For a pleasant and safe vaping experience, it's important to understand the difference and relationship between,
- Voltage/Volts (E)
- Resistance/Ohms (R)
- Amperage/Current (I)
- Watts/Power (P)
With Ohms law there is "nothing gained and nothing lost", if the value of any of these 4 variables change, it will affect the value of all the others to balance out the equation. There are only 2 variables you can change directly, Voltage and Resistance, by changing these you can indirectly affect (change) the others.
You'll first want to figure out your target output power (watts) is, in other words, what wattage you like to vape at? The only real way to do that is to try it, personally I like to vape at ~20Watts and I'll use this target in the examples below. We will see that we can reach our desired power output (watts) with several different voltage, resistance and amperage values, regardless what values are used, if the result of the equation is the same output power, 20Watts in this case, the vape experience should be the same.
I mentioned above that there are only 2 variables you can change directly, Voltage and Resistance, the other 2, Amperage and Watts, are a result of the first 2.
Voltage/Volts: The batteries in your mod output a fixed voltage of 3.7Volts. I say "fixed" because you can't change what the battery outputs but in reality the 3.7Volts they are labelled with is the mid point between 4.2V (full charge) and 3.2V (drained). Now I just said the output voltage of the battery was fixed and also claimed that Voltage is one of the Ohms Law equation variables we could change... which is it?! You can change the output voltage in 2 ways,
- The way the batteries are wired, series vs parallel. In an unregulated mod this is the only control you have over the output voltage, in series the voltage is doubled, in parallel the capacity/battery life is doubled.
- With a Regulated mod. The only thing that is regulated or variable in a regulated mod is the voltage which indirectly affects the output power(watts). Some digital mods will allow you to set the desired output power(watts) but all it is doing is changing the value of the voltage variable in the Ohms Law equation to achieve the desired output power(watts).
Resistance/Ohms: The "coil" in your mod has a fixed resistance (the unit of measure is Ohms), you can change the resistance by using a different coil with a different resistance.
Now for the other 2 variables which can only be changed indirectly, that is by either changing the voltage or resistance,
Amperage/Current: I'm gonna go long here because out of the 4 Ohms Law variables, Amperage is the one I find is hardest to grasp. Amperage is a "draw"... How much current or how many amps a circuit will "draw" depends on the values of the voltage and resistance. Amperage is also the main variable to look at for safety, we must make sure that all the components in the mod (wires, batteries, etc), all the components in a given circuit are "rated" or can handle the amount of current/amps the circuit will draw. If a component exceeds the amount of current it can handle it will fail, typically a high heat/fire/melting type failure.
Watts/Power: Like Amperage the output power(watts) is a variable of the Ohms Law equation that is affected indirectly by changing the values of either the Voltage or Resistance, it is the "result" of the other variables in the Ohms Law equation.
Time for some math! I said I personally aim for a ~20Watt power output, let's look at different ways we can acheive this,
Double 18650 Unregulated Mod, batteries wired in parallel
With the batteries wired in parallel the output voltage remains the same, 3.7Volts. The only variable left I can change is the resistance of my coil, what resistance do I need to achieve my desired 20Watt power output?! One way the Ohms law equation can be presented is,
R = E2 / P
We know that Volts (E) is 3.7 and that Watts (P) is 20 so,
3.72 / 20 = 0.68
I need to make (or purchase) a ~0.68 Ohm coil, What kind of Amperage will this setup draw? Other ways Ohm's law can be represented are,
I = P / E
I = E / R
We know that Volts (E) is 3.7, Watts (P) is 20 and Resistance (R) is 0.68,
20 / 3.7 = 5.4
3.7 / 0.68 = 5.4
This setup will draw ~5.4Amps... Can my components (batteries, wires, etc) handle that? Yup, we're vaping safely!