ve2pid wrote in news:cd290c79-c0c7-4308-af41-
:

Let's suppose that we use to identical length of coax, and on each
one, we connect the shield to the inner conductor at each of its ends.
If we use these two independant (?) sections as a substitute for an
open-wire line, could we use the usual formula Z=276*log(2S/D) to
compute the impedance of that line?

In that case, the characteristic impedance Zo of the coax could be
neglected. And what about the matched losses in dB/100' of the
resultant line? Twice the one of the coax?

Pierre,

The formula you give is an approximation which is not good for low Zo,
and it ignores the effect of the dielectric (coax jacket).

I have seen discussion of this recently in another place.

Yes, you can fabricate a two wire open line like this. You need to find a
way to bind the two cables for consistent physical spacing, if the jacket
is PVC you are using a lossy dielectric, the braided conductor is lossier
at RF than an equivalent solid copper tube. So an expensive low Zo line,
not a low as you might think, with poor performance.

Why would you do this?

Owen

Pierre,

The matched loss of the resulting open-wire line is not related to the
matched loss of the coax cablebecause those cables are not
individually operating in "Transmission Line" mode. For the same
reason, the velocity factor of the open-wire line will not be the same
as the coax.

Here's a write-up of an experiment I did to produce low Zo balanced
line. I didn't measure the loss because my application was for a very
short run:

http://www.karinya.net/g3txq/twin_feed/

73,
Steve G3TXQ

steveeh131047 wrote in news:f6f2c446-5128-4851-aab2-
:

Here's a write-up of an experiment I did to produce low Zo balanced
line. I didn't measure the loss because my application was for a very
short run:

Steve, you haven't let on much detail about the application of your low Zo
line.

I think I suggested elsewhere that you could fabricate such a thing with
parallel coaxes with all of their shields at each end bonded.

I explored using square tubes as a rigid transmission line. This finds
application in a range of areas, eg the boom of a LP. See
http://www.vk1od.net/calc/tstl.htm . A 50 ohm line (if that was your
objective) requires an air spaced line of 1.19 D/d. The article contains a
calculator to solve the curve fit.

Owen

On Feb 24, 8:16*pm, Owen Duffy wrote:
steveeh131047 wrote in news:f6f2c446-5128-4851-aab2-
:

Here's a write-up of an experiment I did to produce low Zo balanced
line. I didn't measure the loss because my application was for a very
short run:

Steve, you haven't let on much detail about the application of your low Zo
line.

I think I suggested elsewhere that you could fabricate such a thing with
parallel coaxes with all of their shields at each end bonded.

I explored using square tubes as a rigid transmission line. This finds
application in a range of areas, eg the boom of a LP. Seehttp://www.vk1od..net/calc/tstl.htm. A 50 ohm line (if that was your
objective) requires an air spaced line of 1.19 D/d. The article contains a
calculator to solve the curve fit.

Owen

Owen.

The application is no secret - the first paragraph in my linked page
explains it.

I was looking for line with a Zo in the range 50ohms-70ohms that could
be used for the band interconnects on the hexbeam. Arguably, balanced
line would be easier to handle, and might have some common-mode
advantages. Total length is about 4ft so loss is not an issue. It
would need to be very easily replicated, using materials readily
available to constructors world-wide.

At the end of the day it became more of a "self learning" exercise
than a serious quest !

73,
Steve G3TXQ

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) ?

ve2pid wrote in news:e44076bb-dfdc-4aa4-8c13-
:

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...

That is not quite what you asked.

However, in the case you described in your first post, the shield of each
cable carries the differential current on the outside of the shield.

In the case of B above, the differential current is carried entirely
inside the shield, no differential current flows on the outside of the
shield.

So, the field structures, and therefore Zo are different in the two
cases.

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 opine that you will have to measure MLL for case A, mainly because the
effective RF resistance of the braid is not easy to estimate.

(B) can be found from tables, or better still,
http://www.vk1od.net/calc/tl/tllc.php , it is the same as the single
cable used to fabricate the pair... think about it.

Owen

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

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.


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

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

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