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Jan 15, 2024 at 15:05 comment added Tim Williams And like I said, inductance is the bigger deal. A 10mm wide trace has about 1/10 the inductance of a 1mm trace. And a 1cm long trace has 1/10 the inductance of a 10cm long trace. Keep it wide and flat, and components close together (short loop length). There aren't design rules for this. Only engineering experience.
Jan 15, 2024 at 9:29 comment added Andr7 Thanks! @jusaca
Jan 15, 2024 at 7:11 comment added jusaca You can use copper sheet resistance for fast approximation. 35 µm copper has 0.5 mOhm / square. So a trace with double the length in comparison to width has 1 mOhm. For 33 mOhm the trace can be 66 times the width. 2 mm width would mean maximum length of 66 mm to stay below 1 V of voltage drop.
Jan 15, 2024 at 7:08 comment added jusaca As I said, I would decide it by specifying a maximum voltage drop. If for example your pulse voltage can not drop by more than 1V over the trace just calculate 1V / 30A = 33 mOhm maximum.
Jan 15, 2024 at 6:51 comment added Andr7 @jusaca, how do I decide the acceptable trace resistance?
Jan 15, 2024 at 6:44 comment added jusaca For µs-long pulses I would design for an acceptable trace resistance and corresponding voltage drop instead of temperature rise. Specify what voltage drop you allow and in combination with your 30A you get a maximum value for the allowed trace resistance. Now you can design trace width too stay below that value.
Jan 15, 2024 at 4:22 comment added Tim Williams It's most likely not an issue.
Jan 15, 2024 at 4:11 comment added Andr7 So how should I deal with the trace widths considering my traces lengths are short only?
Jan 15, 2024 at 4:08 comment added Tim Williams Long traces, where heat only flows laterally out from the trace.
Jan 15, 2024 at 4:06 comment added Andr7 - my double pulses are in us - what do you mean by temp rise is only valid for long traces?
Jan 15, 2024 at 4:03 history answered Tim Williams CC BY-SA 4.0