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According to the NeoPixel best practices (which is the exact product I'm using):

Before connecting NeoPixels to any large power source (DC “wall wart” or even a large battery), add a capacitor (1000 µF, 6.3V or higher) across the + and – terminals as shown above. The capacitor buffers sudden changes in the current drawn by the strip.

Seems to me that the size of the capacitor is linked to the amount of pixels to be used (since more pixels = more max current drawn). If I'm correct, would anyone care to explain how to calculate the capacitor size in relation to the amount of pixels?

Some extra specs:

  • Each individual NeoPixel draws up to 60 milliamps at maximum brightness white (avg 20 milliamps).
  • They require 5V.

Update based on comments:

  • Power supply will be a low quality wall wart that puts out anywhere from 9-15V (but my circuit has a LM7805 voltage regulator that will bring this down to 5V.
  • The distance from the capacitor to the first LED in the strip will be around 1 inch.
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  • \$\begingroup\$ It's not purely dependent on those numbers. It depends on the quality of the PSU and how far you are wiring it, etc. Anecdotally, I designed a PCB with ~100 Anypixels and assembled one of the PCBs without any bypass or bulk capacitor. It worked fine. I still used all bypass and bulk caps on the final design but still, the point stands, YMMV. \$\endgroup\$ – Wesley Lee May 23 '17 at 21:48
  • \$\begingroup\$ @WesleyLee Thanks for commenting. I've updated the question with the info you requested. \$\endgroup\$ – Julian May 23 '17 at 21:57
  • \$\begingroup\$ My point was more like, you might want to determine this experimentally. 1 inch seems quite reasonable though. Another point is, how many LEDs are you powering? Have you considered the power dissipation of the 7805? \$\endgroup\$ – Wesley Lee May 23 '17 at 22:00
  • \$\begingroup\$ @WesleyLee OK, gotcha. My 7805 can handle upto 1.5A which is almost double what I will be using with 10 LEDs (600mA max) so I'm ok on that front. \$\endgroup\$ – Julian May 23 '17 at 22:06
  • \$\begingroup\$ 0.6A * 10V voltage drop (if using 15V supply) = 6W, which is substantial. You will probably not run the LEDs at 100% all the time, but it might become a problem in some situations. \$\endgroup\$ – Wesley Lee May 23 '17 at 22:16
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This will not provide an answer on how to calculate the capacitor size, but I'll use the answer space to draw a small model of a NeoPixel strip + Power supply + cable. This might help you understand the role of the capacitor. I'm sorry if you already knew any of this.

Data lines are not shown, etc.

schematic

simulate this circuit – Schematic created using CircuitLab

Ok, so when the strip draws current, current goes through the cable and with the parasitic R of the cable causes a voltage drop (U = R*i, Ohm's law). The longer and/or thinner the cable, higher the parasitic R and more severe the voltage drop.

During these voltage drops, the capacitor is there as a buffer, as the guide suggests. This also means that if the parasitic R of the cable (and of the power supply) is negligible, the capacitor will make little difference. In the same way, its benefits are limited by the Parasitic R of the Strips itself (measure the voltage of a long strip, it will be far below 5V when on, you might even notice some color change).

So while it may be possible to calculate ripple vs capacitor size, I suggest you determine this empirically, if possible watching the 5V line with a scope as the LEDs light up.

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