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