Higher Resistance, Lower Wattage vs Lower Resistance, Higher Wattage

[Shynobi]

Hey there ECF, so I recently went from a mech mod to an IPV3. Usually when I was using my mech mods I would do sub ohm builds for cloud chasing. I'm now using my IPV3 as my daily driver and I notice with my sub ohm builds I actually have to turn up the wattage to have more volts running which in turn increases the amps quite a bit. But anyways, I noticed that if I go higher resistances, I can maintain the same amount of voltage with less wattage and amp. So my question is that considering two builds: one at .2 ohms and one at 1.2 ohms, if they are both being put at 4.2v, would they perform the same, even with the one with higher resistance running lower watts and putting out less amp to achieve the same volts?

 

[treehead]

Hey Shynobi the answer to your question is when dealing with volts (the 4.2), your coils are going to perform faster/better at the lower resistance (.2ohms), remember the .2ohms will require more amps, in other words using the higher resistance of 1.2ohms at the same 4.2 volts, it should be heating up much slower, because all that extra kanthal is absorbing/eating more power trying to heat up.

I think where you got a bit confused was the fact that you thought the lower res .2ohm setup required more volts to get more wattage, which isn't true. In a nutshell, the lower your resistance the less volts you need to get to the same wattage. So it appeared that your lower res setup needed more power to perform well, when in reality it's just that the smaller coils need to be at higher watts to get the same performance as the larger coils, basically since the coils are smaller, they need to get hotter to perform the same, because there's less surface area (less coil-to-juice contact). Although the .2ohm build needs to get hotter to compete, it actually takes much less power to heat them (that's where it gets confusing lol ).

I'm only giving a long explanation because I was confused about the same thing a while back lol, and I wish someone would've explained it to me .
So,

1.2ohm setup needs 6.9volts for 40watts (and only needs 6 amps to achieve that power)
(Same 40 watts, but .2 needs ALOT less because it's smaller.)
0.2ohm setup needs 2.8volts for 40watts (needs more at 14amps because there's less resistance.

The amps are higher on the low resistance because it's more "wide open". A good way to picture that, is to imagine the less resistance you have, the wider/more open a garden hose is, thus the faster the water comes out, thus the more amps needed. The 1.2ohm setup is resisting 6x more (the hose is tighter, more restraining on the water/amps), the .2ohm setup is getting close to not resisting at all, so it's a 6x wider hose letting the water/power through much much faster, so it needs quite a bit more amps.

Sorry hope this didn't look like a "lecture" lol, it's easy to remember that Watts = same heat on all coils, that IPV3 is made to heat bigger coils up as fast as you could normally only heat sub-ohm coils at, but keep in mind that takes alot more power. Smaller coils require less power for more heat, they're more efficient, but with less surface area. And your amps don't waste battery, you only have to make sure your within your battery's amp limit (which your IPV3 does for you I believe).

This is how I learned volts, amps, watts, and resistance: Ohm's Law Calculator

I'm sure you've already known about this site, but after playing with it for a while it's super easy to catch on. I can pretty much do it in my head now .

Hope this helps brother! Vape on!

 

[Shynobi]

Hey Shynobi the answer to your question is when dealing with volts (the 4.2), your coils are going to perform faster/better at the lower resistance (.2ohms), remember the .2ohms will require more amps, in other words using the higher resistance of 1.2ohms at the same 4.2 volts, it should be heating up much slower, because all that extra kanthal is absorbing/eating more power trying to heat up.

I think where you got a bit confused was the fact that you thought the lower res .2ohm setup required more volts to get more wattage, which isn't true. In a nutshell, the lower your resistance the less volts you need to get to the same wattage. So it appeared that your lower res setup needed more power to perform well, when in reality it's just that the smaller coils need to be at higher watts to get the same performance as the larger coils, basically since the coils are smaller, they need to get hotter to perform the same, because there's less surface area (less coil-to-juice contact). Although the .2ohm build needs to get hotter to compete, it actually takes much less power to heat them (that's where it gets confusing lol ).

I'm only giving a long explanation because I was confused about the same thing a while back lol, and I wish someone would've explained it to me .
So,

1.2ohm setup needs 6.9volts for 40watts (and only needs 6 amps to achieve that power)
(Same 40 watts, but .2 needs ALOT less because it's smaller.)
0.2ohm setup needs 2.8volts for 40watts (needs more at 14amps because there's less resistance.

The amps are higher on the low resistance because it's more "wide open". A good way to picture that, is to imagine the less resistance you have, the wider/more open a garden hose is, thus the faster the water comes out, thus the more amps needed. The 1.2ohm setup is resisting 6x more (the hose is tighter, more restraining on the water/amps), the .2ohm setup is getting close to not resisting at all, so it's a 6x wider hose letting the water/power through much much faster, so it needs quite a bit more amps.

Sorry hope this didn't look like a "lecture" lol, it's easy to remember that Watts = same heat on all coils, that IPV3 is made to heat bigger coils up as fast as you could normally only heat sub-ohm coils at, but keep in mind that takes alot more power. Smaller coils require less power for more heat, they're more efficient, but with less surface area. And your amps don't waste battery, you only have to make sure your within your battery's amp limit (which your IPV3 does for you I believe).

This is how I learned volts, amps, watts, and resistance: Ohm's Law Calculator

I'm sure you've already known about this site, but after playing with it for a while it's super easy to catch on. I can pretty much do it in my head now .

Hope this helps brother! Vape on!

Thanks treehead! The information you gave definitely helped clear things up. After reading what you wrote and taking a look at the formula for ohm's law again I realized that wattage is the result of volts and resistance. I believe I was confused because I was thinking a 1.2ohm coil needed 40 watts to get to 6.9 volts. Thanks for your time and effort in writing all that!

 

[Dristwaz]

TBT, smaller coils are not more efficient and by a significant margin. Using P = I²*R for the same wattage, there is going to be a much higher current draw with a lower resistance; plus it goes up exponentially while resistance goes up linearly.

To show how this works we will say you have a 2000mAh cell that you're using. For 40W of power @ 1.2Ω the current draw will be 5.77A and at 0.2Ω your current draw will 14.14A. The 1.2Ω set-up drawing 5.77A will have about 20 minutes of useful power whereas the 0.2Ω set-up drawing about 14A will only have about 8½ minutes of useful power draw, over half the time! Plus once you add in the power lost due to the internal resistance of the battery and mod the loss is even greater because now you're losing more power there than what you are in the coils where you want it to be.

What does this really mean when using your mod? The higher your current draw, the less efficient your system is going to be, the more stress you are putting on your cell and the shorter amount of time you'll have between changing batteries. The ideal is to find the balance that fits best for you overall and what you're looking for (huge clouds, taste, temp, etc...) while vaping.

 

[treehead]

TBT, smaller coils are not more efficient and by a significant margin. Using P = I²*R for the same wattage, there is going to be a much higher current draw with a lower resistance; plus it goes up exponentially while resistance goes up linearly.

To show how this works we will say you have a 2000mAh cell that you're using. For 40W of power @ 1.2Ω the current draw will be 5.77A and at 0.2Ω your current draw will 14.14A. The 1.2Ω set-up drawing 5.77A will have about 20 minutes of useful power whereas the 0.2Ω set-up drawing about 14A will only have about 8½ minutes of useful power draw, over half the time! Plus once you add in the power lost due to the internal resistance of the battery and mod the loss is even greater because now you're losing more power there than what you are in the coils where you want it to be.

What does this really mean when using your mod? The higher your current draw, the less efficient your system is going to be, the more stress you are putting on your cell and the shorter amount of time you'll have between changing batteries. The ideal is to find the balance that fits best for you overall and what you're looking for (huge clouds, taste, temp, etc...) while vaping.

Interesting, thanks for the info brother! I was always under the impression that the higher the amp draw, the more efficiently a battery puts out its charge. (In other words, with a smaller coil, although it technically "takes" more watts, I always thought you would get up to heat much faster, and generally larger coils end up wasting battery because they are holding down the button for ages trying to get up to heat. I know that larger coils are easier on the current draw, but I thought the general amount of time it takes to get them heated up is what wastes vape time.)

The more you know!****************

 

[mscott]

wow I think I've read this exact discussion at least 15 times here and it just now clicked in my head. Looking at the formulas and the relationship between Ohm's Law and Joule's Law and how they apply to the vaping experience did it!. Now to figure out why cana mod is lying to me...maybe its time to get a real one.

 

[treehead]

wow I think I've read this exact discussion at least 15 times here and it just now clicked in my head. Looking at the formulas and the relationship between Ohm's Law and Joule's Law and how they apply to the vaping experience did it!. Now to figure out why cana mod is lying to me...maybe its time to get a real one.

Right on man! Believe me I didn't pick up on it fast myself lol, it's why electricians are paid so much . It's something that seems truly backwards compared to any other science that we know. I mean *smaller coils = *more watts? Totally mindfckd me when I was trying to understand it lol!

For anybody trying to figure out the same thing, a good metaphor for amps and ohms is a highway.

The wider the highway, the more cars can go through (thicker the gauge wire, the faster/easier the current travels, harder to traffic "jam" the electricity)

The thinner the highway, the less cars can go through (so when they traffic jam, that creates heat, thinner wires make heat easier)

shorter = less room for cars, heats up faster

longer = heats up slower

So longer thicker wire needs more power to heat up, but it also allows for faster current, so it doesn't necessarily take as long as thinner wires.

The shorter thinner wire "pinches" the traffic/electricity, so it heats up faster with slower current.

It's so backwards lmao, one day it just makes sense after playing around with some coils .

 

[Monotremata]

I just noticed this "phenomenon" myself a week ago.. Decided to do a twisted 32ga coil. 8 wraps got me to 1.4 ohms (this would've been close to 3 with a single wire) and yep.. For the first time I'm actually running my MVP at 11w (used to keep it between 7-8), and now I'm already looking for a new mod that'll hit 20 or 30 to see what I've been missing haha.. Although I don't think I'd try it with my old Protanks, those things can't take more than 8 or it just fries your juice..

 

[KenD]

I'm guessing you use thinner wire in the 1.2 ohm coil, that means that the heat-up (and cool-down) time is shorter. That would mean that with the 0.2 ohm coil you either need to hit it longer or apply more power to get the temperature up quicker. As for current draw, regulated kids use more or less the same amps whatever the resistance of the coil, at the same watts of course. With higher resistance coils you up the voltage, which means that you draw more current to reach those higher voltages (check steam engine). Due to the same reason the amp draw increases when the battery charge lowers, dueto the battery needing to work harder to provide the desired voltage (again, check steam engine).

 

[Norrin]

Interesting, thanks for the info brother! I was always under the impression that the higher the amp draw, the more efficiently a battery puts out its charge. (In other words, with a smaller coil, although it technically "takes" more watts, I always thought you would get up to heat much faster, and generally larger coils end up wasting battery because they are holding down the button for ages trying to get up to heat. I know that larger coils are easier on the current draw, but I thought the general amount of time it takes to get them heated up is what wastes vape time.)

The more you know!****************

This is partially true but the problem you have is a battery kicking out 15A will only last about half the time of a battery kicking out 5A when we are talking about vaping when most batteries are swapped at 3.6-3.5V (my opinion not fact) the lower you use your battery the closer this difference is, but you can't do that as a mech starts to under perform and a regulated device cuts out usually at 3.5V IIRK.