Let's say we have a coaxial cable with a thick outer 'screen' and a thin central conductor, separated by a dielectric. We connect a length of it to a signal generator and put a load on the far end.
The skin effect tells us that as the frequency increases, so the bulk of the screen current moves towards the outside. This increases the apparent resistance of the screen in proportion to the square root of the frequency.
Does increasing the frequency also decrease the apparent capacitance as the bulk of the current in the screen moves away from the inner conductor?
High frequency current wants to flow on the inside of the shield and the outside of the centre conductor.
The inductance changes a bit but the capacitance hardly does at all--the E field drops like exp(-z/lambda_D), where lambda_D is the Debye length. It's less than an angstrom for metal, so the capacitance is very close to constant.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
I don't think so. There should be no net (common-mode) current. All the currents are *inside* the coax. The current in the shield actually moves towards the inner conductor at high frequencies; the shield "skin" is inside.
This increases the
I'd expect that most of the skin-effect loss is in the center conductor. It has a lot less surface area than the shield.
Probably not. Different physics.
--
John Larkin Highland Technology, Inc
picosecond timing precision measurement
jlarkin att highlandtechnology dott com
http://www.highlandtechnology.com
What Phil H and John L said, I realized (a few years ago) that if you have a conductive can or cylinder, you can have different currents flowing on each side.. decoupled at some frequency that depends on the thickness of the shield. ~skin depth.. away from DC different 'grounds'. You end up having to hook these two grounds together at some point. and that can cause problems.
The equivalent inductance also decreases. Indeed, this is captured in Coilcraft's SPICE models, which are probably the most accurate SPICE models you can find for any inductors out there.
Spot on. At RF frequencies, the inside and outside are completely* separate; you can only ensure shielding if you fully seal all the internal signals away from external noise (and, for emissions purposes: vice versa).
You're sealing everything inside a Faraday cage, which includes having that cage neck down to cover longer signal runs -- what's otherwise known as a cable.
*Completeness is given by the attenuation (shielding effectiveness), which should be >40dB for most cables at useful frequencies (MHz+), and is easily >100dB for very ordinary enclosures (like 10ga. plate steel, assuming it doesn't have gaps, holes and slots in it of course!). Understandably, getting good shielding at arbitrarily low frequencies is a challenge.
This is the best mindset to have, when designing a product, and considering its EMI performance. Consequently, any connection that must penetrate that Faraday cage, must either be filtered (to reflect or absorb incident radiation), or shielded.
Shielded connectors must be grounded directly to the enclosure, or as close as possible to achieve the same effect. Whether the attached PCB (or wiring harness, or whatever) should also be shielded, or what, depends on the environment within the enclosure. (Hey, no one said Faraday cages can't be recursive.)
This is also why a broken shield is the absolute worst possible option to use -- an anti-option. It makes the EMI situation unconditionally worse. It can only be helpful at low frequencies, where the shield isn't actually a shield anymore -- in other words, not at RF. :)
Hams can have this problem when using a coax with an antenna. They use a choke to stop the currents on the outside of the shield while having no effect on the differential currents on the inside of the shield and the center conductor since they cancel wrt the choke.
I hadn't realised that, I'd thought the current would be nearest the outside surface. AIUI, this would be the case for a single solid conductor. What makes this change for a coax?
Here is one way you can think of it: the opposing currents want to cuddle and close the loop between them. So the current on the coax inner will expand to the outer surface of that conductor and the current in the shield will come inwards to the inner (closest) surface of the shield. With an unshielded single conductor the current's soul-mate it wants to cuddle is the rest of the universe :)
I see, so there must be some intermediate case where the spacings and sizes are such that the current in the the shield flows down the middle of the conductor?
My basic problem is I'm trying to send a tens of kHz signal down a 10mm thick 100mm diameter steel pipe with a coaxial copper inner, and the measured attenuation has taken me by surprise. I've never had to consider skin effect before, but online sources tell me that most of the current is flowing in well under a tenth of the available metal.
If that's the case, then so be it, it can be designed around, but it was a surprise.
Steel? Oh, dear. Not only is its conductivity much less, but unfortunately there's a factor of 1/sqrt(mu) in the formula for skin depth. Thus steel wi th a mu of several thousand is on the order of 100 times lossier than coppe r at high frequency.
If you launch the signal on the shield and not into the coaxial mode, the current will be concentrated on the outside.
The current density and E field in the metal obeys a first-order partial differential equation, which means that the field _diffuses_ in and out of the metal, just like heat. At higher frequency there isn't enough time for the field to make it all the way through before it changes direction and diffuses back out.
So the current is confined to the side facing the nonzero AC E field.
Cheers
Phil Hobbs
--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics
160 North State Road #203
Briarcliff Manor NY 10510
hobbs at electrooptical dot net
http://electrooptical.net
You can look up some skin effect calculators, but at 10's of kHz the skin effect is not appreciable in copper. At 60 kHz a 1 mm copper wire (18 AWG) still has some 60+% of it's DC conductivity (I don't recall the exact number). As has been pointed out the mu of the steel gives it a much thinner skin depth which along with the decreased conductivity makes it a pretty poor conductor even at 10's of kHz. However, you have the large circumference working for you. Use a calculator and see what your actual resistance per foot works out to.
The current concentrates closest to the other conductors in the transmission line. In a single conductor, that's all of the rest of the world.
I'm not sure what happens in a twisted pair or a parallel-line transmission line (like the ones they use to carry power around). I'll be the answers are out there, though.
Yep, I believe a right-hand rule works here - if you point your thumb in the direction of current flow in the conductor the current density will increase in the direction your fingers are pointing if you hold your palm open
Some math should be able to determine the change in capacitance per unit length of a coaxial cable due to skin effect...
The current density at a radius r from the center of a cylindrical conductor carrying an AC current with magnitude equal to Re(z) = C*e^(i*omega*t) is proportional to:
Re(z) = kC/(a*I_0'(ka))*I_0(kr)*e^(i*omagea*t)
where k is a constant depending on the units used, a is the outer radius, and I_0 is the modified Bessel function of the first kind of zero order.
The continuity equation will give you the cross-sectional charge density for a given current, and since you have cylindrical symmetry Gauss' law and another integral should be able to tell you the E field generated by that current in the dielectric. A similar calculation should be possible for the outer conductor.
Your wavelength will be well in excess of 10000 feet. The transmission line model appleis badly here.
My guess is thet you're bitten by ohmic losses in the steel pipe and it creates an RC low-pass with the line capacitance. The situation is the same as with lod baseband telephone lines made of iron wire.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.