How can i generate a 370 A current?

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In summary, high-performance cars and welding machines can provide the necessary currents for magnetizing rare earth magnets, but the wire used must be very thick and the currents must be in the thousands of amperes. DC is generally preferred for welding, but AC can also be used for certain materials. Liquid nitrogen could potentially be used to prevent wire overheating during magnetization.
  • #1
trini
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I need to generate a 20kOe magnetizing field in a 600 turn coil of inductance 0.58181032 microhenries, length 0.166116 m. Given the material I am magnetizing, I need to generate a current greater than 363.095 Amps for a period of about 0.5 s. I know of the use of pulse forming networks in creating large current pulses, but these are very expensive devices. Is there some simpler way to be able to get a single pulse off to meet my needs?
 
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  • #2
According to http://newcarbuyingguide.com/index.php/news/main/3728/event=view:

As a general rule of thumb, under normal load conditions, the starter should draw about one ampere per cubic inch of engine displacement, plus or minus about 25-percent.

If you know someone who's into high-performance cars and they're willing to let you hook up, it might work.

If you can make a .032 Ohm resistor rated for 4400 watts and connect it across a car battery, that'll work too.
 
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  • #3
Find a weld shop with a larger machine. They easily go up to 500A or more. But, think about the fact that people find those currents useful for welding.
 
  • #4
yes clearly there will be a lot of heat generated in my wire, which is thermally rated at 155 C, however my pulse time is going to be less than a second, so i hope that the heat generated during this time will not raise my temperature above this value. Also, do welding machines use AC or DC current, because DC current is what i am after(although i do suppose turning on a 1HZ ac current for half a second will produce a 1 directional current)
 
  • #5
Welders can use either AC or DC depending on the specific weld process being used; some machines can do both. Most basic stick welders will only use 60 Hz AC.
 
  • #6
I need to generate a 20kOe magnetizing field in a 600 turn coil of inductance 0.58181032 microhenries, length 0.166116 m. Given the material I am magnetizing, I need to generate a current greater than 363.095 Amps for a period of about 0.5 s. I know of the use of pulse forming networks in creating large current pulses, but these are very expensive devices. Is there some simpler way to be able to get a single pulse off to meet my needs?

I think you might have the inductance wrong. That is quite a large coil and would likely have a much greater inductance than 0.58uH. Maybe it was MilliHenrys?
A coil of 6 inch length 3 inch diameter and 600 turns would have an inductance of about 11 millihenries or 11000 uH.

Also, the wire in the coil would need to be very thick to carry such currents, even for a short burst.
Could you measure the resistance of the coil ?
If it had a resistance of even ONE ohm, it would need a DC power source of 363 Volts at 363 amps or 131 KW (or about 1200 amps from a 110 volt supply.)
 
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  • #7
From my understanding:

-dΦ/dt = Einduced = ε0LI
L=Area(turns2/length)uo

my field is radial in shape, so I have used the average coil area as a simplification, A=0.03032086m3:

L = 0.03032086(6002/0.166116)(1.25663706 * 10-6)
= 0.082524805H = 83 mH

ah...i took the value from my calculation of the magnetizing field, where i had to multiply by ε0 and not u0.

As for the resistance, I constructed a program which calculated my total length of wire to be 2.170 km. According to typical AWG tables, the resistance per km of 29 gauge wire is 268.4024 ohms, so my total resistance is in the region of 580 ohm.
 
  • #8
To get 370 amps through 580 ohms of wire, you need to apply 214,600 volts. Good luck with that.
 
  • #9
If you have 2170 metres of wire and 600 turns, each turn is 3.616 metres or 11.8 feet circumference or 3.76 feet in diameter. Is this right?

According to a wire table, I found, the maximum current through a 29 AWG wire is 1.2 amps, so it won't cope with 263 amps. It would vapourize immediately.

Seems like you need to rethink this a bit :)

I have remagnetized alnico magnets with 10 amps through a heavy wire coil with about 50 turns on it and the magnet inside it.
 
  • #10
unfortunately for my design i need to use a rare Earth core to achieve my desired B field. How then are typical magnetizing fields achieved for rare Earth magnets? it seems to me that some absurdly high currents would have to be utilized in low resistance wire.

just out of curiosity, if the reason i can't use thin wire is because of overheating, let's say i put my coil and magnet in an open steel drum, and pour liquid nitrogen into it, so that when i flick on my current a large volume of liquid nitrogen would rapidly evaporate, do you think it could stop my wire from vaporizing over that short period of time? coincidentally, using very low temperatures would also improve the remnance of my magnet.
 
  • #11
negitron said:
Welders can use either AC or DC depending on the specific weld process being used; some machines can do both. Most basic stick welders will only use 60 Hz AC.

I didn't know that. Can you elaborate?
 
  • #12
Rare Earth magnets are magnetized commercially using coils with only a few turns, but using very thick wire--really solid copper rods thicker than your thumb bent into a helical coil--and at currents in the several thousands of amperes. These currents are usually generated from a large capacitor bank discharging though the coil.
 
  • #13
dalarev said:
I didn't know that. Can you elaborate?

I'm about half-dead and headed for bed but, briefly, DC is generally preferred for welding since it results in easier arc strikes, smoother arcs, less spatter and deeper penetration. However, AC equipment is generally less expensive and offers higher currents for a given unit rating. So, for various high-temp steels and certain types of refractory metals, AC may be preferred. All things being equal, DC is generally superior but you pay for it in higher equipment cost.
 
  • #14
negitron said:
Rare Earth magnets are magnetized commercially using coils with only a few turns, but using very thick wire--really solid copper rods thicker than your thumb bent into a helical coil--and at currents in the several thousands of amperes. These currents are usually generated from a large capacitor bank discharging though the coil.

When you say a few turns, what kind of range are we looking at? less than 10 turns?

Also if anyone can, please comment on my earlier question regarding liquid nitrogen cooling as well.

I ran some numbers, if i use 5 turns of 0000 gauge copper, thus having a net resistance of 0.0016072 ohm, i would need 15323 A to make my field. Ohm's law suggests that a 24.62757 V source would produce this current in the copper, which seems a little too easy to me, what am i missing?
 
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  • #15
You could have a look at this page:
http://www.dealextreme.com/details.dx/sku.10305
where they have some pretty cheap rare Earth magnets.
Prices are in US dollars, postage paid.

i would need 15323 A to make my field. Ohm's law suggests that a 24.62757 V source would produce this current in the copper, which seems a little too easy to me, what am i missing?
The problem is the 15323 AMPS and the fact it has to be DC. This is an enormous current.
Maybe a large rotating generator could produce that current for a short time. Producing it from the mains supply would not be possible.

If you made the wire out of a superconductor and cooled it below the superconducting temperature, you would have zero resistance, but you still have to get the power and get it to the coil.

Incidentally, a number like 0.58181032 uH could easily be reduced to 0.58 uH because 8 significant figures implies an unlikely precision in the final product. So does 24.62757 Volts.
 
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  • #16
I know that these magnets are very cheap to buy, but the purpose of my project is to make one.

If i understand this correctly, the power of this system would be V2/R = 377.449 kW. Given 5 turns of 0000 gauge copper, the inductance would be 13.31 uH.

The time constant, T=L/R and after 5T, I is essentially completely discharged through my network. 5T = 41.266 ms

Now P = E/t, so 377449 = E / 0.041266 => E = 15.576 kJ

so if i were to build a capacitor bank similar to this one:
http://fastmhz.com/?p=37
it should do the job am i correct? EDIT: i actually was browsing around and spot welding machines do deal with ampages up to 15000A, 300kW do u think this wu be a suitable alternative?

but even with 0000 gauge wire the safe transmission currents are only around 360 A...The larger the wire i use the higher my stable current, but it's still always less than the current I would then require. do u think 0000 gauge would be able to withstand a 15000A pulse?

Incidentally, a number like 0.58181032 uH could easily be reduced to 0.58 uH because 8 significant figures implies an unlikely precision in the final product. So does 24.62757 Volts.

i know i just like to be as precise as possible while still running numbers.
 
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  • #17
4/0 (that's how industry writes 0000) should take a 15,000 A pulse; I work with the stuff daily and we occasionally do short-circuit tests with submillisecond current draws in the 10-30 kA range. There are larger wire diameters but these are sized, not in AWG, but in thousands of circular mils (MCM, or less commonly, KCM). The next few standard sizes up from 4/0 are 250, 350, 500 and 750 MCM. Note, however, that this stuff is THICK and you're going to have a tough time of bending it into a tight spiral, even 4/0.
 
  • #18
trini said:
Incidentally, a number like 0.58181032 uH could easily be reduced to 0.58 uH because 8 significant figures implies an unlikely precision in the final product. So does 24.62757 Volts.

i know i just like to be as precise as possible while still running numbers.

I am facinated by your specification of the length, how did you achieve that level of percision?

My bet is that you didn't.

Fine, carry around all the useless digits you want in your calculations, just do not show them them to us. Save yourself some effort and look more like you know what you are doing, only show us 2 or 3 digits.

I am concerned about the hazards presented by the voltages and currents you want to work with, Please be careful !
 
  • #19
Integral said:
I am facinated by your specification of the length, how did you achieve that level of percision?

My bet is that you didn't.

Fine, carry around all the useless digits you want in your calculations, just do not show them them to us. Save yourself some effort and look more like you know what you are doing, only show us 2 or 3 digits.

I am concerned about the hazards presented by the voltages and currents you want to work with, Please be careful !

It was purely a matter of geometry actually, my coils are not perfectly circular, they are more so rectangular with rounded edges, and were designed to fit as close as possible around my mould. it was the inclusion of pi in my calculations which left such long decimal places, the precision of which was guaranteed by programming a C++ program to calculate length given my dimensions.

And yes i am concerned about these values myself, which is why i will be sure to update this thread for advice as the design process progresses. Thanks for your concern!

Also, to negitron, due to the geometry of my magnetizing field, the rods will only have to be bent to a 5cm turn height, not really that tight. Although your confirmation that it should handle the current is indeed very encouraging.
 
  • #20
There are commercial labs that will provide that kind of current without any problem, except you have to convince them you know what you're doing. There are several here in the Pittsburgh area (or at least they used to be here - manufacturing is slipping away quickly).
 
  • #21
Most welding machines will use AC, so they would be useless.

One thing to note is that air cored coils would be a very inefficient way of magnetising magnets. A large part of the magnetic path (around the outside of the coil) would be in air which is a magnetic equivalent of a high resistance. This is called the permeability of the material and it can be several thousand times better in Iron than in air.

If you had a circle of soft iron with the coil on it and just a gap for the proposed magnet, you would be able to get a much higher magnetic flux in the gap for magnetising the magnet.

This may help to reduce the absurd currents and voltages we are talking about.

Since this is DC there is probably no need for the iron path to be laminated or ferrite. Just as much iron as possible with the coil over the magnet part.
 
  • #22
vk6kro said:
Most welding machines will use AC, so they would be useless.

This is not correct. Most consumer welders use AC, but industrial units generally use DC (and often have the option to switch to AC, if necessary) for the reasons I stated previously.
 
  • #23
vk6kro said:
Most welding machines will use AC, so they would be useless.

One thing to note is that air cored coils would be a very inefficient way of magnetising magnets. A large part of the magnetic path (around the outside of the coil) would be in air which is a magnetic equivalent of a high resistance. This is called the permeability of the material and it can be several thousand times better in Iron than in air.

If you had a circle of soft iron with the coil on it and just a gap for the proposed magnet, you would be able to get a much higher magnetic flux in the gap for magnetising the magnet.

This may help to reduce the absurd currents and voltages we are talking about.

Since this is DC there is probably no need for the iron path to be laminated or ferrite. Just as much iron as possible with the coil over the magnet part.

Because my core is going to be radially shaped, i will be sewing a case out of heat resistant fabric that i will fill with iron powder and pack it around my mold. It will look kind of like an ice cream sandwich, with the 'ice cream' being my iron filled canvas, and the 'cookie' being my coil.
 
  • #24
I didn't want to start a new thread, so I'll ask here, I have to bend some copper rod(5/8" dia), and nowhere near me does the kind of bending i want to do, so i will have to do it myself.

The easiest way i can see to make the bends i have to make(a coil, 4cm turn height) is to have some guides holding the rod, and have the rod pass through an induction heater. One end would be clamped to a steel cylinder, which would move 4cm lengthways when i rotate it once(i will control this with a simple gear system).

To achieve uniform heating, i would have to use an ac supply between 5 and 30 kHZ so that my metal is slowly heated. My question is this, for low frequency induction heating, what kind of special precautions would i have to take when designing my curcuit, or could i just connect the induction coil to an ac power supply(say 24 V 3A) and run an appropriate current.
 
  • #25
Heating the copper over a gas flame would be a lot easier.

The stresses in making such a coil would result in crumpling in the inside and possible rupturing on the outside. It is a very tight turn.
Plumbers have a tool for putting bends in copper pipe but I don't know if it can make a coil that tight.

An alternative would be to start with a copper tube about 4 cm in diameter and cut a slot down it in a spiral so you get a flat coil. This could be done with a small abrasive cutting wheel.
 
  • #26
The stresses in making such a coil would result in crumpling in the inside and possible rupturing on the outside. It is a very tight turn.
Plumbers have a tool for putting bends in copper pipe but I don't know if it can make a coil that tight.


just so that I'm sure we're on the same page, my rod is not hollow, it is solid, also from what i have read slow induction heating of copper makes its ductility properties extremely good. the thing with gas heating is that it does not distribute heat evenly, which is usually why you find internal stresses in the material. i do plan on trying to make at least one coil via my induction method, which is why i would appreciate feedback on the supply setup.
 
  • #27
Copper's excellent heat conductivity would distribute the heat very well if you used a gas flame, like a bunsen burner on a hose, so that you could move it around.

However it is also a very good conductor of electricity which means you would have a lot of trouble trying to get it to heat up with current. After all, that is why you are using copper for your coil.

When you bend anything, there will be areas where the metal has to compress and others where it stretches. I doubt if copper's ductility would help much in bending 5/8 inch copper rod into a circle of 1.6 inches, unless you actually melt it and mould it into the new shape.

But, having never tried it, I'd be interested to see how you go. :)
 
  • #28
I think I should clear up that the radius is 4cm and not the diameter, the vertical distance between the start of a turn and the end is 4cm.

I'm not sure if i understand the part about copper being a good conductor, if I'm not mistaken, isn't it the changing eddy currents which causes heating, in which case wouldn't a good conductor heat up well?

I could set up a heating system with a cross like structure of burners to evenly distribute the main heat sources.
 
  • #29
I think I should clear up that the radius is 4cm and not the diameter, the vertical distance between the start of a turn and the end is 4cm.

That sounds better. Still only 3 inches in diameter though. Is that to the centre of the rod or to the inside edge?





I'm not sure if i understand the part about copper being a good conductor, if I'm not mistaken, isn't it the changing eddy currents which causes heating, in which case wouldn't a good conductor heat up well?

Current is current. To heat up copper rod 5/8ths of an inch in diameter would take eddy currents of hundreds or thousands of amps. I haven't looked it up but it would be a huge current. Gas is a much better way of doing it.

Where are you getting the rare Earth mixture to make these magnets with?
 
  • #30
That sounds better. Still only 3 inches in diameter though. Is that to the centre of the rod or to the inside edge?

it's inside edge.

Where are you getting the rare Earth mixture to make these magnets with?
From an online supplier,magnequench, the powder comes pre-epoxy coated which is very convenient.
 

1. How can I generate a 370 A current using a battery?

Generating a 370 A current using a battery is not possible as most batteries are not designed to produce such a high current. Batteries are typically used to power low-current devices such as flashlights or remote controls. To generate a 370 A current, you would need a power source such as a generator or a power supply that is capable of producing such a high current.

2. Can I use a transformer to generate a 370 A current?

No, a transformer cannot generate a current on its own. It can only transform an existing current to a higher or lower voltage. To generate a 370 A current, you would need a power source that is capable of producing such a high current, and then you could use a transformer to adjust the voltage as needed.

3. Is it possible to generate a 370 A current using solar panels?

Yes, it is possible to generate a 370 A current using solar panels. However, the number of solar panels required would depend on the efficiency of the panels and the amount of sunlight available. It is important to note that solar panels are not designed to produce high currents, so you may need a larger array of panels to generate a 370 A current.

4. How can I increase the current output of a generator to reach 370 A?

To increase the current output of a generator, you can either increase the size of the generator or add more generators in parallel. This will allow for a greater amount of power to be produced, resulting in a higher current output. It is important to ensure that the generator is capable of handling a 370 A current without overloading or damaging it.

5. Can I use a capacitor to generate a 370 A current?

No, a capacitor cannot generate a current on its own. It can only store and release electrical energy. To generate a 370 A current, you would need a power source that is capable of producing such a high current, and then you could use a capacitor to store and release the energy as needed.

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