Mtl tank battery(18650)

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stols001

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May 30, 2017
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LGs are great at first but compared to my 25A Samsungs with 500 less mAh well the Samsungs hold up better. It is rare for me to exceed 14 watts. IDK. It's all a trade off one thing with another.

The LGs are also slightly LONGER. You would think that would not make a difference and in most mods it doesn't but I have a few that have RISEN up and protested disgorging their batteries unexpectedly.

Oh if you go with the chocolate, tissue paper wraps, be prepared to rewrap. I would almost suggest preemptively. At first I was like "Well that's smart, using a poop brown wrap to like show any dings and whatnot. Later, I was like, "They probably asked the wrap makers, what is the cheapest thing you have?" and they replied, "Meh we have this thing the color of dog poop that tears when you look at it," and they were like, "Cool, cool." Occam's razor I was thinking too hard.

Anna
 

Punk In Drublic

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OK, please explain your case.

:pop:

In simple terms – a regulated device is divided into 2 separated parts, an Input (battery) and Output (coil). The input never see’s the coil resistance, only what the regulator circuit is requesting in terms of power in watts. To the battery, 50 watts with a 0.1 ohm coil is the same as 50 watts with a 1 ohm coil (assuming we are working within the voltage and current limitations of the output circuit. But that is another topic).

To determine the current draw on the battery, the math would be as..

Power in watts/# of batteries/battery voltage/ device efficiency. Coil resistance is not part of the equation. And with a regulated device we use the lowest possible battery voltage for that will yield the highest possible current draw for that particular wattage.

So...single cell using the OP’s 25 watts, a typical 3.2 volt cut off, and assuming a device efficiency of 90% we have

25/1/3.2/0.9 = ~8.7 amps

You can read Mooch’s blog on the subject
Calculating battery current draw for a regulated mod | E-Cigarette Forum
 

Punk In Drublic

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mAh is measured at <1amp and down to 2.5 volts. We do not vape at <1 amp and down to 2.5 volts therefore using mAh as a value to determine battery run time can yield false promises.

What we need to look at is Watt Hours. Watt Hours can be measured at any current draw and down to any terminating voltage. WH will display the amount of energy a battery will possess within said perimeters. A 2500 mAh battery could very well yield a higher Watt Hours over a 3000 mAh cell with a 10 amp draw down to a typical 3.2 volt cut off.

Unfortunately we do not always have this specification, Mooch did not measure WH with his earlier tests but has incorporated WH tests with his more current battery measurements.
 

Punk In Drublic

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Oh, but it does. It certainly does.

Coil resistance will only become a factor once we start reaching the limitations of a device in terms of current and voltage output. With the OP vaping at ~25 watts, we are far from those limits.

At 25 watts with a 1.6 ohm coil, Ohms law states we require 6.3 volts to achieve 25 watts which will yield a 3.9 amp current draw.

25 watts with a 0.5 ohm coil would yield less voltage, 3.5 volts, but more current at 7 amps. Regardless of how we cut up resistance, voltage and current the result is 25 watts.

At the battery end the circuit is requesting 25 watts (plus a little for efficiency) from the battery. As the battery voltage depletes, current draw increases in order to meet the power request. But the result is still 25 watts (in this case) - coil resistance is not a factor.
 

AngeNZ

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  • Mar 24, 2018
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    OK, please explain your case.

    :pop:

    Mooch did it for me in his battery blog
    Calculating battery current draw for a regulated mod | E-Cigarette Forum
    One of his comments in that post explains it clearer:
    regbattery.jpg


    For regulated mods I just use this as a guide, from Mooch's recommended batteries
    ampvwattage.jpg
     
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