If I use a trace width calculator to find out the width of copper I would need for a specified current. Now for my situation I need 120-150A of current (peak), 40 - 50A continously. I am using a 4 layer board and using copper planes of 77mm thickness on all of the 4 layers with solder mask removed for the top layer(max width I can go as I am limited by the dimensions of the board and enclosure). Now the internal layers cannot take as much current as there is no way for the heat to escape in the air as it is closed on both sides.
So is it better if i go for 2oz on the outer layers and 0.5 - 1oz on the inner layers ?
For a given current(in the trace calculator kicad or saturn pcb toolkit) if i give 10mm trace width and I get 5A of current is it the continous current that i can take out or is it the peak current for a given incease in temperature ?
If i have 77mm traces on all the 4 layers(46A on a 2oz layer according to saturn pcb toolkit), can i just 2 times it as i have the same copper width and thickness on the bottom layer and say top and bottom layer can carry 92A(only from the top and bottom layer) ?
And saturn pcb toolkit shows that I can have 46A on the internal layer as well for a 77mm and 2oz copper. Is it correct ?
- I am using 0.3/0.7 saturn pcb toolkit says I can have 0.9A per via for a 10 degree delta in temp. So I would need 166 vias for 150A current. Not that the current goes through the vias but they also do heat descipation and distribution among all the layers....how do i determine the optimum number of vias such that all the 4 layers are maintained at equal temperature(good temperature distribution among all the layers) and also make sure that I dont fill up the copper plane with vias that I think can cause decrease in copper mass which would be bad for heat(i assume).
My thinking says(im very bad at thermal) - My mosfets are SMD on the top layer so most of the temp is generated on the top layer and then needs to be transfered to all the 4 layers so that temperature is not conentrated at one spot.
