On Feb 23, 4:41*pm, Jim Lux wrote:
Fred McKenzie wrote:
In article
,
*phaedrus wrote:

Hi again,

Well I have read your informative replies on the problems with making
a low Zo twin feed with much interest. Clearly it's not really a
practical proposition. Shame.

So the obvious question is: why are transmitters normalised to 50 ohms
when clearly 450 ohms would enable us to enjoy cheaper, do-it-
yourself, lower loss feeders? Was this some oversight at the time, or
good practice for some obscure reason that I simply cannot think of?

As others mentioned, the origin is probably based on military standards..

It is my understanding that the lowest loss air dielectric Co-Ax would
have an impedance of around 75 Ohms. *

loss, at constant frequency, is proportional to
1/Z * (1/b + 1/a)
b= diameter of outer , a = diam of inner

This reflects the fact that at lowish frequencies (HF), the loss is
ohmic, so it's the series combination of the resistivity of the center
and outer conductors (the 1/b+1/a term), and the current flowing (the
1/Z term.. higher Z means lower current, so less IR loss)

Z is proportional to log(b/a)
so
loss = k/log(b/a)*(1/b+1/a)

Run the numbers and you see that a ratio of 1:3.6 gets you 50 ohms with
epsilon=2.3 and a ratio of 1:6.7 gets you about 75 ohms.

At 10 MHz loss is about 0.028 dB/meter for the 50 ohm, and 0.017
dB/meter for the 75 ohm.

but you can go higher in Z...and the loss keeps going down, but even at
1:20 diameter ratio, the impedance is 118 ohms, and the loss is 0.010
dB/meter. * So 75 isn't a "lowest loss" frequency. *More likely, 75 ohms
happens to be close to 72 ohms, which is the characteristic impedance of
a dipole in free space, or, more usefully, 1/4 of the 300 ohm impedance
of a folded dipole.

In air, the dimensions and loss a
50 ohm 1:2.3 0.033 dB/meter
75 ohm 1:3.5 0.019 dB/meter

If you use the same mechanical

dimensions but with a polyethylene dielectric, the impedance becomes 52
Ohms.

I think that is just coincidence:

Z = 138/sqrt(epsilon)* log (b/a)

air has epsilon=1 (or very close)
polyethylene has 1/sqrt(epsilon) about 0.66, so, in fact, it happens to
work out (e.g. 50/75 = 0.66)

Agreed that (ignoring loss in the dielectric)

loss(dB) = k * (1/a + 1/b) / log(b/a)

k includes a factor for the permittivity of the dielectric, a factor
for unit length of line, and a factor for the units of a and b.

Now assume a fixed outer diameter b; does this function have a minimum
versus a? Let k=1 and b=1 for convenience (won't affect the ratio b/a
at minimum). So, find "a" to minimize

y = (1/a + 1)/ log(1/a)

I get that to be b/a = 3.5911 or so. It doesn't even matter if you
use log10 or natural log; same result. This says that loss for a line
constructed the same but with a larger b/a and same b will have higher
loss.

With vacuum dielectric, the impedance of coax with that 3.59:1 ratio
is about 76.7 ohms. With solid polyethylene dielectric, it's about
50.6 ohms.

I didn't see anything in your dB/meter numbers about line diameter,
and that's clearly important. You can make the line bigger to get
lower loss, or smaller and get more loss. Clearly if I keep "a"
constant and let "b" grow to get higher impedance, the predicted line
loss will go down monotonically with increasing b/a; things fall apart
when b becomes a significant fraction of a wavelength, of course.

If I calculate the dB loss for air-dielectric lines of several
impedances, all with the same outer conductor inside diameter, and
ratio the results, here's what I get. Note that the attenuation
doesn't really change much with impedance until you get to fairly high
or fairly low impedances. Also...simply multiply the Zo numbers by
0.66 to get the ratios for lines filled with solid polyethylene
dielectric (again assuming no loss in the dielectric itself; generally
accurate up to at least several hundred MHz).

Zo (dB/unit length @ Zo)/(dB/unit length of 76.7ohm line)
25 1.682
30 1.475
35 1.333
40 1.231
45 1.157
50 1.103
60 1.035
70 1.005
75 1.000
80 1.001
90 1.018
100 1.052
120 1.168
140 1.350
160 1.607

One shouldn't read too much into these numbers with several
significant figures; line attenuation also depends on things like
whether the conductors are solid or stranded or braided. Stranding
and braiding adds to the loss.

'Nuff for now...

Cheers,
Tom

On Feb 14, 2:14*pm, Jeff Liebermann wrote:
On Sun, 14 Feb 2010 01:33:58 -0800 (PST), phaedrus

wrote:
why are transmitters normalised to 50 ohms
when clearly 450 ohms would enable us to enjoy cheaper, do-it-
yourself, lower loss feeders?

A 450 ohm coaxial cable would be rather large.
* Zo = 450 = 138 log(b/a) * (where b=OD and a=ID)
For an inner conductor diameter of 1 mm, the outer shield diameter
would need to be 1800 mm or about 71 inches.

--
Jeff Liebermann * *
150 Felker St #D * *http://www.LearnByDestroying.com
Santa Cruz CA 95060http://802.11junk.com
Skype: JeffLiebermann * * AE6KS * *831-336-2558

I understand the Navy used to have some coaxial line you could drive
through. The center conducter was hung from huge ceramic insulators
similar to those used by the power company.

Jimmie