Jeff wrote in
:

Owen Duffy wrote:
Jeff wrote in
:

Owen Duffy wrote:
Jeff wrote in news:hm5dj4$4ls$1
@speranza.aioe.org:

I am sure that in case B it is as for resistors in parallel, ie 2
50ohm cables in parallel give you 25ohms Zo.
Jeff, you you offer more explanation that just that your are
"sure".

If you can't explain it, it speaks of whether you are sure.

Owen
Well having just tried it for real on a network analyser, and
simulated it on Ansoft designer I am now convinced rather than being
sure!!

That is not an explanation at all.

Your confirmation might just be confirmation of a wrong
interpretation of the B configuration.

Owen

and you just might be trolling.

And you just might be wrong on both counts.

I will offer you an explanation of why the ARRL position is IMHO correct.

The configuration is a pair of identical coaxial cables, shields tied
together at each end, and the inner conductors used as a two wire
transmission line (http://www.vk1od.net/transmissionline/stcm/Fig02.png).

Question is what is Zo of the combination?

for each of the coaxial lines, its intinsic Zo defines the ratio of V/I
for a travelling wave at any point. If the two inner conductors are
driven by a differential signal, then I in one conductor equals -I in the
other (defined by differential mode), and V conductor to conductor is
twice V conductor to shield, so V'=2*(I*Zo), and
Zo'=V'/I=2*(I*Zo)/I=2*Zo.

Owen

Owen Duffy wrote in
:

....
....
in the other (defined by differential mode), and V conductor to
conductor is twice V conductor to shield, so V'=2*(I*Zo), and
Zo'=V'/I=2*(I*Zo)/I=2*Zo.

I should have expected that this is not obvious to you Jeff, and expand that
to:

....
in the other (defined by differential mode), and V conductor to
conductor is (by definition of differential mode) V'=(I*Zo-(-I*Zo))=2*I*Zo,
and Zo'=V'/I=2*(I*Zo)/I=2*Zo.

Owen

On Feb 24, 3:36*pm, Cecil Moore wrote:
On Feb 24, 5:21*pm, Owen Duffy wrote:

... mainly because the effective RF resistance of the braid is not easy to estimate.

Is there no data for the RF resistance of the center conductor vs the
RF resistance for the braid? One would think it could be ascertained
by comparing known coax losses to known parallel line losses when the
wires are the same size.
--
73, Cecil, w5dxp.com

There was an article published maybe 15 years ago (RF Design
magazine?? Electronics Design?? EDN??), authored by a fellow from
Andrew as I recall, about some of the finer points of coaxial cable.
I thought it was quite a good article. He had "rules of thumb" for
loss in stranded center conductors versus solid that I remember for
sure, and perhaps for braid as well--I don't recall that for certain.
It wasn't huge, just a few percent, for the stranded center. Of
course in coax, since there's a lot more surface area to the outer
conductor than the inner, the braid would have to be considerably
worse than a solid conductor to significantly add to the total series
RF resistance, so it wouldn't be trivial to resolve by measuring coax
with a solid outer versus a braided outer. You'd have to go to
considerable effort to keep the rest of the construction identical to
nail down the contribution of the braid versus solid outer.

In any event, it seems a reasonable way to get a large diameter RF
conductor that remains flexible...I'd be very surprised if a 5mm
diameter coax braid was a worse two-wire line conductor than 2.5mm
solid, smooth copper. I would want to keep the jacket on it, sealed
against weather at the ends, to keep the copper clean.

Cheers,
Tom

On Feb 24, 3:12*pm, ve2pid wrote:
In the ARRL's *Antenna Book 21st ed page 24-21, we see that if we
connect the two shields of the coax cables together, we obtain
'Shielded parallel Lines' . In that case, the resultant impedance is
simply the sum of the characteristic impedances of each coax.

So, there is quite a difference between the two independent coax I
mentioned in my first message (we connect the shield to the inner
conductor at each of its ends) (A) and the 'Shielded Parallel Lines'
case (B).

*I am trying to understand why and it is the reason I posted my first
message...

In (A), the Z=276*log(2S/D) *applies, so the Zo of each coax does not
matter.. but in (B), Z=Zo1+Zo2, so the value of each Zo matters.

Am I right? And how to compute matched line loss in case (A) and in
case (B) ?

I suppose what follows has already been covered, but perhaps different
words to say it will help...

In the case of the two pieces of coax, used as coax, the fields are
entirely inside the coaxial lines. The center conductors carry
current, and as they have relatively little surface area compared with
the outer conductors, they contribute most of the loss (assuming good
dielectric with negligible loss). Balance depends on the load and how
the lines are fed; there is nothing inherently balanced about the line
itself. In fact, you could make such a line out of two pieces of coax
with different constructions: same outer conductor diameter, but
solid polyethylene dielectric for one and foamed polyethylene for the
other. The result would be different attenuations in the two, and
different propagation velocities, so that for the same physical length
for each, fed 180 degrees out of phase, the signals at the other end
would not be "balanced." Fed as a two-wire line, it would be balanced
(assuming it's kept away from un-balancing structures). The effective
impedance for two identical lines (back to the same dielectric, etc.)
fed that way is just the twice the impedance of a single line.

In the case of the two pieces arranged with constant spacing, fed as a
balanced line, all the fields are external to the lines. There is
essentially no field inside the outer conductor of each line. The
center conductors carry no current, so there is no particular need to
connect them electrically to the outer conductors. The loss is
determined by the RF resistance to the current carried on the outside
of the outer conductor, and by loss in the dielectric between the
wires (or more properly, dielectric in the area around the wires where
the electric field is significant). Since the fields are external to
the coax (not used as coax), the impedance depends only on the
physical configuration outside the coax -- not on the coax inner
conductor diameter, and not on the dielectric inside the coax. Those
are invisible to the fields carrying energy on the two-wire balanced
line.

Note that it's possible to transmit three independent TEM signals on a
couple pieces of coax configured as a two-wire line: one on each coax
and one on the balanced line.

There's a useful formula for estimating the loss of coax or balanced
lines. It includes a term for loss in the resistance of the
conductors, and another term for loss in the dielectric. For coax
using good dielectric, up to a few hundred MHz at least, the second
term can almost always be ignored. In English units (conversion to
metric left as an exercise...), it's:

A100 = 4.34*Rt/Zo + 2.78*f*Fp*sqrt(epsilon)

where
A100 = matched line loss in dB/100 feet
Rt = total conductor RF resistance (see below), ohms
Zo = characteristic impedance of the line
f = operating frequency in MHz
Fp = power factor of dielectric at frequency f
epsilon = permittivity of the dielectric, relative to air
(Note: Fp is essentially the same as the dielectric power factor
for any dielectric you'd care to use in an RF transmission line.)

A good estimate of Rt for copper conductors thicker than a couple of
skin-depths is:

Rt = 0.1*(1/d + 1/D)*sqrt(f)

where
d = diameter of inner coax conductor, or the wire diameter for two-
wire balanced line
D = diameter of the outer coax conductor, or = d for two-wire
balanced line

This formula is from the "Transmission Lines" chapter of Sams'
"Reference Data for Engineers." You can replace 1/d and 1/D in the Rt
formula with Rrf/d and Rrf/D for non-copper conductors, where Rrf is
the RF resistance of the conductor relative to copper. Normally, that
would be sqrt(Rdc(material)/Rdc(copper)), for non-magnetic
conductors. So for aluminum, use 1.23 in the numerator for that
conductor instead of 1.00. It is _probably_ reasonable to use 1.07
instead of 1.00 for stranded copper, and maybe 1.10 for braid, but
"your mileage may vary."

Cheers,
Tom

K7ITM wrote in
:

....
In any event, it seems a reasonable way to get a large diameter RF
conductor that remains flexible...I'd be very surprised if...

Tom,

I have attempted to predict the behaviour of bootstrap coax traps, and an
important factor is the effective RF resistance of the outside of the coax
which forms an inductor.

My measurement capacity is quite limited.

Trying to reconcile my prediction with W8JI's measurements of such traps
makes me think that the effective RF resistance is much poorer than a tube
of the same size. Factors causing this would be braiding, proximity effect
and dielectric losses in PVC jacket.

Owen

On Feb 25, 12:12*pm, Owen Duffy wrote:
K7ITM wrote :

...

In any event, it seems a reasonable way to get a large diameter RF
conductor that remains flexible...I'd be very surprised if...

Tom,

I have attempted to predict the behaviour of bootstrap coax traps, and an
important factor is the effective RF resistance of the outside of the coax
which forms an inductor.

My measurement capacity is quite limited.

Trying to reconcile my prediction with W8JI's measurements of such traps
makes me think that the effective RF resistance is much poorer than a tube
of the same size. Factors causing this would be braiding, proximity effect
and dielectric losses in PVC jacket.

Owen

I think I'll be in good agreement with you, here, Owen, by saying that
the right way to find out about something like this is to actually
make some measurements on the configuration in question. Maybe I can
come up with something... I do have reasonable measurement technology
available, though getting it "right" isn't trivial, especially with
balanced open-wire line. I do have some ideas and if I can find time
will try a couple different ways to see if they agree. Time, right
now, will be the problem.

Cheers,
Tom

In a line made up of two parallel coax cables with connected shields and
with the shields floating at both ends, let's suppose that one center
conductor (cable A) is carrying 1A and the other (cable B) 3A, exactly
out of phase with each other -- that is, the line is not balanced. The
inside of the cable A shield will have 1A and the inside of the cable
shield B will have 3A, again exactly out of phase, and each one out of
phase with its corresponding center conductor current.

Let's consider electrically short cables for simplicity -- short enough
that the currents don't change much along the length.

As Tom said, the fields are contained within the coax cables -- on the
inside of each one is a canceling pair of currents. But at the ends of
the cables, the shield currents can escape to the outside, 1A from cable
A, and an out of phase 3A from cable B. The result is a net 2A flowing
on the outside of the shield. This current will cause the combined cable
to radiate exactly the same amount as if the shield wasn't there at all!
All we've done is to physically separate the differential and common
mode currents, but we haven't changed their values.

So what happens when the conductor currents are balanced, i.e., equal
and opposite? Well, then there's no net current on the outside of the
shield. Now there's no line radiation, again exactly the same as if the
shield wasn't there. The shield likewise has no effect on reception of
unwanted signals -- it provides no shielding at all. The only way to
make it provide shielding is to block the current path from the inside
to the outside of the cables at both ends. In the case of an antenna
feedline, however, this would require putting the antenna into a
shielded box.

The shield has an impact on the differential mode impedance, but it
doesn't help line balance or provide shielding at all.

Roy Lewallen, W7EL

Roy Lewallen wrote in
:

....
The shield has an impact on the differential mode impedance, but it
doesn't help line balance or provide shielding at all.

Roy,

I quite agree with you.

I have received a third hand comment that something in my related article
at http://www.vk1od.net/transmissionline/stcm/index.htm is incorrect. I say
something because there is a language translation involved in the third
hand process.

Have you read the article, and have you any issue with it?

Thanks
Owen