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I have an inductor at the output of my boost converter. My output RMS inductor current is 271mA and the peak switch current is 1.3A. I have selected a 22uH / 2A Inductor as we select the inductor keeping the peak switch current in mind. This is my selected inductor

I dont understand what the below three points in the image are trying to say. Can someone explain me with my case and how to interpret that graph. My Ambient temp of the board is 80degC. So, how should I understand those three points? Please help.

Thank you.

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1) The saturation current value (Idc1) is the DC current having inductance decrease down to 30% (at 20 dec C).

If you exceed this, the inductor will saturate and the current will rise sharply. You want to avoid this or your design won't work as expected.

2) The temperature rise current value (Idc2) is the DC current value having temperature increase by 40 deg C (at 20 deg C).

If you exceed this the i^2 * R copper losses will cause the inductor to get too hot. The manufacturer has decided that more than a 40 deg C rise may stress the part due to temperature differentials.

3) The rated current is the DC current value that satisfies both of the current saturation value and the temperature rise current value.

You shouldn't exceed #1 or #2.

On top of this, there is a derating curve for high ambient temperatures. If your starting temperature is high, then you have less allowable temperature rise before the part exceeds a maximum allowable temperature.

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  • \$\begingroup\$ Thank you for the clear explanation \$\endgroup\$ – Newbie May 22 at 10:32
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When an inductor reaches saturation, its "apparent" inductance value is less than the nominal inductance value. The idea is that a lot of DC current will "fill" the core with magnetic flux, and at a certain point the core will be full, and no longer act as a proper core, thus the inductance starts falling.

Since this inductance lowering is not abrupt, but rather gradual, there is no "saturation current". What they give you there, is the current at which the inductance value is reduced by 30%, so your 22 uH inductor, if you put 2 A into it, will behave just like a 22 uH x 70% = 15.4 uH.

The second point is telling you that if you put 1.6 A (DC) in the inductor, its temperature will increase of 40 degC. With your ambient temp of 80 degC, this means 120 degC for the coil (ouch!).

The third point is telling you that the rated current for your coil is the lower between the two, depending on the ambient and DC bias conditions of your particular application.

In your case, your design parameters are:

  • I_out = 270 mA
  • I_peak = 1300 mA
  • t_amb = 80 degC

I will assume, from your second plot, that the maximum operating temperature for the coil is 125 degC (and you get it to be an ideal wire there BTW).

Since your DC current is well below the 2 A saturation current, your coil will possibly only saturate a bit during the peaks, which is not a problem. Your ambient temperature is very high, and from the second plot you can see that you are right at the knee of the derating curve. My suggestion is to run your converter at room temp, i.e. 20 degC, and measure the coil temperature. If the temprise is within 10 or maybe even 20 degC, which it should be, then your coil is ok for the application.

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  • \$\begingroup\$ Thank you for the very clear and detailed explanation \$\endgroup\$ – Newbie May 22 at 10:32

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