xfmr out of roll of #12 wire

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Just wondering for curiosities sake. I saw a video of mike holt with two rolls of number 12 wires. the bottom roll was hooked to 120 volt line voltage and the top roll was just placed on top (not connected to the bottom roll) but connected to a light bulb. Through induction when turned on power and the bottoe roll was energized the light bulb on top would turn on and if you placed a screw driver through the middle it would magnetize. Just wondering in theory how you would wire that up.
 

ActionDave

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Most reels of THHN have the start of the reel sticking out of a hole near the centre of the reel and the finish under a few wraps of saran-wrap. Hook up a hot and a neutral to either and flip the switch.

If you stick a screwdriver through the hole in the middle of the reel it will jump out of your hand.
 

Sierrasparky

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Somebody is going to try this and cause a big mess.
I remember my chem teacher decided to show us how volital Sodium is. He had done this many a times. Well he droped a chunk in a beaker of water and ..... next thing there is a 1/2 hole in the ceiling and his eye is blood red. Luckily he did not do any permanent damage to his eye.
 
direct short

direct short

So how is that not a direct short, since its essentially putting a piece of wire strait from the hot to the neutral?
 

gar

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jeffrey3001:

Because that coil has an impedance.

Suppose you want the current to be less than 10 A at 120 V 60 Hz, then the impedance has to be greater than 12 ohms. 1000 ft of #14 copper wire is about 2.5 ohms. Thus, we need an inductive reactance of at least 11.7 ohms @ 60 Hz from the coiled wire. This is an inductance of about 0.03 H. I have never measured the inductance of such a coil. If the inductance is not at least this high, then the current would be greater than 10 A, and heating would be more.

If this coil is kept powered and a magnetized screwdriver is inserted and gradually removed, then the screwdriver will be gradually demagnetized as the tool is removed.

Use DC as excitation to the coil and while the tool is in the coil remove excitation to the coil and the tool will be left magnetized, assuming it is magnetizable. Same can be done with the AC coil, but it depends upon when and how excitation is removed.

.
 
thanks

thanks

Thank, not going to try it since I already saw it demonstrated, but just I think that makes since. So basically, if you cut a 4" piece of wire and shove it in there the resistance is so low you will have high amperage that will blow the circuit, but because a roll of wire has a higher impedance it will not draw a high enough current to cause a problem.
 

Besoeker

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Thank, not going to try it since I already saw it demonstrated, but just I think that makes since. So basically, if you cut a 4" piece of wire and shove it in there the resistance is so low you will have high amperage that will blow the circuit, but because a roll of wire has a higher impedance it will not draw a high enough current to cause a problem.
I think it might.
Roughly 12AWG is between 2.5mm^2 and 4mm^2.
Taking the smaller gives a resistance of about 18mohms per metre. It is usually sold in 100m reels. So a reel would be under 2 ohms. With 120V applied that would be 60A. The best continuous rating is about 20A. But much less than that with all the conductors bunched on a reel.
Yes, I know that it ignores the inductance of the coil.
If anyone can come up with a figure for that I'd be happy to review the calculations.
 

realolman

Senior Member
I am not quite so stupid as to argue with gar, but it seems to me that the coil should have some kind of proper "geometry" to create sufficient impedance.

I understand that no one is suggesting that a couple rolls of wire piled on each other is a good transformer, but I think the largest part of the impedance should not be from the resistance of the wire, but from the impedance caused by the winding of the conductor into an inductor.
 
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mivey

Senior Member
I think it might.
Roughly 12AWG is between 2.5mm^2 and 4mm^2.
Taking the smaller gives a resistance of about 18mohms per metre. It is usually sold in 100m reels. So a reel would be under 2 ohms. With 120V applied that would be 60A. The best continuous rating is about 20A. But much less than that with all the conductors bunched on a reel.
Yes, I know that it ignores the inductance of the coil.
If anyone can come up with a figure for that I'd be happy to review the calculations.

How about a rough calculation for a 500 ft roll of #12 CU? Obviously, this is the result of a simplified calculation but:
2" core
4.5" length
0.128" O.D.
15 layers with a 0.866 packing factor
508 ft length

R = 1.75 ohms/kft = 0.889 ohms
X = 20.856 mH = 7.862 ohms at 60 Hz

yielding 15.2 amps at 120 volts
 

hurk27

Senior Member
I am not quite so stupid as to argue with gar, but it seems to me that the coil should have some kind of proper "geometry" to create sufficient impedance.

I understand that no one is suggesting that a couple rolls of wire piled on each other is a good transformer, but I think the largest part of the impedance should not be from the resistance of the wire, but from the impedance caused by the winding of the conductor into an inductor.

Today most rolls of wire come on plastic cores, but in the not to far past they came on metal cores.

To reach the low Henrie's for the impedance required I would be willing to bet that there was some kind of metal core between the two rolls of wire.

air core transformers are more common at higher frequencies but at 60hz it would require many thousands of turns to get a high enough impedance or low enough Henrie's without a ferrous metal core.

There seems to be some missing info in the OP.
 

George Stolz

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Wayne, I've seen the video. It is just two rolls of THHN and a screwdriver. It is a sloppy transformer, and only held for a couple seconds before tripping the breaker. The impedance of the windings is the key, if the same amount of conductor were stretched straight then it would have tripped the breaker instantly.
 

hurk27

Senior Member
Wayne, I've seen the video. It is just two rolls of THHN and a screwdriver. It is a sloppy transformer, and only held for a couple seconds before tripping the breaker. The impedance of the windings is the key, if the same amount of conductor were stretched straight then it would have tripped the breaker instantly.

well the screw driver would explain the metal core? but after looking at Mivey's math I think I'm a little rusty in figuring open air core transformers:ashamed1::lol:
 

gar

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The following site provides an approximate formula for an air core inductor. http://microblog.routed.net/wp-content/uploads/2008/10/pancakewheel.pdf

See multi-layer alternate-expression. Using this formula and values from mivey's calculation I get a lower value, but in the ball park.

Experimentally I have a 500 ft roll of #16 with OD over insulation of 0.104" that measures about 9 mH. This wire is on a spool 3.6" wide and an estimated core of 2" diameter.

How good is the coupling between a primary and secondary of a transformer? This is a function of the magnetic flux that couples both coils. It doesn't much matter whether it is 60 Hz or a 1 MHz. In an air core transformer the geometry of the two coils will be a big factor in the coupling. If you add a high permeability magnetic core between the coils, then geometry is less important.

.
 

kwired

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Location
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So how is that not a direct short, since its essentially putting a piece of wire strait from the hot to the neutral?

Thank, not going to try it since I already saw it demonstrated, but just I think that makes since. So basically, if you cut a 4" piece of wire and shove it in there the resistance is so low you will have high amperage that will blow the circuit, but because a roll of wire has a higher impedance it will not draw a high enough current to cause a problem.

I am not quite so stupid as to argue with gar, but it seems to me that the coil should have some kind of proper "geometry" to create sufficient impedance.

I understand that no one is suggesting that a couple rolls of wire piled on each other is a good transformer, but I think the largest part of the impedance should not be from the resistance of the wire, but from the impedance caused by the winding of the conductor into an inductor.

This is not necessarily a well designed transformer, but the main thing is to remember the screwdriver in the center of the reel is important component of the transformer. Without the steel core the impedance of the coil will not be too significantly different than just a straight run of the same length of conducgtor.

Put any piece of steel in there for a core and you have created an inductor, place a second coil on the same core and you have a transformer.
 

gar

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Try doing calculations on the following problems:

Calculate the inductance of the following round coils. All are #12 solid copper wire 500 ft long.

1. One turn coil.
2. Two turn coil.
3. Four turn coil.
4. 100 turn 1 layer coil. Use 0.1" for the OD of the insulated wire.

5. A single rectangular loop almost 250 ft long with the two parallel wires spaced 1 ft apart.

.
 

mivey

Senior Member
See multi-layer alternate-expression. Using this formula and values from mivey's calculation I get a lower value, but in the ball park.
That Wheeler formula was indeed the one I used. Our differences should just be some assumptions about the winding itself.

I had:
radius = 2.616"
turns = 527.34
length = 4.5"
depth = 1.680"

I may have fat-fingered something but this is what I had for the 15th layer:
base diameter (14th layer) = 5.1037"
wire center diameter = 5.2317"
 

gar

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mivey:

I guessed at the values you might have used.

Mean radius 2"
Approximated mean circumference at 1 ft
Approximated number of turns at 500
Used 4.5" for coil length
Difference between r1 and r2 = 2"

From this I got 6.4 mH.

My post on the calculated inductance of a single straight wire 500 ft long was to show kwired the substantial difference between that and the same length coiled.

The problems I have suggested for the 1, 2, and 4 turn coils using 500 ft of wire are to illustrate how the inductance changes with the geometry.

The long hairpin loop is to give a different perspective.

I think someone other than you and I should do these calculations as a useful exercise.

For those unfamiliar with how inductance changes with turns note that in the equations how N2 appears as a significant factor.

.
 

mivey

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mivey:

I guessed at the values you might have used.

Mean radius 2"
Approximated mean circumference at 1 ft
Approximated number of turns at 500
Used 4.5" for coil length
Difference between r1 and r2 = 2"

From this I got 6.4 mH.

For a 2" radius, 500 turns, 4.5" length, 2" depth I calculate 11.03 mH

I used a 2" spool core. It appears you used a 1" spool core (2" radius - (r1-r2)/2 = 2" - 1" = 1") but I don't think they are normally that small for that size spool.
 

mivey

Senior Member
For a 2" radius, 500 turns, 4.5" length, 2" depth I calculate 11.03 mH
I used a 2" spool core. It appears you used a 1" spool core (2" radius - (r1-r2)/2 = 2" - 1" = 1") but I don't think they are normally that small for that size spool.
Bad math on my part. You had a 1" radius for a 2" dia core: same as mine. Not sure how you got 6 mH.
0.8*(Radius^2 * Turns^2) / (6*Radius + 9*Length + 10*Depth) =
0.8*(2"^2 * 500^2) / (6*2" + 9*4.5" + 10*2") =
0.8*(4*250000) / (12 + 40.5 + 20) =
0.8E6 / 72.5 = 11034.5 uH = 11.03 mH

For mine:
0.8*(Radius^2 * Turns^2) / (6*Radius + 9*Length + 10*Depth) =
0.8*(2.61587"^2 * 527.344^2) / (6*2.61587" + 9*4.5" + 10*1.67987") =
0.8*(6.8428*278092) / (15.69522 + 40.5 + 16.7987) =
1.52234E6 / 72.9939 = 20856 uH = 20.86 mH
 
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