Electronic Cigarette Batteries

E-Cig Batteries and their Voltages

A term you’ll hear allot amongst vapers and electronic cigarette aficionados is “mah” which stands for milliamp hours. This is a value that tells you how long the battery can hold a charge. The higher the “mah” the more puffs you will get between charges

Standard electronic cigarettes are usally about 150mah while the best ones are upwards of 200mah-250mah, however MODS can easliy go over 3000mah

AW’s Lithium Manganese (IMR) are rechargeable batteries that have the benefit of safer chemistry, this removes the need for protection circuits. These batteries offer faster discharge rates and are suitable for high power applications over 3 amps. The safer chemistry over Lithium Ion means no risk of venting, flame or explosion.  Standard e-cigarette batteries usually output at about 3.7 volts (although) the best batteries on the market will have a higher output) and a standard resistance atomizer is about 2.5-3.0 ohms (once again, the best electronic cigarettes will have an atomizer with greater resistance)the best electronic cigarette battery


The power of an electronic cigarette and/or personal vaporizer is determined by the battery output voltage and the resistance of the atomizer. This Output can be varied in more advanced electronic cigarattes and vaporizers to provide more vapor, throat hit and flavor.

In technical terms the overall experience of vaping an electronic cigarette in conjunction with different types of e-liquids, is determined by the output power which is a function of voltage and resistance

Here’s the math:

P = V²/R

Where P = Power, V = Voltage and R = Resistance

So what does this actually mean ?

Lets say you like an electronic cigarette at 4v with a 2 ohm cartomizer, your power output would be:

P= 4²/2 = 8W (Watts)

If you switched to a 4 ohm atomizer the output power would be:

P=4²/4 = 4W

So in order to get your preferred power of 8W, say for a particular e-liquid or e-cig device you would need to increase the voltage

V = √(PxR)
V = √(8×4) = 5.66v You would need to increase the voltage to 5.7v