I am looking at BD9E102FJ buck converter, which has maximum current output of 1A. Reading the datasheet I found this calculation:enter image description here

It shows inductor ripple of 752mA with given values.

I need to convert 24V to 1V, so my ripple calculation yielded me a 247mA ripple. I would be using this buck converter to drive 6 filaments of bulbs, which require 50mA each @ 1V, so a total load of 300mA.

My questions are: How can there be so much ripple? Is it something that Cout takes care of? Would this converter give stable 1V out, or is it not a good converter at all for this task?

EDIT: Added layout and explained application more.

So this is my schematic for this buck: enter image description here

The filament voltage was rated 0.85-1.35 V and these feedback values put it in the middle.


2 Answers 2


You've answered the question yourself - "with given values".

The amount of ripple current is defined by the buck inductor value and the volt-seconds applied to it. If you change operating point (different input voltage, different output voltage, different switching frequency) or change the inductance value, the ripple current will change.

Why is there so much ripple? That's how the circuit was designed. A DCM buck would have even higher ripple, which is a trade-off for simpler loop stabilization. If you want less ripple, make it CCM and design your slightly more complex loop accordingly.

The amount of current ripple applied to Cout is what defines the amount of ripple voltage you get on the output. So yes, in a manner of speaking, Cout "takes care of" the ripple current in so far as it deals with whatever's applied. If Cout is too small, your output won't stay in regulation.

You haven't specified enough about your application, but there is no reason why a 300mA buck operating at 247mA of ripple current couldn't be stable. It would be barely in DCM mode, but that's fine as long as your output capacitor ESR could handle the ripple and your application is OK with the ripple voltage.

  • \$\begingroup\$ I just need to heat the filaments that are rated for 1V and 50mA. Would this be fine for it? \$\endgroup\$ Feb 17, 2018 at 19:40

Adding to Adam's answer:

How can there be so much ripple?

You can lower the ripple current by...

  • using a larger inductor value, but this results in higher losses in the inductor due to more turns, thus a longer wire, core losses also increase, saturation current decreases with more turns, and perhaps you'll need a physically larger and more expensive inductor.
  • increasing the frequency, but this will result in more switching losses.

So, it's a compromise. Your ripple current value will work fine. If it makes you lose sleep at night you can put a 10µH inductor instead.

Depending on the circumstances, you might want to have the buck operate in continuous mode (not skipping cycles) which makes the output voltage more stable and noise easier to filter. If your application is noise sensitive you might want to find the part of the datasheet that says how and when it switches from PWM to "light load" mode.

Is it something that Cout takes care of?

Yes. Cout smoothes the voltage and absorbs the ripple current. The datasheet recommends ceramic caps, which are best for this as they have very low ESR and very low inductance. Pick X5R or X7R dielectric.

Layout is also very important, so feel free to post your layout for a review when it's done.

  • \$\begingroup\$ I added schematic \$\endgroup\$ Feb 17, 2018 at 20:04

Your Answer

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

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