VIC controller

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@sebosfato:
Thank you, quite good explained, this is what I meant. Much better explained than I did...


@All:
I think this is a nice theory you got there about the +-. But I think there's one major flaw in it, which actually IMHO shatters this theory: The WFC was made as a tubular array!
And as I said before, due to the "Faraday-cup"-effect this changes quite thoroughly what is happening in relation to a WFC which would just have 2 plates.


Explanation:
If you have a metal tube, then the water inside will always remain neutral no matter how you change the potential (voltage) of the tube in relation to this water. The water itself would always see a neutral tube from the inside. This is the "Faraday-cup"-effect.


An example for better understanding this effect:
You have a glass bowl, filled with water. Now you put an electrode in it. Up til now, everything is neutral (uncharged, equal pos and minus). Now you put, say  a few thousand volts plus on the electrode. The water will immediately get charged (ionized) by this. Chemically, you will get an anode reaction: 3H20 - 2e -> 2H3O+ + O2 and if some OH- ions happen to be near the electrode 4OH- -4e -> 2H2O + O2
So now, you have more H3O+ ions in the water, than OH- ions. The water is not electrically neutral anymore.


Now you take instead of a glass bowl a metal bowl. And you directly use the metal of the bowl as an electrode. If you now put a few thousand volts plus on the electrode (bowl). Absolutely nothing will happen with the water. No reaction at all. For from the water point of view, the bowl always remains neutral, no matter what charge it has. It could have 100 million volts on it, from the water point of view it would look as if it's neutral. That's the "faraday-cup"-effect.


So from an electrostatic point of view, on which this theory relies, the outer tube is always seen as neutral by the water. At least from the inside of the tube. But as Stan heavily isolated the outside of the outer tube, the outer tube is not able to get any outside contact with the water. So with this information you can easily say, that any water in contact with the outer tube only sees this tube as electrostatically neutral (uncharged).

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Yes your right about that but this is so if you have no applied potential, Is the same principle of the van de graf, when you put a charge inside the ball you will always have 0volts in the center, ok i agree with that. But something you are not accounting is that the charge would remain in the conductor and water in this case become the electrode too, so being the electrical insulated housing the water inside will take on and maintain the electrical charge...


When you apply the voltage between the tubes you will have a differential. For example, you have 40kv between the outer tube and the outside of the container, but now you just put 1000 volts between the tubes, so what happens? Well happens that when you turn off the pulse the voltage between the outer tube and outside will be 41kv for example...  So i think it works somewhat related to the van de graf, of course i'm not sure, is just a theory i've thinking this since a while already... But maybe is just simpler than that..

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so being the electrical insulated housing the water inside will take on and maintain the electrical charge...

I'm not quite sure what you mean by that. If you do not apply any potential across the WFC, but only on the WFC, the water in it has no charge whatsoever. All the charge is on the outer tube.


This is the same about the applied potential. The outside tube will always be seen as electrostatically neutral from the water point of view. Therefore if you say apply a positive 500volts on the inner tube and a positive 1500 volts on the outer tube. Then from an electrostatically point of view of the water inside the tube, the outer tube will be neutral and the inner electrode will have a potential of -1000 volts (in relation to the water itself). Although from the outside point of view it has +500 volts.
This is why the theory of applying + and - relative to the neutral charge of the water goes down the drain, as soon as you have this tubular arrangement of the WFC that Stan used.

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I wanted to mean that if water is inside an insulated place it can take on a net charge just like when you rub a piece of plastic in your hair. Triboelectric effect... Is a propriety of the dielectrics. However the dielectric must be isolated otherwise it would lose this charge for the ambient.


By isolation i mean:


For example the plastic holds its net charge when you simply rub it cause the air surrounding is an insulator.


Water need to be inside an insulator to be able to keep its net charge.


Did you understooD? 



I think is almost that.


I would say if you apply +1000volts to the inner tube and +1500 to the outer tube at the same time you will have a 500v between the tubes. In relation to the water the inner tube will be OV cause it is inside another conductor (the outer tube and water) and as such when we send than the negative voltage to the inner tube restricting the amps (in counter phase) it could always pump charge in the bath cause of the van de graf principle.. but i'm not sure if this is the point we should be discussing. I will work automatically. I mean the water outside of the outer tube will take on an electric charge cause it is part of the positive electrode of the resonant cavity (capacitor)... Have you watched well the meyer cell and did you noticed those micro bubbles everywhere seeming a fog?
I'm not pretty sure on that.. its a confusing analysis...


However is good to try guessing what will happen to predict the functioning.


Check my last post in my thread, i spent the morning producing some knowledge..

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Is a propriety of the dielectrics.

It's actually not only a property of dielectrics. All atoms have this behaviour. But in metals you usually cannot see their effects. This is actually also the basis for electrochemistry.
E.g. if you put a copper metal in contact with air, the metal gets charged negatively and the air positively! This effect is also seen between different metals.


But I'm actually not quite happy with my explanation before.  As I don't like to speak in absolute potentials (voltages), and the example above is also only completely true for point charges (size 0 inner tube).
I will try to explain it, as simple as I can:
You cannot see charges! You can only "see" E-fields caused by charge imbalances. So let's point it that way: If you name the surface e-field of the tubes the current reference potential to the water, with the tubes with such similar diameters as used by Stan, you could say, that the outer tube has maybe +450V and the inner -550V. if you put 1000v across it. But if you now put one tube to 1million volts and the other to 1million+1000volts or not doesn't make any difference to the water INSIDE the tubes. Outside, the story is completely different.

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You are right all atoms have this ability, but dielectrics can hold this charge separation (induced dipole) for longer, in my view..




I think you understood what i mean when i say the water will be an electrode,. a charged electrode with positive, will have missing electrons...


This is the idea behind. So if water in relation to the outside of the container sits at a high positive voltage static, this missing electrons will interact with the bound.


Water molecules links are not covalent but interact by Van der walls interaction (hydrogen bound)... I mean organization.  I need to know more about to explain this better.. but again this is the idea...


Think of the water container and copper capacitor like a high ESR resistance capacitor. One that has a high series resistance if pure water is to be used. This lead us to two problems one is that it affect the Q factor and the other is that it means that a small dc current would give a high voltage.. so the balance of the contaminants and the resonance must be considered.


Is like a balance if you go for one side it become very hard to do such as the other side. So we need to find a balance. 



I would say turn on the thing and add tap water to change the contaminants so the differences can be observed.

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I don't want to sound like a broken record, but, I still feel strongly that the ideal dielectric for the WFC will be air(ionized) mix with water droplets and not bulk liquid water with bubbles. The air adds the capacitance needed and insulation in between the water droplets and exciter plates. Being that the cavity is filled with droplets isolated by air, the conduction path will remain open until the voltage reaches a level to allow arc over. Bulk water will not hold a charge. The container of bulk water can, but, not the water itself. Water droplets mixed with air can take on a huge charge and hold it until a discharge path is provided in some way.

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I dont understand the full depth of the posts in this thread but it gave a thought. As there are ppl here with physics/math/electronic design skill, there is something I would like to ask even if it may not work or be relevant.

On a dual rail powersupply, the secondary has a centre tap. see attached basic diagram.

With the resonant cavity drawing, diagram attached, where is the earth symbolised connection on the water inlet connected to.? Im unclear of its purpose - is it for setting a electrical polarity on the water entering the cavity.?

the question: On the VIC coil driving the cell - has anyone contemplated what would happen in the circuit if the secondary had a centre tap and that centre tap electrically connects to the water in the cell. ?