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

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

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

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

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!!

Jeff