Lr/hr atty and battery life

[LemRox]

Hi guys,
i've been searching the forum long and hard, and 10ml juice after, still no luck. What i've read most with lr/hr attys with Volts are the same experience between them at a certain volt setting. Got that! my question now is, at the same watt output against diff resistance, which of lr/hr extends battery life before charging again? Does it make sense?

 

[Dannation]

Imho I don't think it would make any difference. If you have your PV set at one voltage it should be using the same amount of power whether it's a lr or hr atty. And as such you should get the same battery performance. I'm no expert in these matters though, just my

 

[knivesout]

Given a set wattage, a higher ohm atty will draw more amps than a lower resistance one. I'm not really that knowledgable about how each different mod works, and I'm no electrical engineer, but I'd assume drawing less amps would be easier on the battery.

 

[adami]

Given a set wattage, a higher ohm atty will draw more amps than a lower resistance one. I'm not really that knowledgable about how each different mod works, and I'm no electrical engineer, but I'd assume drawing less amps would be easier on the battery.

That's not really right. If you're changing the resistance, your voltage will have to change to keep the set wattage. 10W@ 1.35 Ohms 3.7V draws 2.5A and at 5.4 Ohms 7.4V it only takes 1.4A.

When you're using an efficient VV device with Buck and/or Boost converters, your input voltage isn't changing so your current draw from the battery will be close to the same if your wattage output is constant. The only variance should be due to the efficiency of the converter under the desired load.

 

[knivesout]

If you're changing the resistance, your voltage will have to change to keep the set wattage. 10W@ 1.35 Ohms 3.7V draws 2.5A and at 5.4 Ohms 7.4V it only takes 1.4A.

That's pretty much what I was trying to say, that you can get the same wattage while drawing less current.

When you're using an efficient VV device with Buck and/or Boost converters, your input voltage isn't changing so your current draw from the battery will be close to the same if your wattage output is constant. The only variance should be due to the efficiency of the converter under the desired load.

It could just be since it's late, but I'm not totally getting the statement in bold. Could you maybe explain it further, and thanks in advance.

 

[adami]

Simply thermodynamics. If you're outputting 10W at any voltage you've got to have 10W + conversion losses coming out of the battery.

 

[LemRox]

That's not really right. If you're changing the resistance, your voltage will have to change to keep the set wattage. 10W@ 1.35 Ohms 3.7V draws 2.5A and at 5.4 Ohms 7.4V it only takes 1.4A.

Hi sir, so is the change from 2.5A to 1.4A would somewhat extend battery life given that all other variables equal? TIA sir!

 

[adami]

so is the change from 2.5A to 1.4A would somewhat extend battery life given that all other variables equal?

Yes, that's basically correct if all other variables are equal. If you want the same "power" of vape at 1.4A your voltage has to be higher. If you're using an efficient VV unit the primary determining factor of your battery life is going to be your output wattage simply because the battery must give up at least that much power.

 

[knivesout]

That makes sense, it's just contrary to what I had seen people post before so I didn't know. A question though, if wattage is really the primary determining factor in battery life (as opposed to amps), why are batteries capacities measured in milliamp hours?

 

[adami]

That makes sense, it's just contrary to what I had seen people post before so I didn't know. A question though, if wattage is really the primary determining factor in battery life (as opposed to amps), why are batteries capacities measured in milliamp hours?

Sometimes you'll see battery packs rated in watt hours, they just multiply the nominal voltage by the amp hours.