I'm working on a custom alarm clock. For the display, I'm using an Arduino Nano to control a 7-segment display made of ninety Neopixels. I have some buttons and an encoder as well.

Based on a 60 mA power draw for each Neopixel and about 40 mA for the Arduino I've calculated a 5.44 ampere current. I want to design the circuit for 6 amperes current (the buck converter I'm using can provide 8, but I shouldn't need that much.) I want to use 1 Oz copper if possible to reduce cost.

The trace width will be 3.56mm for that, which is far too wide. Is there any way I can reduce the trace width?

An idea I had was to split the power input into two 3 ampere power rails, each connected to half the board's components. Would the two rail idea work?

  • \$\begingroup\$ 90*600 is 54A not 5.4A but I suspect the maximum current per neopixel is no6 600mA. \$\endgroup\$ Sep 28, 2021 at 14:16
  • \$\begingroup\$ I think you meant 60mA per neopixel. That is the specification on the LEDs and it also matches your calculation. \$\endgroup\$
    – user253751
    Sep 28, 2021 at 14:31
  • \$\begingroup\$ It is worth noting that it is 60mA at full white maximum brightness which might be too bright to you. Maximum red, maximum green or maximum blue would be about 20mA per neopixel since only one colour channel is activated. \$\endgroup\$
    – user253751
    Sep 28, 2021 at 14:32
  • \$\begingroup\$ Can you show a picture of what you're trying to lay out? I don't see why 3.56mm would be too wide for a reasonable neopixel layout. \$\endgroup\$
    – user253751
    Sep 28, 2021 at 14:33
  • \$\begingroup\$ What's the overall size of the board? Single or double sided? \$\endgroup\$
    – The Photon
    Sep 28, 2021 at 14:39

4 Answers 4


Two-rail for both power and ground should work, but if you're looking for increased ampacity without increasing trace width or PCB dimensions, you can always apply solder to the trace to increase that one's current-handling.

  • \$\begingroup\$ That's what I was always told. But recently I read that adding solder makes hardly any difference to the resistance, because the resistivity of solder is a factor of 10 higher than copper. However, it does help because it increases the thermal mass. At least temporarily: in the long run, though, you're limited by how much heat the solder can dissipate. \$\endgroup\$ Sep 28, 2021 at 19:06
  • \$\begingroup\$ But a fillet of solder will be much thicker than the copper trace it is augmenting, so presumably useful in certain cases? \$\endgroup\$ Sep 29, 2021 at 12:30

I do not find 3.56 mm to wide, so it should not be a problem for the PCB construction. If you want to try to reduce copper width, you can do one (or more of the following):

  • Increase the number of layers, keeping in mind that internal layers dissipate worse the heat than external ones. Connect internal and external layers with via stitching.
  • Use heavy copper PCBs, but that would considerably increase the cost

For the 2 split power rails, it can work, but be sure that the current is well balanced between both rails


To quote the immortal Scotty. "Ye cannae change th' laws o' physics". In this case those are Ohms law (and the resulting power loss) and Onderdonk's fuseing equation.

You will just have to find a supplier that can supply heavier copper at a volume price you can accept.

A 'cheap trick', with diminishing returns, is to use multiple layers for the same trace. e.g. A 6 layer PCB is not much more cost than 4 layers but you'd get 33% more copper.

Another solution (that introduces it's own problems) is to use a thermally conductive PCB, e.g. AluClad but it would probably be cheaper just to use thicker copper.


You can "Load the tracks". This is an old timers term. Sometimes these are called Eltracks. What this entails is running copper wire along the hot track tacking down with the soldering iron. Thru hole componant leads or even fuse wire can be used it a pinch. This is an easy way for 1 or 2 prototype boards but not recommended for volume production.


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