# DJLsb Vapes - How to chose your Batteries



## Pixstar (22/2/16)

Very good battery guide here from _Daniel of DJLsb Vapes:-
*https://www.djlsbvapes.com/chose-batteries/*_

*How to chose your Batteries*
Welcome, here you can find the way of using your devices choosing the right batteries to vape safe and have the more capacity available for the power range usage you are going to use them on your device.

On an unregulated you need to know the resistance of the coil and the nominal voltage of the battery, this makes sense on an unregulated device because the resistance of the coil is what's going to determine how many Amps you are going to be draining from the battery, however, if you are using a regulated mod, this does not apply, resistance of the coil means absolutely nothing on a regulated mod.

Regulated mods use a DC-DC converter, usually a switched-mode converter so the input and output voltage will be separate (in other words, the battery is separated from the atomizer), just because you have 3.6v input from your battery doesn't mean this is what you will have on the output.

From Ohm's law Power (W) = Voltage (V) x Current (A) so as the voltage drops, this necessarily means less power (watts). 

On an unregulated mod, as your battery drains, you have less voltage (and thus less power) hitting the atomizer, this means the vapor production diminishes over time.

On a Regulated mod the regulator will swap voltage (V) for current (A) in order to achieve the power (W) you have your mod set at.

So, because P = I * V. when the voltage drops, the power also must drop then we need to increase the "I" (current) to compensate the battery drain and this will allow you to keep your desired power during the entire battery charge.

The equation for current is I = P / V. That is, Current = Power/Voltage. Voltage will be a fixed parameter on the equation but of course even if you chose to do the math with the nominal voltage of 3.6 Volts that will not apply entirely during the discharge of the battery since most devices are going to drain the batteries down to 3.2, 3.1 or even 3.0 volts. 

With the equation I = P / V you can chose an 20Amp battery to vape at 72Watts an a single battery mod but when the battery drains below the nominal voltage you will get the Power decreasing down to 62Watts and on some mods the message "week battery" display'd on the screen.

With that being said, in my personal opinion you can chose a battery based on the nominal voltage but you would get an more reliable and accurate experience choosing a battery based on the minimum voltage as you can check on the table below.

Some Multiple Battery mods will run multiple batteries in series or parallel. 

A series configuration will treat the two batteries as one and as consequence the double or the triple of the voltage but still the same Amps. As a consequence of that, having more Volts the device will need to drain less Amps from the batteries.

Parallel configurations do not double the voltage, however, the capacity of the cells (amp hours) is doubled and the available Current (A) also. As consequence of that, with the same voltage, now the device is going to drain more Amps from booth batteries.

So, series and parallel devices achieve the same thing but in a different way, series mods are going to do it with more voltage (which decreases the need for more current), while parallel mods are going to do it by doubling the amp hours (increasing the available current).



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https://www.djlsbvapes.com/chose-batteries/
*https://www.djlsbvapes.com/chose-batteries/*​_

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Reactions: Informative 2


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## Kuhlkatz (23/2/16)

Mmm. I tend to disagree with him recommending or using the 'Nominal' or 'Minimum' voltage level for power in any of the above equations.
I do not remove a battery from my charger if it's only halfway there. I leave it in and remove it when it's at 4.2 v.

Any safety calculation you do, should be done using the *maximum *voltage of 4.2 v, *not the nominal* voltage. My battery is not going to explode in my face only at 3.6 or 3.3 volts. It's going to do it when it's fully charged, especially so if I build a coil for 3.6 v.

Looking at the last bit :



A 10A battery with a 0a 0.12 ohm coil.3 ohm coil at 4.2v =
4.2v / 0.3 ohm = 14 A
That is 58W from a 10A battery, nice ! Who needs VTCs or 25Rs.
I would actually be *exceeding *the battery's max rating by 40%.

A 25A rated battery with a 0.12 ohm coil at 4.2 v = 
4.2 v / 0.12 ohm = 35 A
147W from a single 18650 - Winner!

Always do your own calcs and then double check them again. Ask someone else if you are not sure.
Do not trust everything just because it's on the interwebs.


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## Nightwalker (23/2/16)

I tried to follow, but my brain hurts now

Reactions: Funny 1 | Can relate 1


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