I'm not sure how much you know about the Sig150w, I forget if you said you have one, but do you know if the voltage reading/display on the mod is what I should be doubling or do you think it already doubles it for me? For example, the lowest I see it go on the display is somewhere around 2.4V when it stops firing and it displays the whole "Check Battery" thing. Does this mean it's stopping at 2.4V or 2x2.4=4.8V?
On my Sig-- it's an Old Skool without TC-- the voltage display is for the voltage it's putting on the coil, not the battery voltage
per se. The battery voltage is indicated by the battery icon to the side, full = full charge, half = half charge, you know the drill. The voltage display is a different matter. How a regulated mod works is backwards to a mech mod. With a mech, the voltage is (semi) fixed because it's native to the battery. It varies (decreases) from full charge to depleted, but it tends to stay the same for the duration of the puff. Thus, the only way to adjust watts is to change coils. On a mech, at full charge of 4.2 volts across a .5Ω coil, the mod makes 35 watts. As the voltage in the battery decreases due to discharge, down to say 3.5 volts, the watts applied also decreases because the coil stays at .5Ω, down to 24.5 watts.
On a regulated mod, both the coil and the watts are fixed; the coil doesn't change, and the watts you set on the machine. Thus, the mod varies
voltage to change watts, not the coil value. If you put a .5Ω coil on, and tell the machine to make 40 watts, it will hit the coil with 4.47V. If you tell it to make 60 watts on the same coil, it will use 5.47V. That figure is what the voltage reading on the screen is telling you: the mod is using X volts on that particular coil to make whatever watts you set.
None of that has anything to do with the amp draw on the battery. It's helpful to think of it as two separate circuits; a battery circuit to power the board, and a firing circuit to power the atomizer. What I just described above is downstream of the board in the firing circuit. Upstream of the board, in the battery circuit, is where amps are applied to the batteries. All the power to the whole machine is supplied by the batteries; this includes both the power used to fire the coil, as well as the efficiency losses (I^2R losses) on the board itself, and
the board has to take that power at whatever voltage the batteries are then supplying. Therefore, while the power expended to fire the coil stays the same in all cases (so long as the coil and set watts stay the same), the losses can vary on the board, and
the current draw from the battery must increase as battery voltage decreases from discharge to make the required power for the downstream circuit.
Now maybe I am wrong. Perhaps the voltage display on your machine changes, showing applied voltage when the fire button is pressed and snapping back to native battery voltage when the button is released. There's no reason the board couldn't be programmed to do that, although most of them aren't in my experience. In your case, though, I don't think so because 2.4V is lower than it is safe to discharge that type of cell. Generally, 2.7V is considered about the lowest to avoid battery damage, and most mods kick out between 3.5~3.2V. I strongly suspect what you're seeing is the voltage the machine applies to the coil downstream of the board, not the voltage supplied by the batteries upstream of the board. An easy way to check-- I assume you don't have a digital charger that displays battery voltage when you put them in to charge-- is to measure the voltage of the battery with a multimeter when you take it out of the machine. In my Sig, the "discharged" batteries come out within a few hundredths of 3.4V, as measured by my Xtar and Nitecore digital chargers as well as a pair of Fluke multimeters (an 87 and a 287).
