Ionizationx: a clean environment is a human right!
Projects by members => Projects by members => Sebosfato => Topic started by: sebosfato on June 23, 2012, 19:13:27 pm

1° You need an LRC meter to determine the inductance, and a frequency meter to determine the frequency
2° construct a simple frequency generator and an H bridge
3° put a coil a resistor an amp meter and the water capacitor in series as the load, and a frequency metter and voltmeter across the inductor or water.
4° Change the frequency to find where the current and voltage is maximum.
5° calculate from the reactances equations or from the frequency equation, the capacitance, since you have now the inductance and the frequency. and you also know XL=XC at resonance, being XL= 2*pi*L*F the capacitance become C=1/(2*PI*F*XC)
Wasn't it easy?

You all should try it.
There are another methods to determine the Q factor also from the relation between applied voltage and taking a voltage measurement of the tank components.
From my understanding the cell will limit the Q factor, so the higher the frequency the smaller is the capacitiwe reactance, the Q factor for a parallel load acros the component is Q=R/XC so the lower is xc the higher the Q factor.

Make an exel sheet with your measurements ...
Plot graphics with your results...
This will make you understand more about what you are doing.

I just did the test.
I used a 4mh inductor it had half flyback core 1ohm 22awg wire. The resonance happened at the frequency 3,7khz
The capacitance was surprisingly 462nf 100times greater than the calculated value. the XL = 92 ohms
Is consiste of a pair of tubes one inside the other having a gap of only 0,6mm about 4 inches long... and 1cm diameter.. They are inside a fiberglass epoxy resin "encapsulated"... inside a glass of water...
I used filtered water, filtered with resin grain filter. (deionized)
I got a 50volt reading when i pushed the system, across the inductor. I forgot to measure across the water. However i would guess is the same... It was drawing about 1 amp. This tells me that the Q was very law but resonance still clearly happened.
Its pretty shocking!!!
It was pretty easy to conduct the experiment i tried first a 50mh coil than 12mh than 140mh and my readings was that the higher the frequency the lower was the voltage across the capacitor, that was because the frequency of my generator wasn't so low, than i decided to try this smaller coil with only half core than the frequency became to low... i changed the capacitor of the 555 timer and tweeked the knobs and its done.
I took only 20 min to find the resonance.
Hope this inspires you.

I was asking myself if anyone here have ever done this before to determine the capacitance and resonance frequency... ?
I thought of it many times but never realized it because i didn't knew how to construct an H bridge.
Now its clear to me why stan wanted a small size water capacitor... and the gap not too small. The higher the frequency the better ... it seems...

I just did the test.
I used a 4mh inductor it had half flyback core 1ohm 22awg wire. The resonance happened at the frequency 3,7khz
The capacitance was surprisingly 462nf 100times greater than the calculated value. the XL = 92 ohms
Is consiste of a pair of tubes one inside the other having a gap of only 0,6mm about 4 inches long... and 1cm diameter.. They are inside a fiberglass epoxy resin "encapsulated"... inside a glass of water...
I used filtered water, filtered with resin grain filter. (deionized)
I got a 50volt reading when i pushed the system, across the inductor. I forgot to measure across the water. However i would guess is the same... It was drawing about 1 amp. This tells me that the Q was very law but resonance still clearly happened.
Its pretty shocking!!!
It was pretty easy to conduct the experiment i tried first a 50mh coil than 12mh than 140mh and my readings was that the higher the frequency the lower was the voltage across the capacitor, that was because the frequency of my generator wasn't so low, than i decided to try this smaller coil with only half core than the frequency became to low... i changed the capacitor of the 555 timer and tweeked the knobs and its done.
I took only 20 min to find the resonance.
Hope this inspires you.
Fantastic!
Congratulations, Fabio.
Ac resonance is probably step 1, is it.
Now you have more info to play with :)
Steve

Thanks bro.
Really its a good thing, because it changes all the parameters...
This at least tells us that the wfc capacitance calculator and those who are claiming you need to use it for finding the resonant frequency are a big scam.
This also allow us to design a resonant coil based on the water resonance for a desired frequency... Because now we know how to test and find it.
This will allow me to make a further study of the dielectric properties of water as function of applied vt and f. I want now to vary the frequency carefully and try to find a variation on the Q factor of the circuit. Stan said that at 26v his tube resonated at 5khz and would be the frequency related to the back and forth movement of the ions thru the water...
Now i'm pretty sure why meyer used the variable spacing wfc cell as well as the variable inductor... For me theres still another hidden components.
What i also thought is that maybe those all in one vics are kind of non sense. because the ultrahigh inductance values would drop the resonant frequency maybe to 10hz...
I will try to find the sweet spot taking out the inductor and inputing the square wave directly at the water where the current drops would the the sweet spot.
sebos

What i also thought is that maybe those all in one vics are kind of non sense. because the ultrahigh inductance values would drop the resonant frequency maybe to 10hz...
sebos
Or maybe not. The reported thiknes of the core laminat wuld sa the max freq iz 200hz of the all in 1 vic

Well it could be i'm not saying is this or that, but can be.

Took new measurements... at the resonance the voltage is maximum at the inductor only.. at the capacitor the voltage is high at a lower frequency than resonance than at resonance and the further the frequency goes up the lesser is the woltage.
Thus the best way to find the resonance is to measure the voltage across the inductor.

Seems to me the best way to find resonance is to find the inductance value of the coil and the capacitance of the cell and use the standard equation to calculate it. As you stated to me recently, the reactance must be equal between the two. Additionally, there will be a bandwidth where resonance will occur, so it may not be at one single frequency.
TS

Took new measurements... at the resonance the voltage is maximum at the inductor only.. at the capacitor the voltage is high at a lower frequency than resonance than at resonance and the further the frequency goes up the lesser is the woltage.
Thus the best way to find the resonance is to measure the voltage across the inductor.
Thats odd....

Well it might seems strange, i made an excel sheet and ploted a graph, at resonace of course the reactances are the same and thats why current is maximum, however the XC is not linear like the inductive reactance, so at resonance both are the same but as the frequency is increased the higher is the XL and the XC becomes lower as the voltages should be equal to the reactance times the current, the voltages across the inductor and the capacitor become smaller but the capacitance voltage drops more than the inductor.
Also the lower is the Q factor the greater is the difference in frequency between the maximum voltage at the inductor and at the capacitor accoring to the graphs... Actually when i learned about the harmonic oscillator at the physics lecture, i learned there are two types of resonance, one is at the amplitude and the other is at the power. Maybe thats the difference...
Yet I attributed the lower voltage at the capacitor than at the inductor to the fact that although the capacitor holds the charge, it also has the greater loss in the circuit. so or the capacitance change as a function of V(t) and i think it does because i got to change the frequency to find a greater voltage if i raised the Vt applied.
Again at F=0 the capacitor has Vt the inductor has 0v
Closer to resonance the capacitor has greater voltage than the inductor, the current start to raise
At resonance the voltages should be equal but as there are losses in the capacitor it will have a lower voltage than at the inductor. current is maximum
little above the resonance the inductor voltage still greater than the capacitors, the current decreases
way above resonance the voltage at the inductor Vt and capacitor is 0 the current is 0 too
In theory this should be the behavior
In the case of water I think because of the losses, the voltage across it aways decrease from 0 frequency to max frequency. so no voltage peaks... Maybe at resonance it decrease less..
Thats why i said its easier to find the resonance measuring the voltage across the inductor...
Since stanley had the feed back on the inductor, this imply Zero current switch and imply that the frequency is adjusted to max inductor voltages.
Ts as i stated @resonance XL = XC so XLXC=0 condition required for resonance, yes there is a bandwidth not much wide.
What i'm saying is the waters capacitance is not the calculated value because it's a lossy dielectric, this experiment thus allow you to empirically determine the capacitance observing the frequency where resonance happens.
The first graph has 100 ohm resistance and the other 20 ohms
C = 500n
L = 1mh
Fres=7118

If you try this simple circuit you see that the capacitor having a resistance in parallel with it shows a smaller voltage than the inductor.
Now i understand why. The Voltage across the inductor will be = to the current passing thru the capacitance + the current passing thru the resistive path. while the capacitance only the capacitive current...
$ 1 3.4000000000000003E7 61.195984441989225 50 5.0 43
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s 560 128 624 128 0 1 false
c 352 400 560 400 0 5.000000000000001E7 105.48842737546646
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Just did other tests, and i'm not going to share much of the parameters to force you doing this experiment
I tried different inductors, using and not using cores, and got different readings for the capacitances in respect to frequency...
The lowest capacitance I found were at 50khz around 350nf + 20nf

When you did your capacitance calculations for your water cells, did you factor in the dielectric of water into the equation? If factoring in the dielectric into the equations still leads to such a massive error, then either the equation is wrong, or the water dielectric value being assumed is wrong. Either way, it would be more appropriate to correct the equation and dielectric factors and help everyone to correctly calculate the dielectric values. It would save time and help everyone figure out this riddle.
TS

The value of 78.5 is obtained in a different way, it is determined differently...
The problem i see for this is that i think the dielectric constant or the water capacitor behaves different according to applied voltage and frequency.
So the best way is to measure the capacitance for different shapes, sizes, voltages and frequencies, and plot graphs with the results.
I'm using square wave, probably would be better a sine wave...
Yes i used the dielectric constant in the calculation and the calculation is correct because i can measure the capacitance without water inside... but the capacitance don't get anyway close to the capacitance measured on this experiment.
We should be able to create thru calculation a dielectric constant function of voltage and frequency.

Incidentally, when using distilled water, my calculations have usually been within 200Hz of accurate when manually tuning to resonance.
TS

Its know that a RC capacitance resistance series circuit, has a time constant t=R*C
The time constant is the time it takes for a capacitor to charge from zero to 63,2% of the applied voltage. 0,632 * Vo
So If any of you have an oscilloscope and could try some measurements would be a nice.
Some one could Even create a comparator circuit regulated for this parameter and measure the output pulse length. Its easy you just need an opamp and a two resistors divider circuit or a trim pot for the reference. Sharing a common ground the opamp and the cell and having the resistor between the positive and the wfc capacitor the up side of the capacitor is connected to the other input of the opamp... switch on and freeze the scope shot. another way is to record the signal as audio *.*.wav to the computer... and analyze the pulse length with any good audio editing program like sound forge.

Let take out the juice.
just need a battery a resistor a switch a wfc and a potentiometer and a pc. connect the resistor to the cell to the car battery to the switch now across the cell connect the trimpot, and send the signal to the audio in port of the computer... press record in any audio recording software and switch on.
Than inside sound forge or the like measure where the voltage arrived at 63,2%. thats the time constant.
Try different resistors

Connect a battery a resistor and the cell. Measure the current and voltage across the resistor and the cell
The resistance might be a function of voltage.
This is kind of how you measure the internal resistance of a battery (polarization)
You measure the open circuit voltage than connect the battery to a resistor measure again the voltage across the battery and the current. The voltage drop Vd=I*Ri is the current times the internal resistance thus Ri=Vd/I
At the end they are both electrochemical components.
So consider its internal resistances.

If I remember well the relative dielectric constant is determined as the ratio between applied voltage and the induced contrary voltage inside the capacitor and is measured by charging the capacitor and subsequently disconnecting it from the source while still measuring the voltage across the capacitor.
From the document i attached there is an example for the hydrochloric acid which has a dielectric constant of 4.12 acording to the reference...
In the example is not really clear, because they say E0=1,5v and Enet=1,15v being Enet=E0E1 solving E1 becomes 0.35v
Than they say that K=V0/V
however 1,5/1,15=1,3 and the value i found for hydrochloric acid dielectric at the reference is 4,12
so i found that 1,51,15=0,35 that would be the E1 the induced contrary voltage than if i divide 1,5/0,35 than i find a value that is in agreement with the reference 4,3.
this would lead to
For example you charge a water capacitor to 100v And its dielectric constant is 81k Thus the induced contrary voltage equals 100v/81k=1,234v So if a capacitor was charged up to 100v and is disconnected the measured voltage is 100v1,23v=98,76v
If the document was right:
If it was like they say in the document the calculation for water would be 100v/81=1,23v however this time the 1,23v is the remained Enet so the E1 would be 98,76v
If it were vacuum which has 1 as dielectric constant, the induced contrary voltage should be zero than i would guess, so 100/1=100 so Enet=100 and thereto E1 polarization = 0. Which seems to be the case...
So why the contradiction on the hydrochloric acid? is the document wrong only there?
reference:
http://www.asiinstr.com/technical/Dielectric%20Constants.htm#Section%20H

I was thinking about it to try to explain to myself why the voltage across the capacitor has no peak at resonance while resonance still occur clearly at the inductor.

Mystery solved...
At least in part.
The document was right, although the example of the hydrochloric acid is wrong in numbers.
Thats because a dielectric constant affects the ratio of voltage to charge since Q=CV If the capacitance increase the voltage reduces... This reduction is actually this contrary induced voltage (polarization)
So for water if you apply 100v there is 810v induced contrary thereto reducing voltage per unity charge ("increasing capacitance") thus the electric field so the electric field applied is 100v after the capacitor is disconnected from the source.
Or this is correct or the assumption made at the document is wrong... Maybe they wanted to mean that if the dielectric is inserted than the effective electric field reduces by a factor K instead of turning on the switch and turning off while still measuring the voltage.
Because if this is true and the dielectric strength of water is 2,946MV/cm it would become clear that applying 3,637kv the water would breakdown.
reference
http://www.tf.unikiel.de/matwis/amat/elmat_en/kap_3/exercise/s3_2_1.html#_dum_9

One thing that keeps bugging me is the fact that we are dunking water electrodes directly into water. Whever I have done that, I have never seen the voltage across the electrodes go much above a few volts. If we want a very high charge, we need higher resistance than the water is giving. For the electron extraction circuit, we need the electrodes to be in direct contact with the water to pick up the electrons. However, without the electron extraction portion of the circuit, wouldn't it be better to have another dielectric, an insulator, on the electrodes so that there is no loss? I mean, we're not interested in conduction through the water anyway, we don't want the charges of the circuit to directly interact with the water. We want the charges to influence the molecules with atomic forces. So, if the electrodes were completed insulated from the water, they could be charged to higher values with less losses and be able to exert a greater force without the electrochemical reaction with the water molecules.
Does this make sense?
TS

The problem adding insulation the way you described is that if you have two capacitances one big and one small in series the smaller will hold most of the field. For example you have a 100nf and a 100pf in series and charge this to 10kv. It works like a voltage divider, since Q=CV and in series the Q is the same for both. so V/((1/C1)+(1/C2))=Q=> V1=Q/C1 and V2=Q/C2 than C1 will have 9990volts and C2 only 10V.
Tay hee patent explain this.
reference
http://www.google.com/patents/US4427512

Ok, so what about the thin oxide layer that occurs with almost any and every material imaginable? There will always be a secondary capacitive effect.
TS

This capacitance is way too big and too thin. So the strength is low and the capacitance too big if i remember well is 350uf/cm2 theoretically...
That patent also say that if the dielectric constant is huge compared to that of water a high field can be applied to it... in excess of 20kv...
As far as i understand it, the higher the dielectric constant the higher is the conductivity... so i dont know how they did it.. because voltage drop across the water would be much greater...
If your cell had 10kohm of resistance to dc you need to apply 1 amp to raise its voltage to 10kv so you spend 10kw

Hi TS
The reason you not getting high voltage is because you are doing/comparing electrolysis.
If you look close, you notice a 3 seconds timedelay when you apply power to two electrodes, before electrolysis begins.
My suggestion is to have only very short pulses. H2O acts in that case as resistor.
Voltage can be high, as long as amps are restricted to flow....

Hi TS
The reason you not getting high voltage is because you are doing/comparing electrolysis.
If you look close, you notice a 3 seconds timedelay when you apply power to two electrodes, before electrolysis begins.
My suggestion is to have only very short pulses. H2O acts in that case as resistor.
Voltage can be high, as long as amps are restricted to flow....
Ok, makes sense. But isn't that what Meyers PWM was supposed to do, the gated pulse?
Here's where I'm at so far.
1. I have calculated the approximate capacitance of my cell .
2. I have created a VIC coil set with the chokes at approximately 8.78k of resistance.
3. I have calculated my resonant frequency at 6.622kHz and found tuned resonance at around 6.4kHz.
4. Ali has brought my attention to the self resonance properties of coils. This is new to me and is an interesting parameter. Could a coil designed to to self resonate at the same frequency as the circuit resonant frequency improve the effect?
5. My Tony Woodside Vic circuit is presently damaged and I haven't had enough time to figure out how to troubleshoot it yet (the gate pulse isn't working) so I can only work with the resonant frequency ungated until I figure out whats wrong.
6. Now going back to your statement about shorter pulses. I can only assume you mean a lower gated duty cycle as we still want the resonant frequency to exist.
Opinions?
TS

yup.
very short pulses.
some space in between

Hi TS
The reason you not getting high voltage is because you are doing/comparing electrolysis.
If you look close, you notice a 3 seconds timedelay when you apply power to two electrodes, before electrolysis begins.
My suggestion is to have only very short pulses. H2O acts in that case as resistor.
Voltage can be high, as long as amps are restricted to flow....
Ok, makes sense. But isn't that what Meyers PWM was supposed to do, the gated pulse?
Here's where I'm at so far.
1. I have calculated the approximate capacitance of my cell .
2. I have created a VIC coil set with the chokes at approximately 8.78k of resistance.
3. I have calculated my resonant frequency at 6.622kHz and found tuned resonance at around 6.4kHz.
4. Ali has brought my attention to the self resonance properties of coils. This is new to me and is an interesting parameter. Could a coil designed to to self resonate at the same frequency as the circuit resonant frequency improve the effect?
5. My Tony Woodside Vic circuit is presently damaged and I haven't had enough time to figure out how to troubleshoot it yet (the gate pulse isn't working) so I can only work with the resonant frequency ungated until I figure out whats wrong.
6. Now going back to your statement about shorter pulses. I can only assume you mean a lower gated duty cycle as we still want the resonant frequency to exist.
Opinions?
TS
Sorry but seems improbable that resonant frequency... even more improbable that it is the same frequency you calculated, even more using the capacitance found from the capacitance formula. And yet more unlike because of the high resistance.
What you are seeing as peak there, can be impedance matching...
An inductor has a characteristic that it resist to a change in amp flow, using AC because it takes a time t to get charged with current so every time the polarity flips it need to start over again. thus the greater the frequency the lower is the time thus the greater the reactance.
A capacitor impedes Dc current flow when it is charged to dc voltage but allow high frequency to pass.. And its reactance is not linear. A capacitor immediately charges when connected to the source, so every time the polarity of the source flips not only the capacitor gets discharged but also gets charged in the opposite direction so as the frequency increases the more times this happens the lower become the reactance.
Actually the reactance is a measure of how much energy can be transported reactively at a given voltage and frequency applied.

Resonance arises from the fact that when you have an inductor and a capacitor in series, while the capacitor is being charged the current flowing induces a contrary voltage at the inductor, so in series the voltages sums to zero, this let the source to charge the capacitor to higher voltages as it appear like a short circuit for the source. When the current decay the voltage across the inductor becomes smaller while capacitors voltage raises. This is why an inductor has a voltage peak that appear 90° before the capacitor peak voltage.
There are 360° for a complete cycle, so lets examine it:
Consider the capacitor is not charged initially
1° 90° The current is maximum so as voltage across the inductor. The voltage of the inductor is a cosine function, while the capacitors voltage follow a sine function up to its max voltage
2° 90° current stops, and the capacitor completely discharge into the inductor.
3° 90° the inductor charge the capacitor with reversed polarity
4° 90° the capacitor discharge into the inductor.
Summing to that, at the end of the cycle, when current is zero, the capacitor has the full voltage across it, and the polarity is inverted, so you can aways add more energy to it. Because as you apply a contrary voltage to a capacitor charged the, energy at the end will be the sum of the energy you inputed + the energy originally at the capacitor.
Its known that the time constant for a capacitor depends on its capacitance and the resistance in series.
As a Capacitor accumulates a charge Q the peak current will depend on the time constant.
If you discharge the capacitor applying a contrary voltage to it, would the RC time constant be the same?
I think would become smaller, because you are forcing it so the peak current will be greater...
Of course resonance voltage is only limited by the real resistance of the circuit and to the voltage applied...
The resistance will affect the phase limit between voltage peaks therefore it will limit the resonance not only by dissipating power but impeding the voltages in series to completely cancel out.

Thats correct and proven. :)

Meyers indicated that each inductor in the VIC should have a resistance of of 11.6K. How is it you say the resistance of mine is too much?
TS

well, the Q factor of the coil say that it would not resonate. Meyer said this value but probably he was talking about impedance not ohmic resistance. Because than it makes sense if is a resonant choke. Choke.
Take the impedance matrix circuit as reference.

To achieve this resistance I did not use standard copper wire. I used resistor wire. I achieved 8.78k with about 400 turns.
TS

I understood, I tried it also. You wire is very thin isn't.
Try to put this coil and a capacitor in series and find the resonant frequency. You will see that if any peak is observed the band width will be really wide.

I found a thread discussing, dielectrics water...
http://www.overunity.com/9856/relativepermittivityofwater/
Anyway the closer to reality value was found by Steve... 200nf
And they state water is not a dielectric because it conduct dc... many laughs

Here is a link to various methods to test capacitance for unknown capacitors. I will be trying out method 4 soon.
http://www.pupman.com/listarchives/1998/April/msg00625.html (http://www.pupman.com/listarchives/1998/April/msg00625.html)
TS

It seems nice, I wish you luck. Thanks for the link

Another mean for discovering the capacitance is to construct a relaxation oscillator with an opamp. The frequency output ^1 will give you the period of the RC product.
The opamp positive input must be set to 63% of the supply voltage...
As the water is leaky a mean to provide the current needed is to use a low R value. Thus a simple opamp relaxation is not enough, you need a Half bridge to boost the opamp current.
I believe this will give out the best results... if correctly used.
The opamp must be supplied with the same voltage or so going into the bridge.... So one could use two batteries, a full wave bridge rectifier from a variac and using a ground connection to split the DC output into positive and negative between two big capacitors at the dc output...

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Here i used the voltage divider to tune the threshold to satisfy the RC time constant frequency for a known capacitor. Than the only limitation is that as the R in series is = to the R in parallel with water the frequency will become a little smaller, this off set can be interpreted as the ratio between the resistances. In this case the precision will be within around 7%... As the series resistance is preset much smaller than water leak the precision goes up to 1% or less...

So using test 4 from the link in my previous post, I have determined that the actual capacitance of my one test cells is 40nf where I originally calculated it to be 25nf. What is more (and possibly more important) I have also determined that the water cell's internal resistance is 3 ohms.
This also makes the dielectric value of the water I am presently using to be 125. Either that or I am using the wrong constant for permittivity.
In case anyone is interested.
TS

So using test 4 from the link in my previous post, I have determined that the actual capacitance of my one test cells is 40nf where I originally calculated it to be 25nf. What is more (and possibly more important) I have also determined that the water cell's internal resistance is 3 ohms.
In case anyone is interested.
TS
Presumably, to use the fact that the resistance is 3 ohms, to make use of this, I would need to create a pulse that is short enough that 3 ohms is enough to prevent dielectric breakdown by current.
TS

Hi TS
What are the area and separation of your electrodes, if were tubes what are the length and diameters?
How did you made the test? what were the values of the components?
if you connect a battery with 12v across the water the current is 4 amps ?
Is that pure water? tap? distilled?
Thanks

I'm trying some variations to confirm my results. I'll provide more details soon.
TS

Make a tables, take measurements at different frequencies, different voltages and components...
than you can compare and be more sure of your measurement...
good luck