It regards to
vaping, what is the difference between wattage and voltage?
The difference is in which number you want to be displayed in bigger print on the screen. You can adjust in watts (and the applied voltage will be changed to achieve that wattage) or you can adjust in volts (and the derived wattage will change to reflect that voltage) but they are tied to each other by the resistance of your atomizer. The equations that are used are:
Watts = Volts * Volts / Ohms
Volts = sqrt(Watts * Ohms)
So if you have, for example, a 2-ohm atomizer, and you set your device for 8 watts, it's going to apply sqrt(8 * 2) = sqrt(16) = 4 volts to achieve those 8 watts. The device may or may not say "4.0v" somewhere in smaller letters. (The MVP2 doesn't, but some others do.)
On the other hand, if you are setting it in voltage mode and you set for 4v you are going to be getting 4 * 4 / 2 = 8 watts out of it. The device may or may not read "8w" somewhere in smaller letters.
The choice of coils is really going to come down to personal preference. Lower-ohm coils are going to have fatter wire that puts out more vapor (more surface area) but takes longer to heat up (more mass). Higher-ohm coils will be just the opposite, with skinnier wire and a faster ramp-up time but less surface area in contact with the liquid making less vapor output overall. It used to be, with fixed-voltage and unregulated devices, that low ohms were required in order to achieve high power output. These days, with wattage-regulated mods that's no longer the case. You can put on any atomizer that you want, and crank up the voltage as high as it will go. That's not necessarily a good idea, though. If you try to run too much power through the skinny wire of a higher-ohm coil, it might run too hot and burn out the coil. Also you may be limited by the voltage output. A device might have an output limit of say, 8 volts, for example. So if you try to put a 2-ohm atomizer on and set it for 50 watts (your device doesn't do 50 watts, but lots of others do) then it's not going to happen. That would require 50(watts) * 2(ohms) = 10 volts to achieve, but the device can only output 8 volts, so the most you're going to get from 2 ohms is 32 watts (8 * 8 / 2). Put a .5 atomizer on that same device, though, and then 50 watts is only sqrt(50 * .5) = sqrt(25) = 5 volts, well within the available range.
The last (well, first really) thing you need to know about is battery safety. All batteries have an amp limit, and you need to know what that limit is and how to stay underneath it. On a wattage-regulated device, the amp drain is equal to:
Amps = Watts / Volts
Note that this "volts" is the
charge state of the battery, and
not the voltage applied to the atomizer. Note that as the voltage
increases, the amp drain
decreases. That means your amp drain will be highest, when the battery is at its weakest charge state. So for safety's sake, assume a low-voltage cutoff of 3v under load. Take the maximum wattage your device will do, and divide that by 3v. I think the MVP2 does up to 11 watts, so 11 watts / 3 volts = 3.67 amps. This should be plenty low enough for the built-in battery on your device, but on some other devices with higher power outputs and user-replaceable batteries, you need to know your stuff because the amp draw can be up around or even over 20 amps, which is just about the limit for any battery we have available. (*In voltage mode, you will need to calculate the applied wattage at your set voltage, and then divide
that by the battery voltage to find your correct amp draw.)
The last calculation, doesn't really apply to you because it's for mechanical or unregulated devices, is the classic "Ohm's law" equation:
Amps = Volts / Ohms
This is for use with unregulated devices, where raw battery power is just dumped into your atomizer without any circuitry in between. For this notice that the amp draw is higher when the volts are higher and the ohms are lower. In this case you would always use 4.2v as in a fully-charged battery. A .5 ohm coil on a fresh battery would, for example, draw 4.2v / 0.5Ω = 8.4A.
Finally, let's take one last look at that first equation: Watts = Volts * Volts / Ohms. Notice it contains the term "Volts / Ohms," which we've just discovered is also the Amperage. Therefore, another way to write it would be:
Watts = Volts * Amps (Note that this is the same as the "Amps = Watts / Volts" that I said earlier, just rearranged a little bit.)
