J
John Larkin
Your 70K/w for a 20 mil dia hole is the same as my 50C for a 28 mil hole.
500uR per 1 oz square cu works for me too.
Given a fixed heat transfer area and no solder fill, a lot of little vias
are better than one big air hole.
Let's check your 150K k/w with a 20" x 0.010" 1 ohm cu trace:
Rt= 0.1*20/(.010*.0014) = 143K.. not too bad!
Do you use/agree with any of the trace width calculators or even IPC(2222)
that are based on convection cooling? They work for me when the trace is
1.50" long, shorter traces can be thinner, longer traces must be fatter.
Most PCB have internal planes that spread heat and thermal conduction in
traces greatly out performs thermal convection. I strive for <12mV drop in
all traces. This will keep temp rise in all traces to <5C due to conductive
cooling at the trace end points.
Right: usually I care about voltage drop before I care about trace
heating, and if you keep the voltage drop down, the heating is no big
deal. Usually.
John