Can anyone please help me

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wayneh

Joined Sep 9, 2010
17,496
Sorry to offend you before but your statement was retarded...
No problem. You should be aware that the term "perpetual motion" applies to any machine thought to power itself indefinitely or to even produce more power than it consumes. Actual mechanical motion isn't required, at least in the way I know the term.

It's a sad fact of physics that, when you handle energy, the very best you can hope for is mere recovery of what you put in. In real life, it's normal to lose large fractions every time you touch power. The rush towards electric cars, for instance, seems to ignore the fact that you have to make all that electricity somewhere, losing a lot along the way to the pavement underneath you. More coal plants? Nope, too dirty. Nukes? Nope, too dangerous. Homegrown oil? Nope, too messy. I'm befuddled why anyone thinks adding more power conversion steps is an improvement. But I digress.

Here in the world of electricity, we're actually better off and can often achieve >90% conversion efficiencies. That's one reason electricity is considered such a high quality source of energy. You can work on a computer, see the screen and listen to music, all the while warming your room. Compare that to relatively low quality energy like warm air from a solar collector. It can warm the room, but that's about it.

I'm still quite fuzzy on your arrangement, but I don't need to see anything more to know that you cannot power your house without significant power coming from somewhere. Energy is conserved in all cases, and cannot be created out of thin air or made to disappear. We humans can only harness it as it passes from one place to another.
 

Thread Starter

Profiteer

Joined Sep 28, 2011
25
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.
here is a crude line diagram since everyone keeps getting it wrong maybe this will help but remeber what I said earlier about lowering the voltage input to the electromagnet and I thinking now that it will more feasible to build a second electromagnet and just alternate power to each one then the back emf from each one can be fed to the respected feeding battery to charge it.

 

Thread Starter

Profiteer

Joined Sep 28, 2011
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So then what I'm creating could be termed as "perpetual motion" even though there is no motion because I am expecting more out than I am putting in? Then I will agree it is then a perpetual motion machine even though it limited to the life of the battery
 

Thread Starter

Profiteer

Joined Sep 28, 2011
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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
SGT.: could you give me circuit diagram by chance?
 

wayneh

Joined Sep 9, 2010
17,496
So then what I'm creating could be termed as "perpetual motion" even though there is no motion because I am expecting more out than I am putting in? Then I will agree it is then a perpetual motion machine even though it limited to the life of the battery
If you expect more energy out than the amount lost from the battery, then yes, I would call it perpetual motion. I suppose "over unity" is the more accurate term. Whatever you call it, it won't happen.
 

DRock

Joined May 7, 2011
68
Perpetual motion is impossible, therefore this topic should be closed.

You could interchange the word motion with the word energy if it helps you understand it better...you will never put out more energy than you put into the system. This is not the only reason your idea would be unrealistic/impossible but it is perhaps the easiest one to understand.

So with that being said, good luck if you decide to continue.
 

Bernard

Joined Aug 7, 2008
5,784
Thanks for the diagram: With 100 stages of multiplication, fed by a chopped 12V signal, might get around 900V with no load. Connect a load of a few hundred ohms and output goes to about "0". Who said that education was cheap.
 

SgtWookie

Joined Jul 17, 2007
22,230
So you have Battery #1.

It appears that you have a voltage multiplier circuit on there. The type you are showing only works if the input is AC; and you are limited in the number of stages that you can use due to the Vf (forward voltage) of the diodes. Even with low voltage Schottky diodes, you'd lose ~0.4v-0.5v per stage.

Have a look at the attached schematic and simulation. In the schematic, V1 on the far left is your battery; I'm simulating that your battery can output a 60Hz sine wave that is 14v p-p (which it can't by itself of course; this is a simplified schematic).

From there, I have caps and diodes (the 1N5817 diodes are Schottky diodes with a very low Vf) connected to multiply the output voltage. You can see in the plot on the bottom that each successive increase is smaller; after c6 here's hardly any increase in voltage from the prior stage. I have a load on the output; a 1 MEG (one million Ohms) diode - it's only pulling ~16.5uA (microamperes) average.

Also, as the successive stages increase the voltage, the available current decreases by a like amount, less the power losses incurred in the diodes.

At this point, I don't know where you plan on getting the high voltage from, as the voltage multiplier circuit clearly will not do what you want it to do.

Then you show a diode and capacitor somehow siphoning off some power from the electromagnet. Did you know that the way the diode & capacitor are wired in series, that the capacitor might charge to a certain point, but has no discharge path except for leakage back through the diode? That also means the transformer will receive no power, as the diode has too little capacitance to couple AC power through it, and the cap blocks DC - so there's nothing left to go through the transformer.

I'm not sure why you wanted to go with such high voltage in the first place. If it were power transmission over long distances, that would make sense.

You're using an air-wound inductor, which is good from the aspect of being saturation-proof; it won't ever saturate like an inductor with a ferrite core could, for example. However, your inductor is very large physically, yet not able to handle anywhere remotely near the kind of power you are hoping to get out of it.

I'm asking all of these questions and examining every part of the project, as I don't want you to spend a lot of hard-earned money and then have it turn out to be a failure. However
 

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SgtWookie

Joined Jul 17, 2007
22,230
SGT.: could you give me circuit diagram by chance?
Well, if you're going to be below 600V, then you can use parts that are a lot less expensive, and a lot easier to get.

But at this point, I don't see where you're going to even need parts rated for 100V.

Anyway, have a look at the attached; it covers a basic H-bridge using a pair of IR2110 half-H bridge drivers. It is not complete for something like you want to use it for; it's just something to look at for the moment to give you an idea on how these things work.
 

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Thread Starter

Profiteer

Joined Sep 28, 2011
25
SGT. Wookie : When you pulse a HV electromagnet ( or transformer) with HV DC (360V or more)every time you shut off the voltage (like pulsing) you get a backward (reverse polarity) emf. The diode on the negative side of the electromagnet is directing that voltage to a smoothing capacitor that dumps into the in feed of the step-down transformer that steps that down to 120V then on the out feed negative you wire a micro switch triggered by a tiny toy motor with 2 bb's at 90 degree angeles glued to the shaft, spinning at 3600 rpm and you get out pretty near perfect 60hz 120VAC usable power. As you know the back emf is about 1 1/2 times the voltage you put in. So right there is proof that you can get more out than you put in.

Wire this up and test it for yourself if you don't believe me I already know it's fact as I have done it numerous times. I am also building a 400 horse all electric DC powered engine for my truck (waiting on more parts to arrive to finish it) run with opposing electromagnets on the housing and the crank shaft I don't need anyone to "Bless" this project as I really don't care what any "Book Smart" individual believes or doesn't believe.

I've proved it all to myself using miniature motors and transformers and tested it to see the results until someone else follows in those footsteps it is hard to get them to understand what is going on with the device. And anyway I wasn't looking to recruit anyone into my way of thinking I just wanted to know how to reverse polarity automatically on a HV electromagnet and it gotten to this discussion on whether or not I sane or competent to attempt my project which has absolutely nothing to do with my question.

I really appreciate your input SGT. and will use this new found knowledge to make some new calculations. I want you to remember what I said earlier about dropping the input voltage to the electromagnet. I believe that will give me enough of a magnetic field to induce the voltage in the generator coil and I am going to add another electromagnet to the center of the coil and just switch power between them (one wired N/S the other S/N) using the back emf from magnet #1 to feed magnet #2 and the back emf from magnet #2 will be fed to the step-down transformer and used to charge the batteries. Although I'm wondering if I could just use the back emf from magnet #2 to feed back into a super capacitor and then eliminate the batteries altogether? An idea worth looking into, but I need to consider if the voltage will continue to increase and how to control that.

Looking at your simulation I am wondering if it would be better to just use a step-up transformer instead of a voltage multiplier, any thoughts or concerns on that idea?
 

Thread Starter

Profiteer

Joined Sep 28, 2011
25
You're using an air-wound inductor, which is good from the aspect of being saturation-proof; it won't ever saturate like an inductor with a ferrite core could, for example. However, your inductor is very large physically, yet not able to handle anywhere remotely near the kind of power you are hoping to get out of it.
SGT.

I am not getting anything out of the electromagnet the power is coming off the generator coil (12awg wire) I am considering winding another generator coil with 10awg wire though. The back emf that is present in ANY DC motor or magnet is simply being captured so it doesn't destroy the apparatus. I then use that captured energy and step it down to a usable voltage so it can be reused. The magnet is not all that big as you put it 4"x6" the whole device only measures 10"W x 18"L x 6"H :)
 
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