Author Topic: My new approach  (Read 62565 times)

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Re: My new approach
« Reply #504 on: June 08, 2024, 22:18:06 pm »
I forgot to think first how to connect the cell so now I’m wondering how to make it…
The inner electrode is easy but the outer I will need to make some design

I should have made it before all and leave it connected to the copper foil and maybe even apply resin to avoid leak

The problem of the electrodes is that they need to be only where the water generates voltage otherwise they act as a load to the voltage generated

Of course we still get amp restriction but much lower than would be possible

So is important that they are only within the high field

This is absolutely important otherwise you may see very little effect on voltage and amps



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Re: My new approach
« Reply #505 on: June 08, 2024, 23:36:47 pm »

If I have a set of coils with 39,5mh each and when I connect them in series their inductance is 148,6mh what is the mutual inductance and what would be the inductance when I connect them in series with fields opposing ? Their resistance is 18 ohms each and it’s air core

To find the mutual inductance (\( M \)) of the coils and the inductance when connected in series with fields opposing, we can use the following equations.

### Inductance with Fields in Series (Aiding):

When two inductors are connected in series with their fields aiding, the total inductance (\( L_{aiding} \)) is given by:
\[ L_{aiding} = L_1 + L_2 + 2M \]
where:
- \( L_1 \) and \( L_2 \) are the inductances of the individual coils.
- \( M \) is the mutual inductance.

Given:
\[ L_{aiding} = 148.6 \text{ mH} \]
\[ L_1 = L_2 = 39.5 \text{ mH} \]

Plugging these into the formula, we get:
\[ 148.6 \text{ mH} = 39.5 \text{ mH} + 39.5 \text{ mH} + 2M \]
\[ 148.6 \text{ mH} = 79 \text{ mH} + 2M \]
\[ 2M = 148.6 \text{ mH} - 79 \text{ mH} \]
\[ 2M = 69.6 \text{ mH} \]
\[ M = \frac{69.6 \text{ mH}}{2} \]
\[ M = 34.8 \text{ mH} \]

So, the mutual inductance (\( M \)) is 34.8 mH.

### Inductance with Fields Opposing:

When the fields oppose, the total inductance (\( L_{opposing} \)) is given by:
\[ L_{opposing} = L_1 + L_2 - 2M \]

Using the values we have:
\[ L_{opposing} = 39.5 \text{ mH} + 39.5 \text{ mH} - 2 \times 34.8 \text{ mH} \]
\[ L_{opposing} = 79 \text{ mH} - 69.6 \text{ mH} \]
\[ L_{opposing} = 9.4 \text{ mH} \]

Therefore, the inductance when the coils are connected in series with their fields opposing is 9.4 mH.

Each inductance has a value of 86mh and in parallel the 3 have 39,5 what is the mutual inductance please?

Given the following data:

1. Each coil has an inductance \( L = 86 \text{ mH} \).
2. When three identical coils are connected in parallel, the total inductance is \( L_{parallel3} = 39.5 \text{ mH} \).

We need to determine the mutual inductance \( M \) between the coils.

### Inductors in Parallel:

For three identical inductors in parallel, considering mutual inductance \( M \), the formula for the equivalent inductance \( L_{parallel3} \) is:
\[ \frac{1}{L_{parallel3}} = \frac{1}{L + M} + \frac{1}{L + M} + \frac{1}{L + M} \]

Given:
\[ L = 86 \text{ mH} \]
\[ L_{parallel3} = 39.5 \text{ mH} \]

### Mutual Inductance with Parallel Configuration:

\[ \frac{1}{L_{parallel3}} = \frac{3}{L + M} \]

So:
\[ L_{parallel3} = \frac{L + M}{3} \]

Given \( L_{parallel3} = 39.5 \text{ mH} \):

\[ 39.5 = \frac{86 + 2M}{3} \]

Solving for \( M \):

\[ 3 \times 39.5 = 86 + 2M \]

\[ 118.5 = 86 + 2M \]

\[ 118.5 - 86 = 2M \]

\[ 32.5 = 2M \]

\[ M = \frac{32.5}{2} \]

\[ M = 16.25 \text{ mH} \]

Therefore, the mutual inductance \( M \) is 16.25 mH.


The coupling coefficient (\( k \)) is a measure of how effectively two inductors are magnetically coupled. It is defined as:

\[ k = \frac{M}{\sqrt{L_1 \cdot L_2}} \]

Given that each coil has an inductance \( L_1 = L_2 = 86 \text{ mH} \), and the mutual inductance \( M \) has been calculated as \( 16.25 \text{ mH} \):

\[ k = \frac{16.25}{\sqrt{86 \cdot 86}} \]

First, calculate the product of \( L_1 \) and \( L_2 \):

\[ L_1 \cdot L_2 = 86 \cdot 86 = 7396 \text{ mH}^2 \]

Next, take the square root of the product:

\[ \sqrt{7396} = 86 \text{ mH} \]

Now, calculate the coupling coefficient \( k \):

\[ k = \frac{16.25}{86} \]

\[ k \approx 0.189 \]

Therefore, the coupling coefficient \( k \) in this case is approximately 0.189.


And in the case of having them in series 148mh and in parallel 9,4mh what is the coupling being the mutual inductance 34,8

To find the coupling coefficient (\( k \)) given the series and parallel inductance values, we will use the provided mutual inductance (\( M \)).

### Given:
- Series inductance: \( L_{series} = 148 \text{ mH} \)
- Parallel inductance: \( L_{parallel} = 9.4 \text{ mH} \)
- Mutual inductance: \( M = 34.8 \text{ mH} \)

### Inductance of Each Coil:
From the series inductance formula, we know:
\[ L_{series} = L_1 + L_2 + 2M \]

Given \( L_1 = L_2 = L \) (assuming identical coils):
\[ 148 = L + L + 2 \times 34.8 \]
\[ 148 = 2L + 69.6 \]
\[ 2L = 148 - 69.6 \]
\[ 2L = 78.4 \]
\[ L = 39.2 \text{ mH} \]

So, the inductance of each coil is \( L = 39.2 \text{ mH} \).

### Coupling Coefficient:
The coupling coefficient (\( k \)) is given by:
\[ k = \frac{M}{\sqrt{L_1 \cdot L_2}} \]

Since \( L_1 = L_2 = L \):
\[ k = \frac{34.8}{\sqrt{39.2 \cdot 39.2}} \]
\[ k = \frac{34.8}{39.2} \]
\[ k = 0.887 \]

Therefore, the coupling coefficient (\( k \)) is 0.887.

So the result was exact as I designed to be and expected

The coils have a low coupling between them allowing to lower the inductance when connect them in parallel

The coupling between them side by side is high as I expected 88% this couples the resonance together

Is possible that the screw inside holding it may affect some the readings but is just to get the idea..








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Re: My new approach
« Reply #506 on: June 08, 2024, 23:50:59 pm »
So the missing reading is All six in parallel so I can find the capacitor value to be used split between them



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Re: My new approach
« Reply #508 on: June 09, 2024, 10:04:00 am »
So you are really waiting me to show a video of it working?

Who is there reading? Introduce yourself !

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Re: My new approach
« Reply #509 on: June 09, 2024, 14:26:06 pm »
I made a small piece of stainless steel out of a tube and made it very sharp on one side

Than I made at  the outer tube end a small cut to fit the sharp side of the stainless steel metal

The cell will than be pressed and it will be the connection to the outer tube

I’m 3d printing a holder that will avoid the conduction out of the desired location I may glue the metal in it and it will have a 5mm stainless steel screw holding it and make the electrical conector for the output

The inner electrode is simpler

Anyway the big concern is to not alow it leak water

So probably I may need to glue it on

The problem is that it need to be easy for changing the tube inside so I can test different setups

So I may have to solder it






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Re: My new approach
« Reply #510 on: June 09, 2024, 18:56:01 pm »
I found that for calculating the resonant frequency is actually 22mh that is the sum of the coils in parallel with the 16,25mh mutual inductance between them

I got this aproxima-te value from the simulation and tried to figure out what is happening

This is very non intuitive now

Or we can place the capacitor and calculate with only one inductor and apply the coupling factor to the frequency out

Seems it has 88% of the calculated frequency seems could be related

The formula I did for roled capacitors goes like

Length of copper is = d*c / eo* er * width * k * 2

The 2 comes from the fact that they are rolled so the capacitances add on each side considering all full turns

If they are not full tll the end you can still calculate with adjusting this 2 factor excluding the last layer and the portion of the first layer that don’t complete the turn

This will be very small percentage however

To find the correct capacitor for resonance at a specific frequency once you have the inductance I made this formula

C = 1 /  4 * pi^2 * F^2 * L

You can also find L substituting if you know C

This is the easy way to do it



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Re: My new approach
« Reply #511 on: June 09, 2024, 20:29:04 pm »
I found that the variable resistor will be useful to broad the range of the resonance by simply consuming more power and raising the impedance of the resonance

May have around 49k to  499kohm  in parallel with the chokes to consume some watts

This will require a higher turn ratio on the secondaries driving it to keep the same fields of course

After it’s matched will be easy to simply take it out

The secondary loading it also is a good idea however may not be enough

This secondary can also partetipaye on the tuning by adding capacitance to it too and than using the variable inductor connected to it

So from this simulation you see how simple it is to get the pll to tune we just need the scanning circuit to be off from 180degrees lock range and also make the filter and lock in capacitor such that it will show lock only at the required frequency

From this seems the pll actually don’t follow up the frequency in any way it simply locks and when it’s unlock it scan again and lock again to it


To get any let’s say active locking and tracking to maintain the resonance than the ultrasonic sensor need to kick in when it’s locked!


This can be done with simple circuits to combine or switch

After all this sensor is useless until high energy is in the cavity so makes all the sense to me