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.

  • 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 Aug 27 '18 at 16:17
  • If you have 2xAA why don't you connect them directly to 3V3 pin? – Fabio Marzocca Aug 27 '18 at 16:23
  • good idea. so good in fact, that's what i did (VCC == 3.3v) – dandavis Aug 27 '18 at 16:25

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.

  • But I can't connect 4.5v to 3.3 pin, correct? – Fabio Marzocca Aug 25 '18 at 11:35
  • No. That's why I say 2.8 to 3v because the regulator will drop an unknown amount. – Majenko Aug 25 '18 at 11:51
  • In other words: my 3xAA battery pack could work until it supplies at least 3.5-3.6v? – Fabio Marzocca Aug 25 '18 at 11:53
  • 1
    Sounds about right. – Majenko Aug 25 '18 at 11:53

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