What's the battery life like on the Provari Mini???

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klac

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I've been looking into purchasing the Provari. I am pretty much sold on all of the good things that owners have to say about it. However, I am undecided between the mini and full-size models. I know that you can get the extender to have it work with larger batteries, but was wondering what is the battery life like on the Provari Mini with the standard smaller battery?
 

Spazmelda

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I use AW IMR 18650s. I have one 2000 mah and two 1600 mah batteries. The 1600s last me about 1 day + a few hours, the 2000 lasts me about a day and a half; maybe more. I usually run the provari at 4.2-4.5 V and go through about 4 ml a day in a tank.

I need to buy a few more of the 2000 mah batteries. Matter of fact, I think I will go do that right now.
 

DaveP

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Some people like to use low resistance cartos with the Provari. I get better battery life with 3 ohm cartos at 4.2v.

This power chart shows you the wattage draw from various combinations of voltage and resistance on an ecig. Click it for a larger, more readable view. 7 to 8 watts is a reasonable vaping level. Some people say they vape at 12 watts, but that's a battery killer and a hot vape.

Safe Vaping Power Chart
 

njgeek

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Here's a reply I made on he same topic at reddit:

You'll make the best use of your VV by getting SR carts (Boge SR is my favorite), not dual coil and not LR. Fillerup, crank the voltage to about 4.6 to start, and play with it. Trust me, you'll love it, and it helps your battery life too.

I have an 18350 Provari Mini, and the battery life is just fine. I get about 3-4 hours out of each charge with constant vaping. A good AW IMR high drain is recommended, and get about 3 so you can have a rotation.

Remember, 4.7V at 3ohm has better battery than 3.7V at 2ohm.

4.7 * 4.7 / 3 = 7.36 watts

4.7 / 3 = 1.57 amps

3.7 * 3.7 / 2 = 6.845 watts (less power)

3.7 / 2 = 1.85 amps (more battery draw)
 

DaveP

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Using AW 18650 IMR 2000mah (from Provape) in my Provari, I can go from the night before until late the next afternoon before cutoff. I usually charge when the battery hits around 3.4v to 3.5v to reduce battery charging time and stress on the batteries. They don't particularly like being drained to cutoff.

njgeek, I also ran those numbers on 2.0 and 3.0 cartos at my sweet spot and discovered the same thing you did. The amp draw at higher resistance as less while still being in the sweet spot for wattage on the 3 ohm carto. Since wattage is E^2/R and amperage is V/R it works out that the curves are different and you get more battery life at higher resistance with the right voltage.
 
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CyberDj

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Well, if you use an AW IMR 18350, 700mah battery you'll get approximately 7 hours. Of course the voltage you vape at will make a difference. The higher the voltage, the less time you'll get.

Not really. Volts don't have too much to do with the drain. Amps does. A 1.7 ohms dual coil kills may AW 18650 1600 mah battery in about 3 hours. A 3.5 ohm atty on 5.5 runs for 7 hours...
 

Rader2146

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Here's a reply I made on he same topic at reddit:

You'll make the best use of your VV by getting SR carts (Boge SR is my favorite), not dual coil and not LR. Fillerup, crank the voltage to about 4.6 to start, and play with it. Trust me, you'll love it, and it helps your battery life too.

I have an 18350 Provari Mini, and the battery life is just fine. I get about 3-4 hours out of each charge with constant vaping. A good AW IMR high drain is recommended, and get about 3 so you can have a rotation.

Remember, 4.7V at 3ohm has better battery than 3.7V at 2ohm.

4.7 * 4.7 / 3 = 7.36 watts

4.7 / 3 = 1.57 amps

3.7 * 3.7 / 2 = 6.845 watts (less power)

3.7 / 2 = 1.85 amps (more battery draw)

With a boost regulator, amperage at the atty/carto has very little to do with the actual true battery drain. Using the same wattage and comparing low resistance to high resistance, the only difference in battery life is going to come from where the regulator is more efficient. Even then, the difference in efficiency is most likely too small to be noticeable in a scale of hours of battery life. It's more on the scale of minutes of battery life.

Here's a post from another thread explaining input and output current on a boost regulator:

A boost regulator has to transform low voltage into higher voltage. This can only be done by using more current on the input (battery) side. The amount of additional voltage needed is expressed as:

(Volts out - Volts in) / Volts out = Percentage of voltage increase, also know as the switch duty cycle.

Now that we know the duty cycle, we can figure the additional input current required to obtain the desired output voltage.

Amps out / ( 1 - Duty Cycle) = Amps in

Example...

Known factors:
3.7v in
8 watts out
3.0 ohm carto

Ohms Law tells us that we'll need 4.9v and 1.63 amps output to achieve 8 watts.

(4.9-3.7)/4.9 = .24 = 24% increase in voltage = 24% switch duty cycle

1.63 / (1-.24) = 2.16 amps input drawn from the battery.
-----------------------------------------------------------

Now for validation.
As deemed by the Law of Conservation of Energy, power (watts) in must equal power out. (The true statement of the law is power in equals power out + efficiency losses. But for simplicity sake, we'll get to efficiency below.)

Power = Volts * Amps

Input:
3.7 * 2.16 = 8 watts input

Output:
4.9 * 1.93 = 8 watts output
------------------------------------------------------------

But what about efficiency?
Typical efficiency for a boost converter is in the 75-90% range. Efficiency is not constant. It varies with the desired outputs. You can find the efficiency for certain [manufacturer chosen] situations in the regulators data sheet. Using an optimistic value of 90% efficiency we can figure our adjusted input current.

Power out / efficiency = adjusted power in

8 / .9 = 8.89 watts input.

Adjusted power in / Volts in = adjusted amps in

8.89 / 3.7 = 2.4 amps input.
-----------------------------------------------------------
And comparison:
A fixed voltage device @ 3.7v will achieve an 8 watt output using 2.16 amps (Ohms Law)

8 / 3.7 = 2.16 amps input
------------------------------------------------------------
The above calculations explain why I say that boost regulators will get less battery life than a same size fixed volt, and also that you will not achieve better battery life by using a higher resistance coil.

-----------------------------------------------------------
 

Rader2146

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I've certainly had my fair share of those moments.:facepalm: I had the same "low amps equal better battery life" thoughts until about a month ago when I started researching how boost regulators work. Followed by that sinking feeling when I say to myself "every bit of advice that I've given about boost regulators was wrong." :oops:
 
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