Questions about amps and volts

[ltrainer]

I think I know the answer but I'm not sure.

IF you are vaping at a given wattage as you increase the ohms of the coil the voltage goes up and the amps go down. Does this mean that your battery will last longer?

According to ohms law the amps decease so in my head the battery will last longer. Am I thinking about this the right way?

THanks

 

[CATastrophe]

Sounds good to me - logical anyway. But I'd like to hear from the folks who understand this better than I.

 

[NicoHolic]

Remember the principle of conservation of energy. If you are vaping a mod that supports wattage settings, you can change anything you want, but as long as the wattage remains the same, the usage of battery capacity will remain the same (factors such as Peukert being ignored).

eta: if you're vaping a mech mod or a regulated vv mod set at a certain voltage, increasing the resistance won't change the voltage much*, but it will decrease the current (intensity in amps) and the power (in watts). Hence, the battery will last longer, but you'll have a weaker vape.

*only by a portion of that dropped by the internal resistance of the battery due to the reduced current ...and a regulated mod will correct for that.

 

[CATastrophe]

I'll take your word for it, Nico. I don't have a VW mod. (Husband does though, but it's not my thing - I'm a 100% REOnaut.)

 

[NicoHolic]

I'll take your word for it, Nico. I don't have a VW mod. (Husband does though, but it's not my thing - I'm a 100% REOnaut.)

Me too... except for an eVic I got in the big $59.95 sale that's become a resistance checker. It's really fast and shows Ω on boot-up.

 

[inter_ceptor00]

lets say....wattage stays the same 10 watts. Resistance is increased 1ohm. Amperage draw decreases, but since the battery is a fixed voltage, the only variables left to alter are the amperage (lowered) and the wattage(would now be less from amperage drop) Volts x Amps=wattage, Volts/Resistance = amperage, Watts/Amps=Volts longer life but less power, total energy value remains the same.

simple converter Volts/Amps/Watts Converter

 

[ltrainer]

Call me slow. I still don't understand and I certainly am not disputing any information given here. I just want to try to understand this.

Lets say I have a variable wattage pv and I keep my watts at 10 watts. IF I have a coil at 1 ohm I get current and voltage both at 3.16.

When I go to a 3 ohm coil I get voltage at 5.47 and current at 1.82.

Since its putting out less amps(current) why doesn't the battery last longer? A battery only has so many amps and I'm using less of them so shouldn't it last longer?

 

[ChrisEU]

A battery doesn't have "so many amps" - it has an amount of energy stored in it, usually measured in (milli)amp hours.

So, let's take a battery with 3,7 V. It is rated at 3000 mAh, or 3Ah, three Amp hours. All technicalities of chemistry aside, it can either deliver 3 amps for 1 hour (at 3,7 V) or it can deliver 1 amp for 3 hours (at 3,7 Volts).

A battery is like a bucket full of water - you can empty it with a straw and it will take quite long, or you can empty it all at once, quite quickly, it's still the same amount of water.

 

[MamaTried]

>When I go to a 3 ohm coil I get voltage at 5.47 and current at 1.82.

i assume this is what you are seeing at the atty. what is happening at the battery is a little different thanks to the "magic" of the VV or VW circuitry. consider your 18650 battery, which, fully charged will be somewhere near 4.0 volts- it ain't never getting close to 5.47. the circuitry makes the voltage at the atomizer 5.47 by converting some amps into volts. so the amp drain on the battery is actually higher than 1.82, which makes it iffy to predict battery life.

it's been about 4 decades since the last time i designed power supplies or analog circuits, but i believe that's fairly close to the truth.

either that or i'm all wet which is also quite possible

 

[NicoHolic]

There's a heavy price to be paid for that higher voltage, which you're assuming to be a free lunch (looking only at amps and amp-hours). You see or calculate voltage and amps of the coil. What's happening to the battery is hidden from you. For example, if your battery is at 4 volts and your coil voltage at 6 volts, the amps being drawn from the battery by the boost circuit are at least 1.5 times that you calculate for the coil. Most VW mods use pulse width modulation of this boosted voltage, and the higher you set the voltage, the greater the duty cycle (conducting time) of the pulses.

 

[CATastrophe]

ltrainer, you're not slow. I'm fairly lost with all this too. BUT, what I do know, is what the safe limits are for my batts with my coils. And when my batts are getting tired, I swap them.

This is a very interesting discussion. Lots to learn. I feel like you're all talking Martian.

 

[nerak]

ltrainer, you're not slow. I'm fairly lost with all this too. BUT, what I do know, is what the safe limits are for my batts with my coils. And when my batts are getting tired, I swap them.

This is a very interesting discussion. Lots to learn. I feel like you're all talking Martian.

I feel the same way! I just don't care about learning the tech side. I know what I like, and know others run the same, so it is fine!

I do pay attention to the batteries. I change them early enough so they are good!

Now I will sit here like I do when I don't want to hear something, ears plugged and singing lalalalalalalal.............LOL!

 

[ltrainer]

Thanks everyone. I THINK I have it now. I guess what I'm hearing is that if you are using VW, no matter what ohm coil you are using, you'll get pretty much the same battery life. What made me ask this is that I just ordered a Kick 2 for my Grand. So I was playing with the ohms calculator and noticed the relationships in that law. Thanks again one and all.

 

[NicoHolic]

Thanks everyone. I THINK I have it now. I guess what I'm hearing is that if you are using VW, no matter what ohm coil you are using, you'll get pretty much the same battery life.

For the same wattage setting, yes.

What made me ask this is that I just ordered a Kick 2 for my Grand. So I was playing with the ohms calculator and noticed the relationships in that law. Thanks again one and all.

Without the kick, current drawn from the battery decreases as the battery charge and voltage falls. In other words, the weaker the battery gets, the less demand the kickless Grand puts on it, so battery usage decelerates. With the kick, it's all just the opposite. The weaker the battery, the harder the kick works it. And, in this case, you're applying that to a smaller battery. You'll need to watch the battery voltage and not let it get too low, since you won't have the decreasing vapor rate to go by. Eventually, you should get a feel for it. Hope this helps.

 

[nerak]

For the same wattage setting, yes.



Without the kick, current drawn from the battery decreases as the battery charge and voltage falls. In other words, the weaker the battery gets, the less demand the kickless Grand puts on it, so battery usage decelerates. With the kick, it's all just the opposite. The weaker the battery, the harder the kick works it. And, in this case, you're applying that to a smaller battery. You'll need to watch the battery voltage and not let it get too low, since you won't have the decreasing vapor rate to go by. Eventually, you should get a feel for it. Hope this helps.

Lalalalalalalalalalalalal.........................................

I must read every post.......it's an OCD thing.

You all go right ahead........You are very good!

 

[ltrainer]

For the same wattage setting, yes.



Without the kick, current drawn from the battery decreases as the battery charge and voltage falls. In other words, the weaker the battery gets, the less demand the kickless Grand puts on it, so battery usage decelerates. With the kick, it's all just the opposite. The weaker the battery, the harder the kick works it. And, in this case, you're applying that to a smaller battery. You'll need to watch the battery voltage and not let it get too low, since you won't have the decreasing vapor rate to go by. Eventually, you should get a feel for it. Hope this helps.

It has a voltage cut off. I used the original Kick for a long time and enjoyed it. THe new one has some features that I think will make it even better for my usage. Thanks for your help with my thick headedness.

 

[KwhyLE]

For the same wattage setting, yes.



Without the kick, current drawn from the battery decreases as the battery charge and voltage falls. In other words, the weaker the battery gets, the less demand the kickless Grand puts on it, so battery usage decelerates. With the kick, it's all just the opposite. The weaker the battery, the harder the kick works it. And, in this case, you're applying that to a smaller battery. You'll need to watch the battery voltage and not let it get too low, since you won't have the decreasing vapor rate to go by. Eventually, you should get a feel for it. Hope this helps.

If im not mistaken, If the battery drops lower than 3.2V the kick wont fire.

 

[ltrainer]

If im not mistaken, If the battery drops lower than 3.2V the kick wont fire.

The new one cuts off at 2.7 under load. Which will end up about 3.0. Thats one of the things that I think will be sweet.

 

[MamaTried]

Lalalalalalalalalalalalal.........................................

LOL. i finally do a serious post and you do that?

me & Nico just getting our nerd on...

 

[_more_]

Its all geek to me