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I would like to calculate the inductor value for MP1584EN.The datasheet says: $$L= \frac{V_{out}}{f_s \Delta I_L} (1- \frac{V_{out}}{V_{in}})$$

On the first page of the datasheet there is a typical application picture with 12Vin, 3Vout and 10uH inductor.

My calculation for this application is 2.39uH. Why? The 10uH is not too big or the 2.39uH is not too small?

My calculation: $$L= \frac{3.3}{500000\cdot2}\left(1-\frac{3.3}{12}\right)$$

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  • \$\begingroup\$ If you want to do the calculation that Spehro is hinting that you try out, then perhaps it is: 3.3V/(500kHz*15%*4A)*(1-3.3V/28V)=9.7uH. That may help to understand their illustrated 10uH sample design. \$\endgroup\$ – jonk Aug 24 '16 at 22:01
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The minimum inductor value should be calculated at the maximum input voltage.

Ripple current in the inductor (according to the datasheet) should be no more than 1/3 of the 4A switch limit, but 15% is better, in my opinion, as it reduces the stress on the switch and the losses. You used 2A which is much higher than the maximum recommended.

So calculate that with 28V maximum input and I believe you'll get close to 10uH. You may have noticed they show graphs at 12V and 24V input, so they are not assuming the maximum input voltage is only 12V.

Using a larger than required minimum inductance reduces the inductor ripple current and losses in both the inductor and the switch, all other things being equal. It might be larger than otherwise required, but 10uH is a pretty small inductor anyway, due to the relatively high switching frequency of that part.

If your input voltage is limited to (say) half (12V nominal 14V maximum) then you could safely reduce the inductor value to 5uH.

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You can typically use a much larger inductor than minimally necessary to store the energy for the downconversion; this usually only has positive effects.

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