A question about the amp draw in 2 battery regulated mods.

[Sinsa]

First off, new member to this forum, and relatively new to vaping(~5 months).
I like to educate myself before pushing the limits, I have been staying in a comfortable area of around 50-60 watts at ~14 amps on LG HG2s, and Sony VTC4s.

I've heard you don't go by the amp numbers the mod provides, instead you use watts/volts.
I've also heard that the amperage is split between both batteries but the equation I read used some words that I didn't understand.

I noticed the lower ohms(usually run about .3-.4 claptons, tried running them in a dual coil and was reading around .1-.2) my coils would read, the higher my amps would read. If I were to go by my usual wattage/voltage I would be well over 20 amps(more like 22-26)

Can anyone tell me how I can calculate how much stress(amps)each battery is taking ?

 

[Darth Omerta]

Welcome aboard! Good to have you!

For a 2 cell mod you calculate based on the wattage supplied by each battery divided by the voltage at cutoff(how low the battery goes before the mod wont work, usually 3.2V), and then factor in the mod efficiency(I like to keep it on the safe side and assume 90% efficiency for most mods). For example lets say you are vaping at 60W
Each cell is responsible for 30W
30W/3.2V = 9.375A
9.37A/.9 = 10.42A

By this example you are drawing 10.42A at the cutoff.

Might also want to give this a read:
Calculating battery current draw for a regulated mod | E-Cigarette Forum

 

[mongo74]

In regulated mods, the amps are not split (unless you're using a parallel configuration...think there's only one regulated mod out there in parallel) With 100% efficiency at, say for example : 120w, you are drawing roughly 18.75a per cell at the cut off voltage of the mod (in a 2 battery series mod) I don't believe there is any mod with 100% efficency...SO:

A safe vaping formula
W+10%(for inefficiency)/mod cutoff V*number of batteries=A drawn per cell.

Darth pointed you in the right direction...follow that lead.

 

[Sinsa]

Just to make sure I understand.

If I wanted to use my device at 100w I would do this.
50W/3.2v = 15.625
15.625/.9 = 17.361A
So 17 amps per battery ?

 

[mongo74]

17a per battery is correct.

 

[Sinsa]

17a per battery is correct.

Thank you for the clarification, going to read up on that thread as well. I appreciate the help.

 

[Darth Omerta]

Yup! You've got it!

 

[BrotherBob]

First off, new member to this forum, and relatively new to vaping(~5 months).
I like to educate myself before pushing the limits, I have been staying in a comfortable area of around 50-60 watts at ~14 amps on LG HG2s, and Sony VTC4s.
I've heard you don't go by the amp numbers the mod provides, instead you use watts/volts.
I've also heard that the amperage is split between both batteries but the equation I read used some words that I didn't understand.
I noticed the lower ohms(usually run about .3-.4 claptons, tried running them in a dual coil and was reading around .1-.2) my coils would read, the higher my amps would read. If I were to go by my usual wattage/voltage I would be well over 20 amps(more like 22-26)
Can anyone tell me how I can calculate how much stress(amps)each battery is taking ?

Welcome and glad you joined.
Might like to read:
Learn About Vaping Here - Everything Ecigs - From Beginner To Advanced
http://www.ecigarettedirect.co.uk/a...tte-college-guides-tutorials-information.html
One Stop Reference Shop For New and Experienced Vapers
Beginner – Guide To Vaping
May 2016 / Smoking Vapor
Vaping 101: Here's Everything You Need to Know about Vaping
(10) Advancing Up the Vaping Ladder with Egos and Mods | E-Cigarette Forum
(1) Proper Terminology - Is it a carto, a tank, or what? A Guide to Juice Attachments. | E-Cigarette Forum

 

[bwh79]

In regulated mods, the amps are not split (unless you're using a parallel configuration...think there's only one regulated mod out there in parallel)

In a regulated mod, series/parallel doesn't actually make much difference as far as amp drain goes. You multiply in one place, or you divide in another, but the end result ends up being the same either way, for the same wattage.

Mech mods of course are a different story, and it matters very much indeed which way the batteries are wired. One allows you to build at very low resistances, while the other one requires that you build very high instead.

Just to make sure I understand.

If I wanted to use my device at 100w I would do this.
50W/3.2v = 15.625
15.625/.9 = 17.361A
So 17 amps per battery ?

Yes that's right. My "rule of thumb" that I use if I don't know a device's actual cutoff or efficiency, is 3v under load at low-voltage cutoff (same as saying 3.3v then subtracting 10% later, but the numbers are easier to work with if we just call it "voltage under load" and use plain old 3v from the get-go). Conveniently, this works out to a "speed limit" of sorts, of 60W per battery (when using good 20A cells). So 100 watts, spread across 2 batteries, means they're pushing out 50 watts each which is near, but not over, the safe limit, presuming you bought recommended name-brand cells from a trusted vendor.

 

[sonicbomb]

In a regulated mod, series/parallel doesn't actually make much difference as far as amp drain goes. You multiply in one place, or you divide in another, but the end result ends up being the same either way, for the same wattage.

Indeed. Parallel or series battery configuration is a design consideration made by the mod manufacturer. Series is usually used because bucking voltage voltage is slightly more efficient than boosting it from the regulators perspective. In a regulated mod wattage is drawn equally from both batteries, the balance of amps and volts to make that wattage is dependant on the wattage selected and the remaining voltage in the battery.

EG.
I=P/V (-10%)

100w divided by 4.2v equals 23.8 divided by 0.9 = 26.4 amps
100w divided by 3.2v equals 31.2 divided by 0.9 = 34.7 amps


On the atomizer side of the regulator, the balance of amps and volts is dependant on the overall wattage and the resistance of the coil. A higher resistance coil will require more volts, a lower one will require more amps. This is of course on the atomizer side of the equation and has no bearing on the amp draw from the batteries.

EG.
100W @ 1 ohm = 10 amp and 10 volts
100W @ 0.25 ohms = 20 amps and 5 volts

 

[90VG]

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[sonicbomb]

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