Explain Dual Coils and Ohm's Law

[Myk]

I found one very old post that promised to explain in understandable terms this as it works with duals but didn't.

I know two 3Ω coils in parallel reads as a 1.5Ω atty and each coil heats like a 3Ω. I know voltage stays consistent through the circuit.

What I can't find is what goes on with watts to predict what I'll like dripping a certain liquid on a DC (it's seems as though watts may get divided so if I liked one 3Ω coil with a certain liquid at 5w I'll like two 3Ω coils at 10w, but that could be because of the time a lower wattage takes to heat two coils up).

What happens with amps?
What really confuses me about the amps is any way I look at it I see a free lunch with the massive amount of heating area that two 3Ω coils provide but read 1.5Ω. The only payment for lunch I can see right off is it takes longer to heat so there's more firing time. Is that the only cost?

 

[Penn]

I'm actually responding just to keep this on the first page. I will differentiate between what I know and what I am thinking but not sure of.

I know if it is a parallel circuit, you get the same voltage across all components. You would then calculate the current across all components and add those together to get total circuit current.

What I'm not sure of is the wattage but I think each individual coil gets the same wattage as the calculation you see people quoting here. If I remember correct the calculation for circuit watts uses the circuit current and the sum of resistance but I don't remember the formula at all OR the calculation is the same as current (meaning just add all the components). I'm too busy right now to look it up.

ETA: The reason I want to see this answered is for a different question. If a voltage regulation circuit is put in place before the coil, is the amp and watt calculation for the coil based on the output of the regulator voltage only or does the current of the regulator output make a difference?

 

[Myk]

I'm actually responding just to keep this on the first page. I will differentiate between what I know and what I am thinking but not sure of.

I know if it is a parallel circuit, you get the same voltage across all components. You would then calculate the current across all components and add those together to get total circuit current.

What I'm not sure of is the wattage but I think each individual coil gets the same wattage as the calculation you see people quoting here. If I remember correct the calculation for circuit watts uses the circuit current and the sum of resistance but I don't remember the formula at all OR the calculation is the same as current (meaning just add all the components). I'm too busy right now to look it up.

ETA: The reason I want to see this answered is for a different question. If a voltage regulation circuit is put in place before the coil, is the amp and watt calculation for the coil based on the output of the regulator voltage only or does the current of the regulator output make a difference?

You gave me something to think about. If each 3Ω coil is seeing the same Xvolts that equals the wattage going across each coil. I don't think the wattage needs increased compared to what I like with 3Ω but rather the 1.5Ω the atty reads because it's actually powering 3Ω, which I assume would be roughly doubled.

 

[AttyPops]

Take a 1.5 ohm dual coil carto as an example. That's like having two 3.0 ohm atomizers "double barreled" on a shared 510 connector.

So, of course the 1.5 ohms draws more amps than one single 3.0 ohm coil would for a particular voltage. But from the battery's perspective ....ohms are ohms. It doesn't know if there are 1, 2, or 20 coils. It's a 1.5 ohms "load".

So...the question is....how is that different than a 1.5 ohm single coil atomizer? And the answer is that the battery can't tell the difference. However, you can, because two 3.0 ohm coils would benefit from higher voltage. Think about it...what voltage would you vape a single 3.0 ohm coil at as compared to a 1.5 ohm LR? So you turn the voltage up. That also increases the amps.

So you can end up vaping a 1.5 ohm dual coil at 4.8 volts = 3.2 amps (if you PV supports it. It may just object!).

 

[Cool_Breeze]

With half the total resistance and the same voltage, the current doubles. At twice the current for a given voltage, the power doubles and your battery life is cut in half.

 

[Cool_Breeze]

It you want all the sorted details...

Feel free to copy the image and save it to your computer (right-click on the image / select Save picture as...) The saved image will be a little larger than it appears here.

 

[AttyPops]

Well, there is no "free" lunch. However you can get a bit of a boost if you're using a mod made for stacked batteries and bucking (reducing) the voltage. This isn't for the normal booster mods though, since they draw more amps to boost the voltage. However, stacked batteries (in a mod that's safe to use them in...Not suggested in all-mech mods, IMHO) don't trade amps for volts. So the voltage is squared in the watts calculation. This is also why we have "high voltage" lines.

Think about it....You vape a 1.5 ohm single coil at 3.2 volts. Do you vape a 3.0 ohm single coil at 6.4 volts? No! You vape it at 4.8 or 5.0 volts. This is because of the squaring of voltage in the Watts = volts²/ohms calc.

So given how you actually use the stuff, it isn't quite the same for amps because the voltage used is different. Make sense? Another thing you have to consider, of course, is that for the "booster" mods...like most VV and VW mods...they draw more amps to "bump" the voltage above battery voltage with special circuits. And how many amps they need depends on efficiency of the circuit. So there's still no free lunch there.

The benefit of these booster mods is that they are regulated to a consistent voltage. The mech guys have to contend with following battery voltage and they are always swapping batteries. (well, most of em do). lol Sorry, I digress. I've just seen too many posts with "I change my batteries at 3.7 volts".

And before you go and say for certain that stacked batteries are more efficient, you have to know the efficiency of the regulator that's bucking the voltage.

Anyway, theoretically, the stacked (bucked voltage) amps work out like this for a perfectly efficient fictional regulator:
1.5 ohms at 3.4 volts = 2.26667 amps and 7.70667 watts
2x3.0 ohms (a 1.5 ohm DCC) at 4.8 volts = 3.2 amps and 15.36 watts = 7.68 watts x 2 coils

You can see amps don't double in real world use although the watts are close. I chose 4.8v because that's a common top-end range. YMMV depending on preference. So use an ohms law calc.

 

[Steam Turbine]

I don't get why people want more than one coil. But that's just me... I might get it one day

 

[Penn]

I don't get why people want more than one coil. But that's just me... I might get it one day

Wether a person would or wouldn't want to is what is being discussed. Care to elaborate in context of electronic principles?

 

[Penn]

It you want all the sorted details...

Feel free to copy the image and save it to your computer (right-click on the image / select Save picture as...) The saved image will be a little larger than it appears here.

That only addresses effect on one resistance component. How does a more complex circuit affect the calculations? Such as do you just add the watts of 2 resistance components in a parallel circuit to get the full circuit load as with current?

ETA - You aren't always going to get 2 coils with the exact same ohms, this must be considered also.

 

[AttyPops]

That only addresses effect on one resistance component. How does a more complex circuit affect the calculations? Such as do you just add the watts of 2 resistance components in a parallel circuit to get the full circuit load as with current?

ETA - You aren't always going to get 2 coils with the exact same ohms, this must be considered also.

Think of the coil(s)/atty/RBA/Whatever as the single black-box load. It could have 20 coils in it, or one coil, or an electric motor. Ohms are ohms. You could put a meter on it and not even know what's in it and test the ohms.

The real question is...."what voltage works for 3.0 ohm coils?". It varies by preference and juice too. But....see above post #7. I added an example with full #'s and noted that efficiency matters too.

 

[Penn]

Atty, I guess you didn't understand why I was asking. In my initial response to Myk, I told him what I thought and was trying to get the correct answer even though I knew it had been established that it wasn't really relevant. I just finally had time to look it up and I was right, you would add the watts across each component, just like in a series circuit.

 

[AttyPops]

Sorry. Thought it was a general question. I see now.

ETA: The reason I want to see this answered is for a different question. If a voltage regulation circuit is put in place before the coil, is the amp and watt calculation for the coil based on the output of the regulator voltage only or does the current of the regulator output make a difference?

I *think* the answer to your edit on post #2 (quoted above) is....

If you know any two variables, you know them all. So if you know amps and volts, you know watts and ohms. Etc.

Assuming it's a voltage regulator, and assuming that your question assumes a 100% efficient hypothetical regulator (there are none) and assuming that it puts out at least enough amps to power the coil....

Calculate the amp draw from the output side of the regulator based on the coil ohms and output voltage. If amps are insufficient, the voltage will suffer and the coil will not "see" proper voltage.

Of course, no regulator is 100% efficient that I know of. So you have to account for that loss in the battery draw (input side of the regulator).

Best guess. And use a meter to test. Theory is theory. Reality is reality.

 

[Cool_Breeze]

That only addresses effect on one resistance component. How does a more complex circuit affect the calculations? Such as do you just add the watts of 2 resistance components in a parallel circuit to get the full circuit load as with current?

ETA - You aren't always going to get 2 coils with the exact same ohms, this must be considered also.

In effect, it is only the 'overall resistance of the device' that concerns the power supply. To the source, it is one load. I believe the resistance of dual coil devices as sold is given as the overall resistance.

However, for calculating the overall resistance to two coils of differing resistance in parallel...

1/ (1/x + 1/y) = z Ohms
In the case of a 3 Ohm resistance and a 2.5 Ohm resistance...
1/ (1/3 + 1/2.5) = z Ohms
1/ (.333 + .4) = 1 / .733 = 1.363 Ohms overall resistance



You've asked of Wattage...

Given: 4 Volt supply, One coil of 3 Ohm and 1 of 2.5 Ohms.

P = E^2/R
Coil 1 = 4 volts x 4 volts / 3 Ohms = 5.33 Watts
Coil 2 = 4 volts x 4 volts / 2.5 Ohms = 6.4 Watts
Total Wattage = 5.33 Watts + 6.4 Watts = 11.73 Watts

Or, as shown above for adding two unequal resistances...
Atty/Carto Overall Resistance = 1.363 Ohms
4 Volts x 4 Volts / 1.363 Ohms = 11.73 Ohms

The math indicates the power consumed is the same whether for a single coil of 1.363 Ohms or two coils, one at 3 Ohms, one at 2.5 Ohms.

 

[Topwater Elvis]

No matter how complicated a person wants to make it sound, 2 coils in parallel share/split the power being supplied.

 

[Myk]

Well, there is no "free" lunch. However you can get a bit of a boost if you're using a mod made for stacked batteries and bucking (reducing) the voltage. This isn't for the normal booster mods though, since they draw more amps to boost the voltage. However, stacked batteries (in a mod that's safe to use them in...Not suggested in all-mech mods, IMHO) don't trade amps for volts. So the voltage is squared in the watts calculation. This is also why we have "high voltage" lines.

Think about it....You vape a 1.5 ohm single coil at 3.2 volts. Do you vape a 3.0 ohm single coil at 6.4 volts? No! You vape it at 4.8 or 5.0 volts. This is because of the squaring of voltage in the Watts = volts²/ohms calc.

So given how you actually use the stuff, it isn't quite the same for amps because the voltage used is different. Make sense? Another thing you have to consider, of course, is that for the "booster" mods...like most VV and VW mods...they draw more amps to "bump" the voltage above battery voltage with special circuits. And how many amps they need depends on efficiency of the circuit. So there's still no free lunch there.

The benefit of these booster mods is that they are regulated to a consistent voltage. The mech guys have to contend with following battery voltage and they are always swapping batteries. (well, most of em do). lol Sorry, I digress. I've just seen too many posts with "I change my batteries at 3.7 volts".

And before you go and say for certain that stacked batteries are more efficient, you have to know the efficiency of the regulator that's bucking the voltage.

Anyway, theoretically, the stacked (bucked voltage) amps work out like this for a perfectly efficient fictional regulator:
1.5 ohms at 3.4 volts = 2.26667 amps and 7.70667 watts
2x3.0 ohms (a 1.5 ohm DCC) at 4.8 volts = 3.2 amps and 15.36 watts = 7.68 watts x 2 coils

You can see amps don't double in real world use although the watts are close. I chose 4.8v because that's a common top-end range. YMMV depending on preference. So use an ohms law calc.

I think this gets it to sink in, although I'll have to read it again tomorrow.

Knowing what the letters mean in the equations I was looking at trying to figure this out helps too.

I don't get why people want more than one coil. But that's just me... I might get it one day

I'm trying it because I can. I don't know if I particularly like it over a single coil in this RDA because I haven't tried single in it. My other RDA probably won't see duals because I really like it as a single. If I do decide I like dual better than single I want to know what is going on.