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#### SgtWookie

Joined Jul 17, 2007
22,221
OK, so do you have a part number and manufacturer of this electromagnet, or preferably a datasheet for it so that I can see its' specifications?

#### Profiteer

Joined Sep 28, 2011
25
no I built it myself

#### SgtWookie

Joined Jul 17, 2007
22,221
OK, so HOW did you build this electromagnet?
What did you use for a core?
What did you use for insulation?
What did you use for wire?
What is the type of insulation of the wire, and it's voltage rating?

#### t_n_k

Joined Mar 6, 2009
5,455
As they say - "A picture is worth a thousand words". It would be useful to have a sketch of how this proposed system is configured. I'm certainly not confident of what is being proposed.

Is this system really an 'over unity' idea? Keeping in mind the TOS for this site.

#### SgtWookie

Joined Jul 17, 2007
22,221
You may not realize this, but finding H-bridge drivers that will work at 1.2kv isn't trivial, and they won't be cheap. Most of them run out of that proverbial steam by 600v.

#### Profiteer

Joined Sep 28, 2011
25
it is an air core electromagnet wound with triple coated flat magnet wire from www.alphacoredirect.com rated at 2000v, 240c

#### Profiteer

Joined Sep 28, 2011
25
sorry tnk I don't own a camera

#### Profiteer

Joined Sep 28, 2011
25
sorry guys I gotta hit the hay early day tomorrow I'll check this thread before I head out in the morning hopefully I will be able to get an answer to my question tomorrow

#### magnet18

Joined Dec 22, 2010
1,227
You may not realize this, but finding H-bridge drivers that will work at 1.2kv isn't trivial, and they won't be cheap. Most of them run out of that proverbial steam by 600v.
Round a hundred bucks for something that can handle ~500A
(basing off of tesla coil design H-bridge circuits, same principle - different frequency)

#### SgtWookie

Joined Jul 17, 2007
22,221
Here is the datasheet for an IGBT module gate driver:
http://www.pwrx.com/pwrx/docs/m57962.pdf
It's rated for 1.2kv.
Digikey sells them for ~$22: http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=835-1107-ND You'd need four of them for an H-bridge. Then you'll also need four IGBT modules with similar voltage ratings. Haven't looked at those yet. I'm assuming cost will be an issue. However, you need to step back a minute and look at your magnet design a bit more carefully. How many turns of wire are on your electromagnet, what is the inside diameter of the form you wound it on, what is the outside diameter, what is the thickness (length) of the coil, and what is the equivalent AWG rating of your wire? From these numbers I can calculate the approximate inductance of your electromagnet. Thread Starter #### Profiteer Joined Sep 28, 2011 25 3000 turns of 30awg flat magnet wire inside square 2" x 4" outside square 4 1/8" x 6 1/8" wire length 5876' there is absolutely NO cost issue what-so-ever thank you for the concern though. Last edited: #### wayneh Joined Sep 9, 2010 17,133 ...the unit is self powered from then on for as long the batteries continue to take a charge. This thread should be closed immediately. The OP is clearly confused over basic concepts, is pursuing a perpetual motion machine, and despite his assertions of attention to proper safety measures, is planning to build something hugely dangerous. If he comes back with MUCH better documentation (pictures, block diagram, etc.), fine. Thread Starter #### Profiteer Joined Sep 28, 2011 25 This thread should be closed immediately. The OP is clearly confused over basic concepts, is pursuing a perpetual motion machine, and despite his assertions of attention to proper safety measures, is planning to build something hugely dangerous. If he comes back with MUCH better documentation (pictures, block diagram, etc.), fine. There is absolutely no motion to this machine what-so-ever so perpetual motion is a ridiculous statement and your a senior member? What a retard! I had a question on how to reverse polarity on a HV electromagnet and was doing just fine conversing with SGT. Wookie it has blossomed into a discussion of the device I am building and from that you have decided without any understanding of what I am doing is perpetual motion?. I guess I was wrong choosing this forum to get my question answered because clearly you are NOT intelligent enough to grasp the concept and the basic laws of physics and electricity employed here I'll be going to another forum that has people with a brain in it! (That statement does not apply to you SGT. Wookie ) Last edited: #### SgtWookie Joined Jul 17, 2007 22,221 Hmm. That'll be ~1.108H inductance, it'll dissipate about 1.37kW of power due to the resistance of the conductor, RMS current will be around 1.7A. That's about twice the maximum allowed current of 860mA for chassis wiring. Table of AWG sizes vs current limits: http://www.powerstream.com/Wire_Size.htm Do you have a plan for getting rid of the 1.37kWatts of power that the coil will be dissipating? If not, you'll have a very quick melt down - that is, if your supply can keep up with it. That's a bit doubtful, as you'd need more input power to drive it. Here are some power IGBT's that are rated for 1300V Vceo, cheapest Digikey had @$4.41/ea:
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=497-10990-5-ND

#### Profiteer

Joined Sep 28, 2011
25
On the negative side going to the electromagnet I have place a diode and capacitor to collect back emf from the electromagnet then that is fed into a step-down transformer, from there it feeds an outlet that provides outdoor power and a battery charger is plugged into that outlet that charges the second battery (still need to build a circuit that reads each batteries voltage/amperage and switches between the two when the active battery reaches 25% discharge). I believe that will take care of any such melt down unless I am mistaken, I am not perfect and could have missed something though.

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#### Profiteer

Joined Sep 28, 2011
25
Hmm. That'll be ~1.108H inductance, it'll dissipate about 1.37kW of power due to the resistance of the conductor, RMS current will be around 1.7A. That's about twice the maximum allowed current of 860mA for chassis wiring.
Table of AWG sizes vs current limits: http://www.powerstream.com/Wire_Size.htm

Do you have a plan for getting rid of the 1.37kWatts of power that the coil will be dissipating? If not, you'll have a very quick melt down - that is, if your supply can keep up with it. That's a bit doubtful, as you'd need more input power to drive it.

Here are some power IGBT's that are rated for 1300V Vceo, cheapest Digikey had @ \$4.41/ea:
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=497-10990-5-ND
So I need an electromagnet with larger wire size wire on it? but that will consume more amperage from the battery. Also I have been re-thinking the voltage to the electromagnet I am thinking that given the size of the device that I may be able to go as low as 200-300 VDC to feed the electromagnet and still get a strong enough magnetic field to induce voltage in the generator coil. any thoughts on that?

#### wayneh

Joined Sep 9, 2010
17,133
...what is difference between switching polarity and spinning a magnet over a coil of wire to create electricity?
And you're calling ME the retard?

Shaft work provided by an engine or some other energy-consuming device supplies the force to move a magnet through a field. Why is there a field? An offsetting magnetic field sets up in the generator coil once it begins supplying current. No load or current, no field, and the magnet is easy to spin. Once there's current flowing and a field developed, it takes work - energy - to move a magnet through that field. A portion of that shaft work is captured as electrical energy, as output from the generator. The laws of thermodynamics dictate that you cannot get more than unity recovery of the energy. So overall you consume coal, fuel oil, natural gas or whatever and get about half of the chemical energy back as electrical energy.

If you want to use electrical energy make the coils in the generator produce power, instead of spinning a magnet past them, ANY power extracted from the generator (essentially the secondary of a transformer) must first be supplied by the primary - your electromagnet.

Your electromagnet may draw little current when it's not loaded, but once it starts seeing an offsetting field in the secondary - a load - its current draw will increase in proportion. And you'll never get enough juice through it for the amount of power you want out the other end. But that doesn't matter, since the whole thing will consume more electricity than if you just connected your load directly to the source of power.

#### Profiteer

Joined Sep 28, 2011
25
As they say - "A picture is worth a thousand words". It would be useful to have a sketch of how this proposed system is configured. I'm certainly not confident of what is being proposed.

Is this system really an 'over unity' idea? Keeping in mind the TOS for this site.
I am not sure if that is true I am hoping that I can keep the batteries charged enough to keep it going, yes if that is what your asking. If that breaks the code of conduct of here I am sorry I just wanted to know how to reverse polarity on a HV electromagnet automatically

#### Profiteer

Joined Sep 28, 2011
25
And you're calling ME the retard?

Shaft work provided by an engine or some other energy-consuming device supplies the force to move a magnet through a field. Why is there a field? An offsetting magnetic field sets up in the generator coil once it begins supplying current. No load or current, no field, and the magnet is easy to spin. Once there's current flowing and a field developed, it takes work - energy - to move a magnet through that field. A portion of that shaft work is captured as electrical energy, as output from the generator. The laws of thermodynamics dictate that you cannot get more than unity recovery of the energy. So overall you consume coal, fuel oil, natural gas or whatever and get about half of the chemical energy back as electrical energy.

If you want to use electrical energy make the coils in the generator produce power, instead of spinning a magnet past them, ANY power extracted from the generator (essentially the secondary of a transformer) must first be supplied by the primary - your electromagnet.

Your electromagnet may draw little current when it's not loaded, but once it starts seeing an offsetting field in the secondary - a load - its current draw will increase in proportion. And you'll never get enough juice through it for the amount of power you want out the other end. But that doesn't matter, since the whole thing will consume more electricity than if you just connected your load directly to the source of power.
So what you are saying is that the electromagnet will be acting as a generator coil itself and will producing a current in it once a load is connected to the generator? Won't the diode and capacitor used on the negative side of the electromagnet to collect the back emf also capture that induced voltage?

Sorry to offend you before but your statement was retarded, If you can have a rational discussion I would like to hear your input as I stated before I am not perfect and that is why I here to gain knowledge I don't already have.

#### SgtWookie

Joined Jul 17, 2007
22,221
On the negative side going to the electromagnet I have place a diode and capacitor to collect back emf from the electromagnet then that is fed into a step-down transformer, from there it feeds an outlet that provides outdoor power and a battery charger is plugged into that outlet that charges the second battery (still need to build a circuit that reads each batteries voltage/amperage and switches between the two when the active battery reaches 25% discharge). I believe that will take care of any such melt down unless I am mistaken, I am not perfect and could have missed something though.
Let me get this straight...
You're planning on capturing the back EMF from the electromagnet that you're powering from one battery, and using that to charge another battery?

Maybe you don't know this, but charging a lead-acid battery is a "lossy" proposition. For every 100 units of energy that such a lead-acid battery accepts, you get 70 to 80 units back, tops. The rest of the energy is dissipated in the battery itself as heat.

I see that lots of energy is being expended, but I'm wondering what the source of all of this energy is? Surely you must have a power source somewhere? Because a 12v battery won't be able to provide the kind of power output that you probably are expecting, and that power has to come from SOMEwhere.

Your electromagnet will not create power. At best, you might get as high as 99% efficiency at transferring to another inductor (then you will have a transformer, not an electromagnet; transformers are coupled inductors).

So, at this point all I can see is the use of electrical power, and not where it is actually coming from.

Power is not spontaneously created. You have to insert energy somewhere.

As it is, if you try to use one battery to power a theoretically perfect driver to make your electromagnet excite a secondary coil, then rectify that output and use it to charge a 2nd battery, you will have power losses in the 1st battery, the electromagnet, a slight loss in the coupling between the electromagnet and the secondary coil, the copper losses, the rectification and regulation circuit, and the 2nd battery itself. Best case, you'll wind up with perhaps 65%-75% efficiency - which means you'd be better off just leaving your 1st battery fully charged as it is; otherwise you'll wind up with two half-charged batteries.

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