VCC is a power rail.
You can say it's both input and output.
The onboard 5volt regulator supplies power to that rail (you could also do that with an external 5volt supply).
And you draw power from it (to power your sensors).
Leo..
If this is a 5 volt pro mini.
4 x 1.5 = 6v therefore if you use a series silicon diode .7 you can get down to 4.3v
edit, 5.3v
This would work.
Add a schottky diode .2v then you will get to ~5.1v, even better.
However, 3 batteries and a boost converter would be my recommendation.
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4 x 1,5V = 6V is to much for Vcc (and to little for the RAW). 5,5V absolute maximum. But a bit lower is fine. 4,5V, no problem even with a 16MHz crystal. From the top of my mind (but you can look it up in the datasheet of the ATmega328p) its stable at 16MHz up until a little under 4V.
Could probably drop 0.7v through a normal (non-schottky) diode to get 4 AA’s into safe territory. I would not add a schottky diode after that; 5.3 is safe, and that’ll give you more headroom as the batteries run down.
Bypassing the regulator (by supplying 5v directly) will not reduce the current (though since you don’t need to give it an extra 2v for the dropout, the power (ie, volts times amps) is lower)
Removing the regulator from the board entirely will slightly reduce power consumption, but the quiescent current of the regulator isn’t that high - it’s only relevant when the chip is sleeping and you’ve already removed the power LED.
Spec for 16mhz is ya, a little under 4v. In practice, under normal conditions, you can generally run a 16mhz '328p down to 3.6v no problem (which is relevant as that’s the min output voltage for a LiPo battery)
If the AAs are rechargeable (NiMH) then you can probably power directly without using anything to drop the voltage. Even when fully charged, I don't think 4 of them would exceed 5.5V when any appreciable current was drawn.