In message , Owen Duffy
writes
Ian Jackson wrote in
:

...
In my simplistic world, this is how I understand things:

Provided the TX is followed by a tuner/matcher, which matches whatever
is attached to the tuner output to 50 ohms at the tuner input, the TX
will be happy.

"Happey" eh!

Your later correction is noted!

The power loss in the feeder is essentially a function of its inherent
loss (when matched) per unit length, and the SWR on it.

Wrong.

OK, the loss may not be uniform if the SWR is greater at the load end.
But when is the power loss* NOT a function of its matched loss per unit
length, and the SWR on it. [*Should I have said 'in dB'? ]

The SWR is a function of the mismatch between the load on the antenna
end of the feeder, and the feeder characteristic impedance. The

Well, the SWR at the load end is a function of Zl and Zo (both complex
quantities)... but the 'notional' SWR varies along the line as accounted
for by the complex propagation coefficient, in most practical cases at
HF, SWR decreases smoothly from load to source.

greater the SWR and the longer the feeder, the higher will be the loss
on the feeder. 'Lossless' feeder have no loss, regardless of length
and SWR. However, with real-world feeders, the losses rise
increasingly rapidly with increasing SWR.

See your earlier misconception regarding loss being a simple function of
SWR.

I didn't say 'simple'. I said 'essentially'.

The impedance looking into the tuner end of the feeder is the
impedance of the load, transformed by length of the feeder, and is
also modified by the loss of the feeder.

You got that right.

Yes, I know I did.

The higher the feeder loss, the closer the impedance seen at the tuner
end will approach the characteristic impedance of the feeder. [Long
lengths of lossy feeder - maybe with a nominal termination on the far
end - can make good dummy loads at VHF and UHF.]

Yes, but is it of practical application in a transmit scenario?

As a termination, I've used a fairly long piece of coax plus, for good
luck, a low-power termination hung on the end.

If the
input impedance due to a severly mismatched load is at all close to Zo,
then you have lost most of the transmitter power in the line.

Quite.

The "make a good dummy load" recipe doesn't address the power rating,
especially where most of the power is dissipated in a very short length
of cable.

Oh, indeed. But I have to confess that I've realised that I'm suddenly
unsure of where, on a long, lossy, feeder with a mismatched load, most
of the absolute power is actually lost. I'm assuming that it is at the
TX end, where the power is greatest - even though that's where the SWR
is best. Please advise!
--
Ian

On Sep 1, 10:29 am, Ian Jackson
wrote:
Oh, indeed. But I have to confess that I've realised that I'm suddenly
unsure of where, on a long, lossy, feeder with a mismatched load, most
of the absolute power is actually lost. I'm assuming that it is at the
TX end, where the power is greatest - even though that's where the SWR
is best. Please advise!

It seems logical that the highest I^2*R losses would be where the
standing-wave current is maximum and the highest dielectric losses
wold be where the the standing-wave voltage is maximum. As Owen is
fond of pointing out, the locations of those points are very
important. If a current maximum point exists at the load and a voltage
maximum point exists at the source, the losses at the load are
probably higher than the losses at the source on HF. If a voltage
maximum point exists at the load and a current maximum point exists at
the source, the losses at the load are probably lower than the losses
at the source on HF (assuming that losses due to SWR are mostly I^2*R
losses on HF).
--
73, Cecil, w5dxp.com

On 9/1/2010 9:20 AM, Cecil Moore wrote:

...
It seems logical that the highest I^2*R losses would be where the
standing-wave current is maximum and the highest dielectric losses
wold be where the the standing-wave voltage is maximum. As Owen is
fond of pointing out, the locations of those points are very
important. If a current maximum point exists at the load and a voltage
maximum point exists at the source, the losses at the load are
probably higher than the losses at the source on HF. If a voltage
maximum point exists at the load and a current maximum point exists at
the source, the losses at the load are probably lower than the losses
at the source on HF (assuming that losses due to SWR are mostly I^2*R
losses on HF).
--
73, Cecil, w5dxp.com

Uhhh, sorry to pose this question to you here but, doesn't a "tuner"
really just "shorten" and "lengthen" the feed line? I mean, not
physically, of course. But, it would seem to me that, this is exactly
what my xmitter and ant are "seeing."

Regards,
JS

On Sep 1, 11:43*am, John Smith wrote:
Uhhh, sorry to pose this question to you here but, doesn't a "tuner"
really just "shorten" and "lengthen" the feed line? *I mean, not
physically, of course. *But, it would seem to me that, this is exactly
what my xmitter and ant are "seeing."

If all the tuner did was shorten and lengthen the feedline, there
would be only one purely resistive low resistance available on the SWR
circle on the Smith Chart and it wouldn't be exactly 50 ohms. So a
tuner does one other thing - it transforms the non-50 ohm resistive
impedance to a 50 ohm resistive impedance. It is not only the
equivalent of shortening and lengthening the transmission line but
also performs an N:1 transformer function. Of course, it does that
seamlessly, i.e. it is not actually a two-step process.

When I vary the length of my ladder-line to obtain system resonance, I
have to be satisfied with purely resistive impedances between 35 ohms
and 85 ohms. I cannot achieve a perfect 50 ohms on all HF bands with
my matching method. That would require the addition of an actual N:1
transformer which is certainly possible but probably not worth the
effort. Since 35-85 ohms is perfectly acceptable to my SC-500 amp, I
don't need a high-power tuner. And since the SC-500 is spec'ed to
handle an SWR of 6:1, I doubt that an 35-85 ohm load makes it
"unhappy". If my SC-500 has ever been "unhappy", I failed to observe
that "unhappiness" but maybe I am just oblivious to such?
--
73, Cecil, w5dxp.com

On 9/1/2010 11:35 AM, Cecil Moore wrote:

...
When I vary the length of my ladder-line to obtain system resonance, I
have to be satisfied with purely resistive impedances between 35 ohms
and 85 ohms. I cannot achieve a perfect 50 ohms on all HF bands with
my matching method. That would require the addition of an actual N:1
transformer which is certainly possible but probably not worth the
effort. Since 35-85 ohms is perfectly acceptable to my SC-500 amp, I
don't need a high-power tuner. And since the SC-500 is spec'ed to
handle an SWR of 6:1, I doubt that an 35-85 ohm load makes it
"unhappy". If my SC-500 has ever been "unhappy", I failed to observe
that "unhappiness" but maybe I am just oblivious to such?
--
73, Cecil, w5dxp.com

No, I got you. Soon as you stepped though the logic, the match became
quite clear as a "third leg" ... glad you have a "happy amp" ... but if
the amp goes overboard, escalates its' state of happiness and goes into
the "gay zone", DUMP IT at the nearest swap! stoopid grin

Regards,
JS

In message
,
Cecil Moore writes
On Sep 1, 10:29 am, Ian Jackson
wrote:
Oh, indeed. But I have to confess that I've realised that I'm suddenly
unsure of where, on a long, lossy, feeder with a mismatched load, most
of the absolute power is actually lost. I'm assuming that it is at the
TX end, where the power is greatest - even though that's where the SWR
is best. Please advise!

It seems logical that the highest I^2*R losses would be where the
standing-wave current is maximum and the highest dielectric losses
wold be where the the standing-wave voltage is maximum. As Owen is
fond of pointing out, the locations of those points are very
important. If a current maximum point exists at the load and a voltage
maximum point exists at the source, the losses at the load are
probably higher than the losses at the source on HF. If a voltage
maximum point exists at the load and a current maximum point exists at
the source, the losses at the load are probably lower than the losses
at the source on HF (assuming that losses due to SWR are mostly I^2*R
losses on HF).
--
The SWR on a feeder will always be greatest at the load end. However,
because of the feeder losses, the power will also be lowest at the load
end.

So, although the power loss per unit length (in dB) due to SWR will
steadily increase as you approach the load end, because of the feeder
loss, there will be less absolute power (in watts) to lose as you
approach the load end.

The question is, at which end of the feeder is the most absolute power
lost per unit length?
--
Ian

On Sep 1, 1:54*pm, Ian Jackson
wrote:
The question is, at which end of the feeder is the most absolute power
lost per unit length?

I already explained that. On HF, maximum losses depend upon where the
current/voltage-nodes/loops are located on the transmission line. One
cannot answer the above question without knowing that information and
the answer is different depending upon the conditions.

Owen has previously proved that on HF, a higher SWR can result in
lower losses than a flat line when that high SWR is near a current
node and the feedline is less than a quarter wavelength long simply
because the average standing-wave current is *lower* than the flat
line current.
--
73, Cecil, w5dxp.com

In message
,
Cecil Moore writes
On Sep 1, 1:54*pm, Ian Jackson
wrote:
The question is, at which end of the feeder is the most absolute power
lost per unit length?

I already explained that. On HF, maximum losses depend upon where the
current/voltage-nodes/loops are located on the transmission line. One
cannot answer the above question without knowing that information and
the answer is different depending upon the conditions.

Owen has previously proved that on HF, a higher SWR can result in
lower losses than a flat line when that high SWR is near a current
node and the feedline is less than a quarter wavelength long simply
because the average standing-wave current is *lower* than the flat
line current.

Ah, I see what you mean. I was thinking in terms of an 'electrically
long' feeder (many wavelengths long).

I can see that, for a 'short' feeder (for example, if you were feeding a
halfwave dipole via a quarterwave of 300 ohm line), at the antenna
feedpoint you will have a current node (high I^2*R loss), and at the
TX/tuner end, where there is a voltage node, you will have a low I^2*R
loss. It could be that the loss of a matched relatively low-impedance
feeder of 50 or 75 ohms (being inherently higher than that of higher
impedance feeders with similar conductors etc) could be higher than the
average loss of the 300 ohm - even though the antenna will present an
SWR to the feeder of around 4 or 6 to 1.

That still leaves the question of which end of a long feeder has the
greater absolute power loss. Using 'reductio ad absurdum' and 'rule of
thumb' (two very dangerous principles!), I feel that it can only be at
the TX end. If the cable is really long, any power left to be dissipated
at the load end must be negligible compared with that available at the
TX end.
--
Ian

On Sep 1, 3:21*pm, Ian Jackson
wrote:
Ah, I see what you mean. I was thinking in terms of an 'electrically
long' feeder (many wavelengths long).

Even then, the same principle applies, at least on HF. If the standing
wave current maximum point is located near the load and the standing
wave current minimum is located near the source, the maximum
transmission line losses due to SWR will be near the load. If one
reverses those conditions, the results will reverse. As Owen suggests,
plot a graph of I^2 along the line - pick your particular load point
and particular source point and almost any outcome is possible.
--
73, Cecil, w5dxp.com

Ian Jackson wrote in
news

The question is, at which end of the feeder is the most absolute power
lost per unit length?

In response to your two posts...

You might be informed of the traditional graphs (or simple underlying
formulas) of "Additional loss due to VSWR" that appear in some
publications including several ARRL publications.

The formulas on which these graphs usually depend rely on a set of
assumptions that are often, if not usually, not stated.

You could be forgiven for interpreting these things to mean that loss per
unit length (in dB) is constant along a transmission line independent of
VSWR.

The traditional RLGC model of a transmission line does give constant loss
per unit length when VSWR=1.

For most practical lines at HF, most of the loss is due to conductor loss
(the R term in RLGC) and very little due to dielectric loss (G in RLGC).
The result of that is that line loss in any incremental length is very
dependent on current, higher in regions of higher current, and lower in
regions of lower current.

Consider the loss in 1m of RG58 at 3.5MHz with a 500+j0 load. It is a
very short line electrically so current is almost uniform. VSWR=10 and
current is approximately 1/3 of that for a matched line for the same
transferred power.

Why should loss be greater than a matched line when I^2 is one tenth the
matched line case? Well, it isn't. The formulas and graphs do not apply
because the (often unstated) underlying assumptions are not met.

Regarding the dummy load, it might seem intuitive that loss is highest
where the VSWR is worst, but that is not correct. The correct answer is
found by calculating the loss along the line.

If you consider the case of 100m of RG58C/U with O/C termination as a
dummy load for 1296MHz...

Lets section the 100m into 10+90m.

Using TLLC (http://www.vk1od.net/calc/tl/tllc.php), the input impedance
of the last 90m is 50.00-j0.01.

Now lets look at the first 10m of our 100m, it has a termination of
50.00-j0.01 and the loss is 8.5dB, only 14% of the input power flows into
the last 90m, 86% if the input power is lost somewhere in the first 10m.

By the same method, you could find that almost 20% of input power is loss
in the first metre, even though the VSWR in that section is almost
perfect.

Owen