Short duration (10ms) surge rating of 1 oz copper? - Page 1
Forum section: electronics / projects

author: Jester
author_rank: Frequent Contributor
timestamp: March 27, 2021, 10:16:52 pm
content: For continuous 30A on a PCB with 1oz copper (50deg rise) an external trace width of ~500mils is required.Obviously for a short duration pulse of say 10ms and very low duty cycle say << 1%, how does one calculate the minimum trace width required?Trace length about 1.2"

author: TimFox
author_rank: Super Contributor
timestamp: March 27, 2021, 11:03:52 pm
content: The thermal capacity of copper is 0.385 J/(g K),i.e., 1 J of energy dissipated in 1 g of copper will raise its temperature by 1/0.385 = 2.6 K.As a worst-case calculation, ignoring any conduction through the insulating substrate or convection from the uninsulated side, that (along with the mass density and electrical resistivity of copper) gives you enough information to calculate the temperature rise from a single pulse of current through the conductor of given dimensions.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: March 27, 2021, 11:10:28 pm
content: I would expect things get very nonlinear as time reduces.For sub-mS pulses you'd probably be looking at thermal runaway leading to vaporisation type effects.

author: TimFox
author_rank: Super Contributor
timestamp: March 27, 2021, 11:17:54 pm
content: For 50 K rise in 10 ms, I don't think that thermal runaway would be a problem. The tempco of copper is 0.4%/K, or 20% in 50 K.

author: Jester
author_rank: Frequent Contributor
timestamp: March 28, 2021, 12:57:40 am
content: TimFox,Thanks for that.I started with the widest trace I can fit (75 mils) and if my quick calculation is correct I could go a lot narrower for such a short pulse.Area of trace is 90,000 sq mils or 0.09 sq inches, 0.0185 g of copper, R=0.0105 Ohms, P=9.45W, ~0.1J for 10ms, so the temperature rises by about 14.6 degrees.

author: Blinkenlights
author_rank: Contributor
timestamp: April 01, 2021, 07:42:51 am
content: Hi, there is an important concept missing from the thought process:The best way to calculate PCB resistivity is in "ohms-per-square" - take your track and divide it into a whole lot of squares, end-on-end.In this hypothetical case, the squares are 75 mil x 75 mil or 5625 square mil total. You can have as many of those as you want - in other words, the length of the track makes no difference to how hot each square will get - the heat per square is constant, you just have more squares in a longer track.My working, based on 1 oz copper being close to 0.5 milliohms per square:The example used was 30A current, so given i2r the heat produced is 0.45w per square.75 thou track --> 5626 sq mil (I can fit 25,600 of those in a square foot - gotta love our non-si PCB industry units)1 square foot of 1oz copper - thats 28.4g, divide by 25,600 and the weight of copper in our 75x75 is: 1.107mg1.107mg times the specific heat capacity of copper (0.385, see above posts) gives 0.426mj required to raise temperature by 1 degree celsiusGiven the 0.45w per square heat production, the copper will heat up by 1 degree celsius in 0.426 / 0.45 in 0.95msIt might not make any substantial difference to the Original Poster, but I think recording the correct reasoning will assist others following the thread later. Perhaps others with more physics knowledge can correct or improve the numbers. As a PCB designer (for 40 years) I still wouldn't want to put that much current through a track that narrow, but your numbers seem to work out.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: April 01, 2021, 01:48:16 pm
content: Google Onderdonk and related docs.