Thermal Management

[jporter]

Greetings,

I have a 4oz board with a specific trace, designed for a given current at a 10 deg C rise over ambient. A terminal is soldered at a PTH on this trace.

In addition to classical calculations, I have a significant amount of test data to show the temperature induced by the mating device at this interface at various ambient temperatures. There will be no heat sink and I am ignoring convection for the time being. My question is, how can I calculate a size of additional copper at the pad to dissipate the heat induced by the device, given the t-rise and/or power dissipated so that I may (theoretically) maintain the 10 deg rise over the majority of the trace?

All comments and suggestions are appreciated.

Jason

[jporter]

After a bit more thought, this is probably a more complex calculation than I had originally thought. In looking at the thermal capacity of a small volume of copper, while ignoring convection, the amount of energy that it can assimilate is very small. Although, being self-admittedly thermally challenged, to me this means that the object would reach a temperature equilibrium in a very short time and would, therefore, transfer heat to the remaining system accordingly.

[robert Tarzwell]

your question is a good one and onethat most designers and PCB people dont know much about.

There are general heat/temp calculators out there and all are generaly correct and incorrect. The problem you face is one of to many factors to be easy to calculate, when I design for heavy copper I usualy try and get as close as I can then add a "guess" factor, as you increase copper thickness you do not add to the surface area you only decrease the resistance of the trace. And that only reduces heat generated and subsiquently temp by a factor of 1/2.
W=IR2.

To lower the temp around terminals I have had good sucess with lots of vias, they act like thermal heat sinks and increase the convection/radiation off the board. If you need further assistance you can e-mail me direct at rtarzwell@megadawn.com

Not a blatant advertisement but I did write a book on heavy copper and the problems associated with its design and manufature and sell it on the net at cost to cover net fees and printing. Its avalable at megadawn.com
best wishes robert tarzwell

[robert Tarzwell]

sorry I cant type the formula should be w=i2 *r which means the wattage generated halfs with a decrease in the resistance of one half. or a doubling of the copper thickness.
So to go from 4 oz to 8 oz means the wattage generated say 30 watts would decrease to 15.
sorry about that.