Help me understand something about battery use...

[Shortstuff116]

I got my Provari a couple of weeks ago and now that I can adjust my voltage, I'm trying to understand something. I've read so many threads about how if you use a higher ohm carto at the higher voltage, the battery in my Provari will last longer than if I use a lower resistance carto at a lower voltage.

So here are two examples I'd like to work with:

1. Using a 1.6 ohm carto and with my Provari set to 3.7 volts gives me about 8.5 watts. With a fresh battery at approximately 4.2 volts, I'm initially using less volts than the battery has available, so I'm actually using less power until the battery voltage drops below 3.7 and the boost circuit kicks in so now I'm using more power from the battery to keep it at 3.7 volts. So essentially I'm thinking that the battery should last a good while.

2. Using a 3.0 ohm carto and with my Provari set to 5.0 volts gives me about 8.3 watts. With a fresh battery at approximately 4.2 volts, the boost circuit kicks in right away and stays working to keep me at 5.0 volts until the battery gets too low. So here I'm thinking that the battery is not going to last near as long as scenario #1 because it's constantly asking for more volts than the battery has in her.

But, it's been said that by using a higher ohm carto with higher voltage, my battery will last longer. I must be missing something or am I wrong all together?

Thanks for any help.

 

[Frick]

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: http://www.fusionteq.com/html/battery_101_-_the_basics.html


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.

 

[Dying2Live]

Frick hit it on the head

 

[Shortstuff116]

So it's all about the amps. Thanks Frick! See, you really can teach an old dog new tricks!

 

[zoiDman]

So it's all about the amps. Thanks Frick! See, you really can teach an old dog new tricks!

In a Nut Shell, Yes.

Your battery is Rated at mAh. Miliamps per hour. So the More Amps you draw from your Battery, the sooner it will need to be recharged.

 

[Slurp812]

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

Carto current is not the same as battery current in a provari. The boost curcuit needs to take additional current from the battery to achieve higher voltage. 8 watts on ANY carto resistance will draw the same average current from the battery (aside from the boost circuit efficiency) . As outlined here. Please read @ the bottom of the page. The relevant number here is watts, and boost circuit efficiency. The battery basics only applies directly to non VV cases.

 

[Frick]

Carto current is not the same as battery current in a provari. The boost curcuit needs to take additional current from the battery to achieve higher voltage. 8 watts on ANY carto resistance will draw the same average current from the battery (aside from the boost circuit efficiency) . As outlined here. Please read @ the bottom of the page. The relevant number here is watts, and boost circuit efficiency. The battery basics only applies directly to non VV cases.

Very interesting and thanks for the link. I hadn't read that before.

I understand what they're saying about the boost circuit, but even though the circuit needs to take additional current to achieve the higher voltage (and therefore one cannot simply use the math I did above -- in both examples, you're actually pulling more amps from the battery because it's boosted), it doesn't change the fact that at higher resistances you're using higher voltage and therefore draw less current (boosted or not) and your battery will last longer.

Edited my post above to indicate the math is not correct in terms of actual amps drawn.

 

[Slurp812]

Very interesting and thanks for the link. I hadn't read that before.

I understand what they're saying about the boost circuit, but even though the circuit needs to take additional current to achieve the higher voltage (and therefore one cannot simply use the math I did above -- in both examples, you're actually pulling more amps from the battery because it's boosted), it doesn't change the fact that at higher resistances you're using higher voltage and therefore draw less current (boosted or not) and your battery will last longer.

Edited my post above to indicate the math is not correct in terms of actual amps drawn.

Higher voltage @ the carto is generated by drawing more current from the battery even tho the current is less through the carto. There is no free power derived from the boost circuit. A 1.5 ohm carto, and a 3 ohm carto both driven to 8 watts would take the same power, and current from the battery even though the current through the cartos would be different. An HR carto does not make a provai more efficient. It makes it more flexible. Maybe I can ask someone at provape to clarify this?

 

[Frick]

Higher voltage @ the carto is generated by drawing more current from the battery even tho the current is less through the carto. There is no free power derived from the boost circuit. A 1.5 ohm carto, and a 3 ohm carto both driven to 8 watts would take the same power, and current from the battery even though the current through the cartos would be different. An HR carto does not make a provai more efficient. It makes it more flexible. Maybe I can ask someone at provape to clarify this?

I agree that there is no "free power" derived from the boost circuit.

I disagree regarding the amperage draw.

The boost circuit (more commonly known as a boost converter) does its thing to increase voltage. I'll admit that I don't know all the math on that page, but the converter doesn't seem to affect amperage at all. And we know the actual amperage capacity of the battery remains unchanged.

You said: A 1.5 ohm carto, and a 3 ohm carto both driven to 8 watts would take the same power, and current from the battery...

I simply cannot agree with that.

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

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

Again, these numbers are not intended to indicate the draw at the battery (because again, I don't have a full understanding of the boost converter), but the relationship remains the same: you draw more amps to drive a LR carto to a given wattage than you do with a HR carto.

 

[Slurp812]

I agree that there is no "free power" derived from the boost circuit.

I disagree regarding the amperage draw.

The boost circuit (more commonly known as a boost converter) does its thing to increase voltage. I'll admit that I don't know all the math on that page, but the converter doesn't seem to affect amperage at all. And we know the actual amperage capacity of the battery remains unchanged.

You said: A 1.5 ohm carto, and a 3 ohm carto both driven to 8 watts would take the same power, and current from the battery...

I simply cannot agree with that.

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

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

Again, these numbers are not intended to indicate the draw at the battery (because again, I don't have a full understanding of the boost converter), but the relationship remains the same: you draw more amps to drive a LR carto to a given wattage than you do with a HR carto.

In this case the battery would last longer on the 1.5 due to it dissipating slightly less power @ the load. it MUST mean that under these conditions, the battery is supplying slightly LESS current. Simple ohms law. Battery power minus converter efficiency must equal carto power.

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.

In reality the battery would also have to supply the power wasted in the converter circuit. And the battery voltage under load would drop due to these high currents. Also these currents would be averages because the boost converter circuit takes larger pulses from the battery in operation. As stated on the provape faq, battery current is not equal to carto current. The converter circuit generates the higher voltage by drawing more current from the battery. You simply cannot get more power from the battery by using a HR carto. Yes, the theory of operation of a circuit like this is very complicated. As it states on the page you quoted: Since power (P = VI) must be conserved, the output current is lower than the source (battery) current.

 

[zoiDman]

This where I'm confused.

If I take a Freshly charged battery, and say it has 4.2 volts for a round number. I put it in my Provari and I set the Output Voltage to the 3.9 volts to the Carto.

The Output voltage is Less than the Battery Voltage. Is the Booster Circuit Bypassed? Does the Booster Circuit Reduce the Voltage? Or does the Booster Circuit draw Volts/Amps from the Battery at a Lower amount that the Output Volts/Amps?

Then I vape along not thinking of Volts or Watts. But at some point the Battery Voltage drops lower that the Output Voltage. At this point, does the Booster Circuit draw volts/amps from the Battery at the same amount?

 

[AttyPops]

Have you ever heard of people getting 220v service to ...say an air conditioner or a swimming pool? Do you know why? The higher voltage is more efficient. Same reason the high tension wires are HV.

The wattage calculation is: Power = V * V / R (where voltage is squared divided by resistance). It's the squaring of the voltage that gives it the impact. It's exponential rather than linear.

Of course, the device design/efficiency comes into play too. For example, if you want to vape at 3.1 volts from a 7.4 volt input (2 batteries in series) you COULD burn off the difference as heat. Different regulators will handle this differently and I can't really get into the details. However switching regulators are more efficient than linear regulators, for example. The less difference there is from input voltage, the easier it is. Now, buck/boost has it's own efficiency issues....and then I guess I don't know enough about the provari's internals to really say for sure. The specific regulator makes a bid difference in efficiency. The modder's have talked a lot about buck/boost and regulators. Specifics could be found in the modding section.

Basically, buck is where you clip the voltage by using some PWM type of switching circuit to pulse it down... often with a comparator to a reference voltage. Linear regulators bleed the excess off as heat/ground where switching regulators switch off. Boost, OTOH, pulls more amps to up the voltage. You are more or less trading increased amp draw for a bump in voltage. Contrary to some quotes you will find, you are not really trading mAh for voltage, you are trading instantaneous amps for instantaneous voltage (mAh is a capacity rating, not a current measurement). This will use the charge up faster than not, however.

All of this becomes a rather interesting theoretical discussion, but any particular device performance would be dependent on design decisions and would thus have to be tested. However, I personally think it's rather moot in that:
1) You are better off setting your vaping for effect or flavor rather than battery efficiency.
2) You need to have spare batteries ready anyway... so meh.

If it makes a 50% difference in battery life one way or the other, maybe it's worth experimenting with, but I doubt it. I'd rather vape at the higher voltage just because I can use more robust atomizers and they last longer (IMO). Also, the hit has more "oomph".

 

[zoiDman]

...and then I guess I don't know enough about the provari's internals to really say for sure. The specific regulator makes a bid difference in efficiency. The modder's have talked a lot about buck/boost and regulators. Specifics could be found in the modding section.

...

I'm kinda in this same boat also. I just don't know for sure what the Provari does to make Output Voltage at all stages of the Battery's Charge.

I thought I read that the Provari pulls the same voltage from the Battery Regardless of the Battery Voltage. ie: It allways draws 3.2 volts and then kicks up the Voltage to the Output Voltage.

But I could be Completely wrong on this.

 

[AttyPops]

Well, it doesn't really "pull" voltage. It pulls electrons. The rate of electron pull is amps. More or less. Voltage is voltage and changes over time.

All the lithium Ion batteries start out at 4.2 volts on full charge, and the voltage varies as the charge gets used up... down to about 2.x volts at cut-off (assuming a protected battery or whatever the device decides is cut off voltage).

Basically, it must buck for over voltage, and boost for under voltage. Hence buck/boost regulators. IDK how it could do anything else. Who knows, maybe someone that knows the internals will pipe in here.

 

[zoiDman]

Well, it doesn't really "pull" voltage. It pulls electrons. The rate of electron pull is amps. More or less. Voltage is voltage and changes over time.

All the lithium Ion batteries start out at 4.2 volts on full charge, and the voltage varies as the charge gets used up... down to about 2.x volts at cut-off (assuming a protected battery or whatever the device decides is cut off voltage).

Basically, it must buck for over voltage, and boost for under voltage. Hence buck/boost regulators. IDK how it could do anything else. Who knows, maybe someone that knows the internals will pipe in here.

Yeah... My terminology was poor in that post.

I could see it 'Bucking" voltage when the Battery Voltage is More than the desired Output voltage. And "Boosting" when it is less.

But like I said, I thought I read that the Provari Allways Boosts. But like I also said, I could be wrong.

I did a ton of research before I bought my Provari and Batteries. Did hours of reading. But that was a long time ago.

 

[Credo]

In most cases there's not much difference between a LR and HV atty. The HV one just has more winds in the coil (longer wire). That gives more coil to liquid surface area, and a little better lung hit.

Of course there are exceptions to the rule...when getting into custom spec atties they might change everything about the atty design...not simply use a longer coil.

 

[Frick]

In this case the battery would last longer on the 1.5 due to it dissipating slightly less power @ the load. it MUST mean that under these conditions, the battery is supplying slightly LESS current. Simple ohms law. Battery power minus converter efficiency must equal carto power.

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.

You don't cite the math you used to reach those amperage draws. I'm using Volts^2 / Ohms = Watts, and Volts / Ohms = Amps (Ohm's law).

How are you calculating converter efficiency? 90-95%, as stated on the ProVari FAQ?

I'm still not understanding how the basic Ohm's law is violated because of a boost converter. The FAQ does state that the amperage draw at the battery is more than would be predicted by the amperage draw at the coil(s) (because of the boost converter), but that shouldn't change the relationship between current and resistance.

Basically, buck is where you clip the voltage by using some PWM type of switching circuit to pulse it down... often with a comparator to a reference voltage. Linear regulators bleed the excess off as heat/ground where switching regulators switch off. Boost, OTOH, pulls more amps to up the voltage. You are more or less trading increased amp draw for a bump in voltage. Contrary to some quotes you will find, you are not really trading mAh for voltage, you are trading instantaneous amps for instantaneous voltage (mAh is a capacity rating, not a current measurement). This will use the charge up faster than not, however.
.

This may be the answer I'm looking for. The increased voltage is a trade-off, resulting in an increased amperage draw.

Provape should really chime in on this. If the basic "use HR cartos/attys to achieve longer battery life" equation is actually reversed through the use of the boost converter, it would be nice to know this fact.

 

[Slurp812]

This where I'm confused.

If I take a Freshly charged battery, and say it has 4.2 volts for a round number. I put it in my Provari and I set the Output Voltage to the 3.9 volts to the Carto.

The Output voltage is Less than the Battery Voltage. Is the Booster Circuit Bypassed? Does the Booster Circuit Reduce the Voltage? Or does the Booster Circuit draw Volts/Amps from the Battery at a Lower amount that the Output Volts/Amps?

Then I vape along not thinking of Volts or Watts. But at some point the Battery Voltage drops lower that the Output Voltage. At this point, does the Booster Circuit draw volts/amps from the Battery at the same amount?

It is still in action. Its a boost/buck type circuit. So either way larger pulses are drawn from the battery to produce a smooth output @ the carto even if the carto voltage is less. Oh, here is a wiki article. The formulas are nuts, but it is very similar to a boost only type. Sorry for getting all nuts about how they work, I am passionate about anything electronic. That's one of the reasons I picked up E-Cigs. I quit analogs almost 5 years ago!

 

[zoiDman]

... Oh, here is a wiki article. The formulas are nuts, but it is very similar to a boost only type. ...

Thanks for the link.

LOL... I actualy understand the Math much better than I understand what is going on. The Math is just some Basic Calculus.

 

[AttyPops]

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!

Like I said:

All of this becomes a rather interesting theoretical discussion, but any particular device performance would be dependent on design decisions and would thus have to be tested. However, I personally think it's rather moot in that:
1) You are better off setting your vaping for effect or flavor rather than battery efficiency.
2) You need to have spare batteries ready anyway... so meh.

If it makes a 50% difference in battery life one way or the other, maybe it's worth experimenting with, but I doubt it. I'd rather vape at the higher voltage just because I can use more robust atomizers and they last longer (IMO). Also, the hit has more "oomph".

Do you really care if buck is 5% more efficient than boost? If it was a super-major difference, you shouldn't need anyone to chime in. You would know by use. That fact that you ask means that either A) you haven't tried it both ways or B) there isn't that much difference. Besides, in the next revision of the provari, or the foo-e-cig, or whatever, it could be different due to design changes.

So vape at what works best for you. THAT'S the whole goal anyway. Otherwise, you could drive yourself (and Provape) nuts with this stuff. It's a cool discussion, but meh.

And to think this thread isn't even in the modder's section. lol.

/trolling