How to choose a inductor for a buck regulator circuit?

Question

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 1

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.

Answer 2

The big problem I see is you haven't specified any parameters other than "I'm designing a buck".

If you haven't yet figured out these parameters, please do so:

• Input voltage range
• Output voltage
• Maximum output ripple
• Maximum output current
• Size and ESR of output capacitors

These all matter:

• Inductor ripple current is often targeted as a percentage of the total DC output current

• Output ripple voltage is the inductor peak-to-peak current superimposed on the output capacitors' ESR

• The peak-to-peak ripple is also related to the duty cycle, which is

  \$\dfrac{V_{out}}{V_{in}}\$ for CCM mode, less than this for DCM mode (load dependent)

• There are feedback stability implications when operating in CCM (CCM limits the maximum bandwidth you can achieve while maintaining gain and phase margin)

• The inductor has to handle the DC current you want without saturating, and the DC loss in the winding has to be within the limits of the part.

EDIT: Your target is 1A at 5V. If you go with the '10% rule of thumb', the maximum inductor peak-to-peak current should be 100mA.

Duty cycle: \$\dfrac{5V}{15V}=0.333\$

On-time: \$\dfrac{0.333}{2.2MHz}=166.5ns\$

Ripple current: \$V_L = L \dfrac{\Delta I}{\Delta t}\$

\$ L = \dfrac{V_L \cdot \Delta t}{\Delta I} = \dfrac{(15V-5V) \cdot 166.5ns}{100mA} = 16.65 \mu H\$

A larger inductor will give you smaller ripple current. The opposite is true - a smaller inductor will give you larger ripple current.

Generally, ripple should be 1% or less of the DC level, so make sure the output cap ESR is less than \$500m \Omega\$ if you target 100mA of inductor ripple. This should be easy.

Answer 3

For good load regulation and low ripple, you want the series resistance of the inductor and capacitors respectively to be much less than the ON resistance of the switch.

for the MAX16974, RON measured between SUPSW and LX, ILX@ 500mA Ron= 185mΩ typ, 400 mΩ max

So choose Rs of L to be << 185 mΩ such as 10~20% of Ron or Rs of(L)= 19 ~38 mΩ