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Copper heatsink area

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for work

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Does anyone have any equations to calculate the area needed of a copper sheet to dissipate power? I have looked everywhere and cant find any specific formulas.
 

SunnySkyguy

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Since FR4 is epoxy is a thermal insulator or rather the conductance is <<10% of copper, you are basically only using 1 side of copper and as I recall forced air can improve this up to 5x with 1m/s air velocity over the surface depending on geometry of obstructions, lack of eddy currents etc. Vias will improve this but also have low conductance so hundreds of holes per sq.in. to utilize both sides. Whereas solid aluminum or MCPCB is far better.
http://www.daycounter.com/Calculators/Heat-Sink-Temperature-Calculator.phtml

e.g. using 1oz copper 43 deg C/Watt per square inch on both sides or
with FR4 substrate ~80 deg C/(W*sqin)

Lots of sources and it depends on thickness, material, convection resistance, air flow
https://www.google.ca/search?q=copp...X&ved=0CDYQsARqFQoTCKPSs5rXiMYCFVcUkgodu24BtQ

40Deg C may be ok max junction temp rise, 80 deg C rise is bad, and 10deg C rise is an excellent design ( rise above max T ambient)

Tjcn is more important than Tcase. where Tjcn-Tcase = Rjc*Watts ( jcn temp rise)

Also remember that with many nearby devices the local ambient on the board is raised from the outside ambient T.

so consider 2 sqin/Watt for 1 sided exposed copper to free air convection.
or <1 sqin/Watt for MCPCB in free air

PHysical design and materials is critical to thermal design so get used to mechanical design for electronics !!

Then each layer is just a resistor and then you can use Ohm's Law.

Watts* (Rjc+Rc-pcb + Rpcb-air + R1+R2 ....) = Delta C rise

where R is each layer thermal resistance for given size.

 

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