better than alternate ends, but still not as good as using both ends.

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  • \$\begingroup\$ All you need to assume is all loads drawing equal current, right? Not all loads drawing maximum current. \$\endgroup\$
    – Hearth
    May 14 at 5:49
  • \$\begingroup\$ @Hearth all loads max current is worst case, which gives you largest voltage drop and largest differenetial between loads. At a lower, albeit equal current, differential and drop will be lower. It's not just the voltage change with position along the string, it's also the change that occurs between other loads going from zero to full current. At zero current, they all get 5 V. \$\endgroup\$
    – Neil_UK
    May 14 at 6:10

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