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I have a PCB with 10 LEDs (reference : SPHWH2L3D30ED4TPK3) which will each dissipate 3 W (forward voltage of 3 V and 1 A passing through). The PCB is a circle with a diameter of 20 cm and LEDs are evenly distributed.

I need to know, when all LEDs are on, what the temperature of my PCB will be. But how can I approximate it ?

The datasheet of the LEDs is here : http://www.to-nichi.co.jp/pdf/led/LH351B.pdf

The only thing that they mention is the junction to solder thermal resistance, which is max. 6 °C/W. Normally the formula is : T_J = P*R + T_A. But I do not have the junction to ambiant resistance...

If you have any advice it would help. :) Thanks

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  • \$\begingroup\$ I think you're concentrating a bit too much on the LED to PCB interface. What would interest me is how hot would a circle shaped PCB of 20 cm get when 10 x 3 W = 30 W of power is dissipated into it. I expect that there are calculator tools to find this value. Then use the LED to PCB thermal resistance to find the temperature inside the LED. \$\endgroup\$ Commented Jan 14, 2020 at 9:57

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The “junction to ambient” thermal resistance is equivalent to a resistance parallel to the “junction to solder” thermal resistance and, that parallel resistance is usually much greater than the “junction to solder” resistance and can be usually ignored.

So, if you want to prevent your device from overheating, you need to have copper area of some significance around the LED solder pads. The copper area will have a thermal resistance to ambient and that then becomes in series with the junction to solder resistance: -

Total thermal resistance is therefore: -

\$R_{(JUNCTION-SOLDER)} + R_{(CU-AMBIENT)}\$.

You then have to calculate what size copper provides a low enough thermal resistance to prevent your LEDs from overheating when they are dissipating 3 watts. Be also aware that the local ambient temperature around the circuit board will rise as heat is passed to it. By how much is dependent on other factors like forced air cooling or the use of a metal case.

However, there is a good site that will provide useful insights into how big the copper area needs to be and here's a graph that is useful: -

enter image description here

So, if you can get away with your copper to ambient thermal resistance being 40 °C per watt, then about 60 square cm of 1 oz copper is required. I have a feeling that with a total power of 30 watts you will need something substantial.

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  • \$\begingroup\$ Are we done with this question and answer now? If there's something you don't understand, please leave a comment. \$\endgroup\$
    – Andy aka
    Commented Feb 21, 2020 at 12:24

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