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?

73 de Pierre VE2PID

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) ?

steveeh131047 wrote in
:
The application is no secret - the first paragraph in my linked page
explains it.

I missed that!

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

I am not so sure there would be much difference.

would need to be very easily replicated, using materials readily
available to constructors world-wide.

If you tried my calculator for the square tubes, you would have found
that you get lower Zo for the same D/d ration as for round tubes. For
example, a pair of 12mm tubes clamped together with 2mm spacing gives Zo
of around 50 ohms. If such an element could be used to replace one of the
spreaders, then you have neat solution... but it would take some custom
moulding to support such a spreader on the hub.


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

Well, that is a serious quest.

Cheers
Owen


73,
Steve G3TXQ

On Feb 24, 11:28*am, Owen Duffy wrote:
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

I certainly agree with Owen that it's not a particularly good way to
get a low-loss two-wire balanced line. To me, the attraction of a two-
wire balanced line is the low loss when it's properly implemented at
relatively higher impedance. That said, however, one can buy "Siamese
twin" cables with two pieces of coax in the same jacket, arranged to
make it easy to separate them. The web of jacket between the two
lines is relatively thin. Such line is common in CATV installations
in homes.

At low frequencies (HF, especially lower HF frequencies), the
dielectric loss may not be too bad, especially for the low impedance
involved. It may be a cheap way to get an approximately 100 ohm
balanced line. Even though the braided conductors will have higher
loss than equivalent diameter smooth conductors, the copper loss for
typical RG-6-type line used that way (about 5mm OD outer conductor)
will almost certainly be less than that using, say, 2mm diameter solid
conductors. You may be able to find surplus (e.g. reel-ends) of this
sort of line at very attractive prices. Even new, it's not too bad--a
lot cheaper than equivalent diameter solid copper!

Cheers,
Tom

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?

73 de Pierre VE2PID


Sounds kinda messy and unstable, to me.... but do-able I suppoes. I'd
rather search for a few short pieces of 25 ohm coax and do the coaxial
balanced line thing as normally done.

Ed K7AAT