resistance and battery life on VV/VW mods

[ad68]

new ECF member, been vaping for almost a year and a half now. Thought I'd post some info on how the resistance of your atty/carto affects vape time.
Bottom line is that vaping at higher resistance (but fixed power / Watts) should increase vape time on a single charge, in proportion to the square root of the resistance. If you want a formula:
t = Q * sqrt(R/P).
R is resistance in Ohms, P is power in Watts, Q is the charge held by the battery in Ah; the result is the time in hrs. Example: say you're using a 800 mAh battery, which is 0.8 Ah; you're vaping at 8W on a 1.5Ohm atty. You'll find t=0.35 hrs, a little over a third of an hour. If instead you're running a 3Ohm load at the same power of 8W, the time increases by a factor of sqrt(2) = 1.414 (or 41.4%).

That of course assumes a continuous current. You won't vape continuously for a good half hr though, you're gonna take breaks. In effect you may actually fire the device for 10% of the time, say; then multiply the above time by 10. Regardless, the point is that at higher resistance the battery should last longer.

[If you're in to physics etc: you obtain the formula above from t=Q/I, where I is the current in Amps; eliminate it using P = V * I = R * I^2.]

If your mod doesn't maintain fixed power: P = V^2 / R, so you can manually maintain power (Watts) by increasing the voltage in proportion to the square root of the resistance. So, doubling the resistance from 1.5Ohm to 3Ohm for example requires you to bump up the voltage by a factor of sqrt(2) = 1.414

I should also note that the charge Q held by the battery is effectively lower than the mAh rating quoted by the manufacturer. That's because they determine it by discharging down to 2.8V or so. Your mod will shut down with a "low battery" warning though once it's down to ~ 3.4V; the remaining charge goes unused. What fraction that corresponds to depends on the specific battery etc, you'd have to check the battery forums.

Ok, hope this is useful for some.

 

[DaveP]

All you say is true, but the fix is batteries with higher MAH ratings. Kick and boost circuits use some power themselves as they regulate the voltage. Mechs are probably the most efficient devices, but lack the versatility of electronic regulation to keep the vape the same from full charge to depletion.


As you said, we get only a small amount of continuous output into a given load. That's broken up across cumulative vaping time. Chain vapers get less, occasional vapers get lots of vaping time from a charge, making the discussion of vaping time ambiguous unless they divulge their vaping style. Like gas mileage, it depends on style of use ... light vaper or heavy hitter makes a difference in whether a given battery "lasts all day" or "not nearly enough"!

Good information in one spot here, ad68. Welcome to ECF. You will find many threads leaning toward the highly technical side as you move around in the forum threads. We are loaded with technical people who enjoy a good discussion on the various aspects of our habit.

 

[HughDaHand]

You also have to keep in mind the efficiency of the boost circuit used in the VV/VW mod that you are using. The extra voltage above the battery voltage has to come from somewhere. This makes this very tough as we very seldom know these figures. I have a vamo and a mech with a kick, both set at 8 watts using the same battery and tank I get a noticeably longer vape time on the mech with a kick over the vamo. The vamo also runs the battery to a lower voltage in these test, leading me to believe that its boost circuit is horribly inefficient. I did this test about 10 times and got hours more vape time at my pace from the kick every single time before the vamo hit the backup vape box. There are so many variables that have to be taken into account when figuring this out.

 

[Dakota Jim]

Thank you ad68 for this very useful info

so to break it down for beginners with a regular 3.7 volt 900mah ego battery with a 1.8 ohm coil in their mini protank 2

t = Q * sqrt(R/P)

.4379 = .9 * sqrt(1.8/7.6) .4379 = .9 * .4866

so they will get .4379 hours of constant firing or 26.27 minutes of firing

I for example fire mine for 7 seconds on average when I vape so I will get about 225 vapes off of the battery charge.

Please check my math to make sure I did it correctly (I is not colliej edgiecatud)

oh, and welcome to the ECF

 

[ad68]

Thanks for the welcome guys, don't know what took me so long to join, ha ha ha... Great forums.

Dakota Jim: yeah, your math is ok. You'd get about 0.44 hrs of continuous firing. Though I of course agree with the other posts: this is a very optimistic estimate cause it doesn't account for the additional resistance due to the switch, circuitry etc... (and don't forget about the fact that you probably won't be able to actually use all of those 900 mAh for vaping, battery will shut off before it is fully drained of charge.)

I posted this mainly to give you an idea of how vape time would increase with resistance on one and the same mod, battery etc. Besides, I need to accumulate those 5 posts on ECF -

 

[Ryedan]

new ECF member, been vaping for almost a year and a half now. Thought I'd post some info on how the resistance of your atty/carto affects vape time.
Bottom line is that vaping at higher resistance (but fixed power / Watts) should increase vape time on a single charge, in proportion to the square root of the resistance. If you want a formula:
t = Q * sqrt(R/P).
R is resistance in Ohms, P is power in Watts, Q is the charge held by the battery in Ah; the result is the time in hrs. Example: say you're using a 800 mAh battery, which is 0.8 Ah; you're vaping at 8W on a 1.5Ohm atty. You'll find t=0.35 hrs, a little over a third of an hour. If instead you're running a 3Ohm load at the same power of 8W, the time increases by a factor of sqrt(2) = 1.414 (or 41.4%).

That of course assumes a continuous current. You won't vape continuously for a good half hr though, you're gonna take breaks. In effect you may actually fire the device for 10% of the time, say; then multiply the above time by 10. Regardless, the point is that at higher resistance the battery should last longer.

[If you're in to physics etc: you obtain the formula above from t=Q/I, where I is the current in Amps; eliminate it using P = V * I = R * I^2.]

If your mod doesn't maintain fixed power: P = V^2 / R, so you can manually maintain power (Watts) by increasing the voltage in proportion to the square root of the resistance. So, doubling the resistance from 1.5Ohm to 3Ohm for example requires you to bump up the voltage by a factor of sqrt(2) = 1.414

I should also note that the charge Q held by the battery is effectively lower than the mAh rating quoted by the manufacturer. That's because they determine it by discharging down to 2.8V or so. Your mod will shut down with a "low battery" warning though once it's down to ~ 3.4V; the remaining charge goes unused. What fraction that corresponds to depends on the specific battery etc, you'd have to check the battery forums.

Ok, hope this is useful for some.

This is so easy to get wrong ad68. Watts is power, which is the rate of doing work. You're discharging a battery at the same number of watts on heaters of different resistance. Why would your battery last longer on one than the other at the same wattage? The work done is the same in both examples, right?

Rader2146 wrote out the explanation in his blog much better than I can explain it: Battery Life - Low Resistance, High Resistance, and Efficiency.

The crux of the confusion I think is that the battery in the PV always puts out 3.7V nominal. The device however puts out the voltage you ask for via the VV or VW setting. When set to increase voltage output above battery voltage, it does that by drawing more current. The reverse is also true. The battery however always puts out the voltage it's at in its discharge cycle (between 3.3 and 4.2V). That is not adjustable.

The battery in the power supply will last the same amount of time regardless of the ohms of the heater as long as both heaters are run at the same watts. There are other efficiency issues in play, but they are not what we're discussing here. Rader does go through that in his blog also.

 

[ad68]

hi Ryedan,

thanks for the link to Rader's post. Well, sure, he's certainly correct to say that for more realistic estimates you'd need to factor in the efficiency of your device/battery which is less than 100%. (Might even change with voltage.) No argument here.

I don't think there's anything wrong with what I wrote in my first post. But you're right that there's an implicit assumption that the battery can deliver the same charge Q (the same mAh) for both loads. If so, it should indeed last longer, at fixed power, for higher resistance loads. That's because the current (rate of discharge) is less: I = sqrt(P/R), so higher R means lower current I. And yes, you're right: that means that it delivers more electrical energy E = P * t, i.e. more Watts * hrs (Wh). Of course, it can't deliver more energy than it's stored but it won't get there anyway.

You're basically suggesting that the battery would not be able to deliver the same amount of charge Q; rather, that it would deliver a fixed amount of energy E (unit is Wh). If so, it would indeed last the same time, agreed.

I am a physicist but do not deal with batteries and circuits, so I don't claim to have any knowledge of battery discharges and the like. Perhaps there's a better qualified individual around here who could clarify if the battery would deliver a fixed amount of charge (mAh) or a fixed amount of energy (Wh) as the Ohm load varies. Would be interesting to know, for sure.

And those who don't really care about the arguments may just want to try a LR atty/carto vs a SR or HR device and compare. At least you'll know for sure, ha ha... Have fun vaping!

 

[yzer]

LR cartos cost less than SR or HR cartos. That's my deciding factor right there.

 

[DaveP]

I think that to do any meaningful comparison we have to assume a set voltage and recalculate for other common vaping levels. 3.7 is good for mech users. VV and VW is a different story altogether, although wattage drains virtually the same power for all reasonable resistance values if you throw out the unknown efficiency of kick and boost circuits.

If you try to factor in the resistance of switches and the various contacts in a mod, all bets are off. There's no standard to rely on and all we can do is compute the average vaper's time on a charge. It becomes futile when you try to cover all the variables.