# PCB as heat sink - how to calculate the area

I got an Inverter of 6 MOSFETs and I need to estimate the PCB area for keeping the junction temperature below 100 °C. The losses per FET are 0.21 W which leads to a max. Rthja = 143 K/W (junction to ambient). The manufacturer provides Rthjc = 5.3 K/W (junction to case).

Does anyone know an application note or another source to estimate the required Rthca (case to ambient) for passive cooling? e.g. onsemi refers to thermal simulations (like many others). And I think for the first approach it's acceptable to disregard effects caused by adjacent components.

To estimate the thermal capacitance of the PCB heatsink would also be help but the main problem is to find an easy aproach to estimate the thermal resistance/size of PCB.

Many thanks!

• Did you mean something else by the second paragraph? You already have a calculation (not even estimate!) for RthJA, and CA just subtract JC from that. Or did you want to know that this is the correct calculation? Commented Jan 3 at 11:22
• From that calculation I know that I need a heatsink with Rthca = 138 K/W. I have to use PCb copper as heatsink and I want to know which size of copper would fit to the calculated Rthca. sorry for the confusion Commented Jan 3 at 11:40
• It really sounds like you would benefit from external proper heatsinks, like classic TO220 ones. They are superior to PCB. Commented Jan 3 at 15:01

Calculations aren't really possible, more than rough figures or guides, as convection is a complicated boundary-condition problem, and even just conduction and radiation depend on material geometry and property, nearby components, sight lines, etc., to represent fully.

That said, the ballpark figure I internalized is to say: a thermal resistance of 150 °C in2/W. Divide by surface area to get thermal resistance. This is, within 20% or so I think, for modest temperature ranges -- what you would see in commercial equipment, say. Convection is nonlinear, having lower Rth at higher temperature/differential, and especially so when taller (chimney-like) structures are present, or other airflow.

Typical PCB material has a thermal spreading distance of about an inch, give or take number of copper layers and thickness, and how contiguous the pours are. That is, no matter how wide you pour, heat only spreads out a distance proportional to material conductivity. (Or maybe area is proportional, I forget.) Which defines the active dissipating area around a component, and how close components can be before they interact (raise each other's temperature).

Note that PCB has two sides, so the area of a free PCB should be counted twice (in parallel).

If it's in an enclosure, you might consider the area restricted, depending on shape; but for a sealed enclosure, keep in mind it counts double in series, because convection occurs at both inside and outside surfaces of the wall. Even for poorly conductive enclosures (ABS plastic for example), it can pay to use thermal interface material / gap pad between board and enclosure, to short-circuit the internal convection penalty.

Alternately, you can use the compiled data, for a defined footprint on specified material. This is simply how RthJA is measured in datasheets. You might use minimal-footprint data as a worst-case basis, and assume that components placed closer together, but over copper plane, perform about as well together, and double-check with an area dissipation calculation like this. Finally, it can be tested, and verified with a thermal IR camera for example.

• I've used this application note from TI as it has collected much of the relevant info into one note Application Report SNVA419C AN-2020 Thermal Design By Insight, Not Hindsight. ti.com/lit/an/snva419c/snva419c.pdf Commented Jan 3 at 21:48
• Thank you so much, that rules of thumb were what I was looking for. Just a quick and easy approach for the necessary PCB area. And the result matches with the result from this calculator: heatsinkcalculator.com/pcb-temperature-calculator.html Commented Jan 4 at 12:01