High Voltage/High Resistance vs low Voltage/Low Resistance

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sollehahn

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I started vaping since early january this year and got iTaste MVP as my first PV. several day ago i got myself a smok R tank and i notice they provide several coiled wick with different resistance (2, 2.5 and 3 ohm). I dont really notice the difference of using different resistance, except that i need to use higher voltage to get more vapor with higer resistance.

so i really want to know whats the different of using high voltage with high resistance and low voltage with low resistance? :blink:
 

HTX_Vape_Cape

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ditto on this. There are a few advanced topics that I don't understand when it comes to resistance, voltage and things like number of coils. I'm not an electrical engineer, so that might have something to do with it.

I know there is a lot of information on ECF, but some of the advanced topics like this don't have as much i

So, lower resistance on the same battery will give more vapor, because it's hotter? Yes? How can lower resistance be hotter, wouldn't a higher resistance have more electron collisions, and therefore run hotter? Or is it because it sucks out more amperage, and the velocity of the electrons moving faster create more heat?

How does varying the voltage change the vape experience? I assume the same thing, more voltage = more amps over the same resistance, so therefore more heat. Is that correct?

What about coil numbers. How does having two coils change the vape experience?
 

The Ocelot

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Hi - when you first stop smoking cigarettes it takes awhile for your sense of taste to return/improve. As it does, you will be able to better taste the differences of flavor and th you get by varying the voltage. When I first got my Twist, I couldn't figure out what all the hype was about. It took about a month or so to be able to taste the changes.
 

Thompson

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It will have an effect on battery life. I've read high resistance + high voltage = longer battery life, but I can't provide any solid reasoning behind this. Maybe somebody will come along and lay it out.

I'm starting to view LR, SR & HR (possibly incorrectly) in terms of surface area. Thicker, lower resistance wire can be wrapped more times to reach the same ohms as a thinner, higher resistance wire.
 

Cloud Wizard

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It's all about ohm's law V=I*R (v=voltage (volts or v), I=current (amps or A), R=resistance (ohms or Ω). Power P=(V*V)/R

Power roughly = work, which roughly = heat. Heat created by the coil is what vaporizes juice.

The water analogy(An analogy for Ohm's Law : Ohm's Law)
  • Ohm's Law also makes intuitive sense if you apply it to the water-and-pipe analogy. If we have a water pump that exerts pressure (voltage) to push water around a "circuit" (current) through a restriction (resistance), we can model how the three variables interrelate. If the resistance to water flow stays the same and the pump pressure increases, the flow rate must also increase.
  • If the pressure stays the same and the resistance increases (making it more difficult for the water to flow), then the flow rate must decrease.
  • If the flow rate were to stay the same while the resistance to flow decreased, the required pressure from the pump would necessarily decrease.


What it means for vaping is that the higher the power (watts) the hotter your vape. In practice, resistance wire comes in different thicknesses that have different resistance (28awg is fatter than 32awg / 28awg has less resistance than 32awg per length) so when folks are talking about "wraps" or "coils" they are talking about increasing/decreasing resistance by adjusting the length of the resistance wire (more wraps = longer wire = higher resistance).

So, if I am using a battery that is 3.7v I will use a lower resistance to get to what most folks consider the "sweet spot" of ~8watts. If you are using a variable voltage (VV) or variable wattage (VW) device you can change battery output to match any resistance head. With higher resistance heads (cartos, attys, etc) the current (amps or A) required to get the same power (watts) decreases. This is important because any VV/VW devices have a safety limitation on their switches (e.g. Vision Spinner = 2.5A, Provari = 3.5A, ZMax = 5A). So If using a device with a lower amp saftey switch a standard resistance (SR) head will be able to provide more power without tripping the cutoff (e.g. a 1.7ohm carto @3.7v = 8.1 watts at 2.2A, a 2.5ohm carto @4.5v = 8.1watts at 1.8A) The same 2.5ohm carto could run up to 5.4v at 11.2watts at the same 2.2A current (much hotter with out tripping the safety cutoff).

The other thing to think about is the mAh of your battery (storage capacity). It's not safe and could damage your battery if you run low resistance (less than ~2ohm on anything smaller than 650mAh)

A real simple rule of thumb is to set battery voltage to resistance + 2 to get close to 8 watts (like the examples above - 1.7ohm@3.7v=~8watts, 2.5ohm@4.5v=~8watts) and then adjust by .1v to taste. And on 3.7v fixed batts (650mAh or larger) 1.5ohm-2.2ohm depending on how hot you like your vape.
 

Thompson

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Just curious CW, I've read before that LR and small mah batteries are not a good idea. Is it because they are pulling too much amperage?

I like to use an A7 RBA on an SD Keyring 300mah stealth battery, and I wrap it around 2.4 - 2.8Ω to keep it in that 'standard resistance' zone as to not cause it any issues.
 

HTX_Vape_Cape

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It's all about ohm's law V=I*R (v=voltage (volts or v), I=current (amps or A), R=resistance (ohms or Ω). Power P=(V*V)/R

Power roughly = work, which roughly = heat. Heat created by the coil is what vaporizes juice.

The water analogy(An analogy for Ohm's Law : Ohm's Law)
  • Ohm's Law also makes intuitive sense if you apply it to the water-and-pipe analogy. If we have a water pump that exerts pressure (voltage) to push water around a "circuit" (current) through a restriction (resistance), we can model how the three variables interrelate. If the resistance to water flow stays the same and the pump pressure increases, the flow rate must also increase.
  • If the pressure stays the same and the resistance increases (making it more difficult for the water to flow), then the flow rate must decrease.
  • If the flow rate were to stay the same while the resistance to flow decreased, the required pressure from the pump would necessarily decrease.


What it means for vaping is that the higher the power (watts) the hotter your vape. In practice, resistance wire comes in different thicknesses that have different resistance (28awg is fatter than 32awg / 28awg has less resistance than 32awg per length) so when folks are talking about "wraps" or "coils" they are talking about increasing/decreasing resistance by adjusting the length of the resistance wire (more wraps = longer wire = higher resistance).

So, if I am using a battery that is 3.7v I will use a lower resistance to get to what most folks consider the "sweet spot" of ~8watts. If you are using a variable voltage (VV) or variable wattage (VW) device you can change battery output to match any resistance head. With higher resistance heads (cartos, attys, etc) the current (amps or A) required to get the same power (watts) decreases. This is important because any VV/VW devices have a safety limitation on their switches (e.g. Vision Spinner = 2.5A, Provari = 3.5A, ZMax = 5A). So If using a device with a lower amp saftey switch a standard resistance (SR) head will be able to provide more power without tripping the cutoff (e.g. a 1.7ohm carto @3.7v = 8.1 watts at 2.2A, a 2.5ohm carto @4.5v = 8.1watts at 1.8A) The same 2.5ohm carto could run up to 5.4v at 11.2watts at the same 2.2A current (much hotter with out tripping the safety cutoff).

The other thing to think about is the mAh of your battery (storage capacity). It's not safe and could damage your battery if you run low resistance (less than ~2ohm on anything smaller than 650mAh)

A real simple rule of thumb is to set battery voltage to resistance + 2 to get close to 8 watts (like the examples above - 1.7ohm@3.7v=~8watts, 2.5ohm@4.5v=~8watts) and then adjust by .1v to taste. And on 3.7v fixed batts (650mAh or larger) 1.5ohm-2.2ohm depending on how hot you like your vape.

First, huge Thanks for this info, very helpful. Been awhile since I've taken any physics classes, haha. So if I got the math right:

V=I*R
so I = V/R

3.7 volts (~'standard'?) / 2.5 ohm (~'standard'?) = 1.48 amps

so going lower resistance, say 1.7 ohm gives

3.7v / 1.7ohm = 2.176a



In terms of heat (power)

3.7v ^ 2 = 13.69, so divided by different resistances:

1.7 ohm = 8.05 watts
2.5 ohm = 5.478 watts



If I went with a standard 3.7 volt eGO system, using a lower resistance will give me the "sweet spot" 8 watts. If I went with the standard 2.5 ohm, I would get lower power, and not as much vapor. How does this effect taste though, will the 8 watts give a burnt taste? Maybe if the liquid can't wick fast enough?

Also, I'm assuming since batteries are a finite power source, the more amperage you pull from lowering the resistance, the faster the battery will be discharged and you'll have to swap it out. If I like the lower resistance, then I should get a 1000 Mah battery to make it last longer, and the Mah size has nothing to do with voltage output. (pretty sure on that last sentence).

Am I headed in the right direction?
 

Thompson

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Yes, the draw is too much for the battery. There's some math behind it but at the end of the day all batteries are also rated with "C". This site explains it better: CARTO to BATTERY INFO

Here's a chart for quick look ups on power
http://i.imgur.com/Ohn8o.png

OK, thanks :)

That site is actually exactly what I've been looking for. Had a basic understanding of C ratings but I'm seeing how it is applied with this article. Just had one of those "ooooh" moments when things all come together.

Then another wave of variables floods in.
 
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Cloud Wizard

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If I went with a standard 3.7 volt eGO system, using a lower resistance will give me the "sweet spot" 8 watts. If I went with the standard 2.5 ohm, I would get lower power, and not as much vapor. How does this effect taste though, will the 8 watts give a burnt taste? Maybe if the liquid can't wick fast enough?

Also, I'm assuming since batteries are a finite power source, the more amperage you pull from lowering the resistance, the faster the battery will be discharged and you'll have to swap it out. If I like the lower resistance, then I should get a 1000 Mah battery to make it last longer, and the Mah size has nothing to do with voltage output. (pretty sure on that last sentence).

Am I headed in the right direction?

Your math looks right :) Higher resistance on lower voltage gives you lower power - correct

Most of the fixed batts actually run in a range ~4.2v fully charged and ~3.3v when they'll shut-off (LiIon batts should never be over discharged - it shortens the life of the battery). The watts thing is really subjective. When I started vaping I used be very happy with a 2.2ohm carto on a 3.7v "fat" batt (eGo style) around 6ish watts. That pretty quickly wasn't powerful enough so I dropped down to 1.7ohm cartos and 1.5ohm attys.

8 watts does not ever taste burnt to me. Some heads (like Vivi Novas) I run ~8.5 watts, Boge 3.0ohm cartos I run at 8.0watts, Silica wicked rebuildables ~10 watts and genesis style attys around 15 watts all with the same juice. Really depends on your preference, the carto/atty and the juice.

And yes, LR can drain your batteries quicker (higher current) but I've never noticed it enough to steer me away from using them.
 

Cloud Wizard

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OK, thanks :)

That site is actually exactly what I've been looking for. Had a basic understanding of C ratings but I'm seeing how it is applied with this article. Just had one of those "ooooh" moments when things all come together.

Then another wave of variables floods in.

In the infinite set of possibilities, there are infinite solutions. Fortunately for us vaping lives between 3.3v - 6.0v (unless you stack batts to 7.4v which I NEVER do - IMHO stacked batts = unsafe). So, you kinda get a hang for what ranges appeal to you and then the rules pretty much work for any device (PV or head) you may choose. Just a lot of info in a very short amount of time in the beginning.
 

Birchi

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Most of the fixed batts actually run in a range ~4.2v fully charged and ~3.3v when they'll shut-off (LiIon batts should never be over discharged - it shortens the life of the battery). The watts thing is really subjective. When I started vaping I used be very happy with a 2.2ohm carto on a 3.7v "fat" batt (eGo style) around 6ish watts. That pretty quickly wasn't powerful enough so I dropped down to 1.7ohm cartos and 1.5ohm attys.

8 watts does not ever taste burnt to me. Some heads (like Vivi Novas) I run ~8.5 watts, Boge 3.0ohm cartos I run at 8.0watts, Silica wicked rebuildables ~10 watts and genesis style attys around 15 watts all with the same juice. Really depends on your preference, the carto/atty and the juice.

And yes, LR can drain your batteries quicker (higher current) but I've never noticed it enough to steer me away from using them.

This makes a lot of sense to me. My eGo batteries with LR Clearo's were much better when freshly charged and fade a bit after several vapes. Now when I'm using my eVic and LR I run 8.5 Watts to get a steady 4.02-ish volt's with great hits every time.

I must add some Twist's to my arsenal (for out and about).
 

w00st3r

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I always thought of it like a water slide with loops, if you have two slides the same length and one has 10 loops and the other has 0, the slide with 10 loops will be a lot slower because of all the added resistance. It takes less effort to go through the slide with 0 loops so its a free range to dump power from the battery to the device
 

Thompson

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In the infinite set of possibilities, there are infinite solutions. Fortunately for us vaping lives between 3.3v - 6.0v (unless you stack batts to 7.4v which I NEVER do - IMHO stacked batts = unsafe). So, you kinda get a hang for what ranges appeal to you and then the rules pretty much work for any device (PV or head) you may choose. Just a lot of info in a very short amount of time in the beginning.

Would Ohm's law be used to find the amperage draw on a battery from a VV boost circuit? Or would that require extra steps?

Such as having your voltage set to 4.5, using a 2.4Ω coil and the battery is at say 3.7 volts. I can only find the amperage after the increase in voltage right? Would this be the same amperage drain applied to the battery?

Most of the fixed batts actually run in a range ~4.2v fully charged and ~3.3v when they'll shut-off

By fixed batteries do you mean the standard ego? I thought they ran exclusively at 3.2-3.3v even fresh off the charger because they are regulated? Or am I just mixing up terms again?

:p
 
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Cloud Wizard

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This makes a lot of sense to me. My eGo batteries with LR Clearo's were much better when freshly charged and fade a bit after several vapes. Now when I'm using my eVic and LR I run 8.5 Watts to get a steady 4.02-ish volt's with great hits every time.

I must add some Twist's to my arsenal (for out and about).

Yep, that's the beauty of regulated output. Most of the newer VV/VW PVs will give a consistent vape right up until the low battery cut-off. I don't have an eVic but they pretty cool (very futuristic)
 

Regzzz

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ditto on this. There are a few advanced topics that I don't understand when it comes to resistance, voltage and things like number of coils. I'm not an electrical engineer, so that might have something to do with it.

I know there is a lot of information on ECF, but some of the advanced topics like this don't have as much i

So, lower resistance on the same battery will give more vapor, because it's hotter? Yes? How can lower resistance be hotter, wouldn't a higher resistance have more electron collisions, and therefore run hotter? Or is it because it sucks out more amperage, and the velocity of the electrons moving faster create more heat?

How does varying the voltage change the vape experience? I assume the same thing, more voltage = more amps over the same resistance, so therefore more heat. Is that correct?

What about coil numbers. How does having two coils change the vape experience?

Hi there Chry, I used to teach electricity to soldiers a long time ago and at the time I was using this "water hose" comparison to explain volts/amp/and resistance... I'll try to explain here your question about low resistance hotter?

Let pretend that:

Volt is the same as the size of a garden hose, more volts, bigger the hose, ok?

Resistance is like your hand on the tube of the garden hose and the more you squeeze your hand on the hose the more resistance you give it.

Heat (or in our case, coil glow) can be compared to how much water comes out at the end of the hose. (you can ALSO know this as watts)

Now, from here it's easy.

Take 1 inch diameter garden hose (1 volt), open the tap, the water will flow at the end quite good (little resistance from the hose inside = 1 ohms) therefore high heat. Right? Well call this heat 1

Take the same 1 inch hose (1 volt) and squeeze it in the middle about half the size of the hose to create high resistance (5 ohms), therefore less water comes out at the end, less heat. this would be heat 0.2

now, take a 10 inch diameter hose (10 volts) open the tap, the water will flow at the end quite good (little resistance from the hose inside = 1 ohms) therefore high heat. Right? Well call this heat 100

Take the same 10 inch hose (10 volt) and squeeze it in the middle about half the size of the hose to create high resistance (5 ohms), therefore less water comes out at the end, less heat. this would be heat 20.

The formula is really simple.

How do we get from 1 Volt, 1 ohm to 1 watt
or
How do we get from 10 Volts, 5 ohms to 20 watts

easy: multiply volts by itself: 10 x 10 = 100
then divide that by resistance: 100 / 5 = 20

so in the case of our e-cigs, lets see more realistic numbers:


My battery has 2 voltages choice 3.5 volts and 4.0 volts
I have 2 atty heads, one with 1.2 ohms and one with 2.4 ohms.

Lest see what my results would be in some settings possible

my battery at 3.5 volts with 1.2 ohms atty:

3.5 x 3.5 is 12.25
12.25 divided by 1.2 = 10.2 (ish I rounded decimal) Watts

my battery at 3.5 volts with 2.4 ohms atty:

3.5 x 3.5 is 12.25
12.25 divided by 2.4 = 5.1 (ish I rounded decimal) Watts

As you can see, a lower resistance will give you more heat at the same voltage.

My 1.2 ohms atty with my 3.5 volts is 10.2 Watts

if I keep my 1.2 ohms atty and now put the 4.0 volts battery on it:

4.0 x 4.0 = 16
16 divided by 1.2 = 13.3 Watts.

hope you can understand my gribirish...

One thing that I would not want to explain in details, but is good to know, is that with the same coil material, a higher voltage with a resistance that would give you the same watts will fire faster "startup heat"

example:

a 10 volts battery with a 5 ohms atty would give you 20 watts of power

a 100 volts battery with a 500 ohms atty would ALSO give you 20 watts of power, but that red hot coil would be red faster then the lower voltage one.

hope this helps.

cheers
 

Cloud Wizard

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Would Ohm's law be used to find the amperage draw on a battery from a VV boost circuit? Or would that require extra steps?

Such as having your voltage set to 4.5, using a 2.4Ω coil and the battery is at say 3.7 volts. I can only find the amperage after the increase in voltage right? Would this be the same amperage drain applied to the battery?

Good question. I would guess (and I am guessing now) that the actual battery charge becomes somewhat irrelevant because the safety cut-off would be based on the actual current passing through the switch (in your example the resistance of the carto/atty at the set 4.5v). But yes, there is an efficiency loss due to the step-up circuitry.
 

Regzzz

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Would Ohm's law be used to find the amperage draw on a battery from a VV boost circuit? Or would that require extra steps?

Such as having your voltage set to 4.5, using a 2.4Ω coil and the battery is at say 3.7 volts. I can only find the amperage after the increase in voltage right? Would this be the same amperage drain applied to the battery?

No, the amperage draw would not be the same, but the watts output would be same. the electronics of the VV that increases the voltage for you, also draws more amps from the battery then the amps being given to the coil. but in all, the power output (watts) is the same, nothing gained, nothing lost. (except for a very very minimal heat lost in the electronics of the VV)


To find the amperage draw on your battery is simple if you know the watts. ie:

3.0 ohms atty
VV set at 4.5 volts
Amps=1.5 amps

therfore watts = 6.75

your actual battery voltage is 3.5 and watts is 6.75 then Amps draw from battery is 1.93 amps


See: Ohm's Law Calculator
 
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Cloud Wizard

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By fixed batteries do you mean the standard ego? I thought they ran exclusively at 3.2-3.3v even fresh off the charger because they are regulated? Or am I just mixing up terms again?

:p

Fixed batts meaning anything that doesn't have VV/VW.

Didn't think about that but you're right about the regulated output on some eGos@3.4v, SLB@3.7 and others unregulated use the native battery voltage 4.2v/3.3v.
 
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