timeshell

I may provide further information soon.

What make you believe that? and according to your perspective why would current get restricted?

You need to understand the difference between these two attached schematic diagrams and how V=IZ is applied to the VIC at resonance.

You have to understand what the problem actually is before you can understand how to design the VIC to solve it. 

The bottom line is, the resistance value of the chokes and secondary aren't as important as the reactance and impedance values being matched to the WFC at resonance because it is the reactive impedance that solves the equation of high voltage with low current.  So the WFC and chokes need to be designed with high reactive impedance within specific frequency ranges in mind based on the variable range of the capacitance of the WFC.

Quote from: massive on November 16, 2024, 04:10:56 am

Quote from: timeshell on November 15, 2024, 19:20:05 pm

Who said anything about blocking current.

None other than Stan Meyers himself. He said many times ".... yada yada, Blocking diode..." and the diode is in every "circuit" diagram on the net.

This thread is titled "Back to Basics"

Did Stan Meyer show a circuit to anyone?    The answer is no, he did not

V=IR

If you want to measure 20kV across water, which has a low resistance to electric current before electrolysis occurs, what do you suppose the amp level needs to be?

Obviously it can't be zero, can it?  Otherwise voltage would be zero.

But it can't be high current either or you will have electrolysis instead of water splitting.

So how do you get high voltage with low resistance without getting high current?

KISS

As I mentioned earlier, the VIC is a DC circuit due to the diode, but it contains a pseudo-AC circuit formed by the chokes and water capacitor. In this part of the circuit, the relationship follows V=IZ, where Z is the reactive impedance, which limits current.

The VIC leverages reactive impedance through inductors and the water capacitor, rather than relying purely on resistance. The inductive chokes resist rapid changes in current dynamically, and their opposing mutual inductance further suppresses current flow. Meanwhile, resonance between the chokes and the water capacitor allows energy to oscillate efficiently, recycling energy within the circuit to build up voltage across the capacitor over time without requiring continuous high current input. This ensures the voltage rises high enough to induce breakdown without significant energy losses through conduction or electrolysis.

The bottom line is, the resistance value of the chokes and secondary aren't as important as the reactance and impedance values being matched to the WFC at resonance because it is the reactive impedance that solves the equation of high voltage with low current.  So the WFC and chokes need to be designed with high reactive impedance within specific frequency ranges in mind based on the variable range of the capacitance of the WFC.

Quote from: timeshell on November 15, 2024, 19:20:05 pm

Who said anything about blocking current.

None other than Stan Meyers himself. He said many times ".... yada yada, Blocking diode..." and the diode is in every "circuit" diagram on the net.

This thread is titled "Back to Basics"

Did Stan Meyer show a circuit to anyone?    The answer is no, he did not

V=IR

If you want to measure 20kV across water, which has a low resistance to electric current before electrolysis occurs, what do you suppose the amp level needs to be?

Obviously it can't be zero, can it?  Otherwise voltage would be zero.

But it can't be high current either or you will have electrolysis instead of water splitting.

So how do you get high voltage with low resistance without getting high current?

KISS

And yet so often the answer is just under our nose, and we don't see it.

1. Stan's invention wasn't about electrolysis.  It was about using high voltage fields to split the water.
2. Who said anything about blocking current.  The current needs to be limited sufficiently to allow the voltage to climb across the water cell without inducing electrolysis, which would prevent the voltage from building up.

Why do so many deviate from Stan's explanations?  Remember what he said?  KISS

The FBT discontinuous function is available to anyone online

Any yet nobody else has solved Meyer's riddle yet.  Hmmmm....

Again, what you're saying doesn't really make sense.  Hydrogen will only bind to oxygen in gas form after it is ignited to start the reaction, which then creates the water molecule.  Otherwise, once separated, they stay separated.

The VIC's chokes are not a FBT.  The secondary attempts to energize both chokes simultaneously.  If anything the primary and secondary could be FBT.

the oxygen atom wants free electrons, and they have to come from some where

You've mentioned this a couple times.  But you don't really elaborate.  Hydrogen and oxygen don't combine into water just because they are mixed together.  Once separated whether by electrolysis or some other means it would take an ignition source to recombine them.  So please elaborate.

That's rather vague.

Oh, one more thing occurred to me.  I updated the previous post with point 5.

Let's go another step further.  The chokes.  The chokes are intended to be mutually opposing.  We know they were different resistances, just slightly offset.  It just occurred to me why this is (again), but I think this may be more plausible.

We know the current must be miniscule when charging the water, otherwise the current will dissipate into electrolysis.  However, when the water is sufficiently charged, we need it to breakdown across the entire cell's gap and the current will need to be high enough at that point to do that.  Here is what I believe happens.  At the beginning of charging the cell, the voltage is low and the frequency is low.  And since the voltage is low, the current on the VIC should also be low as tuned by the chokes, low enough to prevent electrolysis.  But as the voltage increases at resonance, so will the current passing through the very slightly misaligned chokes.  Eventually, at a high enough voltage field to break the water across the gap, the current through the misalignment will grow enough to trigger the final breakdown and WHOOSH, the gas is produced, the voltage and current drops momentarily and then starts all over again.

LM334 might be a better option.

Just an example.

To set a constant current of 1 mA with an LM317, you can configure it as a constant current source by placing a resistor between its output (OUT) and adjust (ADJ) pins. This configuration uses the LM317's characteristic that it maintains a 1.25V voltage drop between the output and adjust terminals.

Steps to Set Up a 1 mA Constant Current Source

1. Choose the Resistor: To set the current, use the formula:



I = \frac{1.25V}{R}

Rearranging for , we get:

R = \frac{1.25V}{I} = \frac{1.25V}{0.001A} = 1250\ \Omega

So, a 1.25 kΩ resistor will give you 1 mA of constant current.

2. Wiring:

Connect one end of the 1.25 kΩ resistor to the output (OUT) pin.

Connect the other end of the resistor to the adjust (ADJ) pin.

The ADJ pin will then connect to your load.

The input (IN) pin is connected to your input voltage source (ensure it is higher than the combined load voltage and 1.25V for proper operation).



3. Input Voltage Requirements:

Ensure your input voltage is at least 3V above the load voltage to give the LM317 sufficient headroom to regulate.




Example Setup

Suppose you have a 12V input and a load that varies in resistance. With this setup, the LM317 will regulate the current flowing through the load to 1 mA, regardless of the load’s resistance (as long as the LM317 has enough input voltage to supply 1.25V across the resistor and the load’s voltage drop).

Circuit Diagram

Here's a basic representation:

IN pin: Connected to your input voltage (e.g., 12V).

OUT pin: Connected to one side of the 1.25 kΩ resistor.

The other side of the 1.25 kΩ resistor connects to the ADJ pin.

ADJ pin: Connects to the positive terminal of your load.

The negative terminal of the load connects to ground.


This configuration will create a steady 1 mA current through your load, independent of input voltage fluctuations.

If that works for you, sure.  I have found limiting the current on the primary works better for me.

Unfortunately due to significant health issues, I don't really actively work on this project anymore. Every once in a while I tinker with it.

Embora eu tenha aprendido português brasileiro nos últimos anos. 😂

Check my YouTube channel tomorrow you will see hopefully a demonstration of current being restricted and voltage taking over
I hope to be able to show the  stepped up waveform with double frequency just like Stan show

No magic

I’m starting to  think that those tube cells was only for comparison with.electrolysis

Remember who wins the war tell the story

Who's at war?  I've already done the wave form too.

As previously mentioned the water resistance must be equal to or higher than the resistance of the coils.  This must be the case for any amount of voltage applied to it because water resistance changes with voltage applied.

Soooo....

You need to know the resistance of one of your tube pairs in water and what the maximum voltage that can be applied before the water resistance becomes lower than the total resistance of your coils.

And then of course build your coils accordingly.

Well, for mine, with calculations based on Meyers and Ronnie's, the sum of the resistances of the coils was near equal to the resistance of a tube pair in water.  I don't think that's a coincidence. 🤔