I am working with a MAX38909ATD+ voltage regulator and targeting an output voltage of 3.824V. I just plugged in my board and realized when things weren't working that the regulator was only outputting 0.7V. So clearly something is wrong here.

This is a small IC, I did the soldering with a reflow oven, but I thought a short between two pads was likely. But I have gone through and checked each pad so this is not the issue. Does anyone know what my problem might be?

circuit layout IC orientation

  • \$\begingroup\$ did you confirm the IC orientation? \$\endgroup\$
    – jsotola
    Nov 5, 2021 at 1:35
  • \$\begingroup\$ @jsotola yes I believe so but I have added a photo just in case. \$\endgroup\$
    – Feynman137
    Nov 5, 2021 at 1:53

2 Answers 2


Summary: You've got the two resistors in the feedback voltage divider (setting the output voltage) reversed on your schematic compared to the desired values, and presumably also on your PCB.

It's the lower resistor in the feedback voltage divider (R101 on your schematic, R1 on the typical application schematic in the datasheet) which should be 15 kΩ. However you've got that value as the upper resistor (R102 on your schematic).

This is what the relevant part of your original schematic shows:

Relevant part of the original schematic

Here is what your feedback voltage divider resistors should look like, adapted from the typical application circuit in the MAX38909 datasheet:

Adapted typical application circuit from the MAX38909 datasheet

Notice how the upper & lower resistor values are reversed compared to your schematic. With the resistor values shown in my version, you should indeed get your desired 3.82 V.

For those who want to do the calculation, here is the relevant paragraph in the datasheet:

Feedback resistor calculation from MAX38909 datasheet

We can also see in this table of various example output voltages from the datasheet, for an output voltage of 3.82 V, then the lower resistor (R101 on your schematic, R1 on the datasheet schematic) should be 15 kΩ and the upper resistor (R102 on your schematic, R2 on the datasheet schematic) will be between 75.0 kΩ (which would give a 3.6 V output voltage) and 97.6 kΩ (which would give a 4.5 V output voltage). So that also confirms that your 80.6 kΩ as the upper resistor in the feedback voltage divider, will give an output voltage between those two voltages:

Recommended feedback resistor values from MAX38909 datasheet

A quick calculation suggests to me that with your existing feedback voltage divider resistor values (15 kΩ for the upper resistor and 80.6 kΩ for the lower resistor) the output would be around 0.7 V, which matches what you report.

  • 1
    \$\begingroup\$ Switched the resistors as suggested and am getting 3.82V out now. \$\endgroup\$
    – Feynman137
    Nov 5, 2021 at 2:57
  • 1
    \$\begingroup\$ @Feynman137 - That's good news, thanks for the update. \$\endgroup\$
    – SamGibson
    Nov 5, 2021 at 3:02
  • Set the bottom feedback resistor R1 to less than 100kΩ to minimize FB input bias current error
  • R1 of 15kΩ is recommended to optimize noise performance.

\$R2 = R1 \cdot (V_{OUT}/V_{FB} - 1)\$

\$80.6k = R1 \cdot (\frac{3.824}{0.6} - 1)\$

\$80.6k = R1 \cdot (6.373333 - 1)\$

\$80.6k = R1 \cdot 5.373333\$

\$\frac{80.6k}{5.373333} = R1\$

\$R1 = 12,646.4435\Omega\$

Could try this and see if output improves, but it appears that the bottom feedback resistor (R1) should be 15k. Solving for R1=15k:

\$R2 = 15k \cdot 5.373333\$

\$R2 = 80.6k\$

Edit: it looks like R101 and R102 are reversed.


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