3xAA Voltage Drop upon load and voltage regulation question

Hi all, I have a question to which I cannot find answer experimentally at the moment but I am eager to know the theoretical part if you can help me with that.
I have a 3xAA (alkaline) batteries power supply that powers circuit via 12V (pololu) step up regulator. That regulator's minimum input voltage is 2.7V, does this mean that when circuit closes and battery voltage drops below 2.7V due to battery's internal resistance, this will no longer be enough voltage to power voltage regulator and it will quit working?

Yup, that's what it means

IMO, you ought to determine experimentally where it actually cuts out (by powering on the bench and varying the voltage), and how it behaves near that point. Some of them will cut out sharply, some will let the output voltage droop before cutting out entirely, many of them start squealing as they get close to their limits. Behavior also of course depends on output current.

Many people would consider the device cutting out at 2.7 or 3v to be a feature - if you're powering it from LiPo's, you call that "UVLO" (undervoltage lockout) and it's an essential feature to prevent LiPo's from being overdischarged and damaged.

Often boost converters will "cut out" at a lower voltage than they "cut in".

In some cases (typically the ones that work from extremely low voltages - like the 0.9V from half-dead AA cells) they're powering the internal logic and switch from the output, and initially start in a low power mode to get a high enough voltage to run normally, and then go into normal operation from there. I had to deal with one with this behavior the other day, where it caused an issue...

If you knew what that part on the board was (the SOT-23-6 chip - that's the one that's doing all the work), you could pull the datasheet and see what it's specs are and compare with Pololu's specs. Cute part, by the way - you don't see many DC-DC converters with the SEPIC topology (allowing step up and step down in the same converter - that said, if you know it's always going to be operating in boost mode, a straight boost converter will be better and/or cheaper). Shame they don't tell you what part it is, as far as I can tell...

I have a 3xAA (alkaline) batteries power supply that powers circuit via 12V (pololu) step up regulator.

Just in case you're not aware of this, a switchmode DC-DC converter that boosts the voltage consumes more current than it puts-out.

Assuming it's (approximately) 100% efficient, stepping-up from 4.5V to 12V is (approximately) 3 times the voltage. That means you are "pulling" (approximately) 3 times as much current from the batteries as you're getting out of the regulator. And of course, that affects battery life.

...does this mean that when circuit closes and battery voltage drops below 2.7V due to battery's internal resistance...?

If fresh batteries are dropping from 4.5V to 2.7V, you probably need "bigger" batteries.

I have seen boost converters that accept as low as 0.9volts. But they seems to boost to a fixed 5 volts USB.

You can run arduino perfectly fine with such a module.

Something else to consider - the model of a battery is a voltage source in series with a resistor - that resistor is the internal resistance of the battery and it increases significantly as the battery discharges. This limits the current the battery can provide at a given voltage - there may still be significant charge left in the battery, but if you are pulling too much current, the voltage drops to the point where it is not usable. You can sometimes work around that to a degree by putting an electrolytic capacitor across the battery so while the dc resistance goes up, the reactance stays low which can help something that has AC (like a switching supply) and is using the battery as a return path. That was a trick we learned many years ago with the early transistor radios that used the little 9v batteries - you could get significantly longer run time (before the audio sounded really bad) by adding a capacitor across the 9v battery.