On 11 Nov, 14:18, Ian White GM3SEK wrote:
Richard Harrison wrote:
Richard Fry wrote:
"Read Terman`s RADIO ENGINEERS` HANDBOOK, 1943 edition, pp 30-31 for
more on this (or many other sources)."

Amen. Terman doesn`t say different things in different places. He is
consistent. In Terman`s 1955 edition of "Electronic and Radio
Engineering" he writes on page 21: "It is to be noted that some of this
(magnetic) flux exists within the conductor and therefore links with,
i.e., encircles, current near the center of the conductor while not
linking current flowing near the surface. The result is that inductance
of the central part of the conductor is greater than the part of the
conductor nesr the surface; this is because of the greater number of
flux linkages existing in the central region.

What Terman says is true, for the particular example that he chooses.
But it may leave an incorrect impression that the conductor needs to be
completely encircled by flux linkages.

In fact the skin effect will develop on the surface of any conducting
material of any shape, wherever there is RF current flowing.

Here is a link to a detailed mathematical proof, from 'Transmission
Lines for Communications' by C W Davidson (Macmillan Press, 1978, ISBN 0
333 32738 1):http://www.ifwtech.co.uk/g3sek/misc/skin.htm

Davidson's analysis starts with the most general assumption possible:
that RF current is flowing over any small patch of a conductor's
surface. No assumption is required about the reason for the RF current
to be present, only that it is. Likewise no assumption is required about
the cross-section of the conductor, only that it has an exposed surface
(and by implication, that there are no constraints due to a small radius
or insufficient depth). Davidson then derives all the usual equations
for the skin effect. The only drawback of this derivation is that it is
highly mathematical, and difficult to put into words; but it's still
physically correct.

To repeat, I am not saying that Terman's explanation is incorrect; only
that the skin effect is a far more general phenomenon than his
particular examples imply.

This is important because, by taking the existence of the skin effect as
a guaranteed starting-point, the explanations for the behaviour of
coaxial cables,

Ian, I have no disagreement to your reply above other than you are
being to king in your response
I personaly would have put more emphasis on what you stated with
respect
to RF traveling along a path that has no external surface .With
emphasising
where many have about RF travel without which one CANNOT understand
coaxial cables or braid The inside of braid on a coax CAN and DOES
carry
RF current but it does NOT radiate, because it does NOT have an
exposed surface
other than a dielectric interface. The outside surface can and DOES
radiate if a
RF current flows on the outside of the braid. I would also add that
copper/braid
itself does not turn into a dielectric or contain a diode thus it
also WILL
also pass a RF current at its centre but of course does NOT radiate.
This very fact was refuted by popular vote on this newsgroup where
poll
standings always overule science. So yes, without true understandings
errors
are sure to congregate and eventually will create a "fact".
Art KB9MZ...xg





'bazooka' baluns, 'shielded' loops and many other
devices will all fall neatly into place.

--

73 fromIanGM3SEK 'In Practice' columnist for RadCom (RSGB)http://www.ifwtech.co.uk/g3sek

"art" wrote
I would also add that copper/braid itself does not turn into a dielectric
or contain a diode thus it also WILL also pass a RF current at its
centre...
____________

art... so by your post you reject the theory and experience of physical
science?

RF

"art" wrote
I would also add that copper/braid itself does not turn into
a dielectric or contain a diode thus it also WILL also
pass a RF current at its centre but of course does NOT radiate.
_____________

art, you really need to buy and read Terman's RADIO ENGINEERS' HANDBOOK or
similar source, instead of relying on your intuition. Terman provides the
following equation for the r-f attenuation of air-insulated, copper coaxial
transmission line:

a = 0.00362 SQRT(f)*(1+ D/d) / D*log(D/d) dB per 1,000 feet

where f = frequency in MHz, D = inner diameter of outer conductor, d =
outer diameter of inner conductor.

Note that the attenuation is the same whether the inner conductor is solid
or tubular. This is the result of "skin effect," which for r-f frequencies
1.8 MHz and higher confines the r-f current on the inner conductor from its
outer surface to a depth of less than 0.18 mm.

RF

Richard Fry wrote:
"art" wrote

I would also add that copper/braid itself does not turn into
a dielectric or contain a diode thus it also WILL also
pass a RF current at its centre but of course does NOT radiate.

_____________

art, you really need to buy and read Terman's RADIO ENGINEERS' HANDBOOK
or similar source, instead of relying on your intuition. Terman
provides the following equation for the r-f attenuation of
air-insulated, copper coaxial transmission line:

a = 0.00362 SQRT(f)*(1+ D/d) / D*log(D/d) dB per 1,000 feet

where f = frequency in MHz, D = inner diameter of outer conductor, d =
outer diameter of inner conductor.

Note that the attenuation is the same whether the inner conductor is
solid or tubular. This is the result of "skin effect," which for r-f
frequencies 1.8 MHz and higher confines the r-f current on the inner
conductor from its outer surface to a depth of less than 0.18 mm.


One should be aware that this formula applies only to "large" coaxial
transmission lines, where the skin depth is a small fraction of the
conductor thickness.

It's not like the current is confined in a uniform band of the skin
depth, and zero elsewhere. The skin depth is a convenient mathematical
fiction.. it's the depth at which the current density is 1/e, so you can
calculate things like voltage drop by assuming a uniform current density
in a layer that thick, instead of actually integrating it.

On a smallish round conductor, where the circumference isn't many, many
skin depths, there's a broken assumption in the skin depth formula of an
infinite flat plane. Actually solving for the true AC resistance (or
current distribution) involves elliptic integrals which only have
infinite series solutions.


Which is why there are nifty tables and empirical formulas for AC
resistance of round conductors (solid and tubular) that get you
arbitrarily close. See, e.g., NBS Circular 75 or Grover or Reference
Data for Radio Engineers.


Lest you think I am nit picking here.. take a piece of venerable RG-8
style coax, with the AWG13 inner conductor (0.072" diameter, 1.83 mm).
The skin depth at 1.8 MHz (per the above post) is 0.18mm, so the wire is
10 skin depths across, so it's probably a reasonable assumption.

However, let's take something a bit smaller, like RG-8X or RG-58 type
coaxes, which have a inner conductor on the order of 0.9mm. Now, you're
talking only 4-5 skin depths, and the assumption of an infinite plane
probably doesn't hold.


So.. Terman's equation probably holds for coax where the inner conductor
is 20 skin depths, and, as posted, it would make no difference whether
it's a tube (with wall thickness5 skin depth) and a solid conductor.

RF

Jim Lux wrote:
. . .
Lest you think I am nit picking here.. take a piece of venerable RG-8
style coax, with the AWG13 inner conductor (0.072" diameter, 1.83 mm).
The skin depth at 1.8 MHz (per the above post) is 0.18mm, so the wire is
10 skin depths across, so it's probably a reasonable assumption.

However, let's take something a bit smaller, like RG-8X or RG-58 type
coaxes, which have a inner conductor on the order of 0.9mm. Now, you're
talking only 4-5 skin depths, and the assumption of an infinite plane
probably doesn't hold.

That would be nit picking unless very high accuracy is required.

As Jim said, the current density actually decays from the surface in an
exponential manner. The skin depth is the depth at which it's dropped to
1/e its density at the surface. If a conductor is infinitely thick, the
total loss is exactly the same as if the current density was uniform to
the skin depth and zero below. So this approximation is widely used when
it can be assumed that the conductor is at least several skin depths
thick. A rigorous calculation for a round wire really requires a
computer, since it involves evaluating complex Bessel functions, and I
believe that closed form equations for many other wire shapes don't
exist at all.

But there are two levels of approximation you can make with the
assumption that the current is all flowing in a uniform layer. If you
calculate the cross sectional area of the ring of current, you come up
with (from simple geometry)

Area = pi * delta * (OD - delta)

where OD is the outer diameter of the wire, and delta is the skin depth.
The material's bulk resistivity is divided by this area to find the
wire's resistance per unit length.

If the diameter is much greater than the skin depth (OD delta), an
even simpler approximation can be and is often made:

Area ~ pi * delta * OD

I assume this is the infinite diameter assumption Jim mentions.

If you use this infinite diameter assumption, the error in the
calculated resistivity of a copper wire 0.9 mm diameter at 1.8 MHz is
5.4% (compared to a rigorous calculation). This error isn't a big deal
for most purposes. But by simply using the first rather than the second
equation for area, the error drops to less than 0.1%. You're still using
the approximation that the current is flowing in a uniform layer one
skin depth thick, so the entire calculation can easily be done on a
pocket calculator in a minute or so.

Roy Lewallen, W7EL

Jim Lux wrote:
Richard Fry wrote:
[...]
Note that the attenuation is the same whether the inner conductor is
solid or tubular. This is the result of "skin effect," which for r-f
frequencies 1.8 MHz and higher confines the r-f current on the inner
conductor from its outer surface to a depth of less than 0.18 mm.


One should be aware that this formula applies only to "large" coaxial
transmission lines, where the skin depth is a small fraction of the
conductor thickness.

It's not like the current is confined in a uniform band of the skin
depth, and zero elsewhere. The skin depth is a convenient mathematical
fiction.. it's the depth at which the current density is 1/e, so you
can calculate things like voltage drop by assuming a uniform current
density in a layer that thick, instead of actually integrating it.

On a smallish round conductor, where the circumference isn't many, many
skin depths, there's a broken assumption in the skin depth formula of
an infinite flat plane. Actually solving for the true AC resistance (or
current distribution) involves elliptic integrals which only have
infinite series solutions.


Which is why there are nifty tables and empirical formulas for AC
resistance of round conductors (solid and tubular) that get you
arbitrarily close. See, e.g., NBS Circular 75 or Grover or Reference
Data for Radio Engineers.


Lest you think I am nit picking here.. take a piece of venerable RG-8
style coax, with the AWG13 inner conductor (0.072" diameter, 1.83 mm).
The skin depth at 1.8 MHz (per the above post) is 0.18mm, so the wire
is 10 skin depths across, so it's probably a reasonable assumption.


No, that wasn't nit picking; those are all fair points.

The underlying point is that engineering is ultimately about numbers. We
all like to think in words and mental images if we can, but in marginal
cases these simple slogans and cartoons won't work.

On the other hand, the marginal cases don't invalidate the point that
the skin effect *will* be present. If there isn't enough conductor depth
to allow the skin effect to develop unhindered, it only affects our
estimates of the AC/RF resistance.

If the available depth of conductor is too small, the inside boundary
will push the current density profile outward towards the surface. For a
round conductor, we can think of it as 'current crowding' along the
centreline. A closely related case is copper-plated steel, where the
magnetic nature of the steel increases its AC/RF resistance by a further
factor of sqrt(mu), which squeezes a much higher fraction of the total
current into the thin layer of copper.

However, let's take something a bit smaller, like RG-8X or RG-58 type
coaxes, which have a inner conductor on the order of 0.9mm. Now,
you're talking only 4-5 skin depths, and the assumption of an infinite
plane probably doesn't hold.

We can see a little further into this without the need for detailed
math. The radius of the conductor is 2.5 skin depths (again using
0.18mm) so the current density at this depth would normally be 1/e^2.5
or about 1/12 of its surface value. That suggests that the perturbation
in RF resistance due to insufficient depth is only taking place at
around the 10% level. In the context of *estimating* the RF resistance
to help us decide whether to buy a drum of cable, that wouldn't be a
serious error.

However, it warns of a very serious error if the centre conductor was
made of copper-plated steel instead of solid copper.


So.. Terman's equation probably holds for coax where the inner
conductor is 20 skin depths,

Sorry, Jim, you lost me: why such a large number as 20?

At 2.5 skin depths, the current density is 10% of the surface value;
at 5 skin depths, 1%. If at least 5 skin depths are available, we can
be confident in the accuracy of the standard, uncorrected equation for
most purposes.

A more serious effect of insufficient conductor depth may be in
estimating the effectiveness of shielding. The residual fields at the
opposite side of an extremely thin shield can be very significant if
we're looking for attenuations of 40dB, 60dB or more.



--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

Ian White GM3SEK wrote:

So.. Terman's equation probably holds for coax where the inner
conductor is 20 skin depths,


Sorry, Jim, you lost me: why such a large number as 20?

At 2.5 skin depths, the current density is 10% of the surface value; at
5 skin depths, 1%. If at least 5 skin depths are available, we can be
confident in the accuracy of the standard, uncorrected equation for most
purposes.


But it's round... (unless Terman rolled that into his constants)

Consider if you peeled that 2.5 skin depth layer and made it flat. It
would look like a pyramid, not a rectangular bar.

Of course, if you assume that the cross sectional area is an annulus
(the pi*( r^2-(r-skindepth)^2) style calculation) this partially gets
taken into account.

The other factor is that in a wire that is comparable to skin depth in
radius, the current on the far side of the wire also contributes to
squeezing the current towards the near side surface. (and that's why
the actual math gets hairy.. you can't use a simple exponential
approximation for the current density)

At 20 skin depths, the difference is negligble.

As a practical matter, if you have an application that actually cares
about this level of detail, you probably have the resources to deal with
the exact calculations, so the simple "thin layer on the surface" is
close enough.

For what it's worth, this kind of thing is why the loss in coax doesn't
follow a nice k1*f + k2*sqrt(f) characteristic at low frequencies. The
second term is essentially assuming that the skin depth in the
conductors is "small" compared to conductor size.



A more serious effect of insufficient conductor depth may be in
estimating the effectiveness of shielding. The residual fields at the
opposite side of an extremely thin shield can be very significant if
we're looking for attenuations of 40dB, 60dB or more.

Especially at low frequencies... (shielding mains frequency
interference, or PWM switcher noise, for instance)

Jim Lux wrote:
Ian White GM3SEK wrote:

So.. Terman's equation probably holds for coax where the inner
conductor is 20 skin depths,
Sorry, Jim, you lost me: why such a large number as 20?
At 2.5 skin depths, the current density is 10% of the surface
value; at 5 skin depths, 1%. If at least 5 skin depths are
available, we can be confident in the accuracy of the standard,
uncorrected equation for most purposes.


But it's round... (unless Terman rolled that into his constants)

I don't believe so.

Consider if you peeled that 2.5 skin depth layer and made it flat. It
would look like a pyramid, not a rectangular bar.

I see your point, just wouldn't have imagined that the circular geometry
would make such a very large difference compared with a flat surface.


Of course, if you assume that the cross sectional area is an annulus
(the pi*( r^2-(r-skindepth)^2) style calculation) this partially gets
taken into account.

The other factor is that in a wire that is comparable to skin depth in
radius, the current on the far side of the wire also contributes to
squeezing the current towards the near side surface. (and that's why
the actual math gets hairy.. you can't use a simple exponential
approximation for the current density)

Agreed. I do have the more detailed equations involving Bessel
functions, but no time to compute them just now.

At 20 skin depths, the difference is negligble.

No doubt; but it's the journey towards "negligible" that concerns us
more than the destination :-)


--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek