tuner - feedline - antenna question ?
 

Please don't insult our intelligence. If the Bird reads

Of course it is nonsense, but it is a logical development based on Jeff's
words "What you are describing could be called 'transmitted' power or
power delivered into a mismatched load, but that it different from
forward power, or the power delivered by the source" and your words "For
systems


Only a logical development if you selectively snip Owen.

"What you are describing could be called 'transmitted' power or
power delivered into a mismatched load" was referring to "Pload = Pfor -
Pref".

Ok I admit that 'transmitted' power could have been better phrased.

Power may not actually be dissipated in a lossless line but that does not
detract from the fact that there is current flow and a voltage along the
line produced by two distinct and independent waves travelling in opposite
directions. True that the power can only be realised when it encounters a
load, but it is highly pedantic not to regard the reflected signal as having
'power' until it actually encounters such a load. If you extend this theory
to a radiated signal, you could equally say that there is no power
travelling through the aether until it encounters a receiver.

It is naive to believe that reflected power is not dissipated in a matched
source, or partially re-reflected at the source/line interface is not
matched. Again going back to the optics corollary you would not expect a
reflected light signal not to impinge on, and interact with a source.

If you pad out a source with a sufficiently high attenuator such that the
reflected signal will not have significant effect on it, you will see an
increase in dissipation in the attenuators when the load is mis-matched. I
am confident that an attenuator is not having its "load line" changed such
that its dissipation goes up magically just by the same amount as the power
in the reflected wave!! (Of course the dissipation in the load is measurable
as heat).

Adding a circulator to a system will not change "the load line" (if a
transmission line or circulator can have such a thing), but it will cause
the power in the reflected wave to be separated so that it can be monitored
and measured. Surprisingly power monitored in this way ties up with the
notion that power is reflected at a mis-matched load.

73
Jeff

tuner - feedline - antenna question ?
 

Cecil Moore wrote:

My favorite quotation by an antenna guru on this
newsgroup is that "a 50 ohm antenna can be replaced
by a 50 ohm resistor without changing anything".
If that were true, we don't need antennas. :-)


That sounds like a direct misquotation of me.

What I HAVE said - and often - is that if you measure the load impedance
presented by an antenna and feedline at the output socket of the
transmitter, and replace it by the same impedance made from lumped R and
L/C components, then the steady-state operating conditions of the
transmitter will not change.

If the transmitter isn't touched, it will deliver exactly the same power
as before - because that happens to be how much power it can deliver
into that particular load impedance.

That's all the RF power there is. In a lossless system, all of that
power will be radiated from the antenna. With the alternative lumped
load, exactly the same power will be delivered into the resistive part
of the load, and dissipated as heat.

Of course the transmitter is under more stress from voltage, current and
heat when it's operating into an incorrect load impedance (not what it
was designed for) but that's all it is. There is no need to invent
reflected power that is being "dumped" back into the transmitter to
cause this stress.

There is also a strong tendency to invent virtual instruments such as
"directional wattmeters" which do not actually exist. An instrument such
as the Bird 43 is calibrated in watts, but it doesn't actually sample
power. As Owen relates (and so have I) these instrument only sample V
and I on the line - they categorically DO NOT sample power.

The power scale is only a meter calibration - literally, only ink on a
meter scale. It indicates the amount of power delivered into a matched
load, when the "reverse" reading is exactly zero [1]. The instrument was
calibrated under those specific conditions, and the "forward power"
reading is only physically meaningful for that specific case.

For a mismatched load, the meter will read higher in the forward
direction than in the reverse - but that is purely a feature of the
instrument. It all looks so plausible on the meter scale, but those are
not genuine power waves flowing in opposite directions.

Everything that's happening inside the instrument can be completely
explained from the new V and I conditions on the line. Waves of power
simply don't come into it. Most people don't want to go that deep into
the theory... but, regrettably, that may be the only way to understand
that the "power" readings on the meter scale are no longer valid under
these conditions.

What IS physically meaningful is the DIFFERENCE between the forward and
reverse "power" readings. That difference will equal the net power
delivered to the load [1]. But those two readings are only meaningful as
a pair - individually they are only "intermediate results" with no
physical meaning of their own.


[1] Ignoring real-life meter errors such as directivity and scale
accuracy.


I'm going to be away from my computer for 48 hours.

But you'll be back... :-)


--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

tuner - feedline - antenna question ?
 

On Feb 26, 6:02 pm, Owen Duffy wrote:
Cecil Moore wrote et:
Please don't blame me for someone else's words.

A selective partial quotation to misrepesent what was actually written Cecil!

My true statement was combined with someone else's false statement to
make the combination false. I'm asking politely that not be done
again.
--
73, Cecil, w5dxp.com

tuner - feedline - antenna question ?
 

On Feb 26, 6:11 pm, Owen Duffy wrote:
Cecil Moore wrote . net:

Owen Duffy wrote:
I suggest that if a PA / line / load situation transforms the actual
load to some arbitrary impedance Z at the PA end of the line, the PA
will peform exactly as if the PA were directly loaded by a lumped
constant load of Z.

Yes, that is true for the performance of the PA.
Certainly not true for the performance of the
transmission line or antenna.

Please explain?

If the transmission line and antenna are replaced by a lumped constant
load, the transmission line and antenna cease to function. IMO, that's
not "performing exactly as".
--
73, Cecil, w5dxp.com

tuner - feedline - antenna question ?
 

On Tue, 27 Feb 2007 09:10:12 +0000, Ian White GM3SEK
wrote:

As Owen relates (and so have I) these instrument only sample V
and I on the line - they categorically DO NOT sample power.

Hi Ian,

We've been through this before. No instrument operates in the absence
of power. Simply because you and Owen are graced with instruments
that demand so little, does not negate what power they do rob from
what is available. Even the humble electrometer has to overcome the
force of gravity to open its foil leaves, and climbing that potential
energy hill is work over time - power.

Hammer down the directivity as much as you want, and it will still
resolve to some diminution of power available to the load. To wave a
hand and say NOTHING does not make it so.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

On Feb 27, 3:10 am, Ian White GM3SEK wrote:
Cecil Moore wrote:
My favorite quotation by an antenna guru on this
newsgroup is that "a 50 ohm antenna can be replaced
by a 50 ohm resistor without changing anything".
If that were true, we don't need antennas. :-)

That sounds like a direct misquotation of me.

Nope, it wasn't you, Ian. You are usually more careful than that.

For a mismatched load, the meter will read higher in the forward
direction than in the reverse - but that is purely a feature of the
instrument. It all looks so plausible on the meter scale, but those are
not genuine power waves flowing in opposite directions.

But they are genuine energy waves flowing in opposite directions.
Standing waves require two coherent energy waves flowing in opposite
directions. Can you explain how to create a standing wave without
two energy waves flowing in opposite directions? And remember, the
two EM wave components in the standing wave cannot stand still.

There's truth in what you say, Ian, but it is not the whole truth.
There's
no difference except frequency (and all that frequency implies)
between
an RF electromagnetic wave and a visible light electromagnetic wave.
In fact, RF waves are covered in many physics textbooks whose subject
is light. There is a wealth of information available from the field of
optics
that is applicable to RF waves.

Visible light physicists don't have the luxury of measuring voltage or
directly measuring phase. They have to rely on a power measurement
of irradiance. As a result, visible light measurements are actually
power measurements so we indeed do know how EM waves behave
at the joules/second level.

Visible light physicists found that when they superpose two coherent
light waves, Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A)
where 'A' is the angle between the electric fields of the two waves.
That exact same equation applies to coherent RF waves. Phasor
addition is used for the superposition of two coherent RF voltages.
The power equation is used to find out what happens to the power
during that voltage superposition. P1 = V1^2*Z0 and P2 = V2^2*Z0
The last term in the power equation is known as the interference term
and is either constructive, destructive, or zero.

Since antenna radiation patterns depend upon constructive and
destructive interference of EM waves in space, we hams could
learn a lot from the field of physics known as optics.

I'm going to be away from my computer for 48 hours.

But you'll be back... :-)

Yep, I'm posting from my sister's computer through my Google account.
--
73, Cecil, w5dxp.com

tuner - feedline - antenna question ?
 

On Feb 27, 2:53 am, "Jeff" wrote:
Adding a circulator to a system will not change "the load line" (if a
transmission line or circulator can have such a thing), but it will cause
the power in the reflected wave to be separated so that it can be monitored
and measured. Surprisingly power monitored in this way ties up with the
notion that power is reflected at a mis-matched load.

Yes, and a little modulation added to the source signal will prove
that the
signal being dissipated by the circulator resistor has made a round
trip
to the load and back. That's hard to explain if reflected energy
doesn't
actually exist.
--
73, Cecil, w5dxp.com

tuner - feedline - antenna question ?
 

Richard Clark wrote:
On Tue, 27 Feb 2007 09:10:12 +0000, Ian White GM3SEK
wrote:

As Owen relates (and so have I) these instrument only sample V
and I on the line - they categorically DO NOT sample power.

Hi Ian,

We've been through this before. No instrument operates in the absence
of power. Simply because you and Owen are graced with instruments
that demand so little, does not negate what power they do rob from
what is available. Even the humble electrometer has to overcome the
force of gravity to open its foil leaves, and climbing that potential
energy hill is work over time - power.

Hammer down the directivity as much as you want, and it will still
resolve to some diminution of power available to the load.

Richard, that argument is otiose.

What I said (in full context) was that their principle of operation as
measuring instruments does not involve sampling traveling waves of power
from the line.

Of course they must incidentally consume some power to move the meter
needle, but that is not part of their operating principle.

For that matter, almost all measuring instruments abstract some energy
or power from whatever they are measuring - but that is usually
incidental. It certainly does not make every instrument into a power
meter.

Can you not see this?


To wave a
hand and say NOTHING does not make it so.

To wave another hand and say traveling waves of power exist does not
make that so, either.




--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

tuner - feedline - antenna question ?
 

"Cecil Moore" wrote in message
oups.com...
On Feb 27, 2:53 am, "Jeff" wrote:
Adding a circulator to a system will not change "the load line" (if a
transmission line or circulator can have such a thing), but it will cause
the power in the reflected wave to be separated so that it can be
monitored
and measured. Surprisingly power monitored in this way ties up with the
notion that power is reflected at a mis-matched load.

Yes, and a little modulation added to the source signal will prove
that the
signal being dissipated by the circulator resistor has made a round
trip
to the load and back. That's hard to explain if reflected energy
doesn't
actually exist.
--
73, Cecil, w5dxp.com

Indeed; TDR's would have areally hard time (;-))

Jeff

tuner - feedline - antenna question ?
 

Richard Clark wrote in
:

On Tue, 27 Feb 2007 01:07:12 GMT, Owen Duffy wrote:

Richard Clark wrote in
Let's treat this like the Chinese Box problem.

If you didn't know what the load was, could you explain it any
differently? No. Apriori knowledge is not a proof.

Richard, I content that:

Contend or offer in contention.

Richard

Yes, my spelling mistake.


- the power output of the PA; and
- the efficiency of the PA may be (and usually are) sensitive to the
load impedance.

This is not contending nor contention and is content only for a non
sequitur. The line following a tuner exhibits considerable loss (poor
efficiency) that can only occur on the basis of power and mismatch.
You yourself offered in other correspondence that it exceeds cable
attenuation specifications found only in a matching condition. To

I am being picky, but "it *may* exceed cable attenuation specifications
found only in a matching condition, it may also be lower". If I said it
as you stated, I made an error. The common statement (and I have no doubt
made it) that VSWR exacerbates line loss is actually wrong in the general
case. (Having Googled my own web site I see one statement along those
lines which needs further qualification!)

suggest that a PA's sensitivity is somehow exhalted in the face of
identical, ordinary behavior of a passive component is hardly
seperable. Consider the simple substitution to your quote:
- the power output at the terminus of the line; and
- the efficiency at the terminus of the line may be (and usually are)
sensitive to the load impedance.

I meant the output at the PA terminals where an lumped constant load
would be attached for comparison.


....

Though it is often asserted that the PA will get hotter as a result of
"reflected power" being dissipated in the dynamic output impedance of
the PA, and that this may / will damage the PA, the power explanation
doesn't work numerically in the general case.

Heat is the outward proof of power and is always demonstrable in both
specific and general cases. Occurrences of other, significant
radiation from the source (as long as that source physically occupies
a substantially minor region of wavelength) is exceedingly difficult
to achieve.

You don't offer a numerical proof of a general case, and given that
the general case must allow for the specific cases already allowed in
your discussion above - that may be an untenable assertion for you.
Those specific cases are demonstrably caloric and must follow the same
math you suggest.

I suspect you are trying to argue differences by degree (no pun
intended as to heat); but I seriously doubt you can produce the math
to do that. The arguments that flow from that involve what is called
source resistance, and those arguments are legion in this forum (where
naysayers embrace a refusal to accept or name ANY value - a curious
paradox and an engineering nihilism I enjoy to watch).

PAs can be designed to behave as an equivalent fixed voltage or current
source with fixed source impedance of Zo, but HF PAs are not usually
designed in that way.

I know that there is a vein of thought that the process of adjusting a PA
for maximum output always, somewhat magically, creates a match condition
where the source impedance is the conjugate of the load at the PA
terminals, but it is contentious. What of broadband PA designs with no
such adjustment, are they source matched over a broad range of
frequencies? Observations are that experiments to discover the source
impedance by incrementally changing load current can produce a range of
values for the same PA on different frequencies, and at different power
levels. Why do amplifiers with say tetrodes and triodes which exhibit
such different dynamic plate resistance but requiring the same load
impedance deliver the same equivalent source impedance?

I am also aware that supporters of the inherent source match position
assert that you must be selective in choosing tests for source impedance.
It is all rather unconvincing when only some of the implications of a
particular source impedance are effective.

It is my view that modelling the PA as a fixed voltage or current source
with fixed source impedance of Zo, and where reflected waves on a
transmission line are absorbed by the matched source is not a good
general model for HF PAs.

The application of small signal analysis to amplifiers that sweep from
near cutoff to near saturation is suspect.

I believe that it is sound (in the steady state) to resolve the forward
and reflected wave voltages and currents at the source end of the
transmission line, calculate the complex impedance, and predict the
effects of that impedance as a PA load using the same techniques that
were used to design the PA.

Owen

tuner - feedline - antenna question ?
 

On Tue, 27 Feb 2007 16:51:21 +0000, Ian White GM3SEK
wrote:

For that matter, almost all measuring instruments abstract some energy
or power from whatever they are measuring - but that is usually
incidental. It certainly does not make every instrument into a power
meter.

Can you not see this?

Hi Ian,

As a trained metrologist, I've seen it far closer up than you. I've
accurately measured the physics of technology out to more decimal
places than most, and there is nothing incidental about it - it is
quite fundamental. In fact I've encountered a spectrum of conflicting
and interfering physical principles to every measurement that had to
be accounted for to obtain that accuracy.

I dare say none here can equal the accuracy of my bench work at RF
Power measurement here, even those with a shelf full of digital meters
used as bookends for their library.

To wave a
hand and say NOTHING does not make it so.

To wave another hand and say traveling waves of power exist does not
make that so, either.

Your counter is not an argument for their inexistence.

The examples of "traveling waves of power" abound, even to the trade
craft of antennas. Examples of "traveling wave of power" are classic
in whole to the explanation for the operation of the standard
Directional Couplers (to distinguish them from the Bruene design).

There are more such examples, but the Directional Coupler is quite
suitable to their class, and its operation satisfies the proof of
power flow in both directions. There are other examples outside of
this class that also provide proofs; and they are based on physical
principles as well (polarization and magnetic moment).

Traditionally, these points pass in silence until some later time when
we visit them once again.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

Richard Clark wrote in
:

....
We've been through this before. No instrument operates in the absence
of power. Simply because you and Owen are graced with instruments
that demand so little, does not negate what power they do rob from
what is available. Even the humble electrometer has to overcome the
force of gravity to open its foil leaves, and climbing that potential
energy hill is work over time - power.

Hammer down the directivity as much as you want, and it will still
resolve to some diminution of power available to the load. To wave a
hand and say NOTHING does not make it so.

The fact that an instrument may consume power from the circuit does not
imply that it measures power.

For example, a "voltmeter" that samples the voltage and draws a small
current from the cicuit under test does consume some power, but it cannot
"measure" power in the circuit under test without knowledge of the
complex load impedance.

I wrote an analysis of the Breune circuit and show that the meter
deflection is a response to the Vf or Vr component, and I dealt with how
that information can be used (including applicable conditions). If you
(or others) think there are flaws in the article, I welcome constructive
feedback.

Owen

tuner - feedline - antenna question ?
 

Cecil Moore wrote:
The joules/sec are real quantities but whether joules/sec
is power depends upon the definition of "power".

In our case here on the internet, it depends on whether or not you
choose to equate 'units of power' with the definition of power.

73 ac6xg

tuner - feedline - antenna question ?
 

On Tue, 27 Feb 2007 21:14:38 GMT, Owen Duffy wrote:

This is not contending nor contention and is content only for a non
sequitur. The line following a tuner exhibits considerable loss (poor
efficiency) that can only occur on the basis of power and mismatch.
You yourself offered in other correspondence that it exceeds cable
attenuation specifications found only in a matching condition. To

I am being picky, but "it *may* exceed cable attenuation specifications
found only in a matching condition, it may also be lower".

Hi Owen,

Lower? That is rather astonishing in light of responding to my
comment.

If I said it
as you stated, I made an error. The common statement (and I have no doubt
made it) that VSWR exacerbates line loss is actually wrong in the general
case. (Having Googled my own web site I see one statement along those
lines which needs further qualification!)

This is even more astonishing. Irrespective of you being the source,
why inject this confusing comment? SWR always exacerbates line loss!
Give me any normal line attenuation and SWR at the load, and I will
tell you exactly how much additional loss will occur. There's a
general solution for you.

Something tells me that your comments are based on a confusion between
the power loss of a cable, and its mismatch loss. They are not the
same thing although they are usually tightly twined in discussion. You
later exhibit a confusion between a conjugate match and an impedance
match. They are not the same thing either. The confusion on both
these points have abounded in this group in past "debates."

I meant the output at the PA terminals where an lumped constant load
would be attached for comparison.

This then removes the reflection from the argument, doesn't it? It
actually doesn't; but this unwarranted substitution is like Zen
Archery in that the line already demonstrates the validity of
reflected power as distinct from that "power" just being a
mathematical fiction.

Putting the lumped load at the PA terminal merely casts the proof back
into the box, it doesn't negate reflected power. As the proof is
already supported in the line, then removing it is not strictly a
valid counter argument. However, we will explore it further:

PAs can be designed to behave as an equivalent fixed voltage or current
source with fixed source impedance of Zo, but HF PAs are not usually
designed in that way.

OK so we are now in my sidebar of source resistance. Even so, it has
nothing to do with the concept of reflected power except insofar as
that resistance's ability to reveal that power's dissipation.

Other's should ponder how the reflected power has a caloric proof in
the line, and then question why it wouldn't prove out when it arrives
back in the box where the temperature rises on its return.

Same source, same power, same reflection, same loss. The only thing
that varies is the capacity of any point along this signal chain to
support that heat burden. Let's skip these as choices of design.

I know that there is a vein of thought that the process of adjusting a PA
for maximum output always, somewhat magically, creates a match condition
where the source impedance is the conjugate of the load at the PA
terminals, but it is contentious.

That contention arises out of mistaking Z0 Matches with Conjugate
Matches. This is a common affliction among "debaters" here. Let's
skip their prejudices.

What of broadband PA designs with no
such adjustment, are they source matched over a broad range of
frequencies?

Having had designed broadband amps, this is simply accomplished with
the proper feedback such that, yes, they are matched over a broad
range. The math is quite simple, the cost is another matter. Can you
afford one? Probably not. The lack of commercial examples available
to the Ham is not proof they do not exist. Let's pass on from issues
of economy.

On the other side of the aisle, I've worked with active loads that
will absorb as much power (up to a limit) at any frequency (up to a
limit) that you care to throw at it.

Observations are that experiments to discover the source
impedance by incrementally changing load current can produce a range of
values for the same PA on different frequencies, and at different power
levels.

This is called "Load Pulling," and is a classic technique to
demonstrate source Z. Thevenin first described it and Norton followed
suit. I cannot, for the life of me, recall any other intellectual
giants that have pulled these apart.

I have done this with my own gear. The variation from a source Z of
50 Ohms wandered the SWR range of 1.5:1 over all bands and most power
levels. Given this conformed to the manufacturer's specification, I
was not particularly surprised. Where it deviated the most, the rig
also operated the worst. What can we say about experience and
performance design converging?

Why do amplifiers with say tetrodes and triodes which exhibit
such different dynamic plate resistance but requiring the same load
impedance deliver the same equivalent source impedance?

A cable connector instead of binding posts? Let's dismiss this as
being obvious.

I am also aware that supporters of the inherent source match position
assert that you must be selective in choosing tests for source impedance.
It is all rather unconvincing when only some of the implications of a
particular source impedance are effective.

Where is the rig specified to exhibit this condition?

Is your rig a VW that stalls trying to pull a trailer from a stop in
3rd gear? Or is it a Mack truck trying to park in the handicap zone
in an underground mall parking lot? Arguing other's incapabilities is
something I like doing, but with more flair. Let's skip these
Tritonic minnows.

It is my view that modelling the PA as a fixed voltage or current source
with fixed source impedance of Zo, and where reflected waves on a
transmission line are absorbed by the matched source is not a good
general model for HF PAs.

You have already said as much. I see nothing new so far.

The application of small signal analysis to amplifiers that sweep from
near cutoff to near saturation is suspect.

If it is near cutoff or saturation, it is suspect small signal
analysis. Certainly, anyone can conspire to fail gracelessly. Would
you care to elaborate the suspicion beyond the evidence of gross
negligence? Why don't we skip this minor excursion?

I believe that it is sound (in the steady state) to resolve the forward
and reflected wave voltages and currents at the source end of the
transmission line, calculate the complex impedance, and predict the
effects of that impedance as a PA load using the same techniques that
were used to design the PA.

Sound though it may be, if I were to line up another transmitter
boresight down the antenna connector of the first, light it up to
provide power with no equivocation of it being fictional; then yes,
all things may appear to be the same. ...and yet I have just
demonstrated reverse power arriving at the antenna terminal (where did
it go?). My having experience in doing just this (aka active load
already described above) fully conforms to your sound idea, and yet,
as for myself, it is not an idea I would rely on to deny the existence
of reverse power nor its capacity to fry the innards of a transmitter
(active loads are heavily heat-sinked and fan driven).

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

On Tue, 27 Feb 2007 21:37:15 GMT, Owen Duffy wrote:

The fact that an instrument may consume power from the circuit does not
imply that it measures power.

Hi Owen,

If that power is from a reflection, then we've come to the logical
conclusion of the argument.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

Richard Clark wrote in
:

On Tue, 27 Feb 2007 21:14:38 GMT, Owen Duffy wrote:

This is not contending nor contention and is content only for a non
sequitur. The line following a tuner exhibits considerable loss
(poor efficiency) that can only occur on the basis of power and
mismatch. You yourself offered in other correspondence that it
exceeds cable attenuation specifications found only in a matching
condition. To

I am being picky, but "it *may* exceed cable attenuation
specifications found only in a matching condition, it may also be
lower".

Hi Owen,

Lower? That is rather astonishing in light of responding to my
comment.

If I said it
as you stated, I made an error. The common statement (and I have no
doubt made it) that VSWR exacerbates line loss is actually wrong in
the general case. (Having Googled my own web site I see one statement
along those lines which needs further qualification!)

This is even more astonishing. Irrespective of you being the source,
why inject this confusing comment? SWR always exacerbates line loss!
Give me any normal line attenuation and SWR at the load, and I will
tell you exactly how much additional loss will occur. There's a
general solution for you.

Something tells me that your comments are based on a confusion between
the power loss of a cable, and its mismatch loss. They are not the
same thing although they are usually tightly twined in discussion. You
later exhibit a confusion between a conjugate match and an impedance
match. They are not the same thing either. The confusion on both
these points have abounded in this group in past "debates."

Just to deal with this issue, line loss under VSWR, which at first seems
a side issue, but it illustrates one of the problems of a "power
perspective" in analysing a transmission line.

By "line loss" I mean the ratio of power at the load end of the line to
power at the source end of the line, not "forward power" or "reflected
power", but the average rate of flow of energy at those points.

So, to your challenge:

The problem is 1m of Belden 8262 (RG58C/U type) at 3.5MHz with three
loads, 50+j0, 5+j0, and 500+j0.

The loss for 50+j0 load is 0.025dB (equivalent to the Matched Line Loss).

What Line Loss to you get for the other two cases?

(I make it 0.24dB for 5+j0, and 0.014 for 500+j0.)

Owen

tuner - feedline - antenna question ?
 

On Wed, 28 Feb 2007 01:01:55 GMT, Owen Duffy wrote:

By "line loss" I mean the ratio of power at the load end of the line to
power at the source end of the line, not "forward power" or "reflected
power", but the average rate of flow of energy at those points.

Hi Owen,

What's wrong with conventional terms so that we BOTH know what you
mean? The convention would call this Mismatch Loss. If you dispute
this, then it serves my complaint. Further, convention has no
interest in "forward power" nor "reflected power" except as expressed
as SWR. I thought I was quite terse in this regard.

So, to your challenge:
The problem is 1m of Belden 8262 (RG58C/U type) at 3.5MHz with three
loads, 50+j0, 5+j0, and 500+j0.

Well, as I've pointed out, it is not strictly in the terms of my
challenge, is it?

The loss for 50+j0 load is 0.025dB (equivalent to the Matched Line Loss).

Sigh... parentheticals?

What Line Loss to you get for the other two cases?

(I make it 0.24dB for 5+j0, and 0.014 for 500+j0.)

An additional 0.1dB However, this example strains the utility of the
challenge.

Let's try a perverse challenge.

Presume a source of 100+j0 Ohms impedance sees a 50 Ohm line that is
5.35 wavelengths long and is terminated with a load of 200+j0 Ohms.
The normal attenuation of the line is 2.00 dB. What is the loss in
the line?

Can your general solution solve this? It uniquely describes both the
kinetics of reverse power flow AND the impact of source resistance.

No one has every answered this one correctly, by the way (and I can
anticipate you are ready to spring that observation on me with your
source feeding essentialy a voltage oriented high Z load as opposed to
the current oriented low Z load).

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

On Tue, 27 Feb 2007 17:51:37 -0800, Richard Clark
wrote:

Presume a source of 100+j0 Ohms impedance sees a 50 Ohm line that is
5.35 wavelengths long and is terminated with a load of 200+j0 Ohms.
The normal attenuation of the line is 2.00 dB. What is the loss in
the line?

For the others,

Any who complain about their transmitter having:
1. No source resistance;
2. Not this much resistance:
3. Not this little resistance;
4. None of the above (the usual response).
can take heart that if you simply substitute a tuner which presents
the equivalent SWR at the plane of the input to the line, then you can
progress to the solution with equal fluidity (which is to say like
molasses in December).

This particular example has been around for at least 40 years if not
since WWII. No one has rushed to answer it here in at least a quarter
of that time, I don't expect a cascade of guesses soon either.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

Richard Clark wrote in
:

On Wed, 28 Feb 2007 01:01:55 GMT, Owen Duffy wrote:

By "line loss" I mean the ratio of power at the load end of the line to
power at the source end of the line, not "forward power" or "reflected
power", but the average rate of flow of energy at those points.

Hi Owen,

What's wrong with conventional terms so that we BOTH know what you
mean? The convention would call this Mismatch Loss. If you dispute
this, then it serves my complaint. Further, convention has no
interest in "forward power" nor "reflected power" except as expressed
as SWR. I thought I was quite terse in this regard.

So, to your challenge:
The problem is 1m of Belden 8262 (RG58C/U type) at 3.5MHz with three
loads, 50+j0, 5+j0, and 500+j0.

Well, as I've pointed out, it is not strictly in the terms of my
challenge, is it?

Richard,

Your challenge was "Give me any normal line attenuation and SWR at the
load, and I will tell you exactly how much additional loss will occur."

I didn't state the VSWR, but it is 10:1 in both cases. The "normal line
attenation" you refer to is I expect the Matched Line Loss" which I have
given you.


The loss for 50+j0 load is 0.025dB (equivalent to the Matched Line
Loss).

Sigh... parentheticals?

What Line Loss to you get for the other two cases?

(I make it 0.24dB for 5+j0, and 0.014 for 500+j0.)

An additional 0.1dB However, this example strains the utility of the
challenge.

Is that your answer, an additional 0.1db due to the 10:1 VSWR? We do not
agree on either answer.

BTW, my figures were not additional loss, but total Line Loss as I
defined it.

You will note that my calculation for the 5+j0 case is less than the
Matched Line Loss, not higher.

In practical transmission lines, most of the loss is in current flowing
in the R component of an RLGC equivalent of the line, the loss in the
copper conductors forming the line. For VSWR1, the net current varies
along the line forming the classic standing wave pattern, and the loss in
incremental lengths of the line varies approximately with the square of
current in that increment.

So in the two cases above, even though the load VSWR is the same, the
loss is quite different due to the different current distribution in both
cases, one is near a current maximum, and the other is near a current
minimum. Any adjustment of Matched Line Loss for VSWR1 using only the
VSWR cannot take the location of the standing wave pattern into account,
and is an inaccurate approximation in some situations.

Many books showing a VSWR based formula for "additional loss due to
VSWR" don't spell out the assumptions underlying the formula. Phillip
Smith does in his book "Electronic Applications of the Smith Chart", he
says "If a waveguide is one or more wavelengths long, the average loss
due to standing waves in a region extending plus or minus a half
wavelength from the point of observation may be expressed as a
coefficient or factor of the one way transmission loss per unit length."
and he gives the ratio as (1+S^2)/(2*S). Though the ARRL shows graphs and
formulas they don't always (if ever) spell out the assmptions.

So, yes I assert that the Line Loss under mismatch conditions may be less
than the Matched Line Loss.

Owen

PS: I calculated my answers using http://www.vk1od.net/tl/tllc.php ,
Dan's TLDETAILS.EXE and ARRL's TLW3.EXE give similar results.

tuner - feedline - antenna question ?
 

Richard Clark wrote in
:

On Tue, 27 Feb 2007 21:37:15 GMT, Owen Duffy wrote:

The fact that an instrument may consume power from the circuit does not
imply that it measures power.

Hi Owen,

If that power is from a reflection, then we've come to the logical
conclusion of the argument.

Richard,

If you read and understood the article, you would see that the instrument
is based on sampling the V/I ratio at a point, and that being surrounded by
transmission line is not important to the principle of operation, in other
words, it does not directly measure a reflected wave.

Owen

tuner - feedline - antenna question ?
 

On Wed, 28 Feb 2007 03:10:28 GMT, Owen Duffy wrote:

You will note that my calculation for the 5+j0 case is less than the
Matched Line Loss, not higher.

Hi Owen,

I already anticipated that, didn't I? Certainly parsing my last
discussion is hardly necessary.

I note you have no solution that answers for the loss in a fairly
typical instance for a fairly typical line condition. It couldn't
have been any more difficult than your former computations, could it?
(In fact it is, but not conceptually.) And yet the absence of that
effort is notable (OK, so you've been ambushed). I may have stumbled
on a novelty application but I didn't trip over a boulder of a common
usage.

Given this sub-thread flowed from my response that a source does
dissipate a reverse power flow (both of which, the direction and
dissipation, are held in contention); and further given my "perverse"
challenge fully specifies such a condition and has a real solution, it
stands to reason that if your general computation is in fact general,
then it can resolve the contention to one or both of our satisfaction.
All it requires is that your math treatment accepts both directions of
power flow, and loss in the source. This is not unreasonable,
especially when any number of references encompass just such concepts.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

On Wed, 28 Feb 2007 05:05:47 GMT, Owen Duffy wrote:

If you read and understood the article, you would see that the instrument
is based on sampling the V/I ratio at a point, and that being surrounded by
transmission line is not important to the principle of operation, in other
words, it does not directly measure a reflected wave.

Hi Owen,

Can you express it without the presumption and still carry the
argument? I have read and understood many treatments on the topic,
and none claim to be the sole and unadulterated truth as it is
generally understood that many analyses work simultaneously and none
deny the validity of the others.

You have not yet actually offered any treatment that denies the bone
of contention that lies in two subject lines:
1. Reverse power is manifest;
2. The source will absorb and dissipate it.

You may have struggled with others over this in times past, but by
your own descriptions they had little intellectual horsepower, and
less experience in the matter.

I have attended schooling specific to these issues, and have practiced
professionally in their measure to the highest of standards. My peers
have instituted national metrology laboratories in your half of the
planet (OK, so it was Korea).

I have measured SWR with Bruene designs (as vulgar as that is);
Directional Couplers, Slotted Lines; and power with half a dozen
different style of sensors, and as many different methodologies. I
have also calibrated these instruments (all of them including the
vulgar Bruene designs). I can separate out the constituent waves (in
spite of the denial of their existence) by several means - each
appropriate to the problem at hand. I can measure excessively high
SWR precisely where others would shrug and simply call it infinite (it
isn't). I can also reduce residual SWR (anyone know what that is?).
I've done this over a spectrum from nearly D.C. to 12 GHz. And I get
a chuckle out of a claim for 0.014dB loss when I know full well
through experience it is unverifiable, unmeasurable, and hence
unproveable except in a spread sheet as a statistical curiosity.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

Richard Clark wrote in
:

....
I note you have no solution that answers for the loss in a fairly
typical instance for a fairly typical line condition. It couldn't
have been any more difficult than your former computations, could it?
(In fact it is, but not conceptually.) And yet the absence of that
effort is notable (OK, so you've been ambushed). I may have stumbled
on a novelty application but I didn't trip over a boulder of a common
usage.
....

Richard,

No, it is just that I have not posted a solution as yet.

Your problem was described as "Presume a source of 100+j0 Ohms impedance
sees a 50 Ohm line that is 5.35 wavelengths long and is terminated with a
load of 200+j0 Ohms. The normal attenuation of the line is 2.00 dB. What
is the loss in the line?"

Rather than sit down an write a bunch of stuff to calculate it, I took
the lazy way out and played with a scenario using Belden 8262 (RG58C/U)
in my line loss calculator that was pretty close to your scenario (Zo is
a touch different at 50-j0.24, most lines with loss will have a non-zero
jX component in Zo). My result for 15.8m of 8262 at 67MHz with a 200+j0
load is a transmission line loss (power into the load divided by power
into the line at the source end) is 3.3dB. This value of line loss is
independent of the source equivalent impedance.

If I reproduced the algorithms with the exact propagation constant and Zo
for your problem scenario, the answer would be more accurate, but I think
similar, and still obtained independently of the source impedance, and
not worth the time.

The figures from the calculator are below if someone wants to play with
it.

Now, are you prepared to post your solution?

Owen

Parameters
Transmission Line Belden 8262 (RG-58C/U)
Code B8262
Data source Belden
Frequency 67.000 MHz
Length 15.800 metres
Zload 200.00+j0.00 ?
Yload 0.005000+j0.000000 ?
Results
Zo 50.00-j0.24 ?
Velocity Factor 0.660
Length 1924.75 °, 5.347 ?
Line Loss (matched) 1.997 dB
Line Loss 3.279 dB
Efficiency 47.00%
Zin 30.48+j24.99 ?
Yin 0.019622-j0.016087 ?
VSWR(source end) 2.22
VSWR(load end) 4.00
? 1.46e-2+j2.13e+0
k1, k2 1.30e-5, 2.95e-10
Correlation coefficient (r) 0.999884

tuner - feedline - antenna question ?
 

Presume a source of 100+j0 Ohms impedance sees a 50 Ohm line that is
5.35 wavelengths long and is terminated with a load of 200+j0 Ohms.
The normal attenuation of the line is 2.00 dB. What is the loss in
the line?

Can your general solution solve this? It uniquely describes both the
kinetics of reverse power flow AND the impact of source resistance.

No one has every answered this one correctly, by the way (and I can
anticipate you are ready to spring that observation on me with your
source feeding essentialy a voltage oriented high Z load as opposed to
the current oriented low Z load).



VSWR 4.75
Loss 3.55dB

73
Jeff

tuner - feedline - antenna question ?
 

Richard Clark wrote in
:

....

You have not yet actually offered any treatment that denies the bone
of contention that lies in two subject lines:
1. Reverse power is manifest;
2. The source will absorb and dissipate it.

Richard, if you go back over my postings in this thread, I have not denied
either of these things.

I did comment on 2 as an explanation, one which I think is poor because of
the conclusions that might be drawn from it, eg any mismatch creates
reflected power which must be dissipated in the PA.

I did suggest that in the steady state, in a tx-line-load scenario, the
impedance looking into the line can be found, and that equivalent load
adequately explains the PA's behaviour.


You may have struggled with others over this in times past, but by
your own descriptions they had little intellectual horsepower, and
less experience in the matter.

I never said such a thing, if it is your conclusion, I disagree with it.

....

Owen

tuner - feedline - antenna question ?
 

"Jeff" wrote in news:45e53c39$0$26694
:


Presume a source of 100+j0 Ohms impedance sees a 50 Ohm line that is
5.35 wavelengths long and is terminated with a load of 200+j0 Ohms.
The normal attenuation of the line is 2.00 dB. What is the loss in
the line?

Can your general solution solve this? It uniquely describes both the
kinetics of reverse power flow AND the impact of source resistance.

No one has every answered this one correctly, by the way (and I can
anticipate you are ready to spring that observation on me with your
source feeding essentialy a voltage oriented high Z load as opposed to
the current oriented low Z load).



VSWR 4.75

A 200 ohm load on a 50 ohm line is 4:1 at the load end in my view, and it
is lower as you move toward the source.

Owen

Loss 3.55dB

73
Jeff

tuner - feedline - antenna question ?
 

VSWR 4.75

A 200 ohm load on a 50 ohm line is 4:1 at the load end in my view, and it
is lower as you move toward the source.


Indeed; 4:1 decreasing to 3.2:1 at the source, BUT only when the source is
50ohms.

Change the source impedance to 100ohms and the picture changes to 5.5:1 at
the load and 4.75:1 at the source end.

By the way the vswr figures will change cyclically with frequency assuming a
fixed length of coax ( 5.35 wavelengths at a fixed frequency), between about
7.5 and 2.5 at the load and 7 and 1.5 at the source.

73
Jeff

tuner - feedline - antenna question ?
 

Richard Clark wrote:
On Wed, 28 Feb 2007 05:05:47 GMT, Owen Duffy wrote:


If you read and understood the article, you would see that the instrument
is based on sampling the V/I ratio at a point, and that being surrounded by
transmission line is not important to the principle of operation, in other
words, it does not directly measure a reflected wave.


Hi Owen,

Can you express it without the presumption and still carry the
argument? I have read and understood many treatments on the topic,
and none claim to be the sole and unadulterated truth as it is
generally understood that many analyses work simultaneously and none
deny the validity of the others.

You have not yet actually offered any treatment that denies the bone
of contention that lies in two subject lines:
1. Reverse power is manifest;
2. The source will absorb and dissipate it.

You may have struggled with others over this in times past, but by
your own descriptions they had little intellectual horsepower, and
less experience in the matter.

I have attended schooling specific to these issues, and have practiced
professionally in their measure to the highest of standards. My peers
have instituted national metrology laboratories in your half of the
planet (OK, so it was Korea).

I have measured SWR with Bruene designs (as vulgar as that is);
Directional Couplers, Slotted Lines; and power with half a dozen
different style of sensors, and as many different methodologies. I
have also calibrated these instruments (all of them including the
vulgar Bruene designs). I can separate out the constituent waves (in
spite of the denial of their existence) by several means - each
appropriate to the problem at hand. I can measure excessively high
SWR precisely where others would shrug and simply call it infinite (it
isn't). I can also reduce residual SWR (anyone know what that is?).
I've done this over a spectrum from nearly D.C. to 12 GHz. And I get
a chuckle out of a claim for 0.014dB loss when I know full well
through experience it is unverifiable, unmeasurable, and hence
unproveable except in a spread sheet as a statistical curiosity.

73's
Richard Clark, KB7QHC

Hi Richard,

Not that I dispute anything here necessarily, but I would like to know
how you went about measuring the reflected power dissipated within a
source. Also, how the power being dissipated?

Thanks and regards,

Jim, AC6XG

tuner - feedline - antenna question ?
 

On Mon, 26 Feb 2007 20:38:00 GMT, Owen Duffy wrote:

The danger in the "power is refelected at a mismatch" explanation, is
that it follows that power reflected at a mismatched antenna flows back
toward the transmitter and is at least partially absorbed in the PA as
heat. Though that is a popular belief, it is not supported by fact.

This is the complete quote to which I responded in a recent
side-thread. You have lately expressed:

On Wed, 28 Feb 2007 08:55:24 GMT, Owen Duffy wrote:

You have not yet actually offered any treatment that denies the bone
of contention that lies in two subject lines:
1. Reverse power is manifest;
2. The source will absorb and dissipate it.

Richard, if you go back over my postings in this thread, I have not denied
either of these things.

What have I missed about "it is not supported by fact?" What do you
mean by "is a popular belief?" I am swayed by facts and I don't
really like general statements that are couched in belief systems.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

On Wed, 28 Feb 2007 08:55:24 GMT, Owen Duffy wrote:

Richard Clark wrote in
:

...

You have not yet actually offered any treatment that denies the bone
of contention that lies in two subject lines:
1. Reverse power is manifest;
2. The source will absorb and dissipate it.

Richard, if you go back over my postings in this thread, I have not denied
either of these things.

Hi Owen,

It is surprising the conclusions I've drawn from our correspondence
then. As I've steadfastly expressed nearly every posting in these
terms, you have not exactly responded to my misunderstanding in an
uniform manner.

I shall return to those postings to enquire further rather than
laboring the point here.

I did comment on 2 as an explanation, one which I think is poor because of
the conclusions that might be drawn from it, eg any mismatch creates
reflected power which must be dissipated in the PA.

This is not a denial? I see no positive characteristic you have
derived from 2 as allowing it is acceptable.

I did suggest that in the steady state, in a tx-line-load scenario, the
impedance looking into the line can be found, and that equivalent load
adequately explains the PA's behaviour.

Yes, this allowing reflected power in your terms, allowing you to
express it as a fiction suitable to providing a truth in creating the
lumped equivalent. This may have the heavy hand of my
editorialization, but it is forced by the equivocation I find in your
points I am responding to here.

You may have struggled with others over this in times past, but by
your own descriptions they had little intellectual horsepower, and
less experience in the matter.

I never said such a thing, if it is your conclusion, I disagree with it.

As I have never raised the discussion of "others" or how "they"
developed poor explanations or subscribed to faulty premises; then my
perhaps over-arching characterization is what you are rejecting as
your having said. You may note that at that time I explicitly offered
that their contributions were not germane to the facts.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

On Tue, 27 Feb 2007 21:14:38 GMT, Owen Duffy wrote:

I am also aware that supporters of the inherent source match position
assert that you must be selective in choosing tests for source impedance.
It is all rather unconvincing when only some of the implications of a
particular source impedance are effective.

It is my view that modelling the PA as a fixed voltage or current source
with fixed source impedance of Zo, and where reflected waves on a
transmission line are absorbed by the matched source is not a good
general model for HF PAs.

Hi Owen,

This quote gives me no confidence in what you have offered to me
recently:

On Wed, 28 Feb 2007 08:55:24 GMT, Owen Duffy wrote:

You have not yet actually offered any treatment that denies the bone
of contention that lies in two subject lines:
1. Reverse power is manifest;
2. The source will absorb and dissipate it.

Richard, if you go back over my postings in this thread, I have not denied
either of these things.

As to point 1 (or 2 it is difficult to determine what you are
responding to specifically), explicitly stated by me, you have
expressed your self in relation to "supporters of the inherent source
match position" without actually identifying if you stand
1. With them;
2. Against them;
3. Indifferent to them.

As to point 2, explicitly stated by me, you have again described
yourself in a negative relation by discussing a model that does not
work.

Perhaps it is this style of ambivalence that clouded my appreciation
of your statement:
I believe that it is sound (in the steady state) to resolve the forward
and reflected wave voltages and currents at the source end of the
transmission line, calculate the complex impedance, and predict the
effects of that impedance as a PA load using the same techniques that
were used to design the PA.
where you do allow 1 and 2.

However, I could be mistaken again because you don't actually
acknowledge return power impinges upon the final stage, you transform
it into another solution. Note that I accept such a transformation of
the problem. It is common alternative explanation and perfectly
valid. However, that transformation, in and of itself, does not speak
to the issue of reflected power as a physical fact and a separable
entity. In fact, the development of a lumped equivalent doesn't need
to acknowledge SWR either.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

On Wed, 28 Feb 2007 08:11:30 GMT, Owen Duffy wrote:

Now, are you prepared to post your solution?

Hi Owen,


Your quick computation of 3.3 dB is suitably close to my reference's
first pass solution (3.27 dB), but it neglects the contribution of the
source's resistance.

The solution is 4.9 dB.

If we were to revisit your 1 meter long cable used in the 80M band and
force the transmitter to be a voltage source through the common
mechanism of adding a substantial resistor, and mismatch the other end
of that 1 meter long cable to the same degree (each end seeing 10K Ohm
for the purpose of this statistical curiosity); then that same cable
will heat up with its contribution of at least 3dB of ADDITIONAL loss.

The 10K Ohm specification is a forced one, but it responds in kind to
the original forced solution too. In fact, it comes close to the
source resistance found in a tube amplifier (a common voltage source)
driving a halfwave element (a common application for such a source)
and demonstrates the common futility of using coax (that I have
already expressed) to accomplish this.

However, we don't have voltage sources to conveniently solve either of
these statistical curiosities. Both the tube transmitter, and the
solid state transmitter employ impedance matching to either draw down,
or pull up the native source resistance to a level suitable for
applying to a transmission line.

I would again point out that reverse power suitably accounts for the
1.6 dB difference between your answer and the solution, it also
accounts for the 3 dB difference between your short cable's example,
and my twist in its application.

All such differences have been described and used in design for quite
a few decades, and they have been couched in exactly the terms I've
used here.

If anyone wants to challenge the 4.9 dB solution, they can impeach my
reference "Reference Data for Radio Engineers," (various editions). I
can supply other references that have been named in this group too,
but I would suggest with tackling one authority at a time.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

On Wed, 28 Feb 2007 07:15:38 -0800, Jim Kelley
wrote:

Not that I dispute anything here necessarily, but I would like to know
how you went about measuring the reflected power dissipated within a
source. Also, how the power being dissipated?

Hi Jim,

Dissipation is caloric, however it can arrive catastrophically by one
of two mechanisms; and they reflect, no pun here, the two types of
phase sense offered by the random opportunity (being phase adding or
subtracting for current or voltage as the occasion demands).

One caloric method is simple in measuring the heat load expressed by
airflow temperature measurements in a confined volume. When I
designed the Flight Recorder, the FAA mandated a heat budget for its
acceptance. This is certainly far afield from the immediate topic,
but it responds to the attention offered in design to this issue. The
point of this sidebar is that efficiency translated immediately into
temperature and this was rigorously anticipated and tested. The same
design philosophy is mandated in RF final design and considerable
attention has been devoted to it in the trade papers.

Returning to our concerns, for certain phase combinations that caloric
solution can arrive suddenly in the form of an arc. Most operators
will immediately act to correct that situation and the heat build up
may not be great, but the damage may still be irreversible. This
harkens back to my discussion of a kitchen table laser cracking a
window pane. Average power may be unspectacular, but instantaneous
power, localized, can be very dramatic and destructive beyond
expectation (it certainly surprised my friend).

For other phase combinations that caloric solution can arrive
gradually (heat soaking); and catastrophe arrives through thermal
runaway. Operators rarely observe this until it is too late.

I hope that the readers can differentiate between these two, and how
certain designs (eg. solid state, and tube design) respond in these
cases and correlate to experience each to their own characteristic
failure mechanism.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

Expanding generously (gusting on):

When I
designed the Flight Recorder, the FAA mandated a heat budget for its
acceptance.

Aircraft electronics lives with a common airduct. Your design must
not load the cooling air such that it becomes a flame thrower into the
next instrument in the stack. I won't go into issues of crash
survivability.

Returning to our concerns, for certain phase combinations that caloric
solution can arrive suddenly in the form of an arc.

I'm sure most readers who run tube rigs will recognize this situation
immediately. However, there is more than one combination of phases
and currents/voltages. I have also seen heat soaking arrive at a tube
to watch the plates glow cheerily. This, too, is probably an
experience borne by several tube rig operators. In fact, it can be
tolerated far more than a solid state amplifier, and tubes are noted
for their resilience. However, I have also seen the glass envelopes
turned into a taffy consistincy and the vacuum draw them like
heatshrink around the internal structure. Surprisingly, I have also
witnessed that these tubes still worked!

For other phase combinations that caloric solution can arrive
gradually (heat soaking); and catastrophe arrives through thermal
runaway. Operators rarely observe this until it is too late.

The latest generation of solid state components have survivability
design into them such that they are specified to operate into an
infinite mismatch (or some such similar claim). This is suitably
taken care of by being able to withstand more voltage. Other issues
of current crowding, the original thermal disaster for transistors,
has been long solved. That solution revealed how the problem was in
heat confined to a small volume.



Finally, my measurements were never pushed to the point of failure.
All may well anticipate that this sudden arrival would preclude any
accuracy in the heat determination to demonstrate a quid-pro-quo of
returned power. Further, once the failure occured, heat is usually
removed by the very failure it brought - it usually removes the source
too. ;-)

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

On Wed, 28 Feb 2007 09:41:56 -0800, Richard Clark
wrote:

I would again point out that reverse power suitably accounts for the
1.6 dB difference between your answer and the solution, it also
accounts for the 3 dB difference between your short cable's example,
and my twist in its application.

Lest there be any doubt about there being concurrent explanations,
this loss is also expressed in lumped equivalency and circulating
currents. It can be correlated to a very common issue with literal
lumped circuit antenna tuners. It can also be described in terms of Q
and cavities. It can also be correlated to short radiators, radiation
resistance, and Ohmic loss.

Each description is accurate, and as varied as the authors each
offering their interpretations, but no explanation denies the validity
of the other, and reflected power in a line is no exception.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

Richard Clark wrote:
On Wed, 28 Feb 2007 07:15:38 -0800, Jim Kelley
wrote:


Not that I dispute anything here necessarily, but I would like to know
how you went about measuring the reflected power dissipated within a
source. Also, how the power being dissipated?


Hi Jim,

Dissipation is caloric, however it can arrive catastrophically by one
of two mechanisms; and they reflect, no pun here, the two types of
phase sense offered by the random opportunity (being phase adding or
subtracting for current or voltage as the occasion demands).

One caloric method is simple in measuring the heat load expressed by
airflow temperature measurements in a confined volume. When I
designed the Flight Recorder, the FAA mandated a heat budget for its
acceptance. This is certainly far afield from the immediate topic,
but it responds to the attention offered in design to this issue. The
point of this sidebar is that efficiency translated immediately into
temperature and this was rigorously anticipated and tested. The same
design philosophy is mandated in RF final design and considerable
attention has been devoted to it in the trade papers.

What I meant was, in what way were you able to attribute and apportion
this heat to its various sources? What evidence were you able to
obtain to show reflected energy re-entering the source output? What
component in the system in fact dissipated the reflected energy? How
were you able to determine the exact source and amount of energy at
any given location within the source? Or did you just presume that
you understood the underlying mechanisms?

Thanks in advance,

Jim AC6XG

tuner - feedline - antenna question ?
 

Cecil Moore wrote:

On Feb 27, 2:53 am, "Jeff" wrote:

Adding a circulator to a system will not change "the load line" (if a
transmission line or circulator can have such a thing), but it will cause
the power in the reflected wave to be separated so that it can be monitored
and measured. Surprisingly power monitored in this way ties up with the
notion that power is reflected at a mis-matched load.


Yes, and a little modulation added to the source signal will prove
that the
signal being dissipated by the circulator resistor has made a round
trip
to the load and back. That's hard to explain if reflected energy
doesn't
actually exist.
--
73, Cecil, w5dxp.com

Your example is the same as putting a load resistor on an open
transmission line, measuring the dissipated power, and then claiming
the same thing happens without the load resistor there.

ac6xg

tuner - feedline - antenna question ?
 

On Wed, 28 Feb 2007 13:55:47 -0800, Jim Kelley
wrote:

What I meant was, in what way were you able to attribute and apportion
this heat to its various sources? What evidence were you able to
obtain to show reflected energy re-entering the source output? What
component in the system in fact dissipated the reflected energy? How
were you able to determine the exact source and amount of energy at
any given location within the source? Or did you just presume that
you understood the underlying mechanisms?

Hi Jim,

This knowledge arrived by many avenues.

For one, in a heavily heatsinked design, mapping of temperatures
generally reveal a very diffuse origin. That, of course, is the
purpose of the heatsink. So, in that regard the assignment of where
dissipation occurs is done by induction. You can eliminate a lot
circuitry as being incapable of supporting this dissipation, as it is
both remote from the signal path, and remote physically. The
literature of design reveals much of what is discovered in the field.

That literature reveals the dissipation occurs in the
emitter/collector junction of the finals' transistors. Failures have
been confirmed through post-mortem examination by microscope (no, I
have not done this).

Experience with new designs and frequency of failure (those activities
that I have participated in) lead to the same conclusion. In one
particular case it was a manufacturing/assembly problem of mounting
the transistor to the heatsink. A bur was found in many such mounts
that interfered with a complete mating of surfaces. This raised the
thermal resistance in the path from that same junction to the mating
surface, to the heatsink, to the environment. Knowing each thermal
resistance in that path makes it rather simple to forecast the
junction temperature at the time of failure (or rather, to say failure
which occurred was guaranteed a fatal temperature) when you know the
power consumed by the component. All such "resistance" conform to the
simple math of Ohm's law (once you substitute the necessary units for
heat).

When we return to the design guidelines and this junction, almost
every manufacturer of power transistors specifies a junction
resistance value at rated power. Casting this value through the chain
of transformations and to the antenna connector reveals a value very
nearly 50 Ohms. There are newer power amplification designs today,
and yet the market for Ham gear is dominated by the Class AB design
which is exhibits this property nicely.

Inductive logic leads us to this junction as the principle target of
reflected power (the signal path is symmetric, after all). Experience
has supported this logic. Failures are attributable to design flaw
(or assembly flaw), or poor application (driving a mismatch), or both.

As for tubes, I've already testified to the obvious location for
dissipation. It is far easier to see.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

Jim Kelley wrote:

Cecil Moore wrote:
The joules/sec are real quantities but whether joules/sec
is power depends upon the definition of "power".

In our case here on the internet, it depends on whether or not you
choose to equate 'units of power' with the definition of power.

Most engineers equate the units of power to power, i.e.
joules/sec = watts and so does the IEEE dictionary. But I
am content to assert that the joules in the joules per
second of a reflected wave is real energy. Do you disagree?
--
73, Cecil http://www.w5dxp.com

tuner - feedline - antenna question ?
 

Richard Clark wrote:
Any who complain about their transmitter having:
1. No source resistance;
2. Not this much resistance:
3. Not this little resistance;
4. None of the above (the usual response).
can take heart that if you simply substitute a tuner ...

Yep, the great majority of amateur radio antenna systems
are matched by a tuner. That act of matching prohibits
reflected load energy from reaching the PA. Except for
overall efficiency, when an antenna system is matched,
the PA impedance doesn't matter. A 5 ohm PA, a 50 ohm PA,
and a 500 ohm PA all output the same power if the output
voltage is the same into the same load.
--
73, Cecil http://www.w5dxp.com