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I am trying to understand the output voltage of a circuit made using LT3014 LDO regulator.

Datasheet: https://www.analog.com/media/en/technical-documentation/data-sheets/3014fd.pdf

The circuit is pretty simple. Input is 28V, and ADJ pin is connected to GND, which is a little confusing. What will the output voltage be in such a case? On the output is a 7.5 Ohm load resistor through which it is connected to a pressure sensor.

While I am asking for this specific regulator here, I believe there is a fundamental thing that I am perhaps missing.

Could anyone please help?

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This is a typical application diagram of the LT3014:

enter image description here

The regulator regulates the output voltage by regulating its pass transistor (between IN and OUT) such that at the ADJ pin there's a voltage of 1.22 V ("ADJ pin voltage" in table on page 3 in the datasheet). The output voltage gets to the ADJ pin through a voltage divider (the 3.92M and 1.27M resistors).

If you ground the ADJ pin then the regulator will "think" that the output voltage is very low. The regulator will then fully open its pass transistor in an attempt to increase the output voltage. So the output voltage will be only slightly lower than the input voltage. Also no voltage regulation will occur, the regulator will behave more or less like a resistor between IN and OUT.

So grounding the ADJ output isn't very useful as the result will be very similar to just leaving out the regulator.

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  • \$\begingroup\$ Thank you so much. Before I accept this as the answer, one quick clarification: Does it mean output voltage will be Vin - Vdropout? Regarding your point of leaving out the regulator, it makes sense, but I understood that this regulator has been used purposefully as it prevents reverse current flow from output to input as required for this application. \$\endgroup\$
    – LoveEnigma
    Sep 21, 2020 at 6:05
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    \$\begingroup\$ Yes, the output voltage will be somewhat less than Vin - Vdropout. Slightly less because the regulator can still "regulate" when there is Vdropout across it. The regulator must then still be able to decrease and increase its resistance (between IN and OUT) so it cannot work at its lowest resistance. When ADJ is grounded, it will be at its lowest resistance so you get a little less voltage drop. Indeed for reverse current flow protection this could be a solution although this IC would not be my choice as the datasheet doesn't mention that usage case specifically. \$\endgroup\$ Sep 21, 2020 at 6:11
  • \$\begingroup\$ Okay, thank you for the clarification. Unfortunately, I can't change this IC due to various reasons which are out of my control. \$\endgroup\$
    – LoveEnigma
    Sep 21, 2020 at 6:22

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