Coil Question
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0.5Ω (assuming they are wired in parallel).
ETA: Just so you know, when two coils are wired in parallel the resistance is half of one coil, but the amps don't change. The draw on the battery will be that of a 1.0Ω coil twice. Not an issue with your set-up, but it something people should keep in mind when they go really low res with multiple coils.
ETA: Just so you know, when two coils are wired in parallel the resistance is half of one coil, but the amps don't change. The draw on the battery will be that of a 1.0Ω coil twice. Not an issue with your set-up, but it something people should keep in mind when they go really low res with multiple coils.
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So if the coils are wrapped together than it is half. But what if you have a three post RDA, and you put the coils on opposite sides. Is that going to be 1 ohms?
And how do you calculate parallel and non-parallel double coils in the calculator?
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"Parallel" and "Series" (non-parallel) doesn't have to do with how the coils are placed on the deck, it's about how they are wired.
Parallel
Series:
You can just call them dual coils, since you will be wiring them in parallel.
I'm sorry this is long.
Dual Coils: info from tritrash13
[Note: the amp limit of a ProVari v2.5 is actually 3.5, but it doesn’t change his point.]
"...For the sake of this discussion we will use the online Ohm's Law Calculator.
With a parallel dual coil setup (both coils connected directly to the positive and negative post(s) on the atomizer) you divide the ohms of one coil in half. So for example, two 3 ohm coils wouldn't be 6 ohms it would be 1.5 ohms. This is because the power coming to the atomizer is split with half of the power going to each coil. Thus, the results you get from the calculator don't apply to dual coils. Moreover, since the power is split between two atomizers, this means that the resistance is cut in half. The amps, however, would be the amps required to power a 3 ohm coil twice. Now let's assume we set our variable wattage device to 15 watts. So if we enter 3 ohms (the resistance of each individual coil) and 7.5 watts (the amount of power going to each coil) we see that we need 1.58 amps of current to power each coil for a total of 3.16 amps. Therefore, a device with a 3 amp limit would not be able to fully power a 1.5 ohm dual coil atomizer in a 15 watt setting. This equation also gives us the number 4.7 for volts. This is the voltage that a device would be putting out to power a 1.5 ohm dual coil atomizer at 15 watts. This is the clue to tell us how to reverse engineer our voltage numbers for those who don't have a variable watts device.
Using our 1.5 ohm dual coil atomizer, let's figure out what the maximum voltage is that we can vape at on a 3 amp device. Take the overall resistance of each individual coil (3 ohms) and enter that into the ohms section of the calculator. Now, enter the amp limit of your device in the amps section. For the sake of this experiment, we will use 3 amps as this is the limit on the ProVari 2.5 but we will divide the amps in half since that is what the atomizer is going to do with the current when it gets there. So we will enter 3 into the resistance field and 1.5 into the current field. This calculation tells us that we can vape a 1.5 ohm dual coil atomizer at a maximum setting of 4.5 volts on our ProVari.
Now, let's assume you have a 2.5 ohm dual coil and you are going to use it on your provari. By entering 5 (the resistance of each coil) in the ohms and 1.5 (the current traveling to each coil) we see that we would be able to vape this at 7.5 volts without exceeding our amp limit. The device, however, only goes up to 6 volts. So let's figure out the wattage we are vaping at in a setting of 5.5 volts. Enter 5.5 (your device setting) into the volts field and 5 (the resistance of each coil) into the resistance field. This calculation gives us 1.1 amps and 6.05 watts. These numbers need to be doubled, however, as this is a measurement of the current traveling to and the wattage output of each individual coil. Therefore, in a 5.5 volt setting, our ProVari is firing a 2.5 ohm dual coil atomizer at 12.1 watts and using 2.2 amps of current.
You can also use this calculator to help you determine if the device you are using is appropriate for the battery you are placing inside it. Let's say you intend to use a .5 ohm dual coil atomizer on a 4.2 volt battery. How many amps does this battery need to be capable of putting out to make this device fire? Enter 4.2 (the voltage of your fully charged battery) into the voltage field and 1 (the resistance in ohms of each individual coil) into the resistance field. This calculation tells us that we need 4.2 amps of current to power each coil, requiring a total of 8.4 amps of current for this device. I will now pause for you to remove that 18350 battery from your device.
"ETA: There are some better calculators online now, but I will leave it to someone else to provide the links.
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if 2 or more coils share the same positive source and the same negative source, the coils are in parallel....
the ohms drop for each coil added... math 1/((1/R1)+(1/R2)+......)
the ideal situation is to have coils of identical resistance ( makes the math easier)... example 2 coils in parallel each at 1 ohm equates to a load of 1/2 ohm, if it was 3 coils at 1 ohm each, the load would be 1/3 ohm....
the ohms drop for each coil added... math 1/((1/R1)+(1/R2)+......)
the ideal situation is to have coils of identical resistance ( makes the math easier)... example 2 coils in parallel each at 1 ohm equates to a load of 1/2 ohm, if it was 3 coils at 1 ohm each, the load would be 1/3 ohm....
After reading your 2 posts to this thread I suggest you do some Googling on Parallel and Series circuit wiring and for a reference to Ohms Law.
What you are doing can be dangerous to you and people around you if you do not understand how a parallel and series circuits work, how vaping at low ohms (Sub-Ohm) can change the amps drawn on the battery, Ohms Law, and how to be safe.
You really need to understand this stuff to be safe. Once you know it you can then help others on this forum or in real life to understand it.
Your screen name says it all DANGER.
Thank you for posting that! I think I got it. If two separate coils share the positive post it is going to be in parallel.
So when calculating a VW device for more than one coil...
Resistance = ohm of a single coil
Watt = divided by number of coils
Amp = multiply by number of coils
So running a dual coil at .5 ohms (two 1 ohm coils) and 40 watts would need 4.4 volts and 8.8 amps. This would be safe in amps on a Sony 18650, but not safe in volts.
But a .25 dual coil would be safe at 3.2 volts and 12.7 amps on a Sony 18650
On an unregulated device for more than one coil...
Volts = max volts
Resistance = ohm of a single coil
Current = multiply by number of coils
A parallel battery is double the amps and mah
A series battery is double the volts
More resistance = more volts
More watts = more amps
Is all of that right?
So when calculating a VW device for more than one coil...
Resistance = ohm of a single coil
Watt = divided by number of coils
Amp = multiply by number of coils
So running a dual coil at .5 ohms (two 1 ohm coils) and 40 watts would need 4.4 volts and 8.8 amps. This would be safe in amps on a Sony 18650, but not safe in volts.
But a .25 dual coil would be safe at 3.2 volts and 12.7 amps on a Sony 18650
On an unregulated device for more than one coil...
Volts = max volts
Resistance = ohm of a single coil
Current = multiply by number of coils
A parallel battery is double the amps and mah
A series battery is double the volts
More resistance = more volts
More watts = more amps
Is all of that right?
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Build a coil, measure resistance. If you build that same coil again(dual coil) half that resistance, and so on. Whatever your final resistance is, is what you use for ohm's law equation, or any equation
The other posts say each coil will require enough amp as if it was a single coil. So 4 single coils at 1 ohm would need enough amps for 1 ohm x 4 coils.
Ya, like I said final resistance(depending on amount of coils) your amps are directly related to your final resistance(dual, single, and so on)
look up ohm's law, lots of apps out there. Your final resistance is what you use as your equation
There's a slightly cumbersome fractional equation for parallel resistances - but you only need to get into that for combining one or more unequal resistances.
2 equal resistances in parallel halves the value, 3 equal resistance in parallel = 1/3........... and so on.
#1) Forget about series at this point. Parallel is what you will be wiring and the series calculations will likely lead to confusion.
I'm the type of person who gets one way of thinking stuck in my head and that's how I calculate things. While I have variable wattage devices and may use them in that mode, I learned to set power levels using variable voltage, so that's what I think in terms of. I also vape by taste, so I don't aim for a number in watts, I set the power at the low end of an estimated range and slowly dial up until it tastes good.
With a dual coil of 0.5Ω resistance (two 1.0Ω coils) the resistance would be 1/2 of one coil, the amps would what what each coil would draw. With a VW/VV device, the other two factors would depend on what you set the device at.
Using VW if you set the device at 40 watts and it will adjust the volts to produce 40 watts, as long as the required voltage and amps are within the limits of the device.
If you want a specific wattage and want to know what the voltage is, you either have to set the device at the watts you've chosen and read the screen to see what the device has set the voltage at or (without using a device) decide what the wattage will be and calculate what the volts will be (the square root of watts x resistance).
In either case, you would have to calculate the amps based on the other two components (unless there is a VW device out there that shows on the screen how many amps are being used and I don't know of one. There isn't a device that allows you to set the amps).
Look at the Power Wheel. It is a combination of Ohm's Law and Joule's Law that illustrates how to calculate the components:
P = watts (power)
V = volts
I = amps
R = ohms (resistance)
If you notice, you can't calculate the amount of amps being drawn unless you know already know what the wattage and voltage are. There is no way to back into the equation. Amps (I) will always be X (unknown) without the other two, because by itself the only number you have for amps is the battery's potential (limit), not what it will draw.
So while this makes sense:
This doesn't (at least to me) without any values:
Watts has nothing to do with the number of coils. Look at the Power Wheel. Watts are "voltage times voltage divided by total resistance" or "amps times voltage" or "(amps times amps) x resistance."
Using an unregulated device the volts, at least for a fresh battery, are known: ≈ 4.2v.
With a 0.5Ω dual coil (two 1.0Ω coils) the calculation is:
4.2v2/0.5Ω = 35.28 watts
35.28w/4.2v = 8.4 amps
With a 1.0Ω single coil:
4.2v2/1.0Ω = 17.64 watts
17.64w/4.2v = 4.2 amps
If you were to use a 0.25Ω dual coil (two 0.5Ω coils) in an unregulated device:
4.2v2/0.25Ω = 70.56 watts
70.56w/4.2v = 16.8 amps
But with a variable device set a 3.2v:
3.2v2/0.25Ω = 40.96 watts
40.96w/3.2v = 12.8 amps
With the same voltage: the lower the total resistance of the atty, the higher the wattage and amps will be; the higher the resistance of the atty, the lower the wattage and amps will be.
It looks a bit different when you are aiming for the same watts using different voltage and resistance (the example is for a single coil):
2.2Ω atty @ 4.5 volts
4.5v²/2.2Ω = 9.2w
9.2w/4.5v = 2.0 amps
1.5Ω atty @ 3.7 volts
3.7v²/1.5Ω = 9.1w
9.1w/3.7v = 2.5 amps
Even though a 2.2Ω atty at 4.5v and a 1.5Ω atty at 3.7v are producing the same amount of watts (+/- 9), when vaping low-res the amps increase, which draws more from the battery. That's why vaping at higher resistance uses less of the battery charge, increasing your vape time (mAh).
See? Clear as mud.
rawr
ETA: It took me too long to write this (I should have had coffee first). It appears everyone else has answered your question.
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At this point it is wise to suggest you get an Ohm reader for your atty. Do several builds and measure each one as you do them and don't fire any of the builds on a mod. Note the amount of wraps of wire, pin size you are wrapping on, ohm reading, etc. This will help you understand as well as using the calculator. It will also help others on the forum when you have questions about changing the diameter of your coils.
I bought a cheap ciga-like a month ago... now I am learning ohms law... ... lol.
Thanks for your help.
The one thing I still do not get is...
"since the power is split between two atomizers, this means that the resistance is cut in half. The amps, however, would be the amps required to power a 3 ohm coil twice"
So with a 0.5Ω dual coil (two 1.0Ω coils) You would need to find the amps for a 1.0Ω coil twice. So why would you use 0.5Ω in your calculation? Would you not need to use 1Ω in the calculation and then multiply the amps?
Will a DNA 40 do most of this work for me?
But I get it for the most part.... Volts squared / ohms = watts .... Watts/Volts = amps
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This quote, "since the power is split between two atomizers, this means that the resistance is cut in half. The amps, however, would be the amps required to power a 3 ohm coil twice" wasn't written by me. The point he was trying to get across was the concept, as opposed to the calculation.
If the total resistance of your atty is 0.5Ω, that would be what you would use in an equation.
Yes, your DNA40 will do this for you, but it's nice to know what it's doing.
Little did you know that quitting smoking involves learning about electronics, chemistry, biology, metallurgy, and a host of others things. Pay attention, there will be a pop quiz later.
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