Yet another Free energy Awnser...But a really good one ;)

[armagdn03]

A while back, I got interested in the C-Stack http://jnaudin.free.fr/cstack/index.htm . This device represents a 100 percent separation between voltage and current from the source to the load. While this may not mean too much to you at first, it literally translates to free energy. If all you are transferring from the source is voltage with no current you have zero power consumption from the source?..Yet still have a voltage across and a small amperage on the load. Think about that for a second, immediately off the bat, we have free power with this device, yet nobody seems to recognize this.  The importance of separation of voltage (since it is ?Free?) is outlined in detail in Tom Beardens ?The Final Secret Of Free Energy? 1993, which is a must read for anybody shuffling through these forums. The C-Stack represents a solid state replacement for fancy spark gaps in Teslian technology, in Ed Grey technology and in countless other devices that rely on oscillation for the separation or addition of current and voltage.

I would like to expand upon this device and hopefully give people some direction so that useful devices may be built.

First let me re-introduce you to Tank circuits so that we are all on the same page. There are essentially two kinds of tank, or RLC circuits. First we have a series circuit, where an inductor and a capacitor of equal Impedances are hooked in series with a power source. Upon hitting the predefined resonance, an interesting thing happens, we essentially have a short circuit. We achieve resonance, but we are consuming the most power possible. We do have another option. We can use a Parallel setup where we have an inductor, a capacitor, and a power source hooked together in parallel. At reaching the predefined resonant frequency an even more interesting thing happens?. We have an infinite impedance!!! If you haven?t drawn the lines together yet, that means that in an ideal circuit with no resistance, when all components reach oscillation at their resonant frequency, zero amperage is drawn. Zero amperage means zero power is needed to keep the circuit in oscillation!!!!

So what does this mean to us? It means that if we take this setup, we now have a circuit that consumes no power, but yet creates a sinusoidal pressure (voltage) change. Okay?.so what? Well now is where the C-Stack comes into play. Since the C-stack consumes now amperage, only voltage, its ideal power source would be a parallel resonant RLC circuit. In fact, we could replace the capacitor in the circuit with the C-Stack, find its capacitance, and away we go. We now have a voltage source with no power driving a C-Stack that outputs power from its inner plates!

Now since there is no such thing as an ideal circuit because there will always be some resistance, the RLC circuit will dissipate some power, not zero. Well as long as the C-Stack can output more power than is needed to maintain oscillation we have???..OVERUNITY!!!! The power needed to maintain oscillation is very small. The power put out by the C-Stack is directly proportional to the voltage across its plates. This means that if we up the voltage at the source, it will still consume no power, and yet the power at the load (what ever is hooked up to the inner plates of the C-Stack) will increase.

This is a very simple device to test and build. Unfortunately I am in the market for a new signal generator and oscilloscope, so I am limited in my experimentation.

Lets run over what we have again:

1)   RLC circuit, at resonance = infinite impedance = no power consumption = separation of voltage from amperage

2)   The C in the RLC will stand for C-stack (although traditionally it stands for capacitor) The C-Stack takes a voltage (with no amperage) and creates power.

3)   Some of the power can be fed back into the RLC circuit to maintain amplitude lost due to natural resistance.  Any extra is ?Free energy? The amount of free energy is limited only by the amount of voltage we use.

TO do this we need only a signal generator, a high voltage C-stack (build your own! They do not exist on the market, and a little engineering now how and research can help you optimize design, or you can cheat and message me and I will tell you ), an inductor with impedance to match the C-Stack, and some sort of load (I would suggest some sort of step up voltage multiplier circuit, so that we can feed some power back into the RLC to keep the amplitude of the sine wave steady)

Think about this!!!! Don?t just glance over this and say?.hmmmm interesting, I wonder what will ever become of that. This is about as simple as it gets, An AC signal, a couple metal plates, and a trip to radio shack should do the trick!

Now the real trick, is how do we get an AC wave for ?free? We must expend some energy in its creation?.I think I have an answer, or as close as one can get, but I would like to see what other people think first

[tinu]

Very good topic, armagdn0!

My humble suggestion to the members not familiar with C-stack would be to take it one spoon at a time. The C-stack is an excellent learning tool! That?s its main use, imho.

I talk from experience and I found C-stack to be tricky, at a first glance, at least to my way of thinking, which is pretty much common-sense (I hope ). But then, after experimenting and doing a little bit of paper work, well? I will not spoil for now your pleasure of testing. Just ask me if you have any question.


A word of caution: not all ?given truths? on the page are real truths. Take just two of them:

1. ?In a conventional capacitor, if the capacitance is known, and the voltage is known, and the capacitor is allowed to charge to that voltage, then the energy stored must be .5*CV2 . This was not the case.? Really?  Why not?! This is precisely the case, but just not fully understood by the author at the time he wrote that part. Check it out.

2. ?A C-stack can be configured to step-up or step-down (the voltage) simply by which set of leads you select, inner versus outer respectively.? Sorry. It can be used to step the voltage down but not to step it up. Actually the author corrects itself later on: ?This (step-up) appears only to work at its resonant frequency, however. (You can expect that the resonant frequencies of devices on the scale that I built will be in the 3 - 4 MHz range, larger devices will be less).? Now of course the last case is possible at resonance but ask yourself where this resonance really occurs. Is it in the device itself? Well, urrr? 

Tinu

[armagdn03]

 

[tinu]

So, you are a physicist! That makes two of us.
I?m glad for having you here and hopefully so will other members be.

I said I will not spoil everyone?s pleasure for testing and I?ll try keeping that promise but still giving you some hints. Maybe not all of them tonight (it?s already over midnight here and I type slowly in English?) but at least some.

Firstly, consider various equipotential areas in a flat capacitor and define their relative potential difference to the main (outer) plates. Then take two of this imaginary areas at your choice (actually they are planes, parallel to the outer plates, right?) and put metallic inner plates in their place. Now you have a C-stack. But nothing really changed, right? (You can check by various measurements, on the outside plates in cc but the voltmeter needs to be a really good one, with very high impedance ? the cheapometers will be misleading). Do not touch the inner plates as of yet and do not measure the voltage on them because this is going to change your measurement. Just check that the outer plates still keep the same voltage.

So, by inserting parallel inner plates into the stack or by removing them, nothing changes. The outer capacitor is the same, it has the same capacity and it will behave exactly the same as before inserting the plates, in all respects.
The ?effects? appear when you short-circuit two inner plates. I will leave you to do the equations; they are really high-grade, no much painful work here. So, given the outer capacitor initially charged at a specified voltage and then disconnected from the power supply, the main questions you have to deal with are: How much charge is moving between the inner plates when you connect them, using a wire, a load/resistor, a voltmeter (see?) etc.? Why exactly so much and not more or less charge? After the equilibrium is established, what is the voltage to be measured between the outer plates? Why it has decreased?  Ooops, I?ve already said too much.

As results from the above, most of my experiments were conceived and conducted in pure cc and using a bare (but good) voltmeter. (I might have used two of them but that?s all). Why would you need a signal generator and a scope to check the above? It?s fancy, I know, but not really necessary. I?ve tried the C-stack response in ac also but it confirmed what I?ve already found from cc tests. And cc measurements confirmed the known equations (C = Epsi x Surface / Distance). Given the Epsi not well known for various materials one may use, some measurements are still needed to simplify the whole work& calculus but again, the device is 100% obeying the simple equations of capacitance (flat capacitor) and of charge conservation.

For the plates I?ve used 2 regular double-sided circuit boards and one stripped of Cu for the inner insulator. All in all, the whole setup would take minutes before going to take measurements. Yet, the measurements may look very misleading if the above rationale is not properly assimilated. That?s why I said that the main use of C-stack is mainly for learning purposes.

That would be all for now.
If not enough, please ask for more.
Or comment on where I was too laconic.

Tinu

[Unicron]

from what i read in the explanation of the C-stack, the plates not connect to power supply should first (draw on their own) electrons from somewere (maybe to get the EMF in balance) before it can discharge. isn't this correct?