I seem to recall that ladder/open-wire feeders are relatively high
impedance ~500 ohms. Is it possible to construct 50 ohm open wire
feeder and if so, what would the spacing be?

Thanks!

phaedrus wrote:
I seem to recall that ladder/open-wire feeders are relatively high
impedance ~500 ohms. Is it possible to construct 50 ohm open wire
feeder and if so, what would the spacing be?

Thanks!

Possible, perhaps, but not practical. Assuming air insulation, the
spacing between the wires would have to be just under 9% of the wire
diameter, and this spacing would have to be maintained very closely to
keep the impedance constant. I might still have some 72 ohm twinlead in
my junk box, which consisted of wires embedded in molded polyethylene.

With your interest in transmission lines, you'd benefit a great deal
from getting a copy of the _ARRL Antenna Book_.

Roy Lewallen, W7EL

On Thu, 11 Feb 2010 08:02:55 -0800 (PST), phaedrus
wrote:

Is it possible to construct 50 ohm open wire
feeder and if so, what would the spacing be?

Look at your lamp zip cord - close enough.

73's
Richard Clark, KB7QHC

Richard Clark wrote in
:

Look at your lamp zip cord - close enough.

There is an interesting example.

K8ZOA published a characterisation of ZIP line some years ago in this
groupd, and Zo is approximately 100 ohms.

Considering a half wave of ZIP cord with a load of 100+j0 at 14MHz, loss is
around 1dB.

Now, using N7WS's characterisation of Wireman 551 ladder line, again a half
wave at 14MHz, smaller conductors, wider spacing, longer, Zo=400, higher
VSWR, the loss with a load of 100+j0 at 14MHz, loss is around 0.14dB.

Owen

phaedrus wrote in news:fe96541e-75fc-423a-8eb7-
:

I seem to recall that ladder/open-wire feeders are relatively high
impedance ~500 ohms. Is it possible to construct 50 ohm open wire
feeder and if so, what would the spacing be?

Such a low characteristic impedance requires very close spacing of round
conductors. Proximity effect which causes the current to mainly flow in a
small adjacent portion of the conductors at close spacing makes achivement
of very low Zo at low loss very difficult, and not very practical.

TWLLC at http://www.vk1od.net/calc/tl/twllc.htm will perform the
calculations for you, including calculating the matched line loss and loss
under mismatch conditions, but read the comments on the page re proximity
effect.

Owen

On Feb 11, 8:02*am, phaedrus wrote:
I seem to recall that ladder/open-wire feeders are relatively high
impedance ~500 ohms. Is it possible to construct 50 ohm open wire
feeder and if so, what would the spacing be?

Thanks!

As hinted by others, especially Owen, it's quite possible to make
parallel-conductor line of very low impedance. There are two ways
that are easy to describe, and these may well bring others to your
mind:

First, consider a line made of broad, flat conductors: copper
ribbon. If the spacing is close compared with the conductor width,
the impedance will be low. This is not particularly practical for
"open-wire feeders," but is commonly used on printed circuit boards.
In the extreme, you can make transmission lines with impedances under
10 ohms this way, if you're of a mind to. You can use thin, flexible
Kapton tape, perhaps 0.1mm thick or less, and have conductors 10 or
more mm wide. According to a formula in Sams' "Reference Data for
Engineers," the impedance will be approximately 377*s/
(sqrt(epsilon)*w), where w is the conductor width and s is the
spacing. For air dielectric, where epsilon is 1, you can make a 10
ohm line if the width is about 38 times the spacing.

Second, consider a multi-wire line. One common construction is four
wires arranged as in the corners of a square, with diagonally opposite
wires connected together at the ends. That gets you a lower impedance
than open wire with the same spacing as the side of the square and of
course the same wire diameter...it's a bit more than half the
impdeance. You could also have more wires arranged around a circle,
with alternate wires fed in parallel. A variation on this theme has
been used on occasion by rabid stereophiles to feed (~8ohm) speakers:
use a 50-conductor ribbon cable. Mass-terminate with a standard two-
row connector. Tie all the pins on each side of the mating connector
together, so the transmission line is similar to 25 pairs that are
fairly closely spaced. (IMO, there are better ways to accomplish flat
frequency response in a stereo system...) But it does work to get a
low impedance line. (And yes, if you look closely, you'll realize the
transmission line impedance is NOT constant as you go to lower
frequencies...)

On the other hand, there's generally not much advantage to making a
low-impedance balanced line for RF work, so they are pretty uncommon
in practice.

Cheers,
Tom

K7ITM wrote:
On Feb 11, 8:02 am, phaedrus wrote:
I seem to recall that ladder/open-wire feeders are relatively high
impedance ~500 ohms. Is it possible to construct 50 ohm open wire
feeder and if so, what would the spacing be?

Thanks!

As hinted by others, especially Owen, it's quite possible to make
parallel-conductor line of very low impedance. There are two ways
that are easy to describe, and these may well bring others to your
mind:

First, consider a line made of broad, flat conductors: copper
ribbon. If the spacing is close compared with the conductor width,
the impedance will be low. This is not particularly practical for
"open-wire feeders," but is commonly used on printed circuit boards.
In the extreme, you can make transmission lines with impedances under
10 ohms this way, if you're of a mind to. You can use thin, flexible
Kapton tape, perhaps 0.1mm thick or less, and have conductors 10 or
more mm wide. According to a formula in Sams' "Reference Data for
Engineers," the impedance will be approximately 377*s/
(sqrt(epsilon)*w), where w is the conductor width and s is the
spacing. For air dielectric, where epsilon is 1, you can make a 10
ohm line if the width is about 38 times the spacing.


You see this in microstrip construction on various substrates,
particularly where it's part of a matching network for transistor
amplifiers (which can have Zs of a few ohms)


On the other hand, there's generally not much advantage to making a
low-impedance balanced line for RF work, so they are pretty uncommon
in practice.

in RF communications, that's true.

In other high power RF applications, low Z can be desirable (because
you'd rather take the IR hit than deal with the high voltage).

In the pulsed nuclear research business you see lines with Z of a few
ohms in pulse forming networks ften using distilled water as the
dielectric.. that epsilon of 80 gets the C right up there so Sqrt(L/C)
is small.





Cheers,
Tom

On Feb 11, 5:38*pm, Jim Lux wrote:
K7ITM wrote:
On Feb 11, 8:02 am, phaedrus wrote:
I seem to recall that ladder/open-wire feeders are relatively high
impedance ~500 ohms. Is it possible to construct 50 ohm open wire
feeder and if so, what would the spacing be?

Thanks!

As hinted by others, especially Owen, it's quite possible to make
parallel-conductor line of very low impedance. *There are two ways
that are easy to describe, and these may well bring others to your
mind:

First, consider a line made of broad, flat conductors: *copper
ribbon. *If the spacing is close compared with the conductor width,
the impedance will be low. *This is not particularly practical for
"open-wire feeders," but is commonly used on printed circuit boards.
In the extreme, you can make transmission lines with impedances under
10 ohms this way, if you're of a mind to. *You can use thin, flexible
Kapton tape, perhaps 0.1mm thick or less, and have conductors 10 or
more mm wide. *According to a formula in Sams' "Reference Data for
Engineers," the impedance will be approximately 377*s/
(sqrt(epsilon)*w), where w is the conductor width and s is the
spacing. *For air dielectric, where epsilon is 1, you can make a 10
ohm line if the width is about 38 times the spacing.

You see this in microstrip construction on various substrates,
particularly where it's part of a matching network for transistor
amplifiers (which can have Zs of a few ohms)



On the other hand, there's generally not much advantage to making a
low-impedance balanced line for RF work, so they are pretty uncommon
in practice.

in RF communications, that's true.

In other high power RF applications, low Z can be desirable (because
you'd rather take the IR hit than deal with the high voltage).

In the pulsed nuclear research business you see lines with Z of a few
ohms in pulse forming networks ften using distilled water as the
dielectric.. that epsilon of 80 gets the C right up there so Sqrt(L/C)
is small.



Cheers,
Tom



Yes, indeed a low impedance can be helpful sometimes...but the times
I've been involved with low-Z lines, they've almost always been
unbalanced (e.g. coaxial). For a PFN, the high epsilon is also an
advantage in storing a lot of energy. Since even very pure water is
pretty conductive (relative to seriously good dielectrics), I'm a bit
surprised it's used. Are the cases you're thinking of balanced lines,
or coaxial? Is the impedance also chosen to match the load (in some
sense), or is there also a transformer at the output to match to a
higher-impedance load?

It occurs to me that for a PFN, energy storage is a good thing...in
fact, it's half of what a PFN is all about, the other half being
delivering that energy in a controlled manner. But in power
transmission, (excess) energy storage is generally a disadvantage.
Lines with dielectric with high epsilon are lossier, and delay the
signal more, than lines with low epsilon dielectric.

Cheers,
Tom

On Feb 11, 7:21*pm, K7ITM wrote:
On Feb 11, 5:38*pm, Jim Lux wrote:


In other high power RF applications, low Z can be desirable (because
you'd rather take the IR hit than deal with the high voltage).

In the pulsed nuclear research business you see lines with Z of a few
ohms in pulse forming networks ften using distilled water as the
dielectric.. that epsilon of 80 gets the C right up there so Sqrt(L/C)
is small.

Yes, indeed a low impedance can be helpful sometimes...but the times
I've been involved with low-Z lines, they've almost always been
unbalanced (e.g. coaxial). *For a PFN, the high epsilon is also an
advantage in storing a lot of energy. *Since even very pure water is
pretty conductive (relative to seriously good dielectrics), I'm a bit
surprised it's used.

Both coaxial and parallel plate balanced.. For short pulses, the water
isn't all that conductive (because it takes time for the water
molecules to dissociate); you worry more about particulate or gas
bubble contaminants cause a dielectric breakdown.

For a spectacular example, look at the Z-machine.

On Feb 11, 8:02*am, phaedrus wrote:
I seem to recall that ladder/open-wire feeders are relatively high
impedance ~500 ohms. Is it possible to construct 50 ohm open wire
feeder and if so, what would the spacing be?

Thanks!

Oh, and to more directly answer your question, the formula generally
given for accurate description of the impedance of two parallel, round
conductors of the same diameter, in air, is:

Zo = 120 * invcosh(D/d)

where D/d is the center-center distance and d is the conductor
diameter. So...

D/d for 50 ohms = cosh(50/120) = 1.088

For 1" diameter wires, you'd have to space them 0.088 inches apart,
which is tough to maintain unless you embed them in a solid
dielectric.

I suspect the formula above is in error for very close spacings
because of proximity effects...I suspect it assumes a uniform thin
cylinder of current...but I'm not sure about that.

Cheers,
Tom