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I'm trying to build a step down module with a TPS562219: http://www.ti.com/lit/ds/symlink/tps562219.pdf

On page 14 there is the reference design and on page 15 I can calculate the values Il p-p, Il peak and I LO(RMS). Therefore the datasheet shows an example for an 1.05V output:

  • Il Peak = 2.34A
  • I LO(RMS) = 2.01A

With that they recommend an inductor with a peak current rating of 5.5 A and an RMS current rating of 4.3 A. I don't understand why they choose such different values from that what they calculated.

In my case I need to get an 5V output for an 12V input. So my calculations give me the values

  • Il p-p = 0.95A
  • Il Peak = 2.477A
  • I LO(RMS) = 2.0189A

Which leads me nearly to the same thing. So I would choose the same inductor? If this is correct, then I don't understand why to calculate the values, if the result is always the same.

And just by the way: Are R2 und R3 in Table 2 (page 15) named correctly? Shouldn't it be R1 and R2?

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  • \$\begingroup\$ It the inductor saturates, L drops and it turns into a heater resistor and fails. But the threshold is gradual, so safety margins are needed \$\endgroup\$
    – D.A.S.
    Commented Sep 11, 2018 at 15:37

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The inductor current ratings represent minimum requirements. An inductor that can handle more current is fine, so we generally leave some margin.

As far as R2 and R3 in the table, you're right, they correspond to R1 and R2 respectively on the schematic.

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  • \$\begingroup\$ But how much margin should I choose? Double?? \$\endgroup\$
    – user3142695
    Commented Sep 11, 2018 at 15:37
  • \$\begingroup\$ Theoretically, you should be able to use one exactly matching your calculations. However, your calculations probably don't take into account component variation, temperature, etc. I'd be comfortable with anything over 20% more than my calculation gave. \$\endgroup\$
    – Cristobol Polychronopolis
    Commented Sep 11, 2018 at 16:15
  • \$\begingroup\$ Any disadvantage if I choose it too big? Because in the example case they choose more then double value. And that is making me wonder. \$\endgroup\$
    – user3142695
    Commented Sep 11, 2018 at 16:19
  • \$\begingroup\$ It's probably one they were familiar with. Because the current requirement is a lower limit, not a goal, they design inductors for inductance and size; the current limit is a side effect of those primary considerations, so while they come in a fairly continuous array of inductance values (specifically designed in) the operating currents are characterized and specified without a specific target in mind. So you don't always have that many options to choose from, and that's because you don't have to get close, just get one large enough. \$\endgroup\$
    – Cristobol Polychronopolis
    Commented Sep 11, 2018 at 19:41

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