ENERGY AMPLIFICATION

[IWD]

it is instructive, but explanation is easy, on the outuput of the bifiliar every time when you see a spark, is discharged a capacitance which is in bifilar coil very high in contras of inductance, point is: that output have lower discharge frequency than input because charging the capacitance take time, so we can say it is frequency transformer. But power is same or smaler on the output like on input.

[gyulasun]

Hi Mags,

Thanks for the circuit list, it works in the Falstad simulator (I have only the online version but it is ok) and gives indeed higher than 20V in the capacitor after the switch is off at near the 20V charging level.

I would like to suggest to replace the 20V source with a 15uF capacitor, precharged to 20V also and then see how the LC circuit swings energy...
Perhaps the LTSpice circuit simulator used by IWD could be a better choice for simulating it?  I included your Falstad schematic here for IWD in case he could enter it with a 15uF precharged cap instead of the 20V voltage source.  This way can get closer to the real behaviour of the interesting circuit, obviously the final proof is to build it and test.
Nevertheless, if LTSpice simulations give promising results and optimization is also performed in it by adjusting switch on time etc then hands-on tests are done more readily.  Use realistic loss resistor for the 1H or other value coil in the simulator.

Thanks,  Gyula

[pese]

Each

Step-Up-Converter
(Boost Power Stages)
is made this way.

only Choke  Diode Condensor and switch is needed.

(look Wikipediia or electronc forums.

NOT any Overunity.

The step up of volgage , bring the same (as losses) in amperages !!

The wattage (Energy will not amplified ! , tey will
allso lossing 5 to 15% typically)

Pese

www.alt-nrg.de/pppp

alternative link collections

[Magluvin]

Hey Gyula

Ya know, adding that diode just came to me yesterday as we were posting each other. You are right about needing the resistance added for the inductor. Today I dont have much time, but will be back in the bench this week some time. It was just another idea that I have to add to all this.
I will be on the computer over the next few days till I get settled.

Hey Pese

The circuit was put up yesterday in another thread just to demonstrate that in reality, the collapsing field of a coil does not necessarily produce bemf. In order to produce bemf, the coils field would have to change polarity.  Like in an ac generator, as the field builds on the coil, a positive voltage increases to peak, then as the field declines, slow "collapse", the positive voltage declines to zero then when the field reverses, only then will we get a negative voltage. Make sense?   So how does the collapse if the field in a coil produce bemf?  ;]

Then I had the idea of if when we disconnect the source, how could we loop the circuit to let the coil continue to charge the cap further without the source.   Hence the looping diode.

Lets face it, even in a perfect world of inductors and caps, we still dont have overunity, in standard demonstrations, just unity, err infinite running.   So if in a perfect world, where the coils have no resistance, my circuit with the looping diode does show overunity. Thats a good start.

As I explained to Gyula yesterday,  lets say we were to charge the cap to 20v directly from the source, there would in theory be an equal transfer of energy, unity, same out as in.  But in my circuit, if the source is 20v and we cut off the source during the incline at 20v, and the inductor continues to pump the cap to a higher voltage   than the source, with the looping diode, that extra is free. 

I have a lot going on here with the move, but I am always thinking and visualizing as I go, multitasking. =]
As to how to utilize that extra is the key now.  Even if it were only 21v gotten to the cap via the loop, that 1 extra volt is free.  Any thing free is a good thing, no?  Im sure there is room for improvement.  =]

Im not saying this is Titos way, it is a form of energy amplification, even if it is .5% thus far, of which I am lowering it to that value of increase to show an example.

Mags

[IWD]

hm...Mags, in reality the caps wil do not charge more when you open the switch, I test it. (even with another charged capacitor as source) I mean not more, than when you keep it without the diode, and leve the curent stop by himself. The LT spice simulator I am using only to make schematics and to confirm measured data from osciloscope. When you know how to set it, it is very precise simulation. But Reality is always original. So I say it once again, if you have let say batery 12v and conect capacitor in series with induction and diode, the capacitor will charge about 19-21 volt but in case when you use another diode and open the switch early before current stop by himself, the voltage will be smaler than 19v.
Only usefuly point is What to do next whit capacitor easily charged higer than source.

-one more thing, When you use charged capacitor to 20v as source instead using batery. that Lets say if second capacitor will have same capacity, then in best case you will have first capacitor almost empty let say about 1,4v and the second will have 18,4. Of course  you can end wit half votage on both capacitors or any other combination which depednd mostly on size of inductance in clasic case, or on time of opening switch on case presented by you,(in that case you can also end for example: first on -7v and second on 13v) It does not matter, because energy will be always same.In case which first capacitor will be bigger than second you can end with letsay, first on 15v and second on 30v. for example is first 15uf second 100n, but still is same energy with little loos.