Can characteristic impedance of a medium be considered as its resistance?

[Kholdstare]

In this thread Antiphon had written that

In its most general form, the Ohm has nothing to do with materials and everything to do with the scalar ratio of Electric to Magnetic fields over suitable line integrals.

The propagation of energy in free space takes place at 376.7 Ohms.

All material characterizations are specific cases of this general definition.

When I first encountered that, I thought there's something wrong with that, but could not figure it out. In order to create a current the electron must flow from one place to another. But in EM radiation although the electron oscillates or jiggles, it does not flow from transmitting antenna to receiving antenna.

I don't think resistance of a material has anything to do with characteristic impedance of a material.

 

[the_emi_guy]

Circuit impedance (or resistance) is the ratio of voltage and current.

Field impedance is the ratio of E field to H field.

Both have units of ohms which, unfortunately, can lead one to get the impression that they are somehow equivalent. There is obviously a connection between them, but they are not the same.

Consider a 50 ohm coaxial transmission line with free space dielectric. I launch a pulse onto the cable. The ratio of the voltage across the conductors to the current flowing through the conductors is 50.
The ratio of the internal E and H fields is 377 ohms.

Now I increase the diameter of the center conductor and cause the circuit impedance to drop to 40 ohms and launch a pulse. The voltage/current ratio has dropped to 40, but the E/H ratio remains 377.

In other words, I can change the circuit impedance arbitrarily by changing geometry without effecting the E/H ratio.

 

[cabraham]

Circuit impedance (or resistance) is the ratio of voltage and current.

Field impedance is the ratio of E field to H field.

Both have units of ohms which, unfortunately, can lead one to get the impression that they are somehow equivalent. There is obviously a connection between them, but they are not the same.

Consider a 50 ohm coaxial transmission line with free space dielectric. I launch a pulse onto the cable. The ratio of the voltage across the conductors to the current flowing through the conductors is 50.
The ratio of the internal E and H fields is 377 ohms.

Now I increase the diameter of the center conductor and cause the circuit impedance to drop to 40 ohms and launch a pulse. The voltage/current ratio has dropped to 40, but the E/H ratio remains 377.

In other words, I can change the circuit impedance arbitrarily by changing geometry without effecting the E/H ratio.

I don't think so. The V/I ratio & the E/H ratio have to be equal, i.e. 50 ohms in the cable. The 377 ohms is the impedance (E/H ratio) of free space. A 50 ohm coax has a dielectric insulator. It is not free space. Even if it were free space, the Z0 value is influenced by the spacing between conductors as well as diameter. Check any e/m fields reference. Have you ever taken e/m fields classes?

Claude

 

[the_emi_guy]

I don't think so. The V/I ratio & the E/H ratio have to be equal, i.e. 50 ohms in the cable.

A coaxial cable with E/H ratio of 50 ohms? This would require dielectric with DK of 57. I guess if you had a water filled coaxial cable...

The 377 ohms is the impedance (E/H ratio) of free space. A 50 ohm coax has a dielectric insulator. It is not free space

.

I have 50 ohm coaxial transmission lines with free space (air) dielectric in my lab in the form of 2.4mm microwave connectors. Air dielectric means E/H = 377, yet it is a 50 ohm connector.

Have you ever taken e/m fields classes?

Yes. Odd that you would ask. I'll assume that you have as well since I see that you are holding a PhD degree in EE.

I have noticed that this is a commonly misunderstood concept.

 

[yungman]

I don't think so. The V/I ratio & the E/H ratio have to be equal, i.e. 50 ohms in the cable. The 377 ohms is the impedance (E/H ratio) of free space. A 50 ohm coax has a dielectric insulator. It is not free space. Even if it were free space, the Z0 value is influenced by the spacing between conductors as well as diameter. Check any e/m fields reference. Have you ever taken e/m fields classes?

Claude

I just read p125 to 127 of Electromagnetics with Applications 5th edition by Kraus/Fleisch. It seems to support Emi-Guy's assertion.

[tex] Z_{line}=\frac V I = \frac {\int_C \vec E \cdot d \vec l_d}{\oint\vec H\cdot d\vec l_w}[/tex]

And

[tex] Z_{medium}=\frac E H[/tex]

The book explain the intrinsic impedance is like surface impedance that it give the impedance per square. So you have to put in the width w and the distance d to get the final impedance. Obviously the w and d is not like surface impedance, the E and B are contour lines, so as the equation shows, you have to use line integrals to calculate. But basically you need to scale the intrinsic impedance by the two paths to get the line impedance.

 

[Kholdstare]

Guys! let's not divert the topic!

Yeah, radiation resistance/effective resistance of an antenna/transmission line is different from the ratio of internal E field to H field in the medium.

But, my question was a very general one. Is resistance (of anything, say antenna/transmission line/resistor/metal) derived ONLY from the characteristic impedance of the free space?

My answer was no. Resistance of something has no connection with the characteristic impedance UNLESS energy is being dissipated in form of em radiation (antenna/transmission line).

EDIT: corrected , thankx uart for spotting

 

[cabraham]

A coaxial cable with E/H ratio of 50 ohms? This would require dielectric with DK of 57. I guess if you had a water filled coaxial cable...

.

I have 50 ohm coaxial transmission lines with free space (air) dielectric in my lab in the form of 2.4mm microwave connectors. Air dielectric means E/H = 377, yet it is a 50 ohm connector.
Yes. Odd that you would ask. I'll assume that you have as well since I see that you are holding a PhD degree in EE.

I have noticed that this is a commonly misunderstood concept.

I'll double check, I've heard of air-filled lines, but I've never seen one. I'm positive that typical coax like RG-58U, is NOT air-filled, which means that Z0 cannot be 377 ohms. Again, I'll double check, but off the top of my head, most lines, coax, twin-lead, whatever, are not air-filled, not 377 ohm.

Claude

 

[cabraham]

I just read p125 to 127 of Electromagnetics with Applications 5th edition by Kraus/Fleisch. It seems to support Em-Guy's assertion.

[tex] Z_{line}=\frac V I = \frac {\int_C \vec E \cdot d \vec l_d}{\oint\vec H\cdot d\vec l_w}[/tex]

And



The book explain the intrinsic impedance is like surface impedance that it give the impedance per square. So you have to put in the width w and the distance d to get the final impedance. Obviously the w and d is not like surface impedance, the E and B are contour lines, so as the equation shows, you have to use line integrals to calculate. But basically you need to scale the intrinsic impedance by the two paths to get the line impedance.

I don't have my reference books right now, but maybe I spoke too soon. The ratio of the E & H integrals may be what I was thinking, which differs from the pure E/H ratio. So if a line is air filled, then E/H will differ from the integral

[tex] Z_{line}=\frac V I = \frac {\int_C \vec E \cdot d \vec l_d}{\oint\vec H\cdot d\vec l_w}[/tex]

But I still believe that most commercial coax cables have a non-air dielectric with a relative permittivity > 1. So the E/H ratio has to be < 377 ohms. So V/I equals the integral E to integral H ratio, but generally not the pure E/H value ratio. I should not rely on memory, & look these things up. BR.

Claude

 

[yungman]

I don't have my reference books right now, but maybe I spoke too soon. The ratio of the E & H integrals may be what I was thinking, which differs from the pure E/H ratio. So if a line is air filled, then E/H will differ from the integral

[tex] Z_{line}=\frac V I = \frac {\int_C \vec E \cdot d \vec l_d}{\oint\vec H\cdot d\vec l_w}[/tex]

But I still believe that most commercial coax cables have a non-air dielectric with a relative permittivity > 1. So the E/H ratio has to be < 377 ohms. So V/I equals the integral E to integral H ratio, but generally not the pure E/H value ratio. I should not rely on memory, & look these things up. BR.

Claude

Yes for dielectric, the intrinsic impedance is not going to be 377 as

[tex]Z_0=\sqrt {\frac{\epsilon_r \epsilon_0}{\mu_0}}[/tex]

So the intrinsic impedance for scaling is different and is going to be lower with the same tx line dimension.

 

[uart]

One way to think about is this that [itex]Z_o = \sqrt{\frac{\mu}{\epsilon}}[/itex] is only the ratio of E/H for ideal plane wave (far field) radiation. Once you contain and distort the fields in a transmission line then the specific geometry of the line also becomes important.

So with a coaxial line for example,

[tex]Z_o = \sqrt{\frac{\mu}{\epsilon}} \ln \left (\frac{r_2}{r_1} \right) [/tex]

For 50 ohms with an air dielectric you'd need r2/r1 about 1.142. An usual geometry for a regular coaxial line, but obviously it could be done for special purpose as in the case of emi_guy's microwave connectors.

 

[uart]

But, my question was a very general one. Is resistance (of anything, say antenna/transmission line/resistor/metal) derived ONLY from the characteristic impedance of the free space?

My answer was no. Resistance of something has no connection with the characteristic impedance UNLESS energy IS [strike]not[/strike] being dissipated in form of em radiation (antenna/transmission line).

I fixed it for you.

 

[Kholdstare]

Please be on the topic! Here are some of my latest thoughts.In quantum mechanical sense, the electromagnetic radiation is caused by acceleration(deceleration) of electrons in antenna and is emitted as photon particles. The wave being sinusoidal electrons do not get permanently stored/drained in the antenna(dc signal won't radiate). In one cycle in enters one rod(folded dipole) and gets slowed down and in the opposite cycle it accelerates and comes out(Actually this acceleration/decleration is the source of resistance in antenna. Its similar to scattering process in the conductor.).

The intensity of electric or magnetic field is proportional to the number of photon particles. While traveling through a material it polarizes/magnetizes the material. This way the photon separates charge or create current loops, which determines the permittivity or permeability of the material and it affects the speed of photon. Its just coincidental that the ratio of electric to magnetic field has the dimension of ohm and nothing else.

(Here I have some doubts. I think permittiviy and permeability could be related to photon dispersion curve. In that case the E/H ratio could be linked with the electron-photon scattering process or resistance.)

(Otherwise) E/H will not be the cause of Resistance.

 

[yungman]

In this thread Antiphon had written thatWhen I first encountered that, I thought there's something wrong with that, but could not figure it out. In order to create a current the electron must flow from one place to another. But in EM radiation although the electron oscillates or jiggles, it does not flow from transmitting antenna to receiving antenna.

I don't think resistance of a material has anything to do with characteristic impedance of a material.

Emi-Guy answer your question already. That's what we are talking all along. Resistance is R=V/I, that is for the material. For air, vacuum or dielectric, it is very very high. Intrinsic impedance of material is E/H. they are related as I posted. Intrinsic impedance is related to the wave propagation in the medium. We have not gotten off the subject. If you start getting into heavy physics or QM, there are specific sub forums here also.

 

[the_emi_guy]

But I still believe that most commercial coax cables have a non-air dielectric with a relative permittivity > 1. So the E/H ratio has to be < 377 ohms.
Claude

No one has asserted otherwise.

One way to think about is this that [itex]Z_o = \sqrt{\frac{\mu}{\epsilon}}[/itex] is only the ratio of E/H for ideal plane wave (far field) radiation. Once you contain and distort the fields in a transmission line then the specific geometry of the line also becomes important.

So with a coaxial line for example,

[tex]Z_o = \sqrt{\frac{\mu}{\epsilon}} \ln \left (\frac{r_2}{r_1} \right) [/tex]

Uart,
The signal propagating in a transmission line is in fact a far field. Think of it as a guided far field. If I have a transmission line with air dielectric, the ratio of the E and H fields within the cable will be 377 ohms. On the other hand, its circuit characteristic impedance can be whatever we design it to. In other words the cable has simultaneously a wave impedance of 377 ohms within the air dielectric, and a characteristic impedance of 50 ohms across its conductors.

The OP wanted to discuss circuit resistance & impedance, and linked to a thread where someone mentioned free space intrinsic impedance or 377 ohms in the context of circuit resistance.

It is a common misunderstanding that there is some parity between wave impedance and circuit impedance, no doubt because they share the same terminology and the same unit.

For example, I have had folks assert that electromagnetic radiation from a PCB trace is due to 377 ohm free space being in parallel with their 50 ohm microstrip causing some of the energy to bleed off through the 377 ohm path.

 

[yungman]

This subject is not being covered in a lot of textbooks. It is not in Chengs and Ulaby. I don't recall seeing in Griffiths. After I read this post, I had to dig around in my other books that I don't use normally. I found this in Kraus. This is usually the book that I don't like much, but it stepped up to the occasion this time.

This thread is very useful to me also as I never even thought of this. It is really like surface impedance where 377 is the resistance per square, then it's up to how you design the dimension of the tx line to get the tx line impedance. It is like if you have surface impedance of 1Ω per square. If you have a strip that the width is 10 times the length, then the resistance of the strip is 0.1Ω. If you have the length 10 times the width, then the resistance of the strip is 10Ω. This is how the book explained. The difference is the field lines are not linear, so you need the line integral instead of using straight d and w for calculation.

On the side note, I always believe in buying multiple books for each subject. In my experience, there is no one perfect book. Every book presents a little differently and stress in different part of the subject. Not to mention in this kind of upper division books, there usually mistakes inside. If you don't buy what one book said, verify with other books. Believe me, you'll find mistakes. I have 7 or 8 books just in the EM and I end up using every one of them.