In MOSFET datasheets you often see something like junction to ambient thermal resistance of 40K/W being achieved using a 1" x 1" pad on the drain.

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The main problem with this approach is that it takes up a lot of space on the outer layers that could be used for components or routing. I was wondering how much of a performance hit in K/W it would be to put it on an inner layer of a multilayer board. If one looks around in online forums people say things like "it will be much worse", but I have yet to see anyone ever attempt to quantify it with any math, and my intuition says that under the right circumstances it may be only a few percent worse.

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I am considering the case where we put the thermal pad on layer 2, 10 mils down from the surface and join it to the MOSFET drain pad using 9 x 12 mil vias with 1oz plating. The vias would be filled to prevent wicking the solder paste out from under the part.

  • FR4 might have a thermal conductivity of 0.25W/m/K. Though that's relatively high, if the area is wide and the depth is thin the overall thermal resistance can be low. A 1" x 1" square that is 10 mils thick has a thermal resistance of 10 mils / (0.25W/m/K) / 1" / 1" = 1.57K/W.
  • The tables I have say each 12-mil diameter via with 1oz plating and 10 mil length has a thermal resistance of about 24K/W. So 9 of them would have a combined thermal resistance of 2.66K/W.

So (if we ignore the ground plane) and make the worst-case assumption that all heat goes down through the vias to the pad and then back up through the FR4 it looks like the thermal resistance instead of being 40K/W would be 40K/W + 2.66K/W + 1.57K/W = 44.23K/W which isn't actually that much of a difference. And since we ignored other beneficial thermal paths like conduction to the ground plane or direct radiation from the part itself and lateral conduction through the FR4 the real result should actually be better.

Is that calculation roughly correct?

And if I have a ground plane below the thermal pad, things get even better because heat can go from the thermal pad into the ground plane with a thermal resistance of only 1.57K/W.


1 Answer 1


You are absolutely correct. Most of the resistance in the thermal path is in the transition from the PCB through the air to the outside world! If you use enough vias and your thickness is as you claim, it will wok fine. Your calculations are proving it.

A point of caution: you then have more space for other components, but you will of course heat up those other components because they are on top of the heat spreader.


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