Question on SWR
 

Owen Duffy wrote in
:

....
though the conductors might seem relatively thin (RF R is proportional
to diameter for wide spaced line), I^2R loss is lower.

Of course, that should read "RF R is inversely proportional to diameter for
wide spaced line"

Owen

Question on SWR
 

Jim Lux wrote:

144 MHz isn't HF, which is where the original statement is valid. At
frequencies above around 50 MHz, depending on the dielectric, the
dielectric loss starts to be more significant.


The analyses I've done using published and realistic values for
dielectric (PE and PTFE) and conductor loss indicate that dielectric
loss doesn't become significant until the 1 - 10 GHz region, well above VHF.

Another trap for the unwary, when comparing coax losses, has to do with
skin effect and the thickness of the copper or silver cladding on the
center conductor. You could have an air insulated coax with silver
plated over stainless steel where the loss is actually greater at low
frequencies than higher, because the skin depth is greater at low
frequencies and the current is flowing mostly in the SS, rather than the
copper. (such coax is used in cryogenic applications, lest one think
it's overly contrived as an example)

It's unusual to find a plating thickness that's less than several skin
depths thick except at MF and below, or perhaps the low end of HF. It
does happen, though. I have some 0.1 inch diameter 75 ohm cable which
has a very small center conductor which is made of several strands of
even smaller Copperweld wire (thick copper over steel). Even though the
copper is probably a sizable fraction of the total wire diameter, the
wire is so small in diameter that the copper isn't several skin depths
thick at lower HF. It has very noticeably excessive loss at 7 MHz,
something I discovered the hard way one Field Day.

Roy Lewallen, W7EL

Question on SWR
 

In message , Jim Lux
writes
Antonio Vernucci wrote:
since most of the loss in practical coax cables is due to I^2R loss
(compared to V^2G)
A quick question. If most of the the cable loss is due to I^2R, how
can one explain that the foam versions of common coaxial cables show a
much lower loss than versions having solid PE insulation?
For instance RG-213 is rated at 8.5dB loss for 100 meters at 144
MHz, while RG-213 foam at only 4.5 dB. If G is relatively unimportant
with regard to loss, how can one explain that a change of insulation
material yields such a tremendous change in loss?
Thanks and 73
Tiony I0JX

144 MHz isn't HF, which is where the original statement is valid. At
frequencies above around 50 MHz, depending on the dielectric, the
dielectric loss starts to be more significant.

Another trap for the unwary, when comparing coax losses, has to do with
skin effect and the thickness of the copper or silver cladding on the
center conductor. You could have an air insulated coax with silver
plated over stainless steel where the loss is actually greater at low
frequencies than higher, because the skin depth is greater at low
frequencies and the current is flowing mostly in the SS, rather than
the copper. (such coax is used in cryogenic applications, lest one
think it's overly contrived as an example)

Indeed. If you compare a frequency response of a reel of coax with a
plated centre conductor (say, copper on steel) with one with a solid
inner conductor, the former often has a noticeable kink at around 40 or
50MHz. A long time ago, we found this out at work when trying to find
drop cables and miniature cables which we could be used as simulations
of large-diameter, low-loss CATV trunk cables (for use in the design
lab).

At least one major CATV supplier had lots of very large reels of various
lengths the 'real thing' (no pun intended) in a massive trailer parked
immediately on the other side of lab wall. These were patched through
the wall to the test benches. The engineer could then test and adjust
the flatness of the frequency response of wideband amplifiers against
the length of appropriate cable. Thankfully, this technology has largely
been superseded by optical fibre/fiber equipment!
--
Ian

Question on SWR
 

In message , Cecil Moore
writes
Owen Duffy wrote:
The ARRL information on "extra loss due to VSWR" is may be incomplete
in that it may not the assumptions that underly the formula used for
the graphs.

It is possible for a feedline with a high SWR to have
lower loss than the matched-line loss. For instance,
if we have 1/8WL of feedline with a current miminum
in the middle of the line, the losses at HF will be
lower than matched line loss because I^2*R losses tend
to dominate at HF.

I'd never thought of that. I suppose it applies to any situation where
the feeder is electrically short, and the majority of the current is
less than it would be when matched. I presume that the moral is that
formulas only really work when the feeder is electrically long enough
for you to be concerned about what the losses might be.
--
Ian.

Question on SWR
 

I'd never thought of that. I suppose it applies to any situation where the
feeder is electrically short, and the majority of the current is less than it
would be when matched. I presume that the moral is that formulas only really
work when the feeder is electrically long enough for you to be concerned about
what the losses might be.
--

Conversely, for a very short line closed on 5 ohm (instead of 500 ohm), the
extra loss caused by SWR would be higher than that shown on the ARRL graph
(apart from the fact that, when attenuation is so low, the extra attenuation is
generally not of much interest, nor it can be read on the ARRL chart).

Evidently the ARRL chart shows some average between the two cases. On the other
hand they probably had no better way to synthetically illustrate a concept
without giving too many details.

73

Tony I0JX

Question on SWR
 

The 300-ohm TV flat ribbon specifications show an attenuation
generally lower than that of plain RG-8, despite the conductors of the
ribbon are by far thinner than those of RG-8 (especially than the
cable shield).

Under matched line conditions, the 300 ohm line transfers the power at
higher voltage and lower current, one sixth of the current, so even though
the conductors might seem relatively thin (RF R is proportional to diameter
for wide spaced line), I^2R loss is lower.

Thanks, but shouldn't the current ratio be the square root of 6 instead of 6?

73

Tony I0JX

Question on SWR
 

Since IR losses dominate at these frequencies, reducing current reduces loss.
The higher Z means more voltage and less current for the same power. Loss
will be 1/36th, assuming all the conductor sizes are the same.

On another thread I was arguing that, if I am correct, the current ratio should
be the square root of 6, not 6. Accordingly the loss would be 1/6, instead of
1/36.

73

Tony I0JX

Question on SWR
 

Ian Jackson wrote:
In message , Cecil Moore
writes
Owen Duffy wrote:
The ARRL information on "extra loss due to VSWR" is may be
incomplete in that it may not the assumptions that underly the
formula used for the graphs.

It is possible for a feedline with a high SWR to have
lower loss than the matched-line loss. For instance,
if we have 1/8WL of feedline with a current miminum
in the middle of the line, the losses at HF will be
lower than matched line loss because I^2*R losses tend
to dominate at HF.

I'd never thought of that. I suppose it applies to any situation where
the feeder is electrically short, and the majority of the current is
less than it would be when matched. I presume that the moral is that
formulas only really work when the feeder is electrically long enough
for you to be concerned about what the losses might be.

That's a good way of putting it, but it only applies to the generalized
ARRL chart which takes no account of the actual load impedance or the
actual feedline length.

Owen's explicit method should get it right in all cases. If you select
say 0.125 wavelengths of RG213 in Owen's online calculator, the
predicted loss with a 100 ohm load resistance is *less* than the matched
loss. If you change the load to 25 ohms, the predicted loss is *greater*
than the matched loss.

Both of these results make perfect physical sense because the largest
part of the loss is proportional to the square of the current, which
will be greater with the lower-resistance load. The two different
resistances correctly give different results, yet they both have a VSWR
of 2 (based on the 50-ohm system impedance). This shows that VSWR does
not contain sufficient information to give an explicit single-valued
result.

Owen's program will accept a VSWR input, but it correctly posts a bold
red warning that the result is an estimate. If you let the program
select the worst-case load impedance for the supplied value of VSWR,
you're back on track and it can calculate an explicit result.


Although we're debating fractions of a milliBel here, the debate has
shown how often the terms "VSWR" and "load impedance" are used
interchangeably - which they aren't. It isn't a big mistake here, but it
can be in other applications. For example, a solid-state PA designed
for a 50 ohm load will respond very differently to load *impedances* of
100 or 25 ohms, yet the load *VSWR* is the same in both cases.



--

73 from Ian GM3SEK

Question on SWR
 

Ian Jackson wrote:
I'd never thought of that. I suppose it applies to any situation where
the feeder is electrically short, and the majority of the current is
less than it would be when matched. I presume that the moral is that
formulas only really work when the feeder is electrically long enough
for you to be concerned about what the losses might be.

I suspect the ARRL additional loss due to SWR charts
are based on 1/2WL increments of feedlines.
--
73, Cecil http://www.w5dxp.com
"According to the general theory of relativity,
space without ether is unthinkable." Albert Einstein

Question on SWR
 

"Antonio Vernucci" wrote in
:

....
Under matched line conditions, the 300 ohm line transfers the power
at higher voltage and lower current, one sixth of the current, so
even though the conductors might seem relatively thin (RF R is
proportional to diameter for wide spaced line), I^2R loss is lower.

Thanks, but shouldn't the current ratio be the square root of 6
instead of 6?

Yes, of course... my mistake.

Owen