I'm coming to understand VV, but have absolutely no idea about VW. What are the pros and cons? Which is better? I just got an iTaste VTR. I'm running the iClear30S that comes with it and a Protank 2 both with 2.2 resistance at 4.2 volts. What would I or should I set the watts at, or does that only matter if I have it in VW mode? I would really like to understand this better. Please Help!
To begin with, to answer "What should I set the watts at?" question: You adjust it to what you like. Experiment, and when you get the right vape for you, that is where you set it. Same goes if you use the variable voltage setting.
Now the math:
the relationship of Voltage, current, and resistance are described through Ohm's law
Voltage (E or V) = Current (I) X Resistance (R)
Through manipulation of this we have:
I = V/R
R = V/I
So if you have any two of the values, you can find the third.
Power can be calculated through Joule's law:
Power (in Watts) = V X I
Ohm's law can be inserted into this equation:
P( or W) = (I X R) X I = I^2 X R
or
P = V X (V/R) = V^2/R
On our vaping systems, we know that our coil is at a certain resistance... I'll use 2.4 ohms for this explanation.
So we can adjust either voltage, or wattage.
If we want to adjust voltage, then the wattage will change accordingly (by way of the current required to satisfy ohm's law) to provide vapor.
If we want to adjust wattage, then the voltage will change to provide the vapor.
If you use the VV setting with a particular coil resistance, and you change that resistance, you will need to adjust the voltage again to obtain the same level of vapor.
If you use the VW setting with a particular coil resistance, and you change that resistance, the system will adjust the voltage to obtain the same level of vapor.
I have a 2 ohm coil, which is actually measured at 2.3 ohms by my VTR. I set my power to 8.5 watts.
The system is providing V = square root(P X R) = square root(8.5 X 2.3) = square root(19.55) = 4.42 volts
and I = V/R = 4.42/2.3 = 1.92 amps
So my battery is providing 4.42 volts at 1.92 amps each time I vape it.
If I set my VTR voltage setting to 4.4 volts, it should provide the same level of vape using the same coil.
The battery will provide 4.4 volts at 1.91 amps each time I vape it.
Now lets change the coil... To a 1.8 ohm. This time, we will keep the same power setting (8.5 watts) and then the same voltage as before (4.4 volts)
Power setting:
V = square root(P X R) = square root(8.5 X 1.8) = square root(15.3) = 3.91 volts
I = V/R = 3.91/1.8 = 2.17 amps
So the battery is providing 3.91 volts at 2.17 amps, but it is producing the SAME amount of power as before.
Voltage setting:
I = V/R = 4.4/1.8 = 2.44 amps
P = I^2 X R = 2.44 ^ 2 X 1.8 = 5.95 X 1.8 = 10.71 watts
So the battery is providing 4.4 volts at 2.44 amps, but is now producing 10.71 watts.
I have probably just reiterated most of what has been said, but I figured I would provide the math that describes what is going on when you make the adjustments.
If you have a vaping system that provides both adjustments (like the VTR), it is purely up to your desire as to which setting you adjust. Either, in the end, will provide you with a happy vaping exprience. With the VTR, the setting that you adjusted last, is the variable that it uses. So if you set the wattage last, the system will adjust the voltage to provide that power to the coil. If you set voltage last, it will use that voltage to provide power to the coil. You just need to remember that if you use the voltage setting, if you put in a different resistance coil, you will need to make sure you adjust the voltage accordingly.
If that wasn't long winded enough.. I don't know what is... This is what you get when I am waiting on code to finish building.
Thanks for being a good sport about it.
, I just know I often change my watts by .1 on my devices and have liked it that way. As always, our experiences and preferences vary. And, as devices vary also in their accuracy, I may like 6.5 watts on one device and 7 on another.
Clatter box Clown toy... 
