Volt/Watt/Amp???

[UncleChuck]

I can tell you are getting frustrated. I hope you are not taking it any further with trying to ridicule my arguments instead of making sense.I believe you that you have comparable output of vapor on different coils using same watts, just as much as I have similar vapor on different coils using the same amps.



You say you see differences in vapor. And yet the only measurable numbers that change are the amps. The watts are the same.

Household bulbs: Not sure how the difference between AC and DC has an influence on things, but the thing you seem to forget is that one of the main differences between 30 and 60 watt bulbs is the resistance of the filament. And besides changing the power they consume that also changes the current.

I'm not saying that watt has no influence on things. All I'm saying is that in my opinion we change the volt/watt on our mods to adjust the current that flows through the coil. Why not cut out the middle man?

Oh I'm not getting frustrated, I was trying to make a little joke with the beaver thing but I'm not the best at joking

When we adjust the voltage on our APVs, or adjust the resistance on our RBA+Mechs we are doing so to change the power, not the current. While true that current and power are related, power is a far more accurate means of determining vapor than current.

If current determine vapor, than all builds of the same current should yield similar vapor.

4 amps, 6 volts, 1.5 ohms, 24 watts

4 amps, 4 volts, 1 ohm, 16 watts.

The first build, at 24 watts, will produce much more vapor than the second build at 16 watts, because power more or less determines the amount of vapor. If current determine vapor those two builds would put out similar vapor, but they don't.

Hope I don't come off as having an attitude or anything, this is still a friendly conversation as far as I'm concerned

 

[Ryedan]

Ok upon reading a bit on the wiki, it says heat is not JUST current squared, it's current squared times resistance, which is where the power level makes the determining factor.

Yup. A bit further down in that article is also says the following:

"Direct current.

The most general and fundamental formula for Joule heating is:

P=VI "

This stuff can be really confusing for people.

 

[deanthemachine]

I am by no means an expert here and cannot debate whether its current or watts that is the factor for vapor, but I can say one thing, or perhaps more accurately ask one thing from a very high level. The reason sub ohmers get more vapor is because our devices do not have the limits on wattage or current that many PVs have. Whether said PV maker allows the user of a PV to adjust current, or overall power(watts) or the voltage you are still going to hit a defined limit that the PV maker puts in the chip to prevent people from exceeding and since there is a relationship via formula your vapor production is not going to be any better. Am I oversimplifying this? Again my electrical knowledge is confined to what I have learned here.

 

[State O' Flux]

The reason sub ohmers get more vapor is because our devices do not have the limits on wattage or current that many PVs have. Am I oversimplifying this? Again my electrical knowledge is confined to what I have learned here.

Anyone can punch numbers into a ohms law calculator, but understanding - and making proper use of - the numbers generated is something else entirely.

The sub-ohm "equation" is controlled by the resistance itself - which in the case of deep sub-ohm is a near dead short - the batteries potential momentary and continuous output, and ultimately, the intelligence and/or common sense of the user.

An example of this is: How often have you read about someone trying to use a 0.15Ω dual coil RDA on a clone, side button mech with an off-brand 18350 battery (you know, the kind with no amp or 'C' rating on the label)... and questioning why the button gets "warm" with a 5 second lung hit - or why the "best vape" only lasts for about 9 minutes?

 

[Stosh]

......The reason sub ohmers get more vapor is because our devices do not have the limits on wattage or current that many PVs have........ Am I oversimplifying this? Again my electrical knowledge is confined to what I have learned here.

Wattage is always limited, you can't get more energy out of any device than is stored in the device. Current is simply a variable used to calculate wattage or work done. It's the work done that changes the vapor, and that's work as defined in scientific terms, not where you're going tomorrow morning....

Watts, Joule, Horsepower, BTU, Kilo calories, Kilowatt-hours, etc are simply different units of measuring work. For instance 1 hp = 746 watts, the other units of measurement also can be converted.

In physics, energy is one of the basic quantitative properties describing a physical system or object's state. Energy can be transformed (converted) among a number of forms that may each manifest and be measurable in differing ways
https://en.wikipedia.org/wiki/Energy

The work done (in our case heating a coil) is dependent only on the energy supplied, which is determined by the combination of voltage, resistance and current as defined by Ohm's law. If the coil was heated to the same temperature using a butane torch, the amount of work necessary (as measured in watts or any other unit you may choose) would be exactly the same as using a battery or any other power source.

The work, resulting in heat, light, movement, digestion is the end result...how you get there, what type power source is used, what units are used to calculate it only changes the equation used to calculate it, not the amount of the work done.

 

[ckn71nm]

Oh I'm not getting frustrated, I was trying to make a little joke with the beaver thing but I'm not the best at joking

When we adjust the voltage on our APVs, or adjust the resistance on our RBA+Mechs we are doing so to change the power, not the current. While true that current and power are related, power is a far more accurate means of determining vapor than current.

If current determine vapor, than all builds of the same current should yield similar vapor.

4 amps, 6 volts, 1.5 ohms, 24 watts

4 amps, 4 volts, 1 ohm, 16 watts.

The first build, at 24 watts, will produce much more vapor than the second build at 16 watts, because power more or less determines the amount of vapor. If current determine vapor those two builds would put out similar vapor, but they don't.

Hope I don't come off as having an attitude or anything, this is still a friendly conversation as far as I'm concerned

No attitude or offence taken. I'm not sure how to proceed from here. I think we are going in circles. You speak form your experience and I can only speak from mine. I have build many coils and used many different commercial attys before that. Regardless of the resistance (from 0.7 to 2.6 in my builds and 1.8 and 2.1 in store bought units) I have always adjusted the power to a level at which I am running between 2.8 and 3 amp over the coil. Obviously that are very different power levels. I always get my preferred amount of vapor and taste.

I don't deny in any way that you have a different experience and that is absolutely valid. I just don't understand why, and the explanations I get here are not enough to convince me.

1. What I can't get over is Joule heating, describing that current is creating the energy. Power is part of that, but not as the creator of that energy "the watt, is equivalent to one joule per second". You could say that power is used to measure the energy put not to create it.

2. My personal experience as described above.

I have read other posts like this one

http://www.e-cigarette-forum.com/fo...ined-detail-single-dual-coil-atty-cartos.html

But all that is presented as "prove" that the vapor is the same is the author saying it is the same. Obviously I'm doing that too, so no better prove here.

Anyway, since there are no variable current PV's out there, the entire discussion is pretty much just theoretical. I will do what I have been doing so far and enjoy my vape. And so will you.

 

[ckn71nm]

Yup. A bit further down in that article is also says the following:

"Direct current.

The most general and fundamental formula for Joule heating is:

P=VI "

This stuff can be really confusing for people.

It also says even further down "using low current at high voltage reduces the loss in the conductors due to Joule heating" Doesn't that mean lowering the current reduces the heat?

You are right, it's very confusing

 

[WattWick]

We are adjusting amperage... In a way.

Let's distinguish input from output. We input from a battery and output to the coil. The input voltage is set by the battery. The rate at which we pull current (amperage) from the battery is set by the output (coil) resistance.

To adjust output power (watts) we have to adjust either input voltage, output current, or both.

Problem is, we are working with too many constants. Input voltage is a constant. Output current is a constant due to a constant resistance.

We can, however adjust one thing: output voltage; by powering a boost circuit. To do so, we pull a higher amperage from the input battery.

This is variable current at the only end we can have variable current: the battery (input) end. Variable voltage IS variable current.

I am no electrician or physicist, so I may well be mistaken.

 

[ckn71nm]

... Energy transfer can be used to do work, so power is also the rate at which this work is performed. The same amount of work is done when carrying a load up a flight of stairs whether the person carrying it walks or runs, but more power is expended during the running because the work is done in a shorter amount of time. The output power of an electric motor is the product of the torque the motor generates and the angular velocity of its output shaft. The power expended to move a vehicle is the product of the traction force of the wheels and the velocity of the vehicle."...

Not sure how that is valid. The amount of work done (weight ending up the stairs) is still the same regardless of running or walking (more or less power). If you would want to get more work done, you would have to find a way to carry more, like more people carrying stuff. And that sound like more current to me.

 

[WattWick]

Or the same amount of people each carrying more. Which sounds like higher voltage to me

 

[Ryedan]

We are adjusting amperage... In a way.

Let's distinguish input from output. We input from a battery and output to the coil. The input voltage is set by the battery. The rate at which we pull current (amperage) from the battery is set by the output (coil) resistance.

To adjust output power (watts) we have to adjust either input voltage, output current, or both.

Problem is, we are working with too many constants. Input voltage is a constant. Output current is a constant due to a constant resistance.

We can, however adjust one thing: output voltage; by powering a boost circuit. To do so, we pull a higher amperage from the input battery.

This is variable current at the only end we can have variable current: the battery (input) end. Variable voltage IS variable current.

I am no electrician or physicist, so I may well be mistaken.

You got it WattWick . Rader explained it well here.

 

[Ryedan]

Not sure how that is valid. The amount of work done (weight ending up the stairs) is still the same regardless of running or walking (more or less power). If you would want to get more work done, you would have to find a way to carry more, like more people carrying stuff. And that sound like more current to me.

Yes, you are correct that the amount of work done is the same no matter how long it took. But the concept of power is rooted in the rate of doing work. For vaping, one drag of a 15 watt setup will give you more vapor than one drag of a 8 watt setup. Or, you could just take a drag that's twice as long at 8 watts and that is what people do to get a satisfactory vape. But once you're taking 10 second drags, longer is not a good option anymore, so we up the wattage to do more work in the same or less amount of time.

Think of a car. The amount of work done in accelerating from a standstill to 60 mph is the same if you do it in 10 seconds as if you do it in 5 seconds. Doing it faster requires more horsepower, as long as you can connect the power to the road (not spinning the tires).

 

[edyle]

The current flows through the coil wire;
the coil wire has resistance;
because of the resistance the wire gets hot.
if you push more current, the wire gets hotter.
if you increase the resistance, less current can flow and the wire gets cooler; but if you push harder until you get the same current as before, the wire gets hotter than before.

It is the current AND the resistance that produce the heat.

The Power output from the coil wire is Current(squared) x Resistance.

And it is this flow of energy that vaporises the liquid.

A superconducting wire has 0 resistance; it passes high current with no resistance, but it does not get hot, because there is no resistance.
Current alone does not create the heat.

 

[Stosh]

It is the current AND the resistance that produce the heat.

The Power output from the coil wire is Current(squared) x Resistance.

And it is this flow of energy that vaporises the liquid..

The Power output from the coil wire is also Voltage(squared) / Resistance,
simply the same formula using variables to describe the wattage, or work done in heating a coil.

 

[eda123]

Heat is what we are after. Higher heat means more vaporization.

Heat is wattage. It depends on current and the resistive load, the product of the two. A 3A setting will give a hugely different vape with a 1ohm vs 3ohm coil. However an 8W setting would give more similar vapes on those two builds.

Adjusting wattage is direct adjustment of the atomizer heat. Changing only voltage or current is only changing one factor and IMO wouldn't be usable at all.


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[Stosh]

Heat is what we are after. Higher heat means more vaporization.
Heat is wattage.

YES, YES!! the wattage is the measure of the work done by the system, it's the result of any and all other variables.

It depends on current and the resistive load, the product of the two.

OR P = E squared / R all the variables are dependant on each other.

Adjusting wattage is direct adjustment of the atomizer heat.

Any change in wattage, the actual work done, is never directly adjusted, it's by varying the voltage or resistance that the resulting wattage changes. a VW mod is just using the results of the formulas and variances in voltage, resistance and DISPLAYING the results, to make it easier for the vaper.

Changing only voltage or current is only changing one factor and IMO wouldn't be usable at all.

Changing voltage or resistance is the only variables we have easy control over, it results in current draw necessary to satisfy Ohm's Law. With a mech mod the resistance is the only available variable, a change results in higher wattage and a higher current to make the wattage possible. Regulated mods use different methods to increase wattage, some will adjust voltage to increase wattage, some will use duty cycles to approximate a voltage.

Early VV mods used a buck circuit that "regulated" the current / voltage to the coil by drawing some of the power (wattage) off and throwing it off as heat in the actual circuit, not a great solution. But these were the only mods to approximate a variable current control, and are no longer used. Even with these mods, the displayed to the user variable was voltage for simplicity and ease of use.

 

[eda123]

I think this says it all! Agree btw, when setting wattage you are setting the target result the mod will provide by varying the applied voltage.

 

[Mitey F]

That is correct, if you change one of the variables all the others change as well. But current is determined BY voltage and resistance, and it's not a linear relationship.

First of all, this is very wrong. V=IR, thus I=V/R. It is VERY MUCH a linear relationship.

The disagreement is that I'm saying amp is what determines the heat and therefor the vapor and should be adjustable on mods

Quote from ^ "Joule heating, also known as ohmic heating and resistive heating, is the process by which the passage of an electric current through a conductor releases heat. The amount of heat released is proportional to the square of the current"

I know, it's Wikipedia, but a very good explanation why the current is responsible for heating, not the power. And we are not disagreeing about safety. I'm with you regarding that all the way.

You're arguing with yourself here.

I'll try to make this as simple as I can, without going into more math and equations (not to patronize, I want to help people understand without confusing them or muddling the concept).

Let's put it this way. What we're really worried about here is HEAT, right? More heat in our coils will (in general, all else equal) give us more VAPOR, as it is vaporizing our niccotine goo more quickly. Too much heat, you burn your goo/wick. Not enough heat, you don't get any vapor. I think we can all agree that HEAT is our primary concern here. Now how do we relate that to amperage and wattage?

HEAT IS MEASURED IN JOULES (a unit of ENERGY, much like, and directly proportional to calories). THIS IS VERY IMPORTANT!!

1 Watt (power, typically denoted "P") = 1 Joule of energy, transferred in 1 second. If this Joule takes 2 seconds to transfer, you will only have 1/2 Watt (1 Joule / 2 seconds). Likewise, if you transfer 2 Joules in 1 second, you have 2 Watts (2 Joules / 1 second).

WATTAGE is a DIRECT MEASURE OF HEAT TRANSFER!

AMPERAGE is the movement of ELECTRONS that ALLOWS for heat transfer, and while you can CALCULATE wattage (and thus heat transfer) *using* amperage, you must also factor in RESISTANCE. Thus, WATTAGE is much more meaningful to us as vapers, as it measure HEAT TRANSFER DIRECTLY, INDEPENDENT OF AMPERAGE!

 

[ckn71nm]

First of all, this is very wrong. V=IR, thus I=V/R. It is VERY MUCH a linear relationship.



You're arguing with yourself here.

I'll try to make this as simple as I can, without going into more math and equations (not to patronize, I want to help people understand without confusing them or muddling the concept).

Let's put it this way. What we're really worried about here is HEAT, right? More heat in our coils will (in general, all else equal) give us more VAPOR, as it is vaporizing our niccotine goo more quickly. Too much heat, you burn your goo/wick. Not enough heat, you don't get any vapor. I think we can all agree that HEAT is our primary concern here. Now how do we relate that to amperage and wattage?

HEAT IS MEASURED IN JOULES (a unit of ENERGY, much like, and directly proportional to calories). THIS IS VERY IMPORTANT!!

1 Watt (power, typically denoted "P") = 1 Joule of energy, transferred in 1 second. If this Joule takes 2 seconds to transfer, you will only have 1/2 Watt (1 Joule / 2 seconds). Likewise, if you transfer 2 Joules in 1 second, you have 2 Watts (2 Joules / 1 second).

WATTAGE is a DIRECT MEASURE OF HEAT TRANSFER!

AMPERAGE is the movement of ELECTRONS that ALLOWS for heat transfer, and while you can CALCULATE wattage (and thus heat transfer) *using* amperage, you must also factor in RESISTANCE. Thus, WATTAGE is much more meaningful to us as vapers, as it measure HEAT TRANSFER DIRECTLY, INDEPENDENT OF AMPERAGE!

Well, alright. I'm still not entirely convinced, but I'm getting there. What seems to be the case is that everyone has a different opinion than I do and I'm certainly not saying that I'm right and EVERYBODY else is wrong. I'm going to do an experiment.

I marked the fluid level on one of my tanks. for today I will take 8 sec drags from that tank, count the drags and at the end of the day refill it to that line using a syringe giving my the amount of liquid used at a certain wattage and amperage. Then I will change the coil. Right now its a 1.6 ohm. I guess I will make a 2.1-2.5 ohm. Tomorrow I will repeat the 8 sec drag/counting and measure the liquid usage at the same current as today, day after tomorrow, with the same coil, I will repeat the whole thing at the same power level as it is now.

I will post the results here in a few days. If by that time anyone is still reading this and it turns out I have more similar liquid usage with constant wattage, I will humbly accept my defeat.

I know this will not be perfect. I can't really account for surface area of the coil, different wicking or slight differences in the way I drag. I hope that will average out.

 

[edyle]

Well, alright. I'm still not entirely convinced, but I'm getting there. What seems to be the case is that everyone has a different opinion than I do and I'm certainly not saying that I'm right and EVERYBODY else is wrong. I'm going to do an experiment.

I marked the fluid level on one of my tanks. for today I will take 8 sec drags from that tank, count the drags and at the end of the day refill it to that line using a syringe giving my the amount of liquid used at a certain wattage and amperage. Then I will change the coil. Right now its a 1.6 ohm. I guess I will make a 2.1-2.5 ohm. Tomorrow I will repeat the 8 sec drag/counting and measure the liquid usage at the same current as today, day after tomorrow, with the same coil, I will repeat the whole thing at the same power level as it is now.

I will post the results here in a few days. If by that time anyone is still reading this and it turns out I have more similar liquid usage with constant wattage, I will humbly accept my defeat.

I know this will not be perfect. I can't really account for surface area of the coil, different wicking or slight differences in the way I drag. I hope that will average out.

Since you did not say what wattage you have it on,
I am going to assume you are using 6 watts.
so 6watts 1.6ohm; that means sqrt(6/1.6)= 1.94amps.

If you put on a 2.5ohm coil, and try to get the same current,
that would be 1.94(squared)x2.5= 9.4watts.

-corrected-