Battery question

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RamShot Rowdy

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The only plausible explanation I can come up with, is that boosting is more efficient than bucking.

Buck-boost converter


This assumes that the atomizer current is identical to the battery current. That would be correct for a mech mod, but in a regulated mod it is not the case.

50 W at 1 Ohm is 7,07 V.
50 W at 0.5 Ohm is 5 V.

So with a mech mod, in order to run this experiment you'd need two different battery setups, or two different charge states to begin with. So the experiment wouldn't even make sense, it would be apples vs. oragnes. For a more thourough explanation, visit my battery calculator and click the "How it works" link at the bottom left.

Your battery calculator confirms it, if a VW regulated mod is set to 50 watts, the battery will have to produce around 50 watts, regardless of atomizer resistance. So atomizer resistance shouldn't have any significant impact on battery life when using a regulated VW mod. That said, most people run more power to lower ohm builds. If the wattage adjustment on the mod is increased then it will drain the battery faster because the user increased the output wattage, and therefore increased the current drawn from the battery, not because of the atomizer resistance.
 

Rsunderl

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The key point you're missing is that that's all on the output side of the chip. Here's how it was explained to me a few weeks ago when I had this exact same confusion:

The battery itself can only ever output a single voltage (generally somewhere between 3.5 and 4.2v, depending on its charge state -- let's assume for simplicity's sake that it's got a full charge, and so 4.2v). Due to conservation of energy, the power that's delivered to the chip is, must necessarily have to be, the same as the power output by the chip (minus energy lost as heat due to inefficiency -- we'll ignore that for now but it's usually on the order of a few percent.) So in order to get "x-amount" of watts to pump into the atomizer, the chip needs to draw those same "x-amount" of watts from the battery, and it needs to draw them at 4.2v, because that's the only voltage that the battery is capable of putting out. This is done by the chip adjusting its internal resistance, to vary the amp draw from the battery to achieve the selected power level.

No matter what wattage you set your device for, the battery can only output 4.2v. Ever. So, regardless of what's happening on the atomizer side of the chip, in order for it to output 50 watts (on a "whatever-you've-got" atomizer at "whatever-that-works-out-to" voltage), it still needs to draw those 50 watts from the battery at 4.2v, which means it must draw approximately 11.9 amps. You can change the "whatevers" on the atomizer side around however you want (so long as they still work out to 50w) and the board will still draw 11.9 amps from the battery, because that's how much it takes to get 50 watts from 4.2 volts, and the battery will never output anything else besides 4.2v. So really, it doesn't matter what's happening on the atomizer side. As far as the battery (and battery life) is concerned, 50w is 50w, regardless. Power in equals power out. Conservation of energy, and all that cal.

Thanks for the explanation, and you are right, I was looking at the output side. It makes better sense to me now.

Learn something every day in here - that's what's so great about this place :)
 

bwh79

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Also how did you get the Ω symbol in you message?

1: Copy-paste from wikipedia's Ohm or Omega article, or another post that already has it,
2: [ALT]+[234] in DOS and Windows or [OPTION]+[Z] on a Mac, or
3: Add "Greek" to your list of international keyboards on your mobile device and press the "globe" icon to switch between them (this is for Apple iOS devices, but there's probably a similar feature for Android/Windows phones.)
 

ronnbert

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3: Add "Greek" to your list of international keyboards on your mobile device and switch languages on keyboard

Setting up greek totally brought in some needed symbols on my Android! Though I did need to do some trial and error to figure out which one was greek, since it isn't plainly labelled.

New symbols I will use frequently: Ω π Σ θ (and probably in that order too lol)
 
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iTiger

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Interesting read.
I've been trying to figure out if there was any better battery for my IPV2 but looks like the 25R I've been running is the best but that seems crazy as I only run 25-30W max (using with subtanks 0.5-0.7 ohm) and it drains fast.

The OP's Seg50 and my IPV2 both have a 16A max rating and would both require 16A or higher CDR battery. I haven't managed to find anything that's worked better so far but hope to at some point.
 
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