I'm currently designing a buck converter using the AP64501. This issue/question is not specific to that buck however. I have a python script that I wrote that uses the formula's in TI's appnote: AN-1197 Selecting Inductors for Buck Converters. As per their example, you either need to be given the specs for r, or directly given the r value. As per ROHM's appnote: Inductor Calculation for Buck Converter IC (and many other places) r should be around 0.2 -> 0.5.

In CCM mode, valid ranges for r are 0 -> 2 (source)

My design that is using the AP64501 is:

Vin: 24V

Vout: 3.3V

Iout: 200mA

Using the TI appnote, if I use these values (as well as calculating v_sw and v_catch from the rds of both FETs, and an r of 0.3) I get an inductance value of 83.22uH, which is of course not ideal. If instead I use an output current of 5A (which is what the buck is rated for), I get 3.33uH, which is basically the exact value that the AP64501 datasheet uses. So I can confirm that these equations are valid for this converter at the max rated current for the buck.

The only way to have the inductance be near the 3.3uH recommended at 200mA is to increase the value of r to like 7.5, which isn't valid.

So what am I missing with respect to the r and converters at lower currents like this. Also, If I set the inductance to be 3.3uH for the rest of the equations to solve for inductor currents, energy, etc., since I fudged on the inductance, are these values even valid anymore?


1 Answer 1


The formula you're using to calculate inductance is really meant for nominal or max loads.

Taken to the extreme, when your converter is unloaded the ripple current will be 100% of the inductor current (Since it's synchronous and always in CCM) giving an "r" of infinity. Yet the converter will still work fine. ("r" in the TI app note is defined as the ratio of inductor ripple current to output current.)

There's nothing wrong with high ratios of ripple to average current at light loads, so you can go ahead and use your 3.3 uH inductor, or somewhat higher for lower core and AC losses.

Your transient response will be much better than trying to use a higher value inductor as well.

  • \$\begingroup\$ Yeah that's about what I came to assume. But I still have two issues with that conclusion: 1) While you call it a "light" load, it's still my nominal/max load for my application. 2) Since I am kinda fudging my numbers using a lower inductance, I can't really accurately size the inductor for saturation current. \$\endgroup\$ Commented Mar 2, 2022 at 2:09
  • \$\begingroup\$ You can calculate peak inductor current, right? As long as your peak inductor current is well below the saturation current you should have no problem. Your efficiency may not be great, but you picked a converter IC that doesn't have light-load efficiency features. What may drive a larger inductor choice is the short-circuit current limit of your converter. If you don't want the inductor to fry during a short you may need to size it for the full SC current. \$\endgroup\$
    – John D
    Commented Mar 2, 2022 at 3:34
  • \$\begingroup\$ Yeah I can, just wanted to make sure they'd still be valid. Because my ripple current shows higher than the peak inductor current. Also, this buck does have a light load feature. See page 11 "Pulse Frequency Modulation (PFM) Operation" \$\endgroup\$ Commented Mar 2, 2022 at 10:22
  • \$\begingroup\$ @KyleHunter Right, I missed the PFM mode in the controller, so you should be OK for efficiency too. \$\endgroup\$
    – John D
    Commented Mar 2, 2022 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.