Author Topic: Figuring out the Steam Resonator  (Read 43995 times)

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Re: Figuring out the Steam Resonator
« Reply #40 on: January 28, 2012, 05:18:10 am »
Yea see I've had this happen in my 8XA setup. The only difference between my setup and Stan's setup is he had the variable gap with low inductance coils and mine has a fixed gap with high induction coils so to balance the difference. Now my biggest feat is to get the same reaction to take place with the VIC circuit & 5-VIC transformer!

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Re: Figuring out the Steam Resonator
« Reply #41 on: January 28, 2012, 05:46:11 am »
That is awesome,

One thing I have wanted to mention for a while:

In the WFC ions are created. The ions which are created collect near the oppositely charged plates, producing a voltage within the cell which will rise to a higher voltage than the circuit can produce. Under the right conditions this voltage will go into a runaway effect, rising to infinity (The ions will increase the voltage which will in turn produce more ions). This is one of the reasons for gating the circuit, to limit the max voltage inside the cell.
Ref: Delearship sales manual pF7

This further explains the transistor theory. The HV is creating ions.  I know Tony you mentioned before that you were getting much higher voltages near your cell than throughout the rest of the 8XA circuit, this is why.

In order for the ions to become stable atoms they must recieve or give an electron. The Cell forms an NPN transistor with the SS being the N type material. If you look at the atoms which are mixed with the iron to make SS you can see those atoms give the ss material it's symmetry with N type semiconductor material.

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/dope.html
« Last Edit: January 28, 2012, 06:26:14 am by HMS-776 »

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Re: Figuring out the Steam Resonator
« Reply #42 on: January 29, 2012, 03:33:16 am »
If you look at the cell as a NP device, such as a diode in reverse bias, the following will take place.

When contact is made, electrons can lower their energy by flowing from the semiconductor conduction band into the metal. The resulting build-up of charge on the metal-semiconductor interface causes a deformation of the band structure. This continues until the chemical potential in the semiconductor reaches equilibrium with the Fermi energy of the metal. The deformed band structure forms a potential barrier which electrons must overcome in order to flow from the semiconductor into the metal. In the reverse bias situation, the potential barrier is much greater and the depletion region larger. As a result, very little current flows and voltage will build up until dielectric breakdown occurs!

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Re: Figuring out the Steam Resonator
« Reply #43 on: January 29, 2012, 03:51:39 am »
That's a great explanation,

Do you think dielectric breakdown is what is splitting the water though?
Perhaps I'm wrong but I was under the impression that dielectric breakdown results in arcing?
Also, the voltage in the cell will rise toward infinity but it is still limited by the components in the circuit, the coils will see that voltage and if the voltage gets to high the coils will breakdown.

I think the dielectric breakdown of water requires a voltage in the MV range. Perhaps there is something I am not understanding though....

(Ref:http://www.sandia.gov/pulsedpower/prog_cap/pub_papers/water_breakdown_tests_PRSTAB_2009.pdf)


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Re: Figuring out the Steam Resonator
« Reply #44 on: January 29, 2012, 04:13:44 am »
You will still have the pulling force, the depletion layer will act as a barrier between the Water and Plate surface. So you will have a B+ field and a B- field on the surface of the plates. With out this depletion layer forming, you will have a dead short condition and current will flow and voltage will drop. The gate pulse acts as a voltage limiter so that you don't have arcing when breakdown occurs.

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Re: Figuring out the Steam Resonator
« Reply #45 on: January 29, 2012, 04:20:58 am »
I guess I need to go do more studying. Thanks for the explanation!

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Re: Figuring out the Steam Resonator
« Reply #46 on: January 29, 2012, 07:32:42 am »
This is a drawing showing the basics of the depletion layers role in all of this.

(http://www.globalkast.com/images/tonywoodside/B+_B-_Depletion_Layer.png)

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Re: Figuring out the Steam Resonator
« Reply #47 on: January 30, 2012, 08:34:52 am »
ok this is my theory on the purpose of the Gate pulse. When you have the depletion layer form on both electrodes, it will get larger and larger as the  voltage increases. At some point the voltage will make the depletion layer large enough that the two will meet and cause a bridge between the two electrodes. This will cause a dead short condition and current will flow and voltage will drop and you will have a loss of resonance. So this is where the gate comes into play, it limits the depletion layer from getting too big and causing the bridge between the two electrodes.
The B+ & B- depletion layers cause the stripping of the electrons and when the gates switch off, the atoms will link back up as diatomic hydrogen.

The antibonding (top) molecular orbitals of the H2 molecule. (out of phase orbit)
(http://upload.wikimedia.org/wikipedia/commons/e/ea/H2OrbitalsAnimation.gif)
« Last Edit: January 30, 2012, 09:19:17 am by TonyWoodside »