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New to APVs- Battery Drain Question

mkhilario

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I have a new Kbox Mini. With a 1.2 Ohm build it reads 4.2V at 15W, switching to a 0.9 Ohm build it reads 3.7V. Same battery at same state of discharge, I am just switching between tank and my RDA.

My question is: Is the 0.9 ohm coil more energy efficient at the given wattage setting of 15? I know Ohms Law but I'm not sure how the "black box" circuitry actually works.
 

Maverik_X

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I have a new Kbox Mini. With a 1.2 Ohm build it reads 4.2V at 15W, switching to a 0.9 Ohm build it reads 3.7V. Same battery at same state of discharge, I am just switching between tank and my RDA.

My question is: Is the 0.9 ohm coil more energy efficient at the given wattage setting of 15? I know Ohms Law but I'm not sure how the "black box" circuitry actually works.
its reading the resistance of the coil and then figuring out the voltage needed to fire it at with your set wattage. Ohms law. it's normal
 

mkhilario

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Yes, I understand how a VW mod basically adjusts the voltage. Here's a more basic question: at a given wattage of 15 watts, will the battery drain be more with the lower resistance 0.9 coil vs the 1.2 coil?
 

Maverik_X

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Yes, I understand how a VW mod basically adjusts the voltage. Here's a more basic question: at a given wattage of 15 watts, will the battery drain be more with the lower resistance 0.9 coil vs the 1.2 coil?
battery drain would be higher with the lower ohm build.
 

mkhilario

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battery drain would be higher with the lower ohm build.
Thanks.
So there's no magic going on in there. It's like a mech mod in that respect. Steam Engine gave 1.2 ohm as the "sweet spot" so I did that as my first build for the Kbox/ mini tank. The "median resistance" is given as 0.75 which is closer to what I build on my rda /mech mod. Both seem to work well on the Kbox.
 

BoomStick

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battery drain would be higher with the lower ohm build.
That's not true. In a vw device the only thing that affects battery drain is watt setting. The higher the watt setting, the faster the battery drains. Build resistance has nothing to do with battery current in vw mods.
 

Maverik_X

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That's not true. In a vw device the only thing that affects battery drain is watt setting. The higher the watt setting, the faster the battery drains. Build resistance has nothing to do with battery current in vw mods.
Ok for ohms law the lower the resistance the more power needed no? Please explain how a vw device does not drain a battery faster when hitting a lower ohm coil vs a higher ohm coil with same wattage... is it coil size,type,build? thermodynamics? buck/boost? just curious is all.
 

mkhilario

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If I=V/R where I is current in watts, and the mod adjusts automagically the V based on your coil's R, then drain is determined solely by I (wattage)?
Did I get that right?

EDIT: Wait, I is actually current given in Amperes, and DC watts=V x I. Or I= watts/V.
So battery drain in amps is inversely related to voltage. So, assuming constant watt setting, with a lower ohm coil the APV mod selects a lower voltage which actually increases the load (amps)?
I just gave myself a headache. College was a loong time ago..
 
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BoomStick

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The voltage being applied to the coil isn't the same as the voltage being applied to the chip. The current being applied to the coil isn't the same as the current being applied to the chip. The watt setting and battery voltage determine battery current. Coil resistance has nothing to do with the load being placed on the batteries in a vw mod.
 

mkhilario

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So if my display reads 3.7 V @ 0.9ohm and 4.2 V @ 1.2 ohms at 15 watts, what is it telling me? Chip load only? How does that circumnavigate the math if all the vw mod does is vary the voltage based on selected wattage at a fixed resistance?
What else is it doing? Just wondering.

EDIT: If you go to the Battery Drain page of Steam Engine and play around with different resistances on the preset it does NOT affect the battery runtime. So this bears out what Broomstick said.

http://www.steam-engine.org/batt.asp?b=Samsung_INR18650-25R&mah=2500&c=8.8
 
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