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I guess it may be pretty simple thing due to my lack of understanding. I designed a PSU converting 12V to 5V using TPS54202HDDCR. Followed the datasheet and made the schematic as follows:

enter image description here

I used eq. 3 for relationship between, Vout, Vref and the two resistor values of the divider. In the text I saw that Vref is 0.596V. Considered to have a little bit higher voltage than the required.

For example, I have 0.596*(100K/13K+1) which should lead to 5.18V. Also, this is the exact same schematic as the one shown in Section 8.2.1 and Fig. 8-1 for typical application.

Yet, when I made the PCB the output is not 5.18V, but 7.1V.

I tried with other values for 2.5V, and I got 4.5V.

The resistors are 1% so I don't think there is something wrong there. I found I added this divider for EN (R1 and R2) to the PCB just in case. Removed it hoping to get the correct values but had the same high voltage again.

Connected to a load it does not go down also (burnt a chip due to high voltage...). I am curious what am I doing wrong?

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  • \$\begingroup\$ Is the current rating of the inductor adequate? Have you taken into account the DC bias derating for the ceramic caps? PCB layout? \$\endgroup\$
    – Kartman
    Mar 16, 2022 at 12:52
  • \$\begingroup\$ Have you measured that all the feedback resistors have correct value, or that the measured feedback voltage is what is expected? \$\endgroup\$
    – Justme
    Mar 16, 2022 at 14:06
  • \$\begingroup\$ @Kartman inductor is okay, just confirmed again. Not sure about the layout... it is not that big PCB and did not expect problem there. 2 layers, wide traces. DC bias is something I never considered and for X5R, 16V rating it looks bad... I need to think how to buy proper caps to test. Thank you! \$\endgroup\$
    – D. K.
    Mar 17, 2022 at 14:15
  • \$\begingroup\$ @Justme I will measure in the weekend. Resistors are fine, but don't remember checking the feedback voltage. \$\endgroup\$
    – D. K.
    Mar 17, 2022 at 14:16

2 Answers 2

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D.K. You made a mistake in your calculations for the output voltage. The datasheet gives Vout = (Rfb1/Rfb2 +1) * 0.596. This is actually 0.596 * (149.9k/13K + 1) = 7.47V which seems close to your value. So don't forget the 49.9K in your voltage divider calculation. As for the reason(s) for burning out the regulator, get the wanted voltage right first, then double check your absolute limits for the chip (make sure you don't exceed them), and also as some others have mentioned, your inductor value and capacitors being used etc. Use those recommended by the manufacturer. Good luck !

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  • \$\begingroup\$ I just doublechecked that I did not make mistake on the resistor. I basically ignored the 49.9 Ohm one since it is too small in comparison with the 100K. It is not 49.9k but 49.9 Ohm. Or I am missing some other connection? \$\endgroup\$
    – D. K.
    Mar 17, 2022 at 13:18
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TI supplies a handy web tool for implementations using their ICs. It is called TI Web Bench and you will have to create an account to use it. I have shown a screenshot below of their suggested implementation for a 12V to 5V, 2A DC-DC supply using the IC you specified:

Schematic

A cursory glance reveals that your feedback resistors are incorrect, as the upper two resistors in series will provide a different output, as the series resistance to the feedback pin will be different.

If I change the schematic to incorporate a 150 kOhm Resistor on the upper feedback path, as you have done, the tool correctly shows an output of 7.17V, as you have determined: Schematic

The simplest solution to fix the PCB would be to remove the 49.9 Ohm resistor and 75 pF capacitor, and then short the two pads for the 49.9 Ohm resistor, allowing a 100 Ohm trace from the output of the supply to the input of the feedback pin.

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  • \$\begingroup\$ Thank you! I didn't know about the tool you showed. I can try that. Actually, one resistor is 100K, the other 49.9 Ohm so it is negligible more or less. I don't expect big change shorting it but I may be missing something. By the way, you also recommend removing of the 75pF capacitor. Is there a reason for that? \$\endgroup\$
    – D. K.
    Mar 17, 2022 at 13:21
  • \$\begingroup\$ Yes that was my point I guess. Are you sure it is not 50K or there abouts, as even the simulation on teh Ti website seems to suggest the votage divider section is out for whatever reason. The 150K / 13.7K combination appears to give the 7.1V you are measuring. Is this just a conincidence ? \$\endgroup\$
    – citizen
    Mar 17, 2022 at 15:04
  • \$\begingroup\$ Was it ever recommened to have the 75pF capacitor there for whatever reason, it seems you have been advised by the datasheet or otherwise, that it is required. This may not be the case. \$\endgroup\$
    – citizen
    Mar 17, 2022 at 15:06

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