Help me understand something about battery use...

[zoiDman]

Yeah, but you guys are (well, kinda) missing the point. I'm not trying to be a troll, or a smart-... here but seriously ....

Mega Meh!

...

Personally I don’t think the OP’s question, which is a Very Reasonable inquiry, has been answered By Anyone.

I also have found the reply’s in this thread helpful and or informative. So why the Big Meh?

Some of us don’t have the Level of Expertise that others may have but should that stop us from asking questions or making replies?

 

[AttyPops]

Zoid,

The Meh isn't about asking the question, nor the replies, nor the topic. It's about... in the end, even if you get an answer, does it matter much?

So I was not critizing the OP, nor the question. Just saying that after all is said and done, I'd rather set voltage to taste rather than battery efficiency.

Besides, I made a reply too. I think it IS and interesting question. I just don't think the results can be obtained without a real-world test, and then you'd go through a lot of trouble only to find that since you carry a spare battery around anyway, it's not a big deal either way.

That should not be taken to mean we (myself included) can't discuss it.

 

[Sicarius]

1.5 ohm carto at 3.5V = 8.17 watts at 2.33 amps

In this case, to supply 8.17 watts, the battery draw at 3.7 volts would be 2.20811 amps

3 ohm carto at 5V = 8.33 wats at 1.66 amps

In this case, to supply 8.33 watts the battery draw at 3.7 volts would be 2.25135 amps, draining the battery slightly faster due to the higher power output.

Slurp hit on the head on the first page.

Ignoring efficiency ... to supply 8W of power for 1 second you need 8 joules of energy regardless of voltage and current. There is only a finite amount of energy that can be 'held' in the battery and this is it's capacity. Now if you simplify the buck/boost circuit into an impedance the battery will always only see a demand of 8W hence it will supply the load at a fixed Joule rate.

The only thing that may be plausible is that a Li that is subject to a higher discharge current exhibits a lower apparent capacity (i.e. 2A - 2400mAh actual / 1A - 2600mAh actual). For the OP I think this is going to be irrelevant due to the Buck/Boost circuitry with the only variable then being the efficiency.

 

[Slurp812]

Slurp hit on the head on the first page.

Ignoring efficiency ... to supply 8W of power for 1 second you need 8 joules of energy regardless of voltage and current. There is only a finite amount of energy that can be 'held' in the battery and this is it's capacity. Now if you simplify the buck/boost circuit into an impedance the battery will always only see a demand of 8W hence it will supply the load at a fixed Joule rate.

The only thing that may be plausible is that a Li that is subject to a higher discharge current exhibits a lower apparent capacity (i.e. 2A - 2400mAh actual / 1A - 2600mAh actual). For the OP I think this is going to be irrelevant due to the Buck/Boost circuitry with the only variable then being the efficiency.

I adjust my voltage for the best , and vape on! And carry an extra battery with me @ all times!

 

[Shortstuff116]

I am the OP and am very thankful and impressed with so many of the responses I've received so far. After reading them all I have a much better understanding of the technical side of vaping, i.e. what really happens when we push that button. My goal was to gain a better understanding of the comments that so many have posted in so many threads about vaping at a higher voltage with a higher ohms carto/atomizer uses less "energy". Also, why so many say that vaping at higher voltages (with high ohms) provides a better vaping experience. It's not that I'm worried about not having battery power when I need it, I always carry a freshly charged battery with me.

Understanding really helps me enjoy my Provari even more and to reap the benefits of vv.

Thanks!

 

[Slurp812]

Assuming 100% efficiency for simplicity in calculations, watts is voltage (amusing 3.7) times amps. Of course, the battery must actually supply more current in actual operation, and this is just average power/current. The pulses are way higher, and that is why they recommend the high drain IMR batteries. So really what I am saying is to make the battery last longer, you have to use less power (watts). It may be that you find your sweet spot at a lower power on a higher ohm carto. Really, its down to taste and vapor, and different cartos.

 

[Frick]

Assuming 100% efficiency for simplicity in calculations, watts is voltage (amusing 3.7) times amps. Of course, the battery must actually supply more current in actual operation, and this is just average power/current. The pulses are way higher, and that is why they recommend the high drain IMR batteries. So really what I am saying is to make the battery last longer, you have to use less power (watts). It may be that you find your sweet spot at a lower power on a higher ohm carto. Really, its down to taste and vapor, and different cartos.

I still really need to know the math you used to achieve what you wrote above:

1.5 ohm carto at 3.5V = 8.17 watts at 2.33 amps

In this case, to supply 8.17 watts, the battery draw at 3.7 volts would be 2.20811 amps

3 ohm carto at 5V = 8.33 wats at 1.66 amps

In this case, to supply 8.33 watts the battery draw at 3.7 volts would be 2.25135 amps, draining the battery slightly faster due to the higher power output.

None of the standard formulas would get one to those results. I assume you're using some factor for the boost circuit.

Please explain.

 

[Slurp812]

I still really need to know the math you used to achieve what you wrote above:



None of the standard formulas would get one to those results. I assume you're using some factor for the boost circuit.

Please explain.

I was assuming 100% from the circuit. And also assuming the battery would stay @ 3.7 volts. Ohms law says watts divided by volts = amps. so I divide 8.33 watts by 3.7 volts to get the battery current needed to supply the 5 volts @ 1.66 amps at the carto, witch is 2.25135. I use this calculator. So yes, extra current is needed from the battery to supply the higher voltage. So its all about the power. The watts used is what determines the battery life. Taking into account the circuit efficiency, and battery droop makes the calculations much more complex, so this would be in a perfect world. In reality the average current supplied by the battery is higher, and is just the average. The circuit draws much higher current pulses from the battery in operation. This is why they recommend the high drain batteries on the ProVari. I suspect the the same would hold true for other single cell type VV mods as well like the LavaTube.

 

[Frick]

Thanks. I understand Ohm's law and the relevant formulas. What I wasn't understanding is that you used a constant voltage of 3.7V at the battery, whereas I was using the variable voltage output at the coil. That makes absolute sense, although it is definitely made more complicated by the fact that the battery comes off the charger at ~4.2V and runs down to ~3.3V.

I do appreciate the clarification.

For anyone else still reading this (long) thread, my initial math was wrong. I can no longer edit the posts or I would. Slurp is correct: in the calculations, you have to use the battery voltage, not the Provari set voltage, as outlined at the bottom of the page here. I read that the first time Slurp posted it and should have caught the relevant calculation then.

So, use SR or LR cartos at your whim; neither is going to save you battery power nor make it last longer.

 

[bcometa]

I don't understand much of what is being said here, and I'd prefer not to really study this all so I do. I just ordered a Provari today (first PV), my question: can I just set the watts on the Provape and vape without having to really think about all this stuff? Or, does one really need a thorough understanding of all this talk to vape successfully using a Provape?

 

[NickZac]

I don't understand much of what is being said here, and I'd prefer not to really study this all so I do. I just ordered a Provari today (first PV), my question: can I just set the watts on the Provape and vape without having to really think about all this stuff? Or, does one really need a thorough understanding of all this talk to vape successfully using a Provape?

This post below sums it up...it is more complex than just a mathematical calculation of carto ohm, voltage, amps, and battery capacity/drain ability. This is due to the Provari's "boost circuit". Do you need an understanding of that to use the V2? No, not at all. You can function as simple as 1) use 3.0 ohm cartos or atys only, and 2) start the voltage at 3.7. 3) Move up for a stronger hit...4) move down if you get a burnt taste.

However, a basic understanding can help. When you adjust voltage, you are changing the actual wattage. A rough calculation of ideal wattage can help you decide where to set the voltage, which means less experimentation with too weak or too strong/burnt hits. The 'sweet spot' usually rides between 6 and 9 watts. However, what your taste buds tell you is far more important, and you only know if you like something until you try, as taste is subjective. The Provari also has protections on it to prevent you from overloading a carto/aty or damaging the battery...just keep in mind that the resistance (ohms) of the carto or aty will affect where the voltage should be set for best performance, and standard resistance (SR) (about 3 ohms) seems to be the best overall performer in the Provari and is what Provape advises using.

The relevant measurement is how many amps you're using from the battery's mAH capacity.

In your first example, you're applying 8.5 watts to the coils, and pulling 2.3 amps on the battery.

Second example, you're applying 8.3 watts to the coils, but now you're only pulling 1.66 amps. Your battery will last longer.

Some good information here: Battery 101 - the basics


Edit: Read the posts below for clarification. The comparison is valid, but the math here isn't correct because of the boost circuit in the ProVari. You're actually pulling more amps from the battery than this post indicates.

 

[Credo]

I don't understand much of what is being said here, and I'd prefer not to really study this all so I do. I just ordered a Provari today (first PV), my question: can I just set the watts on the Provape and vape without having to really think about all this stuff? Or, does one really need a thorough understanding of all this talk to vape successfully using a Provape?

On the Vari you adjust the voltage and think less about watts. You can figure out the watts if need be, but really it's not 'needed'. Just tune it to taste/feel and enjoy

Charge a battery, but it in. Screw on an atty or carto...juice it up. Turn the Vari on and vape. You can tweak it up and down a bit at the time to find a spot that feels right for you. It's that simple