### Author Topic: Working resonance circuit  (Read 49117 times)

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##### Re: Working resonance circuit
« Reply #80 on: December 22, 2009, 12:27:14 pm »

If you put 12V on your wfc, and it pulls 6amps.....How large is the resistance? 2 ohms!
The higher the voltage, the lower the resistance, and more amps are flowing...

Steve
Thinking out loud:

Am I thinking wrong here? We do not want current to flow through the WFC!  We want to prevent this because we do NOT want dead-short-condition (conventional electrolysis). If we use current we are creating heat (losses).

The only current that flows, is feeding the primary of the transformer to power the charge-pump circuit.

Just as you explained, this circuit is charging itself with BEMF due to the AC resonance (give it a little low power push at a fixed period and amplitude is rising). Just like in nature all is in balance (oscillating circuit). If we do nothing, the system eventually dies.

In the simulation we can see the STEP-CHARGE, first low potentials then rising to a maximum value the components can handle. At low frequency (672Hz) the potential is rising very slow but it is rising, so higher frequency can be used to charge it faster. The component values are much smaller at higher frequencies.

Stan explained as we all know, that he used opposite potentials (of equal value) for PULLING the water molecules at the di-poles.

Yes, there is little current necessary to polarize the water for the fist step, to charge it to a min. of 2V (this value is obvious a fixed threshold) If we do not have current flowing, water is not responding (polarizing). We can polarize water with the CHARGE-PUMP for a fixed time period.

When this threshold is reached, we want to prevent dead-short-condition (gap-sparking). If this occurred, polarization must be done again from the beginning.
He prevented this with a gate-signal. When gating (no pulse) water is regaining its dielectric strength (more/less resistance) when we gate (pulse) there is even more potential because the charge-pump is already charged (EMF is cycling through the circuit). But the dielectric is changing (resistance) due to gas production, so the gating signal must also be compensated to prevent dead-short-condition.

Also this PULLING only works with min. of kV potentials, because the water molecules only wants to interact on this potential level.

Do circuit simulations yourself and see the CHARGE-PUMP STEP-CHARGING...

br,
Webmug

I think you hit the nail right, here, webmug.
Amps work. In this topic, we talk about re-using amps and volts.

The deadshort condition is in case of the resonance circuit not inplace.
You only have a deadshortcondition if you hook a wfc directly to a powersupply.
In case of above mentioned circuit, the current flushes back and forward!

Steve

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##### Re: Working resonance circuit
« Reply #81 on: December 22, 2009, 12:42:39 pm »
when meyer said oh we have to use 5 amps big deal he meant this because of the losses 5 amps flowing tru a resonant circuit of 2 ohms series resistance consumes 50 watts you see?

Hi sebosfato,

Yes, but was he only talking about a transformer (variac)?

Where is the 2 ohms series resistance located? Inputing 50 watts of power is nothing, compared what you get in return (aka Meyer setup).
If you want HV then you must input more current in the transformer.
When I simulate this circuit with input of 110V, you pulling about 8Amps (depending on transformer design) and CHARGE it to 4kV.

I only want to figure out how the CHARGE-PUMP is operating.

br
webmug

• Sr. member
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##### Re: Working resonance circuit
« Reply #82 on: December 22, 2009, 12:51:18 pm »

I think you hit the nail right, here, webmug.
Amps work. In this topic, we talk about re-using amps and volts.

The deadshort condition is in case of the resonance circuit not inplace.
You only have a deadshortcondition if you hook a wfc directly to a powersupply.
In case of above mentioned circuit, the current flushes back and forward!

Steve

You are right, NO DEAD-SHORT-CONDITION.

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##### Re: Working resonance circuit
« Reply #83 on: December 22, 2009, 13:41:16 pm »
Hi webmug

You can't just input 50 watts in a circuit is the circuit that draw the current accordingly with its resistance and supply voltage. The Power you are putting in is the power that is being consumed in the Losses of the circuit. If you higher the voltage obviously than you will allow more current to flow. I found a constant that is simple is like ohms law. If you input lets say 12 volts in a 3 ohms resistor you will draw 4 amps ok. Now dealing will resonant tank circuit i discovered that input voltage will be multiplied by the Q factor witch is The inductance reactance divided by the series resistance of the inductor + capacitor and load. Lets say if we have a reactance of 10.000 our Q factor from the first example will be 10.000/3 ohms  = 3.333 *12v = 40.000v  I discovered from many calculations and experiments than Another way to know what will be the voltage you just have to multiply the reactance by the current. So 10.000 * 4 amps = 40kv and now
Again if you have 4 amps recirculating and 3 ohms of resistance you will dissipate 4^2=16*3 ohms  =48 watts

Ok all meyer did is to create a 12v transformer that transformed the impedance 3 ohms on the primary to 40 megaohms of the secondary with the diode. I mean 40 megaohms right because this is the complex impedance witch you have with the circuit i proposed for inputing the energy with the source in parallel. The impedance is aways relative to " something" the source. In the case of meyer he used the vic and gated the pulses to limit the voltage to go higher than certain top. but any way is all about control the amount of accumulated energy.

IN meyers case he said 40kv 1ma because was what the resonant tank was drawing from the vic but on the tank there was 4 amps recirculating. 40Kv 40mega ohms = 1ma

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##### Re: Working resonance circuit
« Reply #84 on: December 22, 2009, 13:45:26 pm »
Who knows if the voltage theory is right, i built the water fracture apparatus with > 10kV and it didn't work, probably because my gap was to large, but likely because it does not work.
also the test report claims there were several amps for each tube.

• Sr. member
• Posts: 436
##### Re: Working resonance circuit
« Reply #85 on: December 22, 2009, 14:03:22 pm »
Who knows if the voltage theory is right, i built the water fracture apparatus with > 10kV and it didn't work, probably because my gap was to large, but likely because it does not work.
also the test report claims there were several amps for each tube.

Have you used opposite potentials (of equal value) for PULLING the water molecules at the di-poles?

br
webmug

• Sr. member
• Posts: 436
##### Re: Working resonance circuit
« Reply #86 on: December 22, 2009, 14:06:18 pm »
Hi webmug

You can't just input 50 watts in a circuit is the circuit that draw the current accordingly with its resistance and supply voltage. The Power you are putting in is the power that is being consumed in the Losses of the circuit. If you higher the voltage obviously than you will allow more current to flow. I found a constant that is simple is like ohms law. If you input lets say 12 volts in a 3 ohms resistor you will draw 4 amps ok. Now dealing will resonant tank circuit i discovered that input voltage will be multiplied by the Q factor witch is The inductance reactance divided by the series resistance of the inductor + capacitor and load. Lets say if we have a reactance of 10.000 our Q factor from the first example will be 10.000/3 ohms  = 3.333 *12v = 40.000v  I discovered from many calculations and experiments than Another way to know what will be the voltage you just have to multiply the reactance by the current. So 10.000 * 4 amps = 40kv and now
Again if you have 4 amps recirculating and 3 ohms of resistance you will dissipate 4^2=16*3 ohms  =48 watts

Ok all meyer did is to create a 12v transformer that transformed the impedance 3 ohms on the primary to 40 megaohms of the secondary with the diode. I mean 40 megaohms right because this is the complex impedance witch you have with the circuit i proposed for inputing the energy with the source in parallel. The impedance is aways relative to " something" the source. In the case of meyer he used the vic and gated the pulses to limit the voltage to go higher than certain top. but any way is all about control the amount of accumulated energy.

IN meyers case he said 40kv 1ma because was what the resonant tank was drawing from the vic but on the tank there was 4 amps recirculating. 40Kv 40mega ohms = 1ma

sebosfato,
I will take a look at this.

br
webmug