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I am trying to design a PCB which has high current traces on it.

PCB has 1 oz (35 um) trace thickness. Maximum current flow is going to be 12 Amps RMS.

I found this online PCB thickness calculator: https://www.eeweb.com/toolbox/external-pcb-trace-max-current

According to this calculator 5mm width is enough to hold 15 Amps current flow at 70 °C. However, due to some design limitations I need to get the PCB trace narrower as 2mm.

For 2mm width trace, the calculator points that it can hold 11.2 Amps current flow at 125 °C which is too high. To demonstrate the situation please look at the PCB layout below:

PCB layout

My question is, what happens if I put this PCB on work? I tried to keep narrow high current lines as short as possible but they are inevidable. I want to think that the thermal rise on narrow lines will be distributed over wide lines since they are conducted however, I am not really sure what will happen. Will those narrow lines burn out somehow? If so what are your practical advices?

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    \$\begingroup\$ 70°C over ambience or at 70°C then with what ambience? What when the ambience increases? And no, heat will not just flow away, especially not via the already hot traces around it. You can get a bit more capacity by making the traces thicker, either by using more copper, or by tinning them. \$\endgroup\$ – PlasmaHH Aug 31 '17 at 12:21
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    \$\begingroup\$ 4.3 uOhm and 12 amps is a power dissipation of 0.6 mW but I have doubts that the 2mm track that looks to be about and inch long and it doesn't appear to be a through going connection so it's anyone's guess. \$\endgroup\$ – Andy aka Aug 31 '17 at 12:25
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    \$\begingroup\$ @Alper91: You should probably calculate for the max ambient temperature you are expecting. 40°C is not uncommon which would make 45°C rise a whopping 85°C, nothing you want near the average electrolytic cap. What is on the other side? How many layers do you have? Maybe you can add some traces in inner layers or on the other side. \$\endgroup\$ – PlasmaHH Aug 31 '17 at 12:31
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    \$\begingroup\$ @Alper91: stuff inside a car is a few levels outside my experience level. Generally all parts seem to be "automotive grade" that can mean rated for higher temperature, potted (decreasing current carry ability) or conformally coated, more mechanically stable, higher safety margins and a lot more. For quality parts anyways. \$\endgroup\$ – PlasmaHH Aug 31 '17 at 12:44
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    \$\begingroup\$ @Alper91 from the layout I see that in theory you have plenty of space, to increase that 2-mm tracks. If this is not possible, you could put a solder mask opening, and have that track tinned, so that its effective resistance is reduced. I suggest also to perform thermal simulations. The "online" PCB traces calculator assume an "infinite -length" trace approximation, therefore their results are very conservative. I guess in your case, part of the heat (generated where the traces are thin) can be instead partly dissipated where the traces get wider. \$\endgroup\$ – next-hack Aug 31 '17 at 12:52
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I would recommend you to expose solder mask on that trace and use solder to thicken the trace. This is the usual way to increase current capability of such trace.

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    \$\begingroup\$ Doesn't really answer the underlying question though. Why do you think it's not safe to do what he's doing? \$\endgroup\$ – pipe Aug 31 '17 at 17:59
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    \$\begingroup\$ @pipe OP asked for practical advices. It is not very good to stress any part of board/circuit. OP stated that 12A over 2mm trace is a problem due to high temperature. It is obvious that my answer gives solution to that. \$\endgroup\$ – Chupacabras Aug 31 '17 at 18:21
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    \$\begingroup\$ It's not at all obvious that it is being stressed though. \$\endgroup\$ – pipe Aug 31 '17 at 18:28
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    \$\begingroup\$ @pipe Well, it is obvious. There are many trace width calculators. OP used one of those. It gives temperature rise of 125 °C. This is stress IMHO. \$\endgroup\$ – Chupacabras Aug 31 '17 at 18:51
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Copper foil will move the heat vertically into the air or into underlying traces or planes through the FR-4.

And the copper foil will move the heat laterally through the trace. The lateral thermal_resistance of 1 ounce/foot^2 foil is 70 degree Cent per watt per square.

A trace of 20 mil by 200 mils has 10 squares, and end-to-end thermal resistance is 10 * 70 = 700 degree Centigrade per watt.

From middle of the 200 mils to either end will be 350 degree Centigrade per watt.

Assuming heat is uniformly generated (a wrong assumption, because copper has 0.4% per degree Cent temperature coefficient) and that heat only flows along the trace, you'll get a parabolic rise in temperature from either end to middle of that 0.2" trace. For simplicity, assume you have 2 milliiWatts of heat generated in the entire 0.2" long (20 mil wide) trace. Assume 1mW needs to flow through 0.1" of trace, from middle to end. The R_thermal is 350 degree Centigrade per watt, thus only 3.5 degree Cent rise.

Summary: for 20 mil wide trace, 200 mils long (0.2"), with 2 milliWatts dissipated in that trace, the worst case rise will be 350degreeC * 1mW or only 3.5 degree Cent.

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