I have a Raspberry Pi 3 Model B currently controlling and powering a strip of WS2812B LED's. A separate 5V 15A power supply powers the Pi through pins 2 and 6, while the LED's are powered off of pins 4 and 9, receiving data from pin 18.

The only issue I seem to be having with this setup is that on the first 146 of 300 NeoPixels (since I don't plan to use the whole strip, however I have not cut the strip anywhere yet), noticeable brightness drop is evident throughout depending on the color/brightness/etc. By my calculations, each NeoPixel uses up to 60mA and the Pi up to 2A, meaning a 15A power supply should be more than enough to power it. Alas, that seems not to be the case, and my question is why.

I should also mention that the same results are achieved when running power through a breadboard and sending power to the Pi and to the LED's such that the LED's don't directly rely on the Pi for power.

(My knowledge of electrical engineering is very limited, and so comments on whether or not powering the LED's directly from the Pi is discouraged would be appreciated as well.)

  • each GPIO pin can only source a limited amount of current. A much better plan would be include a solid state relay chip (capable of handling the max current) and have the GPIO only turn the relay on/off Mar 18 '17 at 2:06

Powering the Pi using the pins at the GPIO connector and connecting the LEDs at the other pins of the same connector bypasses the Pi's protective circuitry that could not handle that much current. However you need to be aware that the Pi's printed circuit board is likely not designed for that current either.

Using this calculator we find that 15 A requires a trace width of 495 mils (12.5 mm) at the standard copper layer thickness of 1 oz/ft² (35 µm). Even if both +5V and GND have separate ground and power planes this is simply asking too much. You will experience a significant voltage drop and resulting thermal issues this way. Similar restrictions apply to bread boards and the wiring that is used with those.

Simple solution: wire the LED strip directly to the power supply using appropriate cables with sufficient diameter. See AWG for more on the topic: the cross-sectional area of each gauge is an important factor for determining its current-carrying capacity.

  • Yep, that was it. Just powered the strip directly from the power supply without any breadboard jumper wires and it's bright as day. I suppose I'll be looking for real wire for this project.
    – TJ Mazeika
    Mar 12 '17 at 3:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.