I got 30 buck converters on 4 layer pcb (70x130mm). Each will dissipate ~2W for typically 30 seconds followed by a very long break >10min, and I'm worried about overheating. Although it is not a safety issue (the chips are protected) i would like to avoid running into these problems.

My approach:

  • 2 Layers full copper with a lot of thermal vias. The other 2 Layers are ~90% copper.
  • The thermal mass of the large inductors should help
  • A 120mm fan directly above the components
  • The airflow should increase Rca (thermal resistance case to ambient) a lot

If this is not enough to get rid of the 60W of heat, I could add an aluminium block on bottom side.

Question: Do you think this is a viable solution?

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  • \$\begingroup\$ You said: the chips are protected - so what is it that you are trying to achieve? \$\endgroup\$
    – Andy aka
    Commented May 11, 2020 at 8:58
  • 1
    \$\begingroup\$ Blowing air directly down might result in stagnation for the centre few chips. This may not matter if the PCB is strongly heat conductive across the board. However, blowing air across the board would be better. Either that, or if the present fan position is good mechanically, then move the low heat production components to the stagnant centre, and the high heat components towards the edge of the board, where the airflow is good. \$\endgroup\$
    – Neil_UK
    Commented May 11, 2020 at 9:06
  • \$\begingroup\$ Andy: I don't want the protection to become active. Neil: Thx. I can rearrange the fan and add ducts to create a better flow. Maybe I could add walls on 3 sides of the fan so the air can exit only on one side? \$\endgroup\$
    – 1uk3
    Commented May 11, 2020 at 9:09

2 Answers 2


60W x 30 seconds = 1800 joules

That spread over 10 minutes would be an average dissipation of 3W for a rather large board.

If things are not overheating in those 30 seconds you should be fine. Otherwise a heatsink to dump that energy into can help make thigs easier.

How much heatsink? Well to raise a metal a temperature is joules per gram per degree c. For aluminium this is 0.9, so lets say you want the heatsink to not heat more than 40 degrees above ambient for the pulse.

1800 joules / 0.9 / 40 degrees = atleast 50 grams of aluminium.

This does not quite tell you how quickly the heatsinks temperature can dissipate that energy. So it will probably be slightly warmer on subsiquent pulses. But you do not need much metal to absorb thr worst part of the pulse.

  • \$\begingroup\$ Thank you! Didn't think think about calculating the temperature rise. The inductors weigh ~80g (copper + ferrite). If i now subtract the removed energy from the 60W, it should be cool enough ;) \$\endgroup\$
    – 1uk3
    Commented May 11, 2020 at 11:04
  • \$\begingroup\$ Ferrite is 0.8, and copper is 0.39, not quite certain what the ratios will be, but even if all the weight was copper, should only result in a 58 degree rise, so in your application it will be less than that. and you also have the specific heat of the PCB 1.15 \$\endgroup\$
    – Reroute
    Commented May 11, 2020 at 11:23

Your PCB will have about 1/2 second thermal time constant between each regulator. [One cm is 1.14 seconds, 2 cm is 4X longer, 3cm is 9X longer)

The heat rise will be approximately uniform, during your 30 second pulse.

IMHO you need some cooling air.

The central region can dump heat in 4 dirctions, (or even 8 what with corners being used) so dead-spot should not matter.

But with 4 degree per watt thermal spreading resistance out from that central region, and with 30 watts in the central region, you'll have 4 * 30 = 120 degree C rise.

If you redo the PCB, consider 2 ounce/squareFoot for the 2 planes.


By the way, you can model this in SPICE with a 7 by 13 resistor grid (at the 1cm resolution, if you feel that is adequate). Make each resistor be 70 ohms, so you get that 70 degree C per watt thermal response. Then inject 3 amps into each of the interior nodes. Ground the periphery ????? no. That is not valid.

As the other answer suggests, compute the thermal capacity, for each square centimeter and add that to each interior node as a lumped capacitor.

By the way, are your traces wide enough to take the 2,000 amps??? off board?


In the thermal thinking, I use the thermal resistance of standard copper foil (1 ounce / squareFoot) which is 70 degree C per watt per square of foil.

Since there are two inner layers, solid sheets, the lateral Rthermal is 35 degree per watt per square, for any size square.

Now draw a grid, and heat any one of the interior squares. You will see the eight adjacent squares as paths for heat to exit, thus Rthermal drops by 8x, to 4 degree C per watt.

You can include the next set of surrounding squares, eight of them, edges 3x the original size), but the component density suggests the iteration is not needed.

  • \$\begingroup\$ You seem to be quoting information that is not in the question. E.g. the C/ W of the chips and the total current of 2000A, Would you mind sharing where this info might be? \$\endgroup\$
    – Reroute
    Commented May 11, 2020 at 20:42
  • \$\begingroup\$ @Reroute Ask him how long it takes for heat to spread through a silicon crystal cube 1m in dimensions. \$\endgroup\$
    – DKNguyen
    Commented May 11, 2020 at 23:29

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