It seems that the windings I calculated will fit into the core.
Can someone please evaluate this:
12A running through 2oz (70um) of copper tracks with a temp rise of
10degrees C requires 1.2um trace width? I got that info from
http://circuitcalculator.com/wordpress/2006/01/31/pcb-trace-width-calculator/.
It doesn't really match the chart of my PCB manufacturing company.
According to their chart 12A with a temp rise of 100degrees C I
require a 2.5mm track width at 70um?
When designing power electronics you should worry more about power loss and
voltage drop. If this is done the temperature rise will not be significant.
The PCB trace width calculators are designed for single layer boards with
only trace area surface convection cooling. In multi layer PCB much of the
trace heat is conducted to other layers and in shorter traces to their end
points. Those calculators also do not include trace length which is absurd!
I normally allow 12.0mV per trace voltage drop. This yields a A*12mW power
loss and a 12.5mV/amp trace resistance. If this is done then the temp rise
in any trace of any length will be less than 2.5C. In any trace you know how
long it must be so you need only calculate its width. Use this formula:
W=L*A/25*oz where W +L = trace width + length in inches.
A = amps oz = cu thickness 1oz =1.4m"
So for your situation you have 12A and 2oz and I will assign a 1.50" trace
length and not worry about temperature rise, so:
W= 1.5*12/25*2 = 0.36" = 9.14mm Contrary to popular belief, In PCBs,
length is important!
Cheers,
Harry
