I didn´t expect to be having issues with a buck converter, but here I am. My design goal is a simple 15-60 V to 10 V buck converter. I settled with the following schmematic:


I designed a 2-layer board with copper pours on both layers with stitching vias:

KiCad layout

(Ignore the terrible soldering)

actual PCB

During testing, I made temporary changes to component values:
Input capacitance: 11 μF,
Output capacitance: 10 μF,
Feedback resistors: 9.1 kΩ and 1.3 kΩ

I tested the converter with an input voltage of 25 V from a fairly noisy bench supply. I loaded the 10 V rail with 120 Ω. Here are the results:

"DC" results:
Vout = 5.05 V instead of ~10 V, Vfb = 0.63 V instead of ~1.22 V according to the datasheet.

"AC" results (with bad oscilloscope probing):

input ripple measured at the input capacitors (AC-coupled)

Data measured after changing inductor from 22 μF to 100 μF: Vout = f(Rload), parameter: Vin

Weird switch node voltage:

switch node voltage (DC-coupled)

I know that the layout and component selection is not optimal, but I think this should only explain high ripple, low efficiency, etc. but not such a massive regulation issue.

Datasheets: Diode MAX5035

  • \$\begingroup\$ What convinces you that the inductor you chose is suitable? Data sheet please. \$\endgroup\$
    – Andy aka
    Commented Mar 9, 2023 at 19:35
  • \$\begingroup\$ @Andyaka My component selection is highly dependend on what I have available, since shipping takes quite a while or is very expensive. The inductor is a Murata LQH32DN220K53 (0.7Ohms, rated for 250mA, SRF 19MHz). It is undersized for my application but should be fine with the 120 Ohm load which should draw 83mA @ 10V. Link: (murata.com/en-eu/products/productdetail?partno=LQH32DN220K53%23) \$\endgroup\$
    – fakulol
    Commented Mar 9, 2023 at 20:00
  • 2
    \$\begingroup\$ Component selection should be based on what is needed and not what you have available. If you can't do that then you can't expect to take current peaks into your inductor that are bigger than what it's rated at without the obvious problems of failing to meet expectations. Just calculate the peak inductor current in your problematic scenario. \$\endgroup\$
    – Andy aka
    Commented Mar 10, 2023 at 10:36
  • 1
    \$\begingroup\$ Well, it's still feeling like an inductor problem to me. \$\endgroup\$
    – Andy aka
    Commented Mar 10, 2023 at 16:23
  • 1
    \$\begingroup\$ I added a graph that shows 3 different loads for 3 different input voltages. \$\endgroup\$
    – fakulol
    Commented Mar 10, 2023 at 17:58

1 Answer 1


Here's my ideas on the matter:

The inductor value is too small. The equation calls for 44uH at least

vin=[15 25 31]; vout = 10 ((vin-vout).vout./vin)/(0.32*125e3)
ans =
4.4444e-05 8.0000e-05 9.0323e-05

Counter intuitively a lower current needs a higher minimum inductor, so you might be able to put a higher load and get better results.

Not having the feedback at the right point has been a problem for me in the past. One problem is the FB point will see a smoothed version of the inductor switching and may not switch at the right time.

But before you take the effort to do switch the wire, make a differential measurement (with a diff probe at pin2 of the inductor and pin 1 of the C10 pad) if you see a difference more than a few mV than that could be the problem.

enter image description here

  • \$\begingroup\$ I continued testing with 100 uH, forgot to update the question. Changing the feedback point is probably a good idea, might investigate. \$\endgroup\$
    – fakulol
    Commented Mar 10, 2023 at 21:47

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