# Sanity check inductor calculation

• Input: 3.3 V
• Output: 13 V
• Output current: 200 mA (actually less then 60 mA but with some headroom)
• Ambient temp: max 70°C
• Regulator: LMR62421XSDE/NOPB

So my question:

What is the smallest inductor I can use? Page 11 and 12 in the datasheet talks about inductor selection.

$$L= \frac {V_{in}} {2iL}DTs$$

Gives: $$L= \frac {3.3}{2 \times 0.1} \times 0.75 \times 0.000000625 = 0,0000000773 \ \mathrm{H} = 0,0773 \ \mathrm {\mu H}$$

That can´t be right ... or??

To follow what I did here one have to look at the equations in the datasheet (page 11 and 12).

A few comments to my calculation:

• 0.1 I get from 10% ripple current.
• TS I get from: 1/TS = FSW (FSW = 1.6 MHz fixed for this regulator.)
• D = VOUT - VIN/VOUT

The result is 7.7 μH.

Simply, in the equation of inductor, if you put voltages in volts, currents in amps, and frequencies in megahertz then you'll find the result in microhenries because the inverse of the mega is micro:

$$\mathrm{ L=\frac{V_i}{2\ \Delta I_L }\ \frac{D}{F_S} \\ \ \ \ =\frac{3.3}{2\cdot 0.1}\ \frac{0.75}{1.6} \\ \ \ \ =7.7\mu H }$$

• But can that be correct? becuase on page 21 in datasheet they have an example with and 6.8 uH inductor at 12V 500 mAmp as output. 3 V in. So I would expect the result to be under 6.8uH since I use less than 500 mAmp.. Oct 31, 2021 at 20:04
• Also I am a bit unsure of the 10% current ripple. The question if it should be "10" (for 10%) or 0.1 (for 10%) in the calculation... Oct 31, 2021 at 20:07

Went to Webench designer on TI website:

It has calculated 15uH @ 29 mOhm (NPI54C120MTRF). Simple as that.

• 0.1 I get from 10% ripple current.
• TS I get from: 1/TS = FSW (FSW = 1.6 MHz fixed for this regulator.)
• D = VOUT - VIN/VOUT

$$\\Delta I_L\$$ is expresed in amps, not in % or factor, so 0.1 means 0.1 A, that would make your output going from 0.3A to 0A.

• But Webench gives you many diffrent choices for inductor. Going from 10 to 15 uH. Is is really for choosing the inductor? on page 21 in datasheet they have an example with and 6.8 uH inductor at 12V 500 mAmp as output. 3 V in. So I would expect the result to be under 6.8uH since I use less than 500 mAmp.. Oct 31, 2021 at 20:26
• ah ok! It is in amps.. OK, So if my Iout is 0.2 amp and 10% to 30% is recommended I should put 0.2*0.1 = 0,02 if i want 10% ripple? Oct 31, 2021 at 20:29