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I have got a part in a SOIC-8 EP package. The "EP" indicates it is a package with a exposed pad which can transfer heat to the PCB. I would like to have a better understanding how much pcb surface I need to cool the part at different power consumption levels. Lets say 1 Watt 1/2 Watt and 0.1 Watt.

I did read some white papers. They basically tell:

PD = (TJ − T A)/θJA

Where θJA = theta ja (junction to ambient) in C/W TJ = junction temperature rating in C T A = ambient temperature in C PD = power dissipated in watts

θJA can be split into three parts which add up:

θJA = θJC + θCS + θSA

Where: θJC = theta JC (junction to case) °C/W θCS = theta CS (case to heatsink) °C/W θSA = theta SA (heatsink to ambient) °C/W

The datasheet of the part tels me: θJC = 10 °C/W TJ = 150 °C

I can think of a ambient temperature lets say 22°C

But then I am still missing the following: θCS and θSA. I could imagine that θCS is negligible, is this true? θSA I find difficult I plan to use via's to get the heat to the other side of the PCB but I can not find any data giving me a idea what number I can use for θSA. I also find it hard to figure out if I need 35um (1oz) or 70um (2oz) PCB.

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    \$\begingroup\$ The larger the dissipative area, the better (up to a limit). \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 25 '17 at 15:28
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This is actually a pretty deep question. Fortunately, there is an extensive layer of literature on this subject. You are basically on right track, but, for some reason, didn't get to the right articles.

Yes, if the pad is soldered down, you can assume the θCS as zero.

Regarding vias, typical thermal resistance of via over a 1.6mm FR4 PCB is 130 to 250 °C/W, depending on stack-up and via size. So you would need few of them to have any effect. Or you can make one 2-mm diameter via, and fill it with solder. There are plenty of calculators on this subject, Google for "Thermal resistance of via calculator".

All details are perfectly explained with formulas and practical examples in this Application Note AN-2020.

Final results will depend on details of ambient conditions, whether the board is vertically oriented or horizontally, are there obstacles to natural convective airflow, or there might be some forced ventilation around. A thermal image of PCB will help tremendously to evaluate the board thermal condition, and, if necessary, design corrections should be made.

But for a 1W dissipation and 3x3mm thermal pad soldered to 1.5 oz PCB, I wouldn't worry much, given TJ = 150 °C.

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The standard copper foil --- 1 ounce per foot square, 35 micron thick, 1.4 mil thick --- has thermal resistance of 70 degree Centigrade from one edge of the square to the opposite edge of the square.

Thus 0.1 " long trace, of width 0.01", aspect ratio of 10:1 and 10 squares of foil, has thermal resistance along the trace of 10 * 70 = 700 degree Centigrade per watt.

A heat spreader like this has 70/8 = 9 degree Centigrade per watt

schematic

simulate this circuit – Schematic created using CircuitLab

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This is a good resource: HeatSinkCalculator
They do have limited functionality free use.

Also you can do thermal analysis using TI's WebBench.

The LED manufactures have some good documentation on PCB design.
Cree: Optimizing PCB Thermal Performance

Regarding thermal vias I did a lot of research and the consensus was to use 15 mil holes spaced on 35 mils centers and a maximum of 15 holes. And use 2 oz copper.

I found the thermal via method to be inadequate. What I did was extend the thermal pad on the PCB top layer out one end of the chip with a larger pad with a 4/40 or 3mm screw hole to mount a heatsink.

My thinking on this method was the thermal resistance would be lowest on the top layer.

Another thing I did was to use bare copper PCB surface area when a heatsink was not necessary. A oxidized copper pad has much higher thermal emissivity.

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LED

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