I'm making a 12V power distribution box, and it's my first time I work with high currents. This is for a boat, and the idea is to run low-power up to the switch panel and having the high-power cables as short as possible.

Simplified schematic:

Simplified schematic

Input on X1-1, F1 is a 25A fast acting fuse, K1 is a relay switch and finally output X2-1. Ideally the X1-1 should be replaced with a copper bar soldered to the PCB to distribute power to the 10 similar on/off circuits I'm having. 250A is too much for any PCB I suppose... :)

The trace with calculator says I need a 25.5mm trace width. But the trace from the X1-1 to F1 is less than 10mm, and it will be a ugly blob of copper 10mm long and 25.5mm wide.

Hope someone can help me do the math here.

Ok, my math from here assumes a 2oz (\$70\mu m\$) board, single-sided and using through-hole components and still standing air as ambient:

$$R=\frac{\rho \times l}{w \times h} = \frac{1.7 \times 10^{-8}\times 10mm}{12.7mm \times 70\mu m}=191.23\mu\Omega$$ $$U_{trace}=RI=191.23\mu\Omega\times 25A=4.78mV$$ $$P_{trace}=IU=25A\times4.78mV=119.5mW$$ $$\Delta T=\frac{P}{Sa\times h}=\frac{119.5mW}{(1cm\times1.27cm)\times 0.001W/cm^2/^\circ C}=94.1^\circ C$$

It seems my \$10\times 12.7mm\$ blob is too small to dissipate the power adequately. If I use a \$35\mu m\$ copper board and double the area I get the \$\Delta T\$ down to about \$47^\circ C\$.

Are my calculations correct? If yes, will it also perform equally after I build it?

  • \$\begingroup\$ What copper weight are you designing for? If you're using the standard 1 oz, be aware that up to 2 oz copper is readily available, and thicker can be found but might impact lead time and cost. \$\endgroup\$
    – The Photon
    Commented Dec 29, 2012 at 20:32
  • 2
    \$\begingroup\$ What copper thickness are you assuming? High-current designs use 2oz (70 um) or even 3oz (105 um) copper layers. This isn't a project for a dirt-cheap 1oz (35 um)-only prototype service. Also, why use a PCB at all? And why are you distributing so much power at such a low voltage? \$\endgroup\$
    – Dave Tweed
    Commented Dec 29, 2012 at 20:35
  • \$\begingroup\$ @Dave, I added answers to most of your comments in the post. I want to use a PCB because there are other parts of the circuit that doesn't show on the picture and isn't relevant for the question. They could be solved by wires and bars, but it would be cleaner to use a PCB. My PCB service can deliver anything I want (or can afford). :) \$\endgroup\$
    – rozon
    Commented Dec 30, 2012 at 4:21
  • \$\begingroup\$ Where does the 0.001 W / (cm^2 C) term come from? Turned around that's 1000 C / W for a 1 cm^2 or 155 C / W for a 1 in^2 area (just to put it in units I'm more familiar with). That's much worse than the usual values used to figure heat dissipation for surface mount components. Assuming still air may be killing you --- convection (air movement generated by your heat source itself) should be considered as a major contributor to heat flow. \$\endgroup\$
    – The Photon
    Commented Dec 30, 2012 at 4:23
  • \$\begingroup\$ @Photon: I got it from this app note: ti.com/lit/ml/slup230/slup230.pdf \$\endgroup\$
    – rozon
    Commented Dec 30, 2012 at 4:33

2 Answers 2


If F1 and K1 are through hole parts, you could put two tracks in parallel, one on each side of the board. Each track would be roughly 10 - 15 mm wide.

Another option, Don't draw the connection as a track but as a polygon or fill. Most of the current will still follow close to the shortest path, but the added area will help to dissipate heat.

Also, be aware that for such short distances, the current crowding to get to the individual pins of your components may dominate the overall resistance of the connection. Using larger holes (requiring components with larger leads) will help with this.


First, I'm not sure how this plays in, but this ampacity table claims a 0.25 inch trace in 2 oz copper is sufficient to carry 24.5 A with 30 C temperature rise. This is about 1/4 of the number you are coming up with (25 mm). Unfortunately the source is not clear about what assumptions went into their calculations.

Second, the math you added in your edit looks fine.

As you see, increasing the copper area allows you to dissipate more heat. There's no reason you should only increase the area by increasing the trace width. You can simply create a large copper area of whatever shape is convenient the closer to a simple square or circle, the better), so long as it connects your input and output, and the heat will spread readily to allow dissipation over the whole area (with, of course, some concentration in the region where the heat is actually being generated).

Third, no back of the envelope calculation is likely to be especially accurate for these kinds of thermal estimates. If you really want to get a good estimate, you should look into a thermal analysis tool like FloTherm or Ansys IcePak. These tools will both improve the ability to estimate convection effects accurately, and take into account "edge effects" that come into play because your "trace length" is so much shorter than your "trace width".

  • \$\begingroup\$ Yes! But it would require several vias to get the heat transfer adequate and it would complicate the calculations. But you are right, this is a pretty slick way to get the numbers in your favor. I am reluctant to use double sided boards for this project though. \$\endgroup\$
    – rozon
    Commented Dec 30, 2012 at 4:29
  • \$\begingroup\$ Found another ampacity chart that says you need a PWB trace with only about 700 mil^2 cross section area (about 1/2 what you're proposing) to carry 25 A with less than 30 C temperature rise. endless-sphere.com/w/index.php/Ampacity \$\endgroup\$
    – The Photon
    Commented Dec 30, 2012 at 5:17
  • \$\begingroup\$ I really value doing math, even on the back of a envelope, as I learn the reasons behind the simulations and is able to guesstimate a better starting point. In the end I will probably do the calculations for the square needed for 1oz copper and probably use 2oz for the production board. Better safe than sorry... \$\endgroup\$
    – rozon
    Commented Dec 30, 2012 at 5:42

Another option you have is to leave the traces uncovered with soldermask.

Then, you can effectively increase the thickness of the trace with solder when you're assembling the board.

enter image description here
Completely bare traces

Another alternative is to leave the solder mask off the traces, and then solder additional copper wire down onto them during assembly. enter image description here
I've done this with solderwick, myself.

  • 6
    \$\begingroup\$ Keep in mind that the resistivity of solder is about 10x that of copper, so building up with solder alone isn't as effective as you might think. Copper wire or solderwick is a much better idea. \$\endgroup\$
    – Dave Tweed
    Commented Dec 29, 2012 at 22:28
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    \$\begingroup\$ The other downside of using solder as a supplemental conductor is that it, by design, has a low melting point. If you get into thermal issues due to overcurrent or under-design, having solder-as-conductor melt off your board will get you into runaway fast (solder melts, cross sectional area goes down, temp goes up further, more solder melts...). Really high current is best left to laminated bus bar structures (for high frequency switching) or point to point wiring (for normal relays and fuses). \$\endgroup\$
    – HikeOnPast
    Commented Dec 30, 2012 at 0:38
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    \$\begingroup\$ Nice pictures. I do feel strongly on doing the math, and adding solder to the traces is very hard to guesstimate. \$\endgroup\$
    – rozon
    Commented Dec 30, 2012 at 4:24
  • \$\begingroup\$ A lot of this depends on the volume of boards you're making. If you're just making one or two boards, Soldering supplemental wire down on top of the traces is a reasonable cost. On the other hands, if you're doing this in production, really the only option aside from heavier copper is flowing solder onto the traces. \$\endgroup\$ Commented Dec 30, 2012 at 14:33
  • \$\begingroup\$ Anyways, the technique of solder-coating the traces may get you the equivalent of another 1 oz or 1/2 oz of effective copper thickness. If you have a 1 oz copper board, and you really need 4 oz, you're stuffed. However, if you're right at the edge of what's achievable with 1 oz copper, it's a cheap way to improve board performance. \$\endgroup\$ Commented Dec 30, 2012 at 14:35

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