Having designed the PCB for my double pulse test circuit I'm facing problems with the trace widths. Operating current is 30A and I checked on the trace width calculation sites and they are coming up with traces several centimetres wide. However, the MOSFET has a tiny leg that gets soldered into the track that nowhere near matches the required trace thickness. Also I am not able to route with these widths. What am I missing?
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- \$\begingroup\$ I used to wonder this too and still do sometimes. The pin is much thicker than the trace. Several cm wide is too wide though. You entered in something wrong. \$\endgroup\$DKNguyen– DKNguyen2024-01-15 03:54:27 +00:00Commented Jan 15, 2024 at 3:54
- \$\begingroup\$ I am getting traces of around 1 cm that too when I go for 2 oz and 30 degree temp rise \$\endgroup\$Andr7– Andr72024-01-15 03:59:08 +00:00Commented Jan 15, 2024 at 3:59
- 1\$\begingroup\$ This answer might help \$\endgroup\$user319836– user3198362024-01-15 05:12:31 +00:00Commented Jan 15, 2024 at 5:12
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10 (This is a duplicate question, but it's more effort to search for duplicates than to answer, so it's one of those evergreen questions.)
Temp rise is only valid for long traces, where heat only flows laterally out from the trace.
Short traces -- on order of a few cm -- have significant heat flow along their length, i.e. can be heatsunk by component pins/leads, wider sections, etc.
Simply keep the connections short and it doesn't matter.
You need to keep traces short to reduce loop inductance anyway, so I doubt it's a problem (but, I haven't been following your series of questions closely, and maybe you've been given bad advice, I don't know).
On top of that, presumably your double-pulse circuit is just that, pulsed. Traces take seconds and minutes to heat up. If your pulse is that long, I fear for your transistors. For pulses shorter than that, use the RMS current instead.
- \$\begingroup\$ - my double pulses are in us - what do you mean by temp rise is only valid for long traces? \$\endgroup\$Andr7– Andr72024-01-15 04:06:54 +00:00Commented Jan 15, 2024 at 4:06
- 1\$\begingroup\$ 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. \$\endgroup\$jusaca– jusaca2024-01-15 06:44:09 +00:00Commented Jan 15, 2024 at 6:44
- 1\$\begingroup\$ 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. \$\endgroup\$jusaca– jusaca2024-01-15 07:08:26 +00:00Commented Jan 15, 2024 at 7:08
- 1\$\begingroup\$ 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. \$\endgroup\$jusaca– jusaca2024-01-15 07:11:47 +00:00Commented Jan 15, 2024 at 7:11
- 1\$\begingroup\$ 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. \$\endgroup\$Tim Williams– Tim Williams2024-01-15 15:05:04 +00:00Commented Jan 15, 2024 at 15:05