Inductor Saturation Current and Temperature Rise Current
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.