MP1584 inductor calculation

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

• 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.
– jonk
Aug 24 '16 at 22:01