William F. Hagen wrote:
an antenna has an impedence at the frequency it is being used at, and an
impedence at its resonant frequency. If either of these impedences happen to be
50 ohms and the coax being used is 50 ohms, and the transciever is working at
50 ohms, then the swr is 1:1, and the swr is on the transmission line, not on
the antenna.

One source of confusion is, on systems with both coax and ladder-line,
the SWR on the coax Vs the SWR on the ladder-line. A 12:1 SWR on 600
ohm ladder-line can result in a 50 ohm SWR of 1:1 without a tuner.
The ladder-line can be used as an impedance transformer.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

Cecil Moore wrote:

William F. Hagen wrote:
an antenna has an impedence at the frequency it is being used at, and an
impedence at its resonant frequency. If either of these impedences happen to be
50 ohms and the coax being used is 50 ohms, and the transciever is working at
50 ohms, then the swr is 1:1, and the swr is on the transmission line, not on
the antenna.

One source of confusion is, on systems with both coax and ladder-line,
the SWR on the coax Vs the SWR on the ladder-line. A 12:1 SWR on 600
ohm ladder-line can result in a 50 ohm SWR of 1:1 without a tuner.
The ladder-line can be used as an impedance transformer.

That lily really didn't need the gold paint job, Cecil.

But thanks for providing a source of confusion. How could we have a
good argument without one. :-)

73, Jim AC6XG

Jim Kelley wrote:

Cecil Moore wrote:


William F. Hagen wrote:


an antenna has an impedence at the frequency it is being used at, and an
impedance at its resonant frequency. If either of these impedences happen to be
50 ohms and the coax being used is 50 ohms, and the transceiver is working at
50 ohms, then the swr is 1:1, and the swr is on the transmission line, not on
the antenna.





One source of confusion is, on systems with both coax and ladder-line,
the SWR on the coax Vs the SWR on the ladder-line. A 12:1 SWR on 600
ohm ladder-line can result in a 50 ohm SWR of 1:1 without a tuner.
The ladder-line can be used as an impedance transformer.



That lily really didn't need the gold paint job, Cecil.

But thanks for providing a source of confusion. How could we have a
good argument without one. :-)

73, Jim AC6XG


For additional confusion, *IF* your transmission line (coax or ladder
line) is low loss,
and if your rig can load into it, SWR doesn't much matter. Reflected
power will
"bounce" off the rig and go back to the antenna. Our rigs actually
present a very
low impedance to the antenna and transmission line. This is by design;
we want all
of the RF we manufactured to go to the antenna and none wasted as heat
in the
rig. Our rigs don't really look like the Thevinian equivalent (voltage
source with
internal resistor of 50 ohms) feeding a 50 ohm load. For a fixed
voltage and fixed
internal resistance, using a 50 ohm load gets you max power *into the load*,
ignoring the wasted power in the source resistance. Your electric power
company doesn't do that, efficiency would suck. They want all the energy
used to be in paying customer's loads. They do that by keeping their source
impedance very low. Actually, our rigs have a source impedance
of only a few ohms, and are designed to pump power into a 50 ohm load.
There is a delay involved with the reflected power getting to the
antenna, but
for the narrow bandwidth signals we transmit (SSB voice or code) this is not
significant. It will matter for amateur television up on UHF, though.

You can get a lower SWR reading than what your antenna is doing if you have
lossy feedline. The lossy feedline is absorbing some of the reflected
power.
So don't be suprised at the worse SWR if you upgrade your coax. As long as
you can load up into it, it's not a real problem.

=============================================

"What did Santa say at the house of ill repute?"
"Ho ho ho!"

============================================
Keep Santa in Xmas

Standing wave ratios on ordinary 1/2-wave dipoles can soar as high as
10-to-1.


With all the fuss made about excessive SWR everywhere else, why is it the
guru's never mention it, let alone show anxiety about it? What are they
trying to cover up?

Reg Edwards wrote:

Standing wave ratios on ordinary 1/2-wave dipoles can soar as high as
10-to-1.


With all the fuss made about excessive SWR everywhere else, why is it the
guru's never mention it, let alone show anxiety about it? What are they
trying to cover up?




It's a communist plot! ;-) Actually, if your feedline is low loss,
and is
able to handle the higher voltages you'd get with high SWRs, and you're
using a narrowband mode like voice SSB, and your rig can tune load
into it, high SWR is not a problem.

BTW, if you're using coax to feed a dipole, be sure to make a coil of
the coax feedline of several turns at the antenna feedpoint. The object
of this is to keep RF currents from traveling down the *OUTSIDE*
of the coax shield. Otherwise your antenna's radiation pattern will
be goofed up, and also the SWR will get even worse. This also
keeps stray RF out of the shack.

On Wed, 24 Dec 2003 08:09:55 +0000 (UTC), "Reg Edwards"
wrote:

Standing wave ratios on ordinary 1/2-wave dipoles can soar as high as
10-to-1.


With all the fuss made about excessive SWR everywhere else, why is it the
guru's never mention it, let alone show anxiety about it? What are they
trying to cover up?


Ah Punchinello,

You have had your crown returned to you by those same gurus who have
abandoned the field, and who quixotically made just those
protestations you now so miss.

We will leave it in your capable hands to argue both sides of the coin
on this one, in your fulsome, best oratorio.

73's and the best of the season to ye,
Richard Clark, KB7QHC

On Wed, 24 Dec 2003 06:11:10 GMT, Robert Casey
wrote:

Our rigs actually
present a very
low impedance to the antenna and transmission line. This is by design;
we want all
of the RF we manufactured to go to the antenna and none wasted as heat
in the
rig. Our rigs don't really look like the Thevinian equivalent (voltage
source with
internal resistor of 50 ohms) feeding a 50 ohm load.

Hi Robert,

Actually none of this is true. It is the glib explanation that is
bandied about commonly in this forum, but it contains its own internal
inconsistency of logic.

This illogic is present in the single statement:
Reflected power will
"bounce" off the rig and go back to the antenna.
The presumption is that the point of bouncing back, the transistion
point of the so-called low Z transmitter to the high Z line, performs
this action. It contains to howlers:
1. if it were true, no one would ever need a tuner whose sole purpose
is to do exactly that (the bouncing back);
2. if it were true, the original power coming from the transmitter
would see the same reflection and bounce right back in to turn to heat
(which is a fairly true representation of the problem of SWR).

As for the reality of the situation, answer me this:
1. How much power does your rig transmit?
2. How much power does your rig draw?
Correct me if the operation of dividing the first by the second does
not reveal an efficiency of roughly 40% and a power loss to heat of
roughly greater than that transmitted. Your rig has a massive heat
sink with a fan, n'est pas?

Too many of the pundits want to force a literal carbon composition
resistor into the mix so that they can point to its absence proving
Thevenin's Theorem does not apply. The same pundits ignore the fact
that Thevenin did not specify a resistor, he specified an impedance to
satisfy his theorem. It was Edison's pervsion of logic in trying to
persuade the investors that AC distribution was for the birds when it
came time to match loads. He inserted the false claim of resistance
forcing inefficiency. This perversion has been with us ever since and
qualifies such believers only as possible investors in the Edison DC
Power distribution company (which folded immediately due to
inefficiency in the market place).

73's
Richard Clark, KB7QHC

As for the reality of the situation, answer me this:
1. How much power does your rig transmit?
2. How much power does your rig draw?
Correct me if the operation of dividing the first by the second does
not reveal an efficiency of roughly 40% and a power loss to heat of
roughly greater than that transmitted. Your rig has a massive heat
sink with a fan, n'est pas?

n'est pas? ? Anyway, the power lost inside the transmitter I thought
was due to the
inneficiencies of a class A or B amp configuration. That the amplifying
device (tube
or transistor) is metering out varying amounts of current from a
constant DC supply.
And the current has to pass thru the tube or transistor before it gets
to the load.
Assuming a sine wave at RF, zero crossing would be low wasted power, peak
also low wasted power, but at 0.707 peak max wasted power (class B). Class
C is more efficient except the power in the harmonics you get have to be
filtered
out, and you get a heated filter.

Now if the Thevenin impedance of the transmitter was 50 ohms, then half
of the remaining power you successfully converted to RF would heat the
transmitter some more. That 40% would become 20%. But if you design
the output right, making the Thevenin impedance very low, little RF power
is wasted in the transmitter (say 1%) and 39% is pumped out into space
(on the air)
by a matched antenna. Now that low Thevenin impedance will reflect
most of the reflected from the mismatched antenna power back to the antenna.
If the transmitter Thevenin impedance was 50 ohms, it would absorb the
reflections and get even hotter. Oh, this has its uses, like in a
signal coming
from a TTL line driver that passes thru a "source termination" resistor
of 100
ohms, then thru a 100 ohm impedance transmission line (like carefully
designed
traces on circuit boards in that GHz computer) and then to a single
destination
receiving TTL gate. which looks like a high impedance load. The reflected
signal gets absorbed by the source termination. A reason for not
destination
terminations is that this makes the source work harder pumping DC current
when the signal is a "high", vs essentially idling when just source
terminated.
No daisy chaining allowed, the signal looks like crap except at the very
end.

On Thu, 25 Dec 2003 05:47:12 GMT, Robert Casey
wrote:

As for the reality of the situation, answer me this:
1. How much power does your rig transmit?
2. How much power does your rig draw?
Correct me if the operation of dividing the first by the second does
not reveal an efficiency of roughly 40% and a power loss to heat of
roughly greater than that transmitted. Your rig has a massive heat
sink with a fan, n'est pas?

n'est pas? ? Anyway, the power lost inside the transmitter I thought
was due to the
inneficiencies of a class A or B amp configuration.

Hi Robert,

Given you offer no evidence to the contrary, 40% efficiency seems to
be the confirmed rule. Giving it a name does nothing to reduce it or
enhance it, the calories expended rob us of RF output for the power
draw to generate that output. That power loss is confirmed through
everyone's experience as heat. Heat sinks and fans testify to our
acceptance of its loss.

The ONLY advance we can claim in the last half century, is that no
power is lost to lighting up filaments in transistors.

That the amplifying
device (tube
or transistor) is metering out varying amounts of current from a
constant DC supply.
And the current has to pass thru the tube or transistor before it gets
to the load.

All of this goes without saying and changes nothing about efficiency,
or Thevenin issues that isn't already obvious. If it illustrates
anything is that it CONFIRMS the Thevenin mechanism (once you look
under the hood, an argument no-no if you first posit the Thevenin
argument).

Assuming a sine wave at RF, zero crossing would be low wasted power, peak
also low wasted power, but at 0.707 peak max wasted power (class B).

What makes the angle an indicator of waste? The RF wave consists of
360 degrees of variation all of which is propagated. If you have your
head under the hood, playing with the innards of the source, there is
still nothing inherently lossy about the angle of conduction.

It is correlatable to the magnitude of voltage across a source (the
tube/transistor, not the load), or current through the source (same
provisos), but in that respect the least loss occurs at the EXTREME of
the cycle that draws down the voltage differential against the supply
or the current magnitude; certainly not at the zero cross (there is no
such thing at the plate or collector of a powered amplifier) and even
more remotely the 0.707. Again, such discussion simply focuses on the
Thevenin validation for the source Z specification.

Class
C is more efficient except the power in the harmonics you get have to be
filtered
out, and you get a heated filter.

Your logic, as faulty as the presentation may be, simply validates the
Thevenin specification for the Source Z. You exhibit any number of
calorie sinks to satisfy some aspect of the Real R, but I see by the
flawed argument below that you persist in Edison's folly of forcing an
R only into the Z specification of Thevenin.

However, we will next proceed to illustrate that Edison's folly is in
fact demonstrably valid for the present discussion. :-)

Now if the Thevenin impedance of the transmitter was 50 ohms, then half
of the remaining power you successfully converted to RF would heat the
transmitter some more.

With a 40% efficiency, that has been demonstrated without further
analysis.

That 40% would become 20%.

This is ENRON math.

But if you design
the output right, making the Thevenin impedance very low, little RF power
is wasted in the transmitter (say 1%) and 39% is pumped out into space
(on the air)
by a matched antenna.

"IF?" Are we to assume that every transmitter that has come down the
pike has never been designed correctly nor optimized? Forcing such an
argument without showing how to accomplish it presumes you have a
method that transcends all of the current generation of Engineers
capacity to bring it to the market. The force of Capitalism and the
Profit Motive most sincerely invalidates that concept in a heartbeat.

What about the remaining 61%? You have in fact lowered the emitted
power in comparison to the drawn power in your argument above. How
this goes to prove that Thevenin's source is Resistorless is an
argument of the man standing behind the curtains.

Now that low Thevenin impedance will reflect
most of the reflected from the mismatched antenna power back to the antenna.

And how does the low source Z face the prospect of the High load Z
reflecting the power in turn? Thevenin explains the absurdity of this
forced expectation, but you do not. Your argument contains the usual
blindside, to be described below.

If the transmitter Thevenin impedance was 50 ohms, it would absorb the
reflections and get even hotter.

It does get hot, that is demonstrably true to everyone's experience.
That heat has its genesis in the Thevenin model and confirms it.

Oh, this has its uses, like in a
signal coming
from a TTL line driver that passes thru a "source termination" resistor
of 100
ohms, then thru a 100 ohm impedance transmission line (like carefully
designed
traces on circuit boards in that GHz computer) and then to a single
destination
receiving TTL gate. which looks like a high impedance load. The reflected
signal gets absorbed by the source termination. A reason for not
destination
terminations is that this makes the source work harder pumping DC current
when the signal is a "high", vs essentially idling when just source
terminated.
No daisy chaining allowed, the signal looks like crap except at the very
end.


Your final elaborations above, again, go to the proof of Thevenin, how
you expect it to be otherwise has not been demonstrated.

Eventually, these arguments devolve to the myopic observation that
there is no "smoking gun" to be exhibited in the form of a carbon
composition resistor that embodies the forced R argument of Thevenin.
Such posters un/willingly ignore the obvious location found within the
Emitter-Collector junction. When my push comes to shove, like right
now, they generally scrabble around the spec sheets (giving every
evidence of having never built from one) to point out that their
transistor exhibits 2 to 3 Ohms NOT 50! They thump their thin chest
with pride and then proclaim vindication. Again the myopic arguments
collapse in turn when it is pointed out that the Transistor's
Emitter-Collector junction is not the Source Z, but rather the
abstracted or untransformed value that has yet to be observed (hence
the failure of their myopic testimony).

If you are going to look under the hood, you have to look at
everything under the hood. For EVERY finals deck supporting the usual
tandem transistor output, they ALL drive a transformer exhibiting a
1:3 winding ratio. The canonical instruction of Z transformation
reveals that this same series combination of Emitter-Collector Z
presented through the transformer finds at that output their abstract
Source Z transformed to, guess what?, roughly 50 Ohms.

However, that is not ALL of the components under the hood to be
examined. We also have in EVERY finals deck, an output low pass
filter with a characteristic Z of, guess what?, roughly 50 Ohms.

Hence, in the deep interior where we abstract the Thevenin model to
the circuit level, the equivalent R that is demanded, is found in the
several Ohms described above. THAT abstracted Thevenin Source sees
the 50 Ohm Load transformed DOWN to its expected value (the usual, and
expected reverse operation of transformers built with 3:1 ratios, from
the perspective of the load).

Most posters argue from their blind side of wholly ignoring the
MATCHing operation of the finals' transformer to force their low R
argument (which is a corruption of Thevenin's stated Z).

Thus, there is a chain of transformation from the Emitter-Collector
junction through many levels to finally meet a load that exhibits the
confirmation of Thevenin that is consistent with the evidence of heat
dissipated by the physical source.

If that load presents a mismatch, the power returned is delivered to a
MATCHED load: what was formerly the Source. The Source, by design,
reflects NOTHING. When the Source is presented with a mismatched Load
we either perform additional matching operations (a Tuner) or accept
the additional heat burden the mismatch forces upon the Source. This
can be in the form of a long term temperature rise (current phase is
additive) or the sudden spark (heat, n'est pas?) of voltage breakdown
(voltage phase is additive). Contrary to the lore of Transistors
being low voltage devices, failure is found far more frequently in the
current density at the, guess what?, Emitter-Collector junction. The
catastrophic results testify, again, to the validity of the Thevenin
model. As I have repaired more than my share of Electronic equipment,
professional experience has shown the majority of transistor failures
reveal themselves as shorts due to their inability to shed the heat of
that current density (a melt down - calories - heat - R).

The compelling, and obvious proof of this last is found in very few
operators deliberately driving massive mismatches to anything less
than equipment failure as a consequence. Of course, when that occurs,
it validates the reflective mode of the Source for as many cycles of
RF stored in the transmission line returning from the mismatched load;
being generous, say roughly several microseconds?

73's
Richard Clark, KB7QHC

Richard sed,
no power is lost to lighting up filaments in transistors.

=======================

Wrong ! Transistors don't have filaments.