switch mode power supply - How to choose a inductor for a buck regulator circuit?

I am designing a buck-regulator circuit, with possibly the MAX16974 as the regulator. I have never done such a thing before, and actually not too much analog electronics at all. I got stuck at the part where I should select a inductor.

Part of the problem is that there is much to choose from (13000 total from Farnell). I got them filtered down to about a 100. But I am still not completely sure if the values are right, and the how to choose from the rest that are left.

As there will not be made that many copies made, the price is not that big of a concern.

After a bit of googling I found a app note form Texas Instruments concerning the selection of inductors for use with switching regulator, but I have not been able to work out some of the constants used in the equations in it.

UPDATE: The regulator is going to be used on a 10-20 volt input (mostly around 15 volts). The output is going to be 5 volts with the current around 1A.

I don't really now where the other specs should be. I'd like to be able to power different kinds of devices requiring 5VDC, for example a raspberry pi or charge a phone through usb.

Answer

Here is a quick and somewhat dirty way to calculate an inductor value for buck regulators operating in constant conduction mode (CCM). It will result in an inductance that will be close to what you would get with a more exact calculation, and will not get you into trouble.

What you need to know to calculate inductance:

• Output Voltage, \$V_o\$


• Output Current, \$I_o\$


• Switching Frequency, \$F_{\text{sw}}\$


• L = \$\frac{\text{$\Delta $t}V_o}{\text{$\Delta $I}}\$

Make a couple of assumptions:

• \$\text{$\Delta $I} = \frac{I_o}{10}\$


• \$\text{$\Delta $t} = \frac{1}{F_{\text{sw}}}\$

so

L = \$\frac{10 V_o}{I_o F_{\text{sw}}}\$

for \$I_o\$ = 1A and \$F_{\text{sw}}\$ = 2.2 MHz

L = 22.7 \$\text{$\mu $H}\$

When choosing the inductor:

• 
  

  Find one that is rated for 1.4 to 2 times the output current. In this case 1.4A to 2A. Most standard inductors are specified for 40C heat rise with rated current, which is kind of hot. Conductive losses scale by the square of the current. Using a current rating of 1.4 \$I_o\$ will reduce that heat rise by half, and a current rating of 2 \$I_o\$ will reduce heat rise to 1/4.

  
  



• 
  

  Make sure the series resonant frequency (SRF) is at least a decade higher than the switching frequency.