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Square coil

Why is only the first turn in the coil hotter compared to other two turns?

Spiral coil

In this, only the turns of the top part seem to be hotter compared to the bottom.

Current through the coils is 2 A. It is connected to a half-bridge operating in buck mode at 1 MHz.

I don't know if I am making any mistakes in capturing the images, but I expected a more uniform distribution of the heat.


Here are some details about the second coil:

Top Layer

Bottom Layer

and Coil connected to the board

The board is FR4 material and turns are copper.

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Why is only the first turn in the coil hotter compared to other two turns?

The inner turn will get hotter because it has less surface area to dissipate heat produced and, it is enclosed by outer turns that themselves will push heat towards that inner turn.

In this, only the turns of the top part seem to be hotter compared to the bottom.

It looks like the 2nd picture has an inner turn that is connected to a central copper circle (I could be wrong about this of course) and that will act as a heat sink for the inner turn at short distances from the central copper.

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  • \$\begingroup\$ You should also consider the emissivity of the surfaces your are looking at with your IR cam. The IR cam should be adjusted accordingly. \$\endgroup\$ – stowoda Jun 7 at 10:44
  • \$\begingroup\$ @stowoda It is FR4 which has an emissivity of 1. So will it make any difference or do I need to adjust this setting in the camera? \$\endgroup\$ – Stoner Jun 7 at 11:24
  • \$\begingroup\$ @Andy aka It looks like the 2nd picture has an inner turn that is connected to a central copper circle I don't understand this point. But I will add that it is a two layer spiral coil (there are same number of turns on the bottom layer as well). There is via at the center which connects the top layer to the bottom. \$\endgroup\$ – Stoner Jun 7 at 11:28
  • \$\begingroup\$ @stoner, just wanted to make you aware that the surfaces could have different emissivities, so you would read errenous temperatures. From the photo you have posted I see,two materials which are radiating: Copper and epoxy, which emit different amount of IR. The copper trace you could paint black with a magic marker in order to rise its emissivity. \$\endgroup\$ – stowoda Jun 7 at 11:51
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I expected a more uniform distribution of the heat.

You're making a common mistake. The heat IS evenly distributed — all parts of the coil have the same resistance and the same current, and therefore, they convert electricity into heat at the same rate.

But that tells you nothing about the distribution of temperature, which is what the FLIR is measuring. Temperature is a function of thermal capacity, time, heat flow and thermal resistance.

In your case, while all segments of the coil have essentially identical heat input, they have different abilities to get rid of heat, and so their temperature varies.


Based on the pictures you added, it is clear that the "glow" is coming from the FR4, not the copper, whose emissivity is much lower. The parts that have more heat input relative to their ability to get rid of it are much hotter.

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  • \$\begingroup\$ So with time the temperature of the parts also increase. Then how is this test useful to us when designing something? Is there a point beyond which the temperature stops increasing? \$\endgroup\$ – Stoner Jun 7 at 12:17
  • \$\begingroup\$ It depends on whether the heat flows ever reach equilibrium. I have no idea about the "usefulness" of the test, since I don't know what you were testing for. \$\endgroup\$ – Dave Tweed Jun 7 at 12:18

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