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I am designing a board with 9 SK6812 LEDs. On the back of the board I want to place the Atmega328P microcontroller.

I have doubts whether the heating of the LED can affect the operation of the microcontroller?

Board Thickness: 1.6mm Two layers

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    \$\begingroup\$ You could use big copper areas for your LED traces and (where possible) thin copper traces to the CPU . That would improve getting rid of the heat and keeping it away from your CPU. Why is your CPU not further away from the LEDs e.g. in the upper right corner? \$\endgroup\$ – Oldfart Sep 8 at 7:24
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The data sheet of the SK6812 that I was able to find made no mention of the amount of current required to the device when the three LED are lit. As such it is not possible to directly calculate the net power consumption of nine (9) of these devices on your circuit board. You may have to setup an experiment to measure the current draw of one of the SK6812's using a breakout board that includes a chip of that type.

Once you know the worst case current draw of a SK6812 with all three LEDs lit (propose that the RGB color would be "white") you can proceed to compute the total power of your LEDs. Only then can you start to consider whether the MCU on the back side of the board would be exposed to excessive temperatures.

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    \$\begingroup\$ It's 40mA at full brightness of all three. Now what? \$\endgroup\$ – Soldersmoke Sep 8 at 7:35
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    \$\begingroup\$ Maximum LED power would be 9 * 5V * 0.04A = 1.8W. That amount of power spread across one side of your circuit board is lot likely to upset your MCU on the other side of the board as long as there is free air circulation available. If you stuff everything inside a tight enclosure then the total temperature rise inside the enclosure will depend upon how much heat transfer can occur through the sides of the enclosure to free air. Easiest thing to do if this raises a concern is to setup an experiment with a power resistor dissipating the 1.8W inside (continued) \$\endgroup\$ – Michael Karas Sep 8 at 7:51
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    \$\begingroup\$ (continued from above) the enclosure and a thermal probe inserted to measure the air temperature inside the box. From this you can deduce the temperature range that your MCU and its support components would be exposed to. It that temperature exceeds the data sheet temperature ratings for any of these components then you could have a problem. \$\endgroup\$ – Michael Karas Sep 8 at 7:54
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    \$\begingroup\$ Doesn't part of that 1.8w get converted to light instead of heat? \$\endgroup\$ – Soldersmoke Sep 8 at 7:55
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    \$\begingroup\$ Let me answer that simply by posing something else....If you decide to run the experiment that I proposed to obtain empirical data then it would be well to have the power resistor setup to dissipate something more like 2.5W or 3W to give you some margin to work with. \$\endgroup\$ – Michael Karas Sep 8 at 7:59

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