Volt/Watt/Amp???

[eda123]

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

Its not an opinion in my case, what i post is fact. I have a B.S.E.E., I design circuits for a living and chances are you are using one of my patents right now as you type

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.

You said "amps are what determines heat". No. thats half of it. Its current AND the resistance its travelling through. Power is the product of current * resistance. The same current can deliver more heat/wattage if you lower the resistance its travelling through.


For a FIXED LOAD, yes, changing current will vary your wattage in a 100% linear fashion. But if you change resistance, that same current is NOT going to give you the same power as you had with another coil with different resistance.

For your experiment, if you swap in a 2.1-2.5 ohm and dont adjust your voltage applied, you will be increasing resistance and therefore lowering the power dissipated by the coil. This is not opinion.

 

[ckn71nm]

You said "amps are what determines heat". No. thats half of it. Its current AND the resistance its travelling through. Power is the product of current * resistance. The same current can deliver more heat/wattage if you lower the resistance its travelling through.


For a FIXED LOAD, yes, changing current will vary your wattage in a 100% linear fashion. But if you change resistance, that same current is NOT going to give you the same power as you had with another coil with different resistance.

For your experiment, if you swap in a 2.1-2.5 ohm and dont adjust your voltage applied, you will be increasing resistance and therefore lowering the power dissipated by the coil. This is not opinion.

Not sure what you are trying to say with this post. Are you saying the experimental setup is flawed? What I want to test is the claim (or fact if you want), made by almost everyone in this thread that regardless of coil resistance, I will get the same liquid consumption when I use the same power. Or my claim that I will get the same liquid consumption when I use the same current.

 

[ckn71nm]

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 6/1.6= 3.75amps.

If you put on a 2.5ohm coil, and try to get the same current,
that would be 3.75(squared)x2.5= 35watts.
No, Mr ckn71nm. Please do not do that.

Fortunately at least, I expect your mod won't allow you to try.

You are right, and I have thought about it.
Today: 1.6 ohm @ 10 watt = 2.5 amp
Tomorrow: 2.3 ohm @ 2.5 amp = 14 watt
Day after: 2.3 ohm @ 10 watt = 2 amp

I know that you guys don't want to sound patronizing, but I have been building coils for about a year. Sub-ohm and all above. I know about safety and I won't do anything unsafe (or impossible like 35 watt)

Edit: btw 1.6 ohm @ 6 watt = 1.9 amp not 3.75

 

[eda123]

Not sure what you are trying to say with this post. Are you saying the experimental setup is flawed? What I want to test is the claim (or fact if you want), made by almost everyone in this thread that regardless of coil resistance, I will get the same liquid consumption when I use the same power. Or my claim that I will get the same liquid consumption when I use the same current.

I guess i missed the part where you said you were targetting the same wattage. I was probably getting confused with the posts where you were saying you were after targetting current, not wattage. Yeah, if you are going to target a certain wattage for each setup, i would think these end up very close. I suppose # of wraps and surface area of the coil will change things a bit, so this will be an interesting experiment!

 

[ckn71nm]

I guess i missed the part where you said you were targetting the same wattage. I was probably getting confused with the posts where you were saying you were after targetting current, not wattage. Yeah, if you are going to target a certain wattage for each setup, i would think these end up very close. I suppose # of wraps and surface area of the coil will change things a bit, so this will be an interesting experiment!

I agree with the number of wraps having an influence. I hope I will see a significant difference in liquid usage in one setup and a insignificant difference in the other. And even that might be very optimistic. But still interesting.

 

[Stosh]

If you do the experiment using the same wattage, the measured results are showing the efficiency of a particular wire gauge, coil shape or number of wraps. These variables will make a difference in vapor produced, liquid used, flavor, heat of the vapor, all good things to consider in vaping.

 

[ValHeli]

 

[sahsah]

Amperage is somewhat of a byproduct figure, useful for checking to see if your build is safe to run with a given battery, but it's a terrible way to judge performance or anything else.

Power, or wattage, is what really matters. Although even wattage figures aren't very useful on their own. Coil temperature depends on how much power you are pumping into it AND how much surface area the coil has. Coil temperature affects flavor, but you shouldn't use coil temp the adjust vapor and TH. Juice vaporizes best at a certain temperature, go hotter and you just burn juice, cooler and you aren't vaporizing properly.

Once that coil temp is reached, if you want to adjust vapor you adjust your surface area. But if you increase surface area, you need to increase power as well to maintain the same coil temp. So ideally, with high-power sub-ohm builds, your coil isn't getting any hotter, there is just far more coil there, at the same temp, so you get far more vapor. Like boiling two pots of water instead of one, you get far more water vapor, but the temp doesn't go up.

If you had a variable current mod, since you cannot adjust the resistance of the coil, the device is simply going to adjust voltage. So it's literally just a VV mod, nothing different, only difference is the units or letter put after the number on your settings screen.

In short, I'm afraid your looking down the wrong path. Current figures don't mean what you think they mean, and aren't useful to us much beyond checking safety.

I figure I'll throw in a real world example:

I currently have a 1.5 ohm coil build sitting on top of my Zmax, which I run with dual 18350s at 6v.

That's 24 Watts of power, drawing only 4 amps. It performs like you would imagine 24 watts would perform, tons of good vapor.

I also have a .5 ohm build on one of my mechs, lets use 4v as the voltage value for calculation:

That's 32 watts of power, drawing 8 amps. As you can see my amperage draw has DOUBLED, but the power has only gone up by about 25% I do get more vapor from this build, but that's because of the increase in power, not current.

Subohm builds perform well because they put out obscene amounts of power, not because of their current draw. Using a high voltage/resistance setup, you can get the same obscene power, but with a relatively tiny current draw. If current was the deciding factor in vapor, then high power, high voltage, high resistance setups wouldn't perform well, and they do. Power is what matters, but it's not the only thing.

Stated perfectly, bravo!

 

[sahsah]

So, both, power and current are higher. How do you know it's the power and not the current that gives you more vapor?



Same thing here, Both power and current go up. I think it's the increase in current that gives more vapor. Nothing in your post is a convincing argument that it's not.

You are right that high power/voltage high resistance setups perform well. And that's because:

4 Volt / 0.5 ohm = 8 amp
6 Volt / 0.75 ohm = 8 amp
12 Volt / 1.5 ohm = 8 amp

Obliviously 12 V is ludicrous. That's the reason why sub ohm will always have more vapor than 1.5 ohm. You can't get the current you need in a 1.5 ohm coil.

The thing is, if you take ohm and power/volt as your starting point, when power/volt goes up so will amp. When power/volt goes down, so will amp. I don't think there is a mathematical way to prove which one is responsible for the coil heat. Leaves current rated fuses in households to prevent wires from overheating.

Edit: From your post: ".. Using a high voltage/resistance setup, you can get the same obscene power, but with a relatively tiny current draw.. " No, you can't. High voltage combined with high resistance is high current draw. Unless you build something ridiculous like a 5 ohm coil you run at 10 volts. In that case you have 2 amp. But according to your line of thinking that' also only 20 watts.

What do you mean only 20 watts? That's much more than most people Vape at like ever. The dna20, a recent vaping achievement mind you, was such a huge deal because it allowed you to Vape all the way up to, you guessed it, 20 watts. But anyways, bro, it's about the wattage ie power, think about it, the heat which is determined by the power is what is making the juice change to vapor, not the current, the current is just affecting the electrical side of things and only factors into how high a strain the power cell you are utilizing can endure.

 

[edyle]

Its not an opinion in my case, what i post is fact. I have a B.S.E.E., I design circuits for a living and chances are you are using one of my patents right now as you type




You said "amps are what determines heat". No. thats half of it. Its current AND the resistance its travelling through. Power is the product of current * resistance. The same current can deliver more heat/wattage if you lower the resistance its travelling through.


For a FIXED LOAD, yes, changing current will vary your wattage in a 100% linear fashion. But if you change resistance, that same current is NOT going to give you the same power as you had with another coil with different resistance.

For your experiment, if you swap in a 2.1-2.5 ohm and dont adjust your voltage applied, you will be increasing resistance and therefore lowering the power dissipated by the coil. This is not opinion.

you got confused a minute there mate; should still be able to edit it.
p=vi and
lower --> increase

 

[gdeal]

You need to also consider watt density. ie: how power is distributed along the coil. For example a 28g A-1 round wire 1.5mm coil at .8ohm has will have 9 coils and a surface area of ~42mm sq. For comparison a 30g A-1 round wire 1.5mm coil at .8ohm will have 6 coils and a surface area of ~21.5mm sq.

So you have for this example ~2x the surface with 28g. If you have a wick that can deliver juice consistently to the coil surface there would be no difference, because temperature would stay at the heat of vaporization of your ejuice . But in reality, our wicks dry out at higher power with thinner wires because they can not keep up with the focused transfer of energy.

 

[edyle]

Not sure what you are trying to say with this post. Are you saying the experimental setup is flawed? What I want to test is the claim (or fact if you want), made by almost everyone in this thread that regardless of coil resistance, I will get the same liquid consumption when I use the same power. Or my claim that I will get the same liquid consumption when I use the same current.

Oh, yes that would be a good test; set power, use different coils, you'll get the approx the same consumption.

 

[edyle]

You are right, and I have thought about it.
Today: 1.6 ohm @ 10 watt = 2.5 amp
Tomorrow: 2.3 ohm @ 2.5 amp = 14 watt
Day after: 2.3 ohm @ 10 watt = 2 amp

I know that you guys don't want to sound patronizing, but I have been building coils for about a year. Sub-ohm and all above. I know about safety and I won't do anything unsafe (or impossible like 35 watt)

Edit: btw 1.6 ohm @ 6 watt = 1.9 amp not 3.75

Thanks; corrected; came out to under 10 watts actually! Thats workable; but's its gonna be a HOT VAPE!

 

[sahsah]

Take it one step further. Watts (power) = volts * amps. When working with a power supply with a constant voltage, if you double the amps (by changing the resistance of the coil) you double the watts. The problem is that when you change both the voltage and the amperage, amps stop having meaning in terms of defining power. Then you have to go to the formula to compare different combinations by using wattage.

Volts define electromotive force . Amps define the flow of electric charges through a conductor (wire in this case).

Watts defines the amount of work done. Higher voltage with lower current can do the same amount of work as lower voltage with higher current.

Exactly, think of it this way: volts are the calories you need to fuel your muscles to run a marathon, while amps are the time you burnt those calories in (volts) to create the force (watts) to push off the ground after each step in resisting gravity (resistance).
I just came up with that on the fly so if someone sees a comparison that doesn't fit please point it out, I won't take it personal none =] Vape on my cloud-taming brethren!

 

[sahsah]

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?

No the coil surface area changes and thus heat bc if I have one match at 100 degrees Fahrenheit that's two inches tall or a forty foot tall log at 100 degrees you can vaporize more juice with the log, hence more vapor more flavor more throat hit. So what have we learned from all this? We need to start vaping forty foot logs =]

 

[ckn71nm]

What do you mean only 20 watts? That's much more than most people Vape at like ever. The dna20, a recent vaping achievement mind you, was such a huge deal because it allowed you to Vape all the way up to, you guessed it, 20 watts. But anyways, bro, it's about the wattage ie power, think about it, the heat which is determined by the power is what is making the juice change to vapor, not the current, the current is just affecting the electrical side of things and only factors into how high a strain the power cell you are utilizing can endure.

If I remember correctly, I was referring to the fact that given the same resistance, If you increase the power you also increase the current and if you decrease the power you decrease the current. In contrast to the post where it was stated that you can have huge power and small current. Obviously that's possible, but go up in power you go up in current.

Edit: I'm aware of the power limitations current mods have. And btw, I want the Futura. 0.7 ohm limit, DNA30 board. Sub-ohm in a regulated PV. Sweet!!!

 

[WattWick]

If I remember correctly, I was referring to the fact that given the same resistance, If you increase the power you also increase the current and if you decrease the power you decrease the current. In contrast to the post where it was stated that you can have huge power and small current. Obviously that's possible, but go up in power you go up in current.

Edit: I'm aware of the power limitations current mods have. And btw, I want the Futura. 0.7 ohm limit, DNA30 board. Sub-ohm in a regulated PV. Sweet!!!

There is a major difference between an Ohms law calculator and an actual PV. In the calculator, we are free to put any number we want anywhere. This is not the case with a PV.

For sake of example, forget VW. It only adds to the confusion. We cannot really increase power or decrease current.

In the calculator, we have 4 values we can adjust at will: voltage, resistance, current and power.
In a PV, we only have two values we can adjust: Resistance and voltage. As you know, resistance is not re-adjustable. Once it's set, it's set. It's only adjustable in the way that it is us who decide what value it has by making a coil.

Forget power. Forget current. These are not for us to decide. We get what we get. The value for current is given to us through the relationship between resistance and voltage. We now have current. (yey!) The relationship between this current and voltage gives us the value for power.

So... for sake of making adjustments on the fly... it boils down to voltage. Current is based on voltage. Power is based on voltage and current. A VW mod is just a VV mod capable of doing math.

 

[ckn71nm]

... Forget power. Forget current. These are not for us to decide. We get what we get. The value for current is given to us through the relationship between resistance and voltage. We now have current. (yey!) The relationship between this current and voltage gives us the value for power.

So... for sake of making adjustments on the fly... it boils down to voltage. Current is based on voltage. Power is based on voltage and current. A VW mod is just a VV mod capable of doing math.

I actually agree with that. I know that I'm beating a dead horse and we are going in circles again. I interpret what you are saying as: The two intrinsic properties of our setups are the resistance of the coil and the voltage of the battery. We can't change that (once the coil is mounted) and I 100% agree with that, I never disputed that and was saying that from the beginning.

Current and power are calculated from those intrinsic properties. Like someone before (forgot who) pointed out power is happening at the level of input into the coil. Current is happening at the level of the coil itself. That sounds to me like, on the coil level, what in the end determines the heat is the current. Obviously we can adjust the power in order to change the current. But that's all it is. Power is the means to adjust the current in order to produce heat.

 

[eda123]

Exactly, think of it this way: volts are the calories you need to fuel your muscles to run a marathon, while amps are the time you burnt those calories in (volts) to create the force (watts) to push off the ground after each step in resisting gravity (resistance).
I just came up with that on the fly so if someone sees a comparison that doesn't fit please point it out, I won't take it personal none =] Vape on my cloud-taming brethren!

Nice.

Also think in terms of a stream going from the top of a hill to the bottom in a trench.

Voltage is analogous to how high the hill is from the bottom
Current is analogous to how fast the water is flowing
Resistance is analogous to how narrow the trench is

Here's a good web site: Ohm's Law - The basics - Ohms Law Formulas Explained to save you time

What is voltage? An anology would be a huge water tank filled with thousands of gallons of water high on a hill.
The difference between the pressure of water in the tank and the water that comes out of a pipe connected at the bottom leading to a faucet is determined by the size of the pipe and the size of the outlet of the faucet. This difference of pressure between the two can be thought of as potential Voltage.

What is current? An analogy would be the amount of flow determined by the pressure (voltage) of the water thru the pipes leading to a faucet. The term current refers to the quantity, volume or intensity of electrical flow, as opposed to voltage, which refers to the force or "pressure" causing the current flow.

What is resistance? An analogy would be the size of the water pipes and the size of the faucet. The larger the pipe and the faucet (less resistance), the more water thatcomes out! The smaller the pipe and faucet, (more resistance), the less water that comes out! This can be thought of as resistance to the flow of the water current.
All three of these: voltage, current and resistance directly interact in Ohm's law.
Change any two of them and you effect the third.

 

[eda123]

Current and power are calculated from those intrinsic properties. Like someone before (forgot who) pointed out power is happening at the level of input into the coil. Current is happening at the level of the coil itself.

Not sure what that means. current is present in every part of the circuit. Power is dissipated across any resistive element.

That sounds to me like, on the coil level, what in the end determines the heat is the current.

Current AND voltage determine power. P= V * I.

Obviously we can adjust the power in order to change the current. But that's all it is. Power is the means to adjust the current in order to produce heat.

Other way; voltage and resistance determine the current through a circuit. You apply a voltage to a resistor, and current is what results. You can change either of these two to increase current. If you are only changing voltage, then that is changing current as well.

With a FIXED resistance, higher voltage == higher current == higher power. But what you are adjusting is the voltage output. current and wattage are not the cause, they are the effect.