Could someone explain the relationship between wire gauge, number of wraps, etc and the resulting resistance (Ohm)?

[guruguhan]

Hi,

I'm wondering if someone could please explain the relationship between wire gauge, the number of wraps around the wick and whatever else contributes to the final resistance of a coil. Unfortunately, I dont have a multimeter/ohm meter with me (its coming in the mail but may take a while) so can't experiment myself.

Thanks

 

[custom-classic]

This might help: http://www.e-cigarette-forum.com/forum/reos-mods/446263-micro-coils-ohms-description-12.html

 

[opuscroakus]

The easy answer is that there are only 2 things that contribute to the resistance of the coil, the gauge of the wire and the overall length of the resistance wire between termination points. I say resistance wire because there are some devices that require non-resistance - resistance - non-resistance setups.

For our purposes, the thicker the wire the less resistance there is in the wire meaning more wire is needed to obtain the desired resistance of the coil. This means for a 1.5 ohm coil..

28 gauge Kanthal A-1 needs 3.25 inches of wire
30 gauge Kanthal A-1 needs 2.0625 inches of wire
32 gauge Kanthal A-1 needs 1.3125 inches of wire

From those lengths you would wrap your coil, whether it is nano, micro, big huge hole, whatever..

 

[Seebs]

It's rather simple really.

Resistance increases as distance increases. So the more wraps you do around the wick; the higher your resistance will be.

Resistance is also dictated by wire gauge (thickness) in the sense that electricity (electrical current) is the movement of electrons on the wire from point A to B... So thinner (higher Gauge) wire will have more resistance than thicker wire (lower Gauge) as it has less material (electrons).

Final resistance of a coil can be "sort of predicted" with a bit of practice and some math.

Here's a chart of the resistances of each of the Gauges of wire per "lenght".



With that chart and a bit of ingenuity one can predict what the resistance on any given coil would be. Just take your wick; you should know what diameter it is (or at least an approximation of it... When in doubt go with the wickhole diameter as that is known).
With that diamter; you can calculate the circumference.

Warning: Math below
Area of a Circle
A = πr²

Circumference of a Circle
C = 2πr or πd -> Where "d" is the diameter of the circle (wick hole diameter).

Now that you know the circumference of the wick hole; you then figure out how many wraps you plan on doing and multiply that number by said circumference. That gives you a length that you can then look up on the chart and approximate the expected resistance.

Example:
- Kanthal A1 - 28 Gauge Wire
- 2.5mm diameter wick hole.
- 5 wraps coil.

Circumference of the wick hole
Cw = π * 2.5mm => 7.85mm

Number of wraps * Cw
5 * 7.85mm = 39.27mm

Add 5mm on each end for connection to the posts on the atty.
Total lenght = 49.27mm (Round this up to 50mm)

We now have the length of wire that will be used. Look it up on the chart.
28 Gauge Kanthal A1 is 0.017Ω/mm
50mm * 0.017Ω/mm = 0.85Ω

And there you have it. A 5 wrap coil of 28 Gauge Kanthal A1, done around a 2.5mm wick should come out at 0.85Ω. Give or take a few tenths of an Ω.



Hope this helps.

Seebs

 

[dr g]

If you have an android device you could download the "vaper's toolbox"

 

[guruguhan]

Thanks a lot all of you, every answer in this thread was useful. From the general to more in depth...and that app looks good too!

 

[guruguhan]

It's rather simple really.

Resistance increases as distance increases. So the more wraps you do around the wick; the higher your resistance will be.

Resistance is also dictated by wire gauge (thickness) in the sense that electricity (electrical current) is the movement of electrons on the wire from point A to B... So thinner (higher Gauge) wire will have more resistance than thicker wire (lower Gauge) as it has less material (electrons).

Final resistance of a coil can be "sort of predicted" with a bit of practice and some math.

Here's a chart of the resistances of each of the Gauges of wire per "lenght".

View attachment 258701

With that chart and a bit of ingenuity one can predict what the resistance on any given coil would be. Just take your wick; you should know what diameter it is (or at least an approximation of it... When in doubt go with the wickhole diameter as that is known).
With that diamter; you can calculate the circumference.

Warning: Math below
Area of a Circle
A = πr²

Circumference of a Circle
C = 2πr or πd -> Where "d" is the diameter of the circle (wick hole diameter).

Now that you know the circumference of the wick hole; you then figure out how many wraps you plan on doing and multiply that number by said circumference. That gives you a length that you can then look up on the chart and approximate the expected resistance.

Example:
- Kanthal A1 - 28 Gauge Wire
- 2.5mm diameter wick hole.
- 5 wraps coil.

Circumference of the wick hole
Cw = π * 2.5mm => 7.85mm

Number of wraps * Cw
5 * 7.85mm = 39.27mm

Add 5mm on each end for connection to the posts on the atty.
Total lenght = 49.27mm (Round this up to 50mm)

We now have the length of wire that will be used. Look it up on the chart.
28 Gauge Kanthal A1 is 0.017Ω/mm
50mm * 0.017Ω/mm = 0.85Ω

And there you have it. A 5 wrap coil of 28 Gauge Kanthal A1, done around a 2.5mm wick should come out at 0.85Ω. Give or take a few tenths of an Ω.



Hope this helps.

Seebs

Awesome - thanks so much - just awesome!