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I have read several articles an datasheets, but I couldn't find an affordable answer...

This is my problem: I need to power a NodeMCU (on Vin pin) with 3xAA batteries. Which is the minimum operating voltage at Vin accepted by NodeMCU? In other words, which voltage drop from 4.5 will let NodeMCU still working? thanks.

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  • i run several simple devices on 2xAA batteries connected to VCC. Only uses about 2/3rds of the battery, but for me, that's years of operation before they drop down to under 2.5v...
    – dandavis
    Commented Aug 27, 2018 at 16:17
  • If you have 2xAA why don't you connect them directly to 3V3 pin? Commented Aug 27, 2018 at 16:23
  • good idea. so good in fact, that's what i did (VCC == 3.3v)
    – dandavis
    Commented Aug 27, 2018 at 16:25

1 Answer 1

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The NodeMCU uses the SPX3819 3.3v voltage regulator.

That regulator, according to the datasheet has a dropout of maximum 700mV, but much less at lower currents.

Taking the maximum, that means that you need to supply 3.3 + 0.7 = 4V to the board for the regulator to regulate properly. However, it is doubtful you will be using that much current, so probably nearer a 350mA dropout - which means 3.3 + 0.35 = 3.65V.

However, the ESP8266 itself will work down to 2.5V - so if you operate the board with the regulator out of regulation (not recommended, but it should work fine), you could probably go as low as 2.8 to 3V without problems.

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  • But I can't connect 4.5v to 3.3 pin, correct? Commented Aug 25, 2018 at 11:35
  • No. That's why I say 2.8 to 3v because the regulator will drop an unknown amount.
    – Majenko
    Commented Aug 25, 2018 at 11:51
  • In other words: my 3xAA battery pack could work until it supplies at least 3.5-3.6v? Commented Aug 25, 2018 at 11:53
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    Sounds about right.
    – Majenko
    Commented Aug 25, 2018 at 11:53

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