### Author Topic: My new approach  (Read 62567 times)

• Global Moderator
• Hero member
• Posts: 4181
##### GPT talks
« Reply #32 on: February 16, 2023, 19:33:44 pm »
What is the electric charge at the tip of a 10mm diameter wire of copper having 10m of length and a voltage of 1000v across its length

To calculate the electric charge at the tip of a copper wire, we need to use the formula:

Q = CV

where Q is the charge, C is the capacitance, and V is the voltage across the wire.

The capacitance of a wire is given by:

C = (2πεL) / ln(b/a)

where ε is the permittivity of free space, L is the length of the wire, and a and b are the radii of the wire and a surrounding conductor, respectively. Since the wire is assumed to be isolated, b can be taken to be infinity, so we can simplify the formula to:

C = (2πεL) / ln(b/a) = (2πεL) / ln(infinity/a) = (2πεL) / ln(infinity) = (2πεL) / infinity = 0

This means that the capacitance of the wire is effectively zero, so the charge at the tip of the wire will also be zero. The voltage across the wire does not produce any significant charge accumulation due to the low capacitance of the wire.

Therefore, in this case, the electric charge at the tip of the copper wire is essentially zero.

Fabio Tortorella
But as copper has a resistance it would never go to infinity
You are correct that the resistance of the copper wire will limit the current flow and prevent the voltage from going to infinity. However, even with a finite resistance, the capacitance of the wire is still very small, which means that the charge that accumulates on the wire is negligible compared to the amount of charge flowing through it.

To calculate the charge on the wire, we need to use the formula:

Q = CV

where Q is the charge, C is the capacitance, and V is the voltage across the wire. As I mentioned earlier, the capacitance of the wire is very small, and we can assume it to be approximately zero. Therefore, the charge on the wire can be calculated as:

Q = CV ≈ 0

So, even though the wire has a resistance and the voltage will not go to infinity, the charge on the wire is still negligible due to the small capacitance. The resistance of the wire will affect the current flow, and the voltage drop across the wire, but it does not significantly affect the charge accumulation on the wire.

Fabio Tortorella
What would be that small capacitance and charge?
To calculate the capacitance of the wire, we can use the formula:

C = 2πεL / ln(b/a)

where ε is the permittivity of free space, L is the length of the wire, and a and b are the radii of the wire and a surrounding conductor, respectively. Since the wire is assumed to be isolated, we can take b to be very large compared to a, so we can simplify the formula to:

C = 2πεL / ln(b/a) ≈ 2πεL / ln(b)

The permittivity of free space is approximately 8.854 × 10^-12 F/m, and the length of the wire is given as 10 m. If we assume the wire has a radius of 5 mm (which corresponds to a diameter of 10 mm), then we can estimate the capacitance of the wire as:

C ≈ 2π × 8.854 × 10^-12 F/m × 10 m / ln(10 m / 5 mm) ≈ 1.16 × 10^-13 F

To calculate the charge on the wire, we can use the formula:

Q = CV

where Q is the charge, C is the capacitance, and V is the voltage across the wire. The voltage across the wire is given as 1000 V. Using the capacitance we calculated above, we get:

Q = CV ≈ (1.16 × 10^-13 F) × (1000 V) ≈ 1.16 × 10^-10 C

So, the capacitance of the wire is approximately 1.16 × 10^-13 F, and the charge on the wire is approximately 1.16 × 10^-10 C. While this charge is small, it is still non-zero and can be measured in certain circumstances.

Fabio Tortorella
What if instead of copper it was stainless steel 430 ?
The capacitance of a stainless steel wire would depend on the same factors as a copper wire, namely its length, radius, and the permittivity of free space. However, the resistivity of stainless steel is different from that of copper, which would affect the voltage drop across the wire.

To calculate the capacitance of a stainless steel wire, we can use the same formula as before:

C = 2πεL / ln(b/a)

However, the permittivity of free space and the length of the wire would be the same as before. The radius of the wire would also be 5 mm, assuming the same diameter of 10 mm. The resistivity of stainless steel 430 is approximately 6.9 × 10^-7 Ωm, which is higher than that of copper. This means that for the same current, a higher voltage drop would occur across the stainless steel wire compared to a copper wire of the same length and radius.

Assuming a voltage of 1000 V across the wire, we can estimate the charge on the wire using the capacitance we calculated earlier, which gives:

Q = CV ≈ (1.16 × 10^-13 F) × (1000 V) ≈ 1.16 × 10^-10 C

So, the charge on the stainless steel wire would be approximately the same as for the copper wire, assuming the same dimensions and voltage. However, the voltage drop across the wire would be higher due to its higher resistivity.

Fabio Tortorella
So it don’t depend much on the material? Or if it was like water in place of that wire?
The capacitance of a wire or any conductor depends on the geometry of the conductor, which is determined by its length, radius, and the permittivity of the surrounding medium. The permittivity of the surrounding medium is usually the permittivity of free space for most practical purposes, as the effects of other surrounding materials are usually negligible.

The resistivity of the material used for the wire affects the amount of current that can flow through the wire and the voltage drop across the wire. However, for the purpose of calculating the charge on the wire, the capacitance is the most important factor.

If we consider a wire made of water, its capacitance would depend on its length, radius, and the permittivity of water, which is much higher than that of air or most other non-polar materials. As a result, the capacitance of a water wire would be higher than that of a copper or stainless steel wire of the same length and radius.

However, it is important to note that the concept of a wire made of water is not physically possible, as water is a liquid and cannot be formed into a wire shape without the aid of a solid structure. Nonetheless, the capacitance of a wire-like structure made of water would depend on the same factors as a solid wire, and its charge would be calculated using the same formula as before.

Fabio Tortorella
Calculate for me assuming it would be possible to make water into a wire form
Assuming it were possible to create a water wire with a diameter of 10 mm and a length of 10 m, we can calculate the capacitance of the wire using the same formula as before:

C = 2πεL / ln(b/a)

where ε is the permittivity of water, L is the length of the wire, and a and b are the radii of the wire and a surrounding conductor, respectively. Since the wire is assumed to be isolated, we can take b to be very large compared to a, so we can simplify the formula to:

C = 2πεL / ln(b/a) ≈ 2πεL / ln(b)

The permittivity of water is approximately 80 times higher than the permittivity of free space, or 80ε_0, where ε_0 is the permittivity of free space. Therefore, we can estimate the capacitance of the water wire as:

C ≈ 2π × 80ε_0 × 10 m / ln(10 m / 5 mm) ≈ 2.93 × 10^-10 F

To calculate the charge on the water wire, we can use the same formula as before:

Q = CV

where Q is the charge, C is the capacitance, and V is the voltage across the wire. If the voltage across the wire is 1000 V, we get:

Q = CV ≈ (2.93 × 10^-10 F) × (1000 V) ≈ 2.93 × 10^-7 C

So, the charge on a water wire with a diameter of 10 mm and a length of 10 m and a voltage of 1000 V across its length would be approximately 2.93 × 10^-7 C. However, it is important to note that creating a wire-like structure out of water is not a practical or feasible solution due to the fluid nature of water.

Fabio Tortorella
If we had half the length would be double? Right?
Yes, that's correct. If we halve the length of the water wire to 5 m, the capacitance of the wire would be half of what it was before, since capacitance is directly proportional to the length of the wire. Therefore, the new capacitance would be:

C ≈ 2π × 80ε_0 × 5 m / ln(10 m / 5 mm) ≈ 1.46 × 10^-10 F

Similarly, if we assume the same voltage of 1000 V across the wire, the charge on the wire would be double what it was before, since charge is directly proportional to the capacitance of the wire. Therefore, the new charge on the wire would be:

Q = CV ≈ (1.46 × 10^-10 F) × (1000 V) ≈ 1.46 × 10^-7 C

So, if we halve the length of the water wire, the capacitance would be halved and the charge on the wire would be doubled, assuming the same voltage across the wire.

Fabio Tortorella
You are assuming a round water wire?
Yes, in the calculations I assumed that the water wire had a circular cross-section, as this is the most common and simple shape for wires. However, the formula for capacitance that I used applies to any cylindrical conductor, regardless of its shape, as long as its radius and length are known. In practice, the capacitance of a non-cylindrical water wire with a different cross-sectional area and shape would have to be calculated using different methods, but the same principles of capacitance calculation would still apply.

Fabio Tortorella
Could you please do the calculation for 10mm square shaper wire ?
Sure! To calculate the capacitance of a 10 mm square wire, we can use the formula for capacitance of a rectangular parallel-plate capacitor:

C = εA/d

where ε is the permittivity of the medium between the plates, A is the area of one plate, and d is the distance between the plates.

Assuming that the wire is isolated, we can treat it as one plate of the capacitor, and the surrounding medium as the other plate. If we assume that the wire is a perfect square, its area would be:

A = (10 mm)^2 = 100 mm^2

The distance between the plates would be equal to the width of the wire, which is 10 mm. The permittivity of free space, ε_0, can be used as the permittivity of the surrounding medium, since air is a good insulator. Therefore, the capacitance of the wire would be:

C = ε_0A/d = ε_0(10 mm^2)/(10 mm) = ε_0 × 10 mm

Using the value of the permittivity of free space, ε_0, which is approximately 8.85 x 10^-12 F/m, we can convert the capacitance to farads:

C ≈ 8.85 × 10^-12 F/m × 10 mm ≈ 8.85 × 10^-14 F

So, the capacitance of a 10 mm square wire would be approximately 8.85 x 10^-14 F.

To calculate the charge on the wire if a voltage of 1000 V is applied across it, we can use the formula:

Q = CV

where Q is the charge, C is the capacitance, and V is the voltage across the wire. Assuming a voltage of 1000 V, we get:

Q = CV ≈ (8.85 × 10^-14 F) × (1000 V) ≈ 8.85 × 10^-11 C

Therefore, the charge on a 10 mm square wire made of a perfect electrical conductor, assuming a voltage of 1000 V across its length, would be approximately 8.85 × 10^-11 C.

Fabio Tortorella
And if it was made of water?
To calculate the capacitance of a 10 mm square water wire, we can use the same formula as before:

C = εA/d

where ε is the permittivity of the medium between the plates, A is the area of one plate, and d is the distance between the plates.

For a water wire, the permittivity would be the permittivity of water, which is about 80 times larger than the permittivity of free space. The area of the wire would be the same as before:

A = (10 mm)^2 = 100 mm^2

The distance between the plates would also be 10 mm, since this is the width of the wire. Therefore, the capacitance of the water wire would be:

C = εA/d = (80ε_0)(10 mm^2)/(10 mm) = 800ε_0 × 10^-6 F

Using the value of the permittivity of free space, ε_0, which is approximately 8.85 x 10^-12 F/m, we can convert the capacitance to farads:

C ≈ 800ε_0 × 10^-6 F/m × 10 mm ≈ 7.08 × 10^-12 F

So, the capacitance of a 10 mm square water wire would be approximately 7.08 x 10^-12 F.

To calculate the charge on the water wire if a voltage of 1000 V is applied across it, we can use the same formula as before:

Q = CV

where Q is the charge, C is the capacitance, and V is the voltage across the wire. Assuming a voltage of 1000 V, we get:

Q = CV ≈ (7.08 × 10^-12 F) × (1000 V) ≈ 7.08 × 10^-9 C

Therefore, the charge on a 10 mm square water wire, assuming a voltage of 1000 V across its length, would be approximately 7.08 × 10^-9 C.

Fabio Tortorella
What is the impedance of a two copper foils having 3m length and 50mm width having 0,3mm polietileno plastic isolation between them
To calculate the impedance of two copper foils with plastic insulation between them, we need to take into account the capacitance and inductance of the structure, as well as the resistance of the copper foils.

Assuming that the copper foils are ideal conductors, the resistance can be calculated using the formula:

R = ρL/A

where ρ is the resistivity of copper, L is the length of the foil, and A is the cross-sectional area of the foil. Using the resistivity of copper, which is approximately 1.68 x 10^-8 Ωm, and the dimensions given, we get:

A = (50 mm) x (0.3 mm) = 15 mm^2 = 1.5 x 10^-5 m^2

R = (1.68 x 10^-8 Ωm) x (3 m) / (1.5 x 10^-5 m^2) ≈ 0.00336 Ω

To calculate the capacitance of the structure, we can use the formula:

C = εA/d

where ε is the permittivity of the plastic insulation, A is the area of one of the copper foils, and d is the thickness of the plastic insulation. Using the permittivity of polyethylene, which is approximately 2.25 x 10^-11 F/m, we get:

C = (2.25 x 10^-11 F/m) x (15 mm^2) / (0.3 mm) ≈ 1.125 x 10^-12 F

To calculate the inductance of the structure, we can use the formula:

L = μL/π * ln(4h/d)

where μ is the permeability of free space, L is the length of the structure, h is the height of the structure (which we can assume is negligible), and d is the distance between the copper foils. Using the permeability of free space, which is approximately 4π x 10^-7 H/m, we get:

L = (4π x 10^-7 H/m) x (3 m) / (π ln(4(0.3 mm)/(1 μm))) ≈ 8.74 x 10^-7 H

Now, we can use these values to calculate the impedance of the structure using the formula for the impedance of a parallel RLC circuit:

Z = R + j(ωL - 1/ωC)

where ω is the angular frequency, which is equal to 2π times the frequency of the signal passing through the structure. Assuming a frequency of 1 kHz, we get:

ω = 2π x 1 kHz = 6.28 x 10^3 rad/s

Substituting the values we calculated, we get:

Z = 0.00336 Ω + j(6.28 x 10^3 rad/s x 8.74 x 10^-7 H - 1/(6.28 x 10^3 rad/s x 1.125 x 10^-12 F))

Z ≈ 0.00336 Ω + j(7.07 Ω - 14.12 Ω)

Z ≈ 0.00336 Ω - j 7.05 Ω

So the impedance of the structure is approximately 0.00336 Ω - j 7.05 Ω.

Fabio Tortorella
I mean like if it was a transmission line
If we consider the two copper foils with plastic insulation between them as a transmission line, then we can use the transmission line equations to calculate its impedance.

Assuming a coaxial transmission line configuration, where the two copper foils are concentric cylinders with the plastic insulation in between, the characteristic impedance of the transmission line can be calculated using the following formula:

Z0 = 276 log10(b/a) / sqrt(εr)

where a is the inner radius of the inner copper foil, b is the outer radius of the outer copper foil, and εr is the relative permittivity of the plastic insulation.

Assuming the same dimensions as before (3m length, 50mm width, and 0.3mm plastic insulation thickness), we can calculate the inner and outer radii of the copper foils as:

a = 0.05 m - 0.0003 m = 0.0497 m
b = 0.05 m

The relative permittivity of polyethylene is approximately 2.25, so we can substitute these values into the formula to get:

Z0 = 276 log10(0.05 m/0.0497 m) / sqrt(2.25) ≈ 69.2 Ω

So the characteristic impedance of the transmission line is approximately 69.2 Ω.

Note that the actual impedance of the transmission line may be different due to the effects of frequency, length, and other factors.

Fabio Tortorella
Yes, it is possible to calculate the impedance of a flat two-wire transmission line with a dielectric insulator between them. In this case, we can use the following formula to calculate the characteristic impedance:

Z0 = (276 / sqrt(εr)) * ln(2h/w)

where h is the distance between the two wires (the thickness of the dielectric insulator), w is the width of the transmission line, and εr is the relative permittivity of the dielectric.

Assuming the same dimensions as before (3m length, 50mm width, and 0.3mm plastic insulation thickness), we can calculate the distance between the two wires as:

h = 0.3 mm = 0.0003 m

Substituting this and the other values into the formula, we get:

Z0 = (276 / sqrt(2.25)) * ln(2*0.0003 m/0.05 m) ≈ 72.5 Ω

So the characteristic impedance of the flat two-wire transmission line is approximately 72.5 Ω.

Note that the actual impedance of the transmission line may be different due to the effects of frequency, length, and other factors.

Fabio Tortorella
If we had a 0.5mm plastic what would be the impedance?
If the thickness of the plastic insulation between the two wires is changed from 0.3mm to 0.5mm, the characteristic impedance of the transmission line will change. Using the same formula as before:

Z0 = (276 / sqrt(εr)) * ln(2h/w)

and assuming the same dimensions as before (3m length, 50mm width), but with a plastic insulation thickness of 0.5mm, we can calculate the characteristic impedance as follows:

h = 0.5mm = 0.0005m

Z0 = (276 / sqrt(2.25)) * ln(2 * 0.0005m / 0.05m) ≈ 62.8 Ω

So with a 0.5mm plastic insulation, the characteristic impedance of the transmission line would be approximately 62.8 Ω.

Note that changing the thickness of the plastic insulation can have a significant impact on the characteristic impedance of the transmission line, and this will affect the performance of the line in terms of signal propagation and reflection.

Fabio Tortorella
Só tô increase the impedance I need to increase the capacitance?
Increasing the capacitance between the two conductors in a transmission line can increase the characteristic impedance of the line. This is because the characteristic impedance of a transmission line is determined by the ratio of the line's inductance to its capacitance.

In a two-wire transmission line, increasing the capacitance between the two conductors can be achieved by increasing the surface area of the conductors or by decreasing the distance between them. This can be accomplished by using a dielectric material with a higher relative permittivity (also known as dielectric constant), which will increase the capacitance for a given separation distance. Alternatively, you can increase the separation distance between the conductors or use conductors with a smaller surface area to decrease the capacitance.

However, it's important to note that changing the capacitance in a transmission line can also affect other important characteristics of the line, such as its propagation velocity and attenuation. So any changes to the line's capacitance should be carefully considered in the context of the desired performance characteristics.

Fabio Tortorella
Does the lenght affect the impedance?
Yes, the length of a transmission line can affect its characteristic impedance. The characteristic impedance of a transmission line is determined by its distributed inductance and capacitance, and these characteristics are affected by the length of the line.

In general, for a given line geometry and dielectric material, the characteristic impedance will be proportional to the square root of the line's inductance per unit length and inversely proportional to the square root of its capacitance per unit length. So if the length of the line is increased, the inductance and capacitance per unit length will both increase, which can result in a change in the characteristic impedance.

However, the effect of length on the characteristic impedance may be negligible for relatively short lines, such as those used in many electronic circuits. For longer lines, the effect of length on the characteristic impedance can become more significant and must be taken into consideration when designing the line.

In addition to the length of the line, other factors such as the dielectric constant and the distance between the conductors can also affect the characteristic impedance of a transmission line.

Fabio Tortorella
What would be the impedance if it were 10m
If the length of the transmission line were increased from 3m to 10m, the distributed capacitance and inductance of the line would increase. This would result in a change in the characteristic impedance of the line.

Assuming the same line dimensions as before (50mm width and 0.3mm plastic insulation thickness), and a dielectric constant of 2.25 for the plastic insulation, we can calculate the characteristic impedance of the 10m line as follows:

Inductance per unit length:
The inductance per unit length of the line can be approximated by:
L' = (μ/π) * ln(4h/d)

where μ is the permeability of free space, h is the distance between the conductors, and d is the diameter of the conductors (assuming circular cross-sections). For a two-wire transmission line, we can use the width of the line (50mm) as the diameter of the conductors. Therefore:

L' = (4π * 10^-7 H/m / π) * ln(4 * 0.3mm / 50mm) = 1.89 nH/m

Capacitance per unit length:
The capacitance per unit length of the line can be approximated by:
C' = εr * ε0 / (ln(D/d) - 1.0)

where εr is the relative permittivity (dielectric constant) of the plastic insulation, ε0 is the permittivity of free space, D is the distance between the centers of the conductors, and d is the diameter of the conductors. Assuming a distance of 50.3mm between the centers of the conductors (50mm width plus two 0.15mm thicknesses of plastic insulation), we get:

C' = 2.25 * 8.85 pF/m / (ln(50.3mm / 50mm) - 1.0) = 67.3 pF/m

Characteristic impedance:
The characteristic impedance of the line can then be calculated as:
Z0 = sqrt(L' / C') = sqrt(1.89 nH/m / 67.3 pF/m) = 40.8 Ω

So, for a 10m transmission line with the given dimensions and dielectric constant, the characteristic impedance would be approximately 40.8 Ω.

Fabio Tortorella
And if it were flat plates?
If the two conductors of the transmission line were flat plates with a separation distance of h and a width of w, the capacitance per unit length would be given by:

C' = εr * ε0 * w / h

where εr is the relative permittivity (dielectric constant) of the material between the plates, and ε0 is the permittivity of free space.

The inductance per unit length would be given by:

L' = μ * w / (2 * h)

where μ is the permeability of free space.

Using the same parameters as before (dielectric constant of 2.25 for the plastic insulation, and assuming a separation distance of 0.3 mm between the plates), the capacitance per unit length would be:

C' = 2.25 * 8.85 pF/m * 50 mm / 0.3 mm = 3.74 nF/m

The inductance per unit length would be:

L' = 4π * 10^-7 H/m * 50 mm / (2 * 0.3 mm) = 0.104 µH/m

The characteristic impedance of the line would then be:

Z0 = sqrt(L' / C') = sqrt(0.104 µH/m / 3.74 nF/m) = 18.8 Ω

So for flat plate conductors with the given dimensions and dielectric constant, the characteristic impedance of the transmission line would be approximately 18.8 Ω.

Fabio Tortorella
How long it should be to have 180ohms
To determine the length of a transmission line required to have a characteristic impedance of 180 ohms, we need to use the equation for characteristic impedance:

Z0 = sqrt(L' / C')

Rearranging this equation, we get:

L' = C' * Z0^2

Substituting the values for capacitance per unit length (C') and characteristic impedance (Z0) that we calculated earlier for the flat plate conductors with plastic insulation, we get:

L' = 3.74 nF/m * (180 Ω)^2 = 120.5 µH/m

To find the length of the transmission line required to achieve a characteristic impedance of 180 ohms, we need to solve for the length that would give us an inductance per unit length of 120.5 µH/m. Using the equation for inductance per unit length that we derived earlier for flat plate conductors, we get:

L' = μ * w / (2 * h)

Solving for the length (l) in terms of inductance per unit length (L'), we get:

l = 2 * h * L' / μ * w

Substituting the values for separation distance (h), inductance per unit length (L'), and width (w) that we used earlier, and assuming the permeability of free space (μ), we get:

l = 2 * 0.3 mm * 120.5 µH/m / (4π * 10^-7 H/m) * 50 mm = 109 m

Therefore, a transmission line with flat plate conductors separated by 0.3 mm of plastic insulation and a width of 50 mm would need to be approximately 109 meters long to have a characteristic impedance of 180 ohms.

Fabio Tortorella
this inductance per unit length can be changed by adding a core around the flat plate conductors?
Yes, it is possible to change the inductance per unit length of a transmission line by adding a magnetic core around the conductors. When a magnetic core is added, it increases the effective permeability of the medium surrounding the conductors, which in turn increases the inductance per unit length of the transmission line. The amount of inductance increase depends on the permeability of the core material and the geometry of the core.

The use of a magnetic core is a common technique for designing transmission lines with specific characteristic impedances and inductances. The choice of core material, size, and shape can be optimized to achieve the desired electrical characteristics of the transmission line.

Fabio Tortorella
I coil this transmission line with another plastic of same thickness as a separator how would the impedance be?
Coiling a transmission line around a plastic separator can affect its impedance, depending on the geometry of the coil and the separation distance between the conductors.

If the coiled transmission line has a similar geometry to the original flat transmission line, and the separation distance between the conductors is the same as in the flat line, then the impedance of the coiled line would be similar to the flat line. This is because the characteristic impedance of a transmission line is determined by the geometry of the conductors and the separation distance between them, and these factors are preserved in the coiled line.

However, if the coiled transmission line has a different geometry than the flat line, or if the separation distance between the conductors changes due to the coiling, then the impedance of the coiled line could be different than the flat line.

In general, coiling a transmission line around a plastic separator could introduce additional parasitic capacitance and inductance due to the proximity of the conductors and the dielectric material. These parasitic components could affect the impedance and other electrical characteristics of the transmission line, depending on their magnitude and frequency dependence.

Fabio Tortorella
How would it be affected?
Coiling a transmission line around a plastic separator can affect its electrical characteristics in a few ways, depending on the geometry of the coil and the properties of the dielectric material. Some possible effects are:

Parasitic capacitance: The close proximity of the conductors and the dielectric material in the coiled transmission line can increase the parasitic capacitance between the conductors. This extra capacitance can reduce the overall impedance of the transmission line, and also affect its frequency response.

Parasitic inductance: The coiling of the transmission line can also introduce additional parasitic inductance due to the magnetic fields generated by the current flowing through the conductors. This inductance can also affect the overall impedance and frequency response of the transmission line.

Crosstalk: If multiple coiled transmission lines are placed close together, they can also interfere with each other due to electromagnetic coupling between the conductors. This effect is known as crosstalk and can cause additional signal degradation and noise.

Resonances: The coiling of the transmission line can also create additional resonant frequencies due to the complex interactions between the conductors and the dielectric material. These resonances can affect the frequency response of the transmission line and cause additional signal distortion.

Overall, the effect of coiling a transmission line around a plastic separator on its electrical characteristics depends on the specific geometry of the coil and the properties of the dielectric material. Careful design and modeling of the coiled transmission line can help minimize these effects and ensure the desired electrical performance.

Fabio Tortorella
Design a coiled transmission line made of copper foil with 50mm width, 0,3 mm thickness and a initial diameter of 30mm  having a 0,3 mm polietileno separator and wound over a polietileno that act as separator with same thickness… what is the length required to match 78 ohms?
To design a coiled transmission line with a characteristic impedance of 78 ohms, we need to calculate the inductance per unit length of the line, and then use this value to determine the length required for a specific impedance.

Assuming the transmission line is wound into a tight, uniform coil, the inductance per unit length can be approximated using the formula:

L' = (0.00508 * ((d+D)/2) * (ln((4*(d+D))/(D-d)) - 1)) / (pi * sqrt(er))

Where:

L' is the inductance per unit length in henries per meter (H/m)
d is the diameter of each copper foil strip (0.05 m)
D is the diameter of the coil (0.06 m, based on an initial diameter of 30 mm)
er is the relative permittivity of the polietileno separator (assumed to be 2.25)
Plugging in these values, we get:

L' = (0.00508 * ((0.05+0.06)/2) * (ln((4*(0.05+0.06))/(0.06-0.05)) - 1)) / (pi * sqrt(2.25))
L' = 2.309e-7 H/m

Next, we can use the formula for the characteristic impedance of a transmission line:

Z0 = sqrt((L' / C')

Where:

Z0 is the characteristic impedance in ohms
L' is the inductance per unit length in H/m
C' is the capacitance per unit length in F/m
Assuming the polietileno separator has a relative permittivity of 2.25 and a thickness of 0.3 mm, the capacitance per unit length can be calculated using the formula:

C' = (2 * pi * er * eps0) / ln(b/a)

Where:

C' is the capacitance per unit length in farads per meter (F/m)
er is the relative permittivity of the polietileno separator
eps0 is the permittivity of free space (8.854e-12 F/m)
b is the outside radius of the coiled transmission line (0.075 m, based on an initial diameter of 30 mm)
a is the inside radius of the coiled transmission line (0.045 m, assuming a 0.3 mm separation between the copper foils)
Plugging in these values, we get:

C' = (2 * pi * 2.25 * 8.854e-12) / ln(0.075/0.045)
C' = 7.434e-11 F/m

Finally, we can solve for the length required to achieve a characteristic impedance of 78 ohms:

Z0 = sqrt((L' / C') * tanh(gamma * L))
78 = sqrt((2.309e-7 / 7.434e-11) * tanh(gamma * L))

Solving for L, we get:

L = (1 / gamma) * (acosh(sqrt((78 * 7.434e-11) / (2.309e-7))))

Where gamma is the attenuation constant, which can be calculated using the formula:

gamma = sqrt((R' + jwL') * (G' + jwC'))

Assuming a frequency of 1 GHz and a loss tangent of 0.001 for the polietileno separator, we can calculate the resistance and conductance per unit length using the formulas:

R' = sqrt

Fabio Tortorella
?
Based on the given specifications, the capacitance per unit length of the coiled transmission line can be calculated as:

C = (2πεr) / ln(D/d)

where:

εr is the relative permittivity of the polyethylene separator (assumed to be 2.25 for polyethylene)
D is the outer diameter of the coil (equal to the circumference of the initial diameter plus the width of the copper foil)
d is the diameter of the copper foil (assumed to be the same as the initial diameter of 30 mm)
C = (2π * 2.25 * 8.4e-3) / ln((2π30mm + 50mm)/(230mm))
C = 9.41 pF/m

The inductance per unit length of the coiled transmission line can be estimated using the Wheeler's formula:

L = μrμ0 / 8 * (ln(2h/w) - 0.75)

where:

μr is the relative permeability of the medium (assumed to be 1 for air)
μ0 is the permeability of free space (4π × 10^-7 H/m)
h is the height of the coil (assumed to be the same as the thickness of the copper foil and the polyethylene separator, which is 0.3 mm)
w is the width of the coil (50 mm)
L = (1 * 4πe-7) / (8 * (ln(2*0.3mm/50mm) - 0.75))
L = 0.47 μH/m

The characteristic impedance of the coiled transmission line can be calculated as:

Z0 = sqrt(L/C)
Z0 = sqrt(0.47e-6 / 9.41e-12)
Z0 = 220 ohms

To match the desired characteristic impedance of 78 ohms, a length of the coiled transmission line can be calculated as:

l = Z0 / (2π * sqrt(εr)) * tanh^-1(ZL / Z0)

where:

ZL is the desired load impedance (78 ohms)
l = 220 / (2π * sqrt(2.25)) * tanh^-1(78 / 220)
l = 0.577 m or 57.7 cm

Therefore, a length of approximately 57.7 cm of the coiled transmission line would be required to match a characteristic impedance of 78 ohms.

Hope you like it

• Member
• Posts: 102
##### Re: My new approach
« Reply #33 on: February 18, 2023, 15:50:05 pm »
Do you use the American formula notation or the European one?
A large amount of calculations.
I usually write the physical parameters multiplied by the geometric coefficient K when deriving formulas.
And then I output the geometric coefficient separately, let's say K.
Then you just need to replace the geometric coefficient K with the value of what K is equal to.

Take a closer look at the article.
Is this your correspondence with someone?
By deciphering the letters of the formula, I understood what was at stake.
I was surprised that you call E_0 - the permittivity of Free Space.
We call it the permittivity of VACUUM
« Last Edit: February 18, 2023, 16:18:48 pm by tur55 »

• Global Moderator
• Hero member
• Posts: 4181
##### Re: My new approach
« Reply #34 on: February 19, 2023, 12:58:46 pm »
I use the metric standards always… pounds and inches and miles is only in United States and maybe UK and Australia …
This is the chat gpt have you heard about it? It’s a artificial intelligence that is free to use and can make wonderful stuff. This week I made a skill for Alexa for my irrigation controller using it… it can code for you it can make very complex calculations it’s fucking amazing! I’m afraid of it!!! What if this could have hands? What it could build? You can ask many things and it will give nice answers hope you like it! Elon musk did well in this… hope it don’t kill us hahaha
I think that was the answer from the artificial intelligence calling free space… we also here call the the permititivity of Vacumm here in Brazil

In this chat I didn’t ask what would be the charge if the wire was subjected by an electromagnetic field … like if it was a coil… I’m not sure how it would come

As you see in the answer it’s considering 1000 volts applied to it because it talks about a voltage drop and that implies that there is a current flowing

• Member
• Posts: 102
##### Re: My new approach
« Reply #35 on: February 19, 2023, 13:10:34 pm »
Thanks now I understand.
Need a phone number outside of Russia.

• Global Moderator
• Hero member
• Posts: 4181
##### Re: My new approach
« Reply #36 on: February 20, 2023, 05:58:41 am »
I keep discovering stuff using the chat look at this

I calculated with it the electric field strength of the water molecule it’s about 258 million volts per meter per hydrogen atom….

This makes a mm spaced electrodes require to get 258kv to achieve the break of the bond! It’s also possible that if the molecule start to polarize the time share ratio of the electrons may change and reduce the force required somehow…

Meyer says is a resonance effect in the water and may be the only explanation possible… to achieve it with much lower voltage that that

« Last Edit: February 20, 2023, 07:18:37 am by sebosfato »

• Member
• Posts: 102
##### Re: My new approach
« Reply #37 on: February 20, 2023, 11:57:55 am »
According to my calculations, in order to break the covalent bond between oxygen and hydrogen atoms H2O (one molecule of water), you need to spend 1.07 attoJoules.
Or 1.07 attoVolts squared

• Global Moderator
• Hero member
• Posts: 4181
##### Re: My new approach
« Reply #38 on: February 20, 2023, 12:29:27 pm »
According to my calculations, in order to break the covalent bond between oxygen and hydrogen atoms H2O (one molecule of water), you need to spend 1.07 attoJoules.
Or 1.07 attoVolts squared

If the field is 258MV/m holding the water molecule… than it would need a voltage that goes beyond this…

If you have a book on the ground and you want to lift it you need to apply a force greater than gravity… similarly with the molecule if we don’t apply a field capable of starting the split it will never split…

I guess this is very fast action and after the split start the fields decrease exponentially so maybe a very short pulse combined with a longer dc field would be able to break the bonds and keep it apart enough time to become hydrogen

• Global Moderator
• Hero member
• Posts: 4181
##### Re: My new approach
« Reply #39 on: February 21, 2023, 22:50:43 pm »
When I say that the water molecule may have a way to get exited but for that we would have to assume the electrons are not always at the minimum energy level or that there is a lower energy level

Like in the example I gave of the book, if the ground was flexible elastic it could be excited little by little and as such get higher in amplitude

In the case of the water is possible that this can be done by maintaining a dc polarizing voltage and than apply to it a very high pulse of inverted polarity but very short just to tickle it

Another thing to consider is that

When we imagine the water molecule polarized we do it wrong… this cause to blind us about how things could really be

I was considering the other day that if you have a angled molecule it will not allign instead it will align one of its poles and the others will keep agitating… this makes the frequency close to the microwave that is already high

But is a manner to get high field strength if you think about it!

Water is going to get hot but also it will produce hydrogen on demand… just need to get around with the vapor

I mean that if we indeed had a microwave cavity with water the fields inside being only electromagnetic in nature may be able to break the water molecule if it’s polarized or if electrodes are provided as to extract the microwave energy out of the system.