# How to choose trace width for very high current PCB?

I am currently designing a PCB layout for my team's high current, high voltage project and am having trouble with choosing my trace width.

The circuit is composed of 12 inductors placed in series with short traces connecting them together (about 5mm max length). We are feeding 10μs pulses at ~300A/1000V at a maximum rate of 10 pulses per second by discharching capacitor banks into the coils. The capacitors are recharged by an external source.

I have looked for resources on the subject, but nothing I could find applies to the level of current and voltage we are applying.

Is there a rule for determining trace width for very high current pulses?

• By googling up a sufficiently feature rich trace size calculator Jul 15, 2016 at 15:19
• IPC-2222 and IPC-2152. Jul 15, 2016 at 15:23
• Could you use solderable (e.g. brass) standoffs screwed into the board and thick copper wire between them instead of using a PCB as a conductor? That way you can still monitor the voltages between the inductors if desired, and it might be easier to replace an exploded inductor. Jul 15, 2016 at 17:05
• @AndrewMorton Sadly, that's not possible as we need it to be as small as possible. We actually used that technique for our initial developpement. Jul 15, 2016 at 17:07

It's not the current that matters, it is heat. The current will heat up your copper on top of the FR4.

The heat generated is current squared time resistance of (trace + solder joints + component leads). The square in there is what makes the high currents so scary, even ath the 1:10k duty cycle you have on the positive side of the balance sheet.

So I agree with Dans answer very much: a trace suitable for a 10A continous current should do the trick. "Suitable" menas that the heat does not increase your PCB temperature more than your specifications allow for. Which as usul depends on many factors like type of PCB, overall heat sources on it, hos the heat is dissipated to the environment etc.

hase

10us pulses at 10 Hz, so a duty cycle of 0.0001?

300A, so I^2T is 90000 * 0.0001 = 90.

Square root of 90 = ~9.5, so you need at a minimum a track that can handle 10A continuous (I would use something good for a few times that, maybe base on the size of the inductor pads), job done.

Now finding inductors to take that abuse. good luck.

Regards, Dan.

• Thanks for the info! As for the inductors, we are using custom coils made with #30 AWG, we've had no problem so far. Jul 15, 2016 at 20:35

I would think about using PCBs with thick copper, 70 or even 105 µm instead of 35 µm. Some manufacturers offer boards with 210 or 400 µm copper. If a trace 20 mm wide and 35 µm thick is too large, think about 2 mm wide traces with 400 µm.