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We have a power connector, which is capable of delivering 20Amps to a PCB. On the PCB, one layer is dedicated to Vdd, which is 5V. The board is quite large, 200mmx300mm.

The power connector is through-hole. In order to solder the connector, we assumed we would need thermal reliefs on the pads so that during soldering, we can heat up the pads properly. But then on the other hand, thermal relieving the pad reduces its current carrying capacity, increases resistance, and increases power consumption. Is this a correct assumption?

What are the issues that should be taken into account when designing for high current power inputs to large power planes?

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I would definitely use thermal reliefs. The resistance increases, but a thermal spoke's resistance will only be 1m\$\Omega\$ for a spoke 1 mm wide and 2 mm long. And you'll have 4 of them in parallel. All in all the increased resistance is negligible.

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  • \$\begingroup\$ your assumption being the board is 1oz copper? <strike>Also your count is off, there are 8 in parallel assuming top and bottom.</strike>oops, that assumes both top and bottom connect somewhere. \$\endgroup\$ – Jason S Sep 14 '11 at 12:34
  • \$\begingroup\$ @Jason - Yes, that's for 1 oz (35\$\mu\$) copper thickness. It's easy to calculate with: a 1mm wide trace has 0.5\$\Omega\$ resistance per meter length. \$\endgroup\$ – stevenvh Sep 14 '11 at 12:41
  • \$\begingroup\$ I've heard this before (that reliefs are short so they don't matter), but I don't understand it. After all, something like a 1/16W 0402 0 ohm resistor is extremely small and extremely low resistance... but I don't think it's going to handle much more than it's rated for. I can see how thermals are going to have a thermal sink into the nearby copper.. but any current imbalances would also be magnified by heat. Each individual spoke if treated as a trace would have about a 2.5A capacity (and only 1.25A on internal layers)... that would make me quite nervous. \$\endgroup\$ – darron Sep 14 '11 at 15:36
  • \$\begingroup\$ @darron - According to this calculator 2.5A through a 1mm wide trace causes a temperature increase of 50°C, which in practice will be a lot less, thanks to the thermal sink into the plane. I'm not so nervous about it. \$\endgroup\$ – stevenvh Sep 14 '11 at 15:46
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    \$\begingroup\$ @darron - the temperature coefficient of copper is positive (0.39%/°C), so a higher temperature due to a higher current should reduce this current again due to the higher resistance. System should keep itself balanced IMO. \$\endgroup\$ – stevenvh Sep 14 '11 at 16:51

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