Howto make a 200/300 volt coil??

[ray0energy]

Hello to all
I am going to build a magnet motor
Whit coils that need to give me abaut
200/300 volt my core is ferrit of 2 in diameter
by 2 in long
And I want to use 0.800mm coil wire.

this is my 1e time that I make this kind of

coils
so I don't no how many turns I need to get 200/300 volt.

Can anybody help me on this

And like to no also how many turns I need if I do this in Teslas befilare coil.

[EgmQC]

Hello to all
I am going to build a magnet motor
Whit coils that need to give me abaut
200/300 volt my core is ferrit of 2 in diameter
by 2 in long
And I want to use 0.800mm coil wire.

this is my 1e time that I make this kind of

coils
so I don't no how many turns I need to get 200/300 volt.

Can anybody help me on this

And like to no also how many turns I need if I do this in Teslas befilare coil.

Its the ratio in number of turn who will tell you if its a step up or down or isolation coil. if you have 10 turn primary and 100 turn secondary with 12v sine wave input , the output will be 120v. For the energy output you need to know at which frequency you will work to calculate accurately the impedance. There alot of information about transformer on the net with all the math for it.

Best Regards,
EgmQC

[Sephiroth]

do you mean for a generator coil...

Couldn't tell you the precise spec you would need... that depends on too many factors, like the stength of your magnets, their speed, their size, but I don't think you will be able to get that kind of voltage with 0.8mm wire... I would go for something much finer....

[tjnlsn255]

L = (d2n2) / (1 + 0.45d)

Inductance of a single layer coil...so...

if d is the diameter of the coil and n is the number of turns ...

If I take a coil with 100 turns with 20 guage wire and replace the wire with 18 guage wire the 18 guage wire is fatter so there will be a larger diameter but the number of turns will be less.... so...

If the toroid is the same and we increase wire size or decrease wire size what happens to the inductance in Henrys.....?

Hopes and Dreams....

Tj

[ray0energy]

Generator coil

do you mean for a generator coil...

Couldn't tell you the precise spec you would need... that depends on too many factors, like the stength of your magnets, their speed, their size, but I don't think you will be able to get that kind of voltage with 0.8mm wire... I would go for something much finer....

Well I order magnets for a Muller generator of 2ins wide by 1ins long
de have about 250 pounds pull power.

So the plan is to have like 30 2 times 15 coil that will
go to a diode to make DC of it that will go in to a big cap bank
of 200/450 vdc.
so I will not use direct the HF AC comming from the coils
I find a way to make from DC 200/450 caps agen ac 220vac
at 60hz but that come later 1e find my self a laser cutter;-)

[juju]

hello ray of light

i will tell you the values of my smaller motor, i can get 6 VAC out of 3 pound force magnets ... the resistance of my coil is 200ohm, but is like 0,2mm thin. Here is my calculation, just to have an idea.

3 pound - 6 Volts
250 pound - ?

250P * 6V / 3 = 500 VAC

But im not shore if its the right way to do the calculation, and im talking of values without load, but my conclusion is that you should have 100/150 ohm coil resistance, to get the 200/300VAC. Someone can tell me if this calculation can be applyed?

The current is irrelevant?

I can get 60/80VAC of 3.4pound magnets, with a 1.6K ohm coil but very low current, or 2 Volts from a 10 ohm coil but with very much high current on load.

but that something that you may be already aware.



king regards

[Harvey]

The voltage you induce in your coil will be dependent on how fast the magnetic flux of your permanent magnet passes through it. The faster the motion, the greater the voltage.

So your first step will be to identify the geometry involved and determine the angular velocity of your moving magnets when the device is at operating speed. From these two, you can determine the change in flux relative to the stationary coil.

Once you know the change of magnetic flux over a specified period of time you can derive the voltage from the Law of Induction. From that we find the equation relating to Volts = Number of Turns times (change in flux divided by change in time) V = N (d Phi B / dt )
(you may see E instead of V, which stands for Electromotive force which is voltage)

As you may imagine, your problem becomes slightly complex because you need to know how much your strong magnet's flux changes for a given distance. A Gauss-meter would be good to have in that case. Then you could actually measure the Phi B (flux density) where the field passes through the coil as the rotor turns. That would give us the numerator - d Phi B.

Once you know the numerator, and you know the denominator ( how fast the rotor turns - dt) we can rewrite the equation as V = (N d Phi B) / ( dt ) which mathematically is the same as (N d Phi B) = Vdt and that is the same as N = Vdt / (d Phi B). So, if you know your V (300) and you know your delta t (change in time) and you know your delta Phi B (change in flux), then you can determine your N (Number of turns) for your coil.

You will notice from these calculations that the voltage produced is not affected by the diameter of the wire or the size of the coil - but there is some practicality that must be observed insomuch that the change in flux must encompass the entire coil throughout the transaction. So naturally, the geometry does play a small factor here. You simply cannot take a Double-Ought Jumper Cable (00 = 9.266mm) coiled up and move a tiny 1mm magnet inside the coil real fast and expect a lot of voltage. There needs to be a difference, a change in flux in the entire winding for the math to be accurate.

If you cannot get a way to measure the flux density, you may be able to approximate it using the information provided by the magnet manufacturer for your magnets. They will give you the flux density at the surface of the pole. This is considered 1r from the center of the magnet. If you move a distance of 1r away from the pole, you will be at a distance of 2r, etc. The magnetic flux density (Phi B) diminishes over distance by a value of 1 / r². For example, let us say, hypothetically, that your magnets are 2 inches Wide and One inch Long (or thick) like the DYOXO and have a value of 4667 Gauss at the surface. This means that 1r = 0.5". So let us say your coil is centered 2" away from the magnet as it passes. This would be 4r. So we would take 4667 / 4² to find the Gauss at that distance. So the Gauss would be about 292 two inches away. These are only approximations, as the true nature of the magnetic field is really 1 / r³ as we get farther from the pole surfaces. So you can expect a change in flux between near zero and 292 Guass but we would be looking for Tesla's to plug into our equation above. 10,000 Gauss equals 1 Tesla.

So, lets say you had a rotor that turned 3600 RPM. That would be 1/60 of a second per revolution. And let us say the rotor is large enough, that it has 30 coils around it and each of those is 2 inches in diameter. So our circumference is 60 inches with the coils touching. This would make your rotor about 19 inches. So we have 1/60th of a second divided by 30 coils and that tells us our dt value which would be 556 µs or 5.56 x 10^-4 (0.000556) seconds. So if N = Vdt / (d Phi B) let's see what N would be.
(300 x 0.000556) / (292 / 10,000) = 5.7 Turns

So you could probably do 6 turns and be within your tolerance at those RPM's That would produce 300V AC on each coil, phased 12° apart. As you said, you can run that through a bridge rectifier on each coil and charge up some caps.

Next we would need to see if the geometry is practical for those RPM because of the inductance of the coils in those dimensions.

[ray0energy]

thanks guys

The voltage you induce in your coil will be dependent on how fast the magnetic flux of your permanent magnet passes through it. The faster the motion, the greater the voltage.

So your first step will be to identify the geometry involved and determine the angular velocity of your moving magnets when the device is at operating speed. From these two, you can determine the change in flux relative to the stationary coil.

Once you know the change of magnetic flux over a specified period of time you can derive the voltage from the Law of Induction. From that we find the equation relating to Volts = Number of Turns times (change in flux divided by change in time) V = N (d Phi B / dt )
(you may see E instead of V, which stands for Electromotive force which is voltage)

As you may imagine, your problem becomes slightly complex because you need to know how much your strong magnet's flux changes for a given distance. A Gauss-meter would be good to have in that case. Then you could actually measure the Phi B (flux density) where the field passes through the coil as the rotor turns. That would give us the numerator - d Phi B.

Once you know the numerator, and you know the denominator ( how fast the rotor turns - dt) we can rewrite the equation as V = (N d Phi B) / ( dt ) which mathematically is the same as (N d Phi B) = Vdt and that is the same as N = Vdt / (d Phi B). So, if you know your V (300) and you know your delta t (change in time) and you know your delta Phi B (change in flux), then you can determine your N (Number of turns) for your coil.

You will notice from these calculations that the voltage produced is not affected by the diameter of the wire or the size of the coil - but there is some practicality that must be observed insomuch that the change in flux must encompass the entire coil throughout the transaction. So naturally, the geometry does play a small factor here. You simply cannot take a Double-Ought Jumper Cable (00 = 9.266mm) coiled up and move a tiny 1mm magnet inside the coil real fast and expect a lot of voltage. There needs to be a difference, a change in flux in the entire winding for the math to be accurate.

If you cannot get a way to measure the flux density, you may be able to approximate it using the information provided by the magnet manufacturer for your magnets. They will give you the flux density at the surface of the pole. This is considered 1r from the center of the magnet. If you move a distance of 1r away from the pole, you will be at a distance of 2r, etc. The magnetic flux density (Phi B) diminishes over distance by a value of 1 / r². For example, let us say, hypothetically, that your magnets are 2 inches Wide and One inch Long (or thick) like the DYOXO and have a value of 4667 Gauss at the surface. This means that 1r = 0.5". So let us say your coil is centered 2" away from the magnet as it passes. This would be 4r. So we would take 4667 / 4² to find the Gauss at that distance. So the Gauss would be about 292 two inches away. These are only approximations, as the true nature of the magnetic field is really 1 / r³ as we get farther from the pole surfaces. So you can expect a change in flux between near zero and 292 Guass but we would be looking for Tesla's to plug into our equation above. 10,000 Gauss equals 1 Tesla.

So, lets say you had a rotor that turned 3600 RPM. That would be 1/60 of a second per revolution. And let us say the rotor is large enough, that it has 30 coils around it and each of those is 2 inches in diameter. So our circumference is 60 inches with the coils touching. This would make your rotor about 19 inches. So we have 1/60th of a second divided by 30 coils and that tells us our dt value which would be 556 µs or 5.56 x 10^-4 (0.000556) seconds. So if N = Vdt / (d Phi B) let's see what N would be.
(300 x 0.000556) / (292 / 10,000) = 5.7 Turns

So you could probably do 6 turns and be within your tolerance at those RPM's That would produce 300V AC on each coil, phased 12° apart. As you said, you can run that through a bridge rectifier on each coil and charge up some caps.

Next we would need to see if the geometry is practical for those RPM because of the inductance of the coils in those dimensions.

hello guys tanks a lot for the replay

my internet is off at home so i had to do it from my ipod.

any way i find the detail of my magnets
http://

http://

my mathematics ware never that good so agen thanks for the help
wel i will study this tekst a bit more but now u guys have the ditails of my magnets. and for the people that like to no wat i am buiding have a look at this site it was ofline for sam time but its back on (i dont no for how long) i am buildind dubbel this size Magnetricity.com ... NEOGEN Dynamo Project

agen thanks for the help

PS: the hole buiding proces will be on video

[juju]

uauu

harvey, you r the man


only with 6 turns you will get 300V from 250pound magnets? im very impressed...

thats good news for Ray, your coils will not give any trouble to wind... jejeje

lets connect soon you have internet.. have some pics to show you!!


hugs

[ray0energy]

Ferrit cores

Cool so now I need to find my ferrit cores or pouder
any one no ware I can order this (in EU )

I can make them my self the coils I have like 14kg for 11€

[Harvey]

harvey, you r the man


only with 6 turns you will get 300V from 250pound magnets? im very impressed...

thats good news for Ray, your coils will not give any trouble to wind... jejeje

lets connect soon you have internet.. have some pics to show you!!


hugs

Keep in mind that those are hypothetical values based on 3600 RPM and close coil proximity with a 19" rotor and manufacturers best Gauss readings.

In real world experience the magnets may not give a near zero flux midpoint between them when placed that close to each other. And the flux may distribute differently as well depending on how the magnetic poles are oriented on the rotor. Also, the real world Gauss may be less, and the flux could fall off along the inverse cube rule rather than the inverse square rule.

Using the above equations, you should be able to find a lower RPM and more turns that produce a similar result.

That's why everybody builds prototypes before ever attempting to manufacture a product in production - there are always little things that change the theoretical perfection and bring it 'down to Earth' so to speak

Cheers!