I am tired of seeing this error being made around here. Most people still seem to be living in the mech mod days, where in order to calculate the current (amps) being drawn from the battery, you only need to know the resistance of the coil and the nominal voltage of the battery. This makes sense on a mech because the resistance of the coil is the ONLY thing you have control over (well that and the charge of the battery). However, if you are using a regulated mod, this does not apply. Resistance of the coil means absolutely zero on a regulated mod.
Let me give a brief explanation on why this is. (NOTE: I am not an electrical engineer, but I did stay in the honeymoon suite of a Holiday Inn Express last night).
Regulated mods typically use a DC-DC converter (probably in most cases a switched-mode converter). That is, they separate the input and output voltage (in other words, they separate the battery from the atomizer). So, just because you have 3.7v going in from your battery doesn't mean this is what will be hitting the atomizer. On a mech mod, yes, that is what happens because there is nothing in between the atomizer and the battery. On a regulated mod, there is a voltage regulator in between the battery and the atomizer.
On a mech mod, as your battery drains, you have less voltage (and thus less power) hitting the atty. This means the vapor production diminishes over time. We know from Ohm's law that Power = voltage X current. As you can see from this simple equation, as the voltage drops, this necessarily means less power (watts).
The circuitry in a regulated mod stops this from happening. The regulator will swap voltage for current in order to achieve the power (watts) you have your mod set at. Again, P = I * V. As the voltage on the right side of the equation drops, the power also must drop. So, looking at this equation, how can we keep the power constant throughout the charge of the battery? Yep, we need to increase the "I" (current) to compensate for the battery being drained. This will allow you to keep your desired power setting all the way through the battery's charge.
I think the biggest bit of confusion comes from the fact that people don't know the difference in input voltage (what comes directly from the battery and varies based on charge level) and output voltage (what the regulator puts to the atomizer to achieve your desired power level). And the mods we use don't help the matter. A lot of these mods show the power, the resistance of the coil and the applied voltage on the screen at the same time. So, a lot of people assume this is the voltage value you plug into an Ohm's law equation. Using our .5Ω example, if you had it set to 50 watts, the screen would show 5v. So if you plugged in 5v and 0.5Ω into a calculator, you would once again get back the incorrect result of 10 amps.
I saw a guy earlier who was talking about his build on a regulated mod. He said "I am running 0.3Ω at 80 watts. This equals 16 amps, so I am well within the safe limits of my 20 amp battery." Well, he committed the cardinal sin of using the output voltage in his calculation (or using Power and atomizer resistance to calculate current -- neither are correct). Let's do his calculation properly. If his battery is fully charged (around 4v) and he is running at 80 watts, then:
I = P/V
80/4 = 20 amps.
In reality he is pulling 20 amps on a full charge, not 16 as he thinks. But that's not all. Since the battery voltage drops during use, the regulator will have to increase the amperage drawn to keep him at 80 watts. So, let's say his battery is near dead and is at 3.2v.
80/3.2 = 25 amps
Now, since most batteries are 20 amps continuous, we might be getting into some danger territory (possibly). Meanwhile this guy is vaping happy thinking he is still only pulling 16 amps from the battery.
Multiple Battery mods:
Example: You are running your Sigelei 150 at 150 watts and you want to know the amp draw on your batteries.
I = 150/ 7.4v = 20 amps
Since your two batteries become a single battery in series, the nominal voltage effectively doubles. This means you don't need as much current to hit that 150 watts.
TL;DR: Don't confuse output voltage for input voltage on regulated mods. If you want to determine your current draw from the battery on a regulated mod, here is the only correct way to do it: I = P/V. That means your current will equal your watt level divided by how much charge you have on your battery. If you don't know the charge, then just plug in 3.7v (as that's the nominal rating). Atomizer resistance has nothing to do with the current being drawn from your battery on a regulated mod.
Also, you need to factor in efficiency (~10% overhead) as well depending on the board (check your board's specs for efficiency).
