Hello, currently Im using the Evic VTC Mini with Sony VTC6 battery. According to Mooch the maximum amp draw of the VTC6 is 19Amps. Currently with my build it is 19.2Amps and around 4-5 hours battery life. Since I started to carry few extra batteries with me and looking forward to even lower builds I need a bigger mod. I really like the Wismec RX2/3 and plan to use it with two VTC6 batteries. My questions are? Is this mod capable of splitting the amp draw between two batteries? In other words am I going to have a limit of 38 Amps? Edint: Already found an answer. The amp limit in series mode is not doubled but since I will have more voltage, the amp draw will decrease. Correct? Is this mod going to have longer battery life than my EVIC if I use the same resistance? I often read the following statement: Atomizer resistance has nothing to do with the current being drawn from your battery on a regulated mod. Why is that? If Im correct on mech mod this is very important.

For battery life comparisons, I prefer to use watt hours. I use a rating of 3.7 volts in my calculations. For an 18650 of 2000mah capacity at 3.7 volts, 2 amp hours (2000 converted to ah) x 3.7 = 7.4 watt hours. For 2 in series, 2 amp hours x 7.4 (2 batteries in series double) = 14.8 watt hours. For 2 in parallel, 4 amp hours (2 in parallel double) x 3.7 = 14.8 watt hours. As you can see, the battery life in either series or parallel mode will double over one battery. For 2 batteries in parallel, available current draw is double. For 2 batteries in series, available current draw is equal to 1 battery's rating. When it comes to battery life related to atomizer resistance, just divide the watt hours of your battery setup by the watts you are using for your atomizer and the answer will be the portion of an hour your batteries will be able to power your coil. As far as your bolded statement: Right now I am using a 0.2 ohm atomizer at 50 watts. Calculated, this means I am powering the coil with 3.16 volts. This is below normal battery voltage of 3.7 volts, being regulated down by the chip. Current draw from my battery will also be lower than in a mech. At 3.7 volts in a mech, current draw across a 0.2 ohm coil is 18.5 amps, 68.5 watts. At the 3.16 volts so that I have 50 watts, current draw in the system is only 15.8 amps. Confused yet?

2 sides to the equation! Coil side and battery side. Usually they are different. So lets work backwards from this 20 amp deal. Assuming maybe 3.5 volts or so at the coil. That indicates a 0.175 ohm coil at round 70 watts. Lets do math for a 2 battery series mod. We set the mod for our desired 70 watts, and realize that we need a bit more from the batteries for the regulator circuit. So for simplicity's sake, lats say we need 75 watts from the batteries to get 70 at the coil. Also we assume the batteries are in good shape, nearly full charge. When we press the button, they do 3.7 volts per cell. so 7.4 volts total. 7.4 volts at 75 watts is 10.13 amps from the batteries. Also we see that it looks like a 0.73 ohm load on the battery side of things. I hope this makes sense?

Your amp draw will actually decrease on the input side with a regulated mod at a higher voltage- 50W with a single 18650 mod would be 16.12A, 50W with a dual 18650 mod would be 8.06A, then 50W with a triple 18650 mod would be just 5.37A. This is with a cut off of 3.1V per cell, so 3.1V, 6.2V, and 9.3V respectively. You will get longer battery life at the same wattage setting- resistance won't play much of a role apart from bucking vs boosting. On a mech mod, resistance is very important. On a regulated mod, the calculation is watts used/min input voltage. You can also divide by mod efficiency too for a more accurate result. The resistance on a regulated mod only matters in regards to not hitting amp output and voltage limits- which are different to the input voltage. Also efficiency comes into play but is nothing to do with amp load on your batteries.

It's because the whole purpose of getting a regulated mod is to let the chip of the board automatically calculate how much current draw will be required in order to achieve the desired power output level.

Boards with batteries in series tend to be more efficient than the ones with batteries in parallel. Just extra info

Thanks so much for all the replies. I understand almost everythink right now except one thing. Lets assume that Im vaping with 0.1 coil at 75 watts and its pulling 20 amps. If the resistance is not important on regulated mod, what will happen with the amp draw if I change the 0.1 coil with 0.05 at 75 watts? Sent from my iPhone using Tapatalk

The output amp load will increase- Ohms law applies as normal to the output side, and if your mod can fire that low (many regulated mods can't, some can), the output amps going to your coil will increase to give the 75W. What happens at the input side doesn't change though. 75W is 75W with a regulated mod and resistance does not change the amp load on the input side, only watts and voltage matter for amp load on your batteries, on the input side.

Just to be on the safe side... I just made some SS Stagerred staple fused clapton. It is 0.051 after heating with a torch. My mod is capleble of firing everything greater than 0.05 on TC. The problem is that now the mod VTC Mini 2 (Vtwo) is getting pretty hot when chain vaping. Is it a problem for the battery (VTC6) or the mod to be this close to the limit? It is showing more than 25 Amps on the output side, but as you thought me this is not important. Still dont know how to xalculate the amp draw on the input side. I was thinking about connecting multimeter. Edit: I think I got it: 75w/3.2=23Amps on the input side when battery almost dead. Is this too much for VTC6 battery? I very rarely vape below 20% battery life. Is it safe?