Where does it go? (mismatched power)
 

On Sun, 13 Jun 2010 03:38:48 -0700 (PDT), K1TTT
wrote:

infinitely long of course. i think it was you who added the infinite
part.

Hi David,

It's a staple of the catechism.

i don't care as long as it is not losing cecil's precious
energy.

Now talk about entropy.

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

On Jun 13, 9:35*am, Cecil Moore wrote:
On Jun 12, 8:52*pm, Owen Duffy wrote:

If there was a valid Thevenin equivalent circuit for a transmitter (and
that is questionable), then you can not use that equivalent circuit to
make any inference about the internal dissipation of the source (the
transmitter in this case), or its efficiency. Nevertheless, I see people
trying to do this one way or another in the various threads here.

In his food-for-thought article on forward and reflected power, Roy
(w7el) says: "So we can model a 100 watt forward, 50 ohm nominal
transmitter as a 141.4 volt (100 * sqrt(2)) RMS voltage source in
series with a 50 ohm resistance." He goes on to calculate power
dissipation in the source resistor.
--
73, Cecil, w5dxp.com

And a mere two sentences back one find Roy writing:
"I make no claim that the model circuit represents what’s going on
inside the transmitter. For one thing, a real transmitter will
typically be more efficient than the model. However, if measurements
and observations are limited to the outside of the transmitter,
the model is very good (within the non-shutdown range). Although
a real transmitter won’t contain the model’s resistance as a
resistor, we will take a look at the model resistor’s dissipation
to see how it interacts with the “reverse” power."

Was your disingenuity deliberate?

....Keith

Where does it go? (mismatched power)
 

On Jun 13, 2:42*pm, Cecil Moore wrote:
On Jun 13, 6:16*am, K1TTT wrote:

just for fun... explain why i can see that discontinuity between z01
and z02 when i hook up my tdr.

That's easy, because it *physically exists in reality* with a voltage
reflection coefficient of (Z02-Z01)/(Z02+Z01) = 0.5. It is related to
the indexes of refraction in the field of optics from which the same
reflection coefficient can be calculated.

The difficult question is: Exactly why doesn't that same physical
reflection coefficient reflect half of the forward voltage when it is
Z0-matched during steady-state?

The answer is that it indeed does reflect 1/2 of the forward voltage
during steady-state but that wavefront interacts with 1/2 of the
reflected voltage returning from the mismatched load which is equal in
magnitude and 180 degrees out of phase. In this case, superposition of
the two waves results in wave cancellation (total destructive
interference). The energy components in those two waves are combined
and redistributed back toward the load as constructive interference in
phase with the forward wave from the source. That's where the
reflected energy goes.

That's why the s-parameter analysis theory could be important to hams.
By merely measuring the four reflection/transmission coefficients
(s11, s12, s21, s22) one learns the basics of superposition. S-
parameter analysis was covered in my 1950's college textbook, "Fields
and Waves in Modern Radio", by Ramo and Whinnery (c)1944. I don't know
how or why the younger generation missed it.
--
73, Cecil, w5dxp.com

because they use the newer 'fields and waves in communication
electronics' by ramo, whinnery, and van duzer that uses reflection
coefficients and calculates voltages, currents, electric and magnetic
fields.

Where does it go? (mismatched power)
 

On Jun 13, 3:02*pm, Cecil Moore wrote:
On Jun 13, 9:34*am, K1TTT wrote:

why are they (voltages) indeterminate? *i can calculate them, why can't you?

Purposefully, the numerical values of Z01 and Z02 are not given and
unknown. The answer to the problem would be the same if Z01=50 and
Z02=150, or if Z01=100 and Z02=300, or an infinite number of other
combinations. Please tell me how you can calculate an absolute voltage
when Z0 is an unknown variable?
--
73, Cecil, w5dxp.com

the same way you calculate the power to be zero watts. no need to
know the z0 if the voltage is zero.

Where does it go? (mismatched power)
 

On Jun 13, 4:10*pm, Richard Clark wrote:
On Sun, 13 Jun 2010 03:38:48 -0700 (PDT), K1TTT
wrote:

infinitely long of course. *i think it was you who added the infinite
part. *

Hi David,

It's a staple of the catechism.

i don't care as long as it is not losing cecil's precious
energy.

Now talk about entropy.

73's
Richard Clark, KB7QHC

ooooooh, entropy... i hated thermodynamics.

Where does it go? (mismatched power)
 

Richard Clark wrote in
:

On Sat, 12 Jun 2010 19:52:15 GMT, Owen Duffy wrote:
....
My recollection of Walt's tests were that they tested at points other
than Zl=0 and Zl=infinity.

Steps 1 and 2 are quite explicit.

I have reviewed Walt's additional material at
http://www.w2du.com/r3ch19a.pdf .

Walt's load at step 1 is stated as 50+j0 without tolerance, but I accept
that Walt is thorough and would have confidence that it is low in error.
He reports Pf=100W and Pr=0, again accepted as indicating that the load
and calibration impedance of the '43 are similar, and probably quite
close to the nominal impedance, close enough for the purpose at hand.

At Step 8, he substitutes a load with measured impedance and calculated
rho^2 as 0.235, VSWR(50) is approximately 3. Reported Pf=95W, Pr=25W, and
on that basis rho^2=0.263. The discrepency in rho is not discussed in the
paper.

A key observation between Step 1 and Step 8 is that changing from a VSWR
(50)=1 load to a VSWR(50)=3 load resulted in a 5% change in observed Pf.
This observation depends on the difference between two direct readings
from a single instrument (the Bird 43), and the accuracy of calibration
of both the Bird 43 and the 50 ohm load to their nominal 50+j0 impedance.

I have given a mathematical development at http://vk1od.net/blog/?p=1028
of why Pf should be independent of load impedance.

The observed change in Pf at 5% is rather small considering the load
change, and suggests that Zs in this case is probably quite close to 50
+j0 although the experiment was limited to just one load case.

I make the "just one load case" condition because in my experience, a
transmitter might exhibit insignificant change in Pf for some load
variations, but for others it is signficant. The case I documented
earlier is one such example.

Owen

Where does it go? (mismatched power)
 

Owen Duffy wrote in
:

....
At Step 8, he substitutes a load with measured impedance and
calculated rho^2 as 0.235, VSWR(50) is approximately 3. Reported
Pf=95W, Pr=25W, and on that basis rho^2=0.263. The discrepency in rho
is not discussed in the paper.

I have made an error here, Walt reports Pf=95W, Pr=20W, and on that basis
rho^2=0.211. The discrepency in rho is not discussed in the paper.

The error has no impact on the rest of my post.

My apologies.

Owen

Where does it go? (mismatched power)
 

On 13 jun, 13:19, Keith Dysart wrote:
On Jun 13, 9:35*am, Cecil Moore wrote:

On Jun 12, 8:52*pm, Owen Duffy wrote:

If there was a valid Thevenin equivalent circuit for a transmitter (and
that is questionable), then you can not use that equivalent circuit to
make any inference about the internal dissipation of the source (the
transmitter in this case), or its efficiency. Nevertheless, I see people
trying to do this one way or another in the various threads here.

In his food-for-thought article on forward and reflected power, Roy
(w7el) says: "So we can model a 100 watt forward, 50 ohm nominal
transmitter as a 141.4 volt (100 * sqrt(2)) RMS voltage source in
series with a 50 ohm resistance." He goes on to calculate power
dissipation in the source resistor.
--
73, Cecil, w5dxp.com

And a mere two sentences back one find Roy writing:
"I make no claim that the model circuit represents what’s going on
inside the transmitter. For one thing, a real transmitter will
typically be more efficient than the model. However, if measurements
and observations are limited to the outside of the transmitter,
the model is very good (within the non-shutdown range). Although
a real transmitter won’t contain the model’s resistance as a
resistor, we will take a look at the model resistor’s dissipation
to see how it interacts with the “reverse” power."

Was your disingenuity deliberate?

...Keith

Hello

Last night I was a little tired when I send my answers to Owen
stinging my eyes. It was too late and time to go to sleep :)
I should tell owen he was inadvertently moving towards a logical
bifurcation's fallacy because is not absolutely true that Thevenin
theorem can not be used to calculate -for example- circuit efficience.
There are an exception to this limitation = It can be used to
efficience calculations when disconnecting the load in the original
circuit the dissipated power is null.

Miguel LU6ETJ

Where does it go? (mismatched power)
 

lu6etj wrote in
:

....
I should tell owen he was inadvertently moving towards a logical
bifurcation's fallacy because is not absolutely true that Thevenin
theorem can not be used to calculate -for example- circuit efficience.
There are an exception to this limitation = It can be used to
efficience calculations when disconnecting the load in the original
circuit the dissipated power is null.

Miguel, when I said "cannot" I was of course meaning in the general case,
as most readers would understand.

Obviously there are cases where the efficiency calculations will be
correct, the obvious one being where the Thevnin equivalent is identical
to the original network.

If you want to make inferences about a source solely on the basis of its
Thevenin equivalent, you are on dangerous ground because you will not
always be correct.

People making the inference for example, that Zeq of a certain source
cannot be 50+j0 because the conversion efficiency of that source with a
50+j0 load is greater than 50% are wrong.

Owen

Where does it go? (mismatched power)
 

On Sun, 13 Jun 2010 21:48:58 GMT, Owen Duffy wrote:

Steps 1 and 2 are quite explicit.

I have reviewed Walt's additional material at
http://www.w2du.com/r3ch19a.pdf .

Walt's load at step 1 is stated as 50+j0 without tolerance, but I accept
that Walt is thorough and would have confidence that it is low in error.

Convention has it that any report is understood to vary by 50% of the
least significant digit. There are other standards that do not wander
far from this simple expectation.

Hence 50±0.5+j0±0.5

Walt has an excellent GR 1606-A Bridge - same model as mine.

To give everyone some idea of what "Qualified" means in my former
trade, my bridge came with a traceable certificate of measurement that
tells me that me my accuracy:

" Resistance:
±[3.0 + 0.0024 · f² · (1 + R/1000)]% ± [(X/f · 10000) + 0.1] Ohms
" Reactance:
±4.0% ±(1.0 + 0.0008 R · f) Ohms
"This certificate is valid only when the bridge is balanced at the
connector reference plane and when all measurements made at this plane
are corrected as outlined in para 4.4 and 4.5 in the manufacturer's
operating manual wit the exception that the capacitive reactance to
ground (Fig. 6) is computed from a value of 2.2pF. Fig. 8 should be
used for the resistance dial multiplying factor."

The correction factors easily leave behind the nominal value of 1.0
for high resistance (such as Walt's 1400 Ohms) with 5% error in the
20M band, and 60% error in the 6M band. Unless, of course, you apply
all corrections (the whole point of having the bridge calibrated).

He reports Pf=100W and Pr=0, again accepted as indicating that the load
and calibration impedance of the '43 are similar, and probably quite
close to the nominal impedance, close enough for the purpose at hand.

Understanding what is available at Walt's bench, Pf could be off by
5W or 5%. The 0 reading also suffers the same 5W indeterminacy if it
is measured on the same scale (assuming he is using a Bird wattmeter
with a 100W slug).

My bridge was not remarkably far off, roughly only 1% more error in
indicating the true value for resistance following corrections.

If I were to try to measure 1400 Ohms in the 20M band, I could expect
it might read 1333 Ohms by dial with an error of roughly ±50 Ohms.

So, between a Bird wattmeter, and it measuring power into a load with
corresponding percentages of error, the we have 5% for the Bird and 3%
for determining the load.

A metrologist can report these two (and there are more) sources of
error by one of two ways.
1. Absolute worst case = 8% error;
2. RSS = 3.9% error
Which one picked is a function of other factors.

It doesn't take long to accumulate error and driving out variables is
found only in the skill of the practice.

At Step 8

Is a step I have no current interest in.

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

On Sun, 13 Jun 2010 21:53:35 GMT, Owen Duffy wrote:

I have made an error here, Walt reports Pf=95W, Pr=20W, and on that basis
rho^2=0.211. The discrepency in rho is not discussed in the paper.

This needs to be put into the perspective of my last post too, as it
will impact your rho.
Pf = 90 to 100W,
Pr = 15 to 25W

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

Richard Clark wrote in
:

Understanding what is available at Walt's bench, Pf could be off by
5W or 5%. The 0 reading also suffers the same 5W indeterminacy if it
is measured on the same scale (assuming he is using a Bird wattmeter
with a 100W slug).


Richard,

You can wax on all you like, and try as hard as you like, but if the
source was exactly Zs=50+j0, then the Bird 43 reading for Pf should be
exactly the same.

RF measurement is as you note, fraught with problems, but here is a
relatively simple case. Walt reported that he adjusted the radio on a 50
ohm load, measured Pf and Pr, then substituted another load, no other
changes and reported a 5% reduction in Pf.

I will leave you to arguing wether Walt's observed 5% reduction is of any
value.

Anyone who accepts that a competent measurement technician, competently
using the Bird 43, can reasonably discern a 5% reduction in Pf when it
should be 0% if Zs=50+j0, might question whether that observation
supports the proposition that Zs=50+j0.

Owen

Where does it go? (mismatched power)
 

On Jun 13, 10:04*pm, Owen Duffy wrote:
Richard Clark wrote :

Understanding what is available at *Walt's bench, Pf could be off by
5W or 5%. *The 0 reading also suffers the same 5W indeterminacy if it
is measured on the same scale (assuming he is using a Bird wattmeter
with a 100W slug).

Richard,

You can wax on all you like, and try as hard as you like, but if the
source was exactly Zs=50+j0, then the Bird 43 reading for Pf should be
exactly the same.

RF measurement is as you note, fraught with problems, but here is a
relatively simple case. Walt reported that he adjusted the radio on a 50
ohm load, measured Pf and Pr, then substituted another load, no other
changes and reported a 5% reduction in Pf.

I will leave you to arguing wether Walt's observed 5% reduction is of any
value.

Anyone who accepts that a competent measurement technician, competently
using the Bird 43, can reasonably discern a 5% reduction in Pf when it
should be 0% if Zs=50+j0, might question whether that observation
supports the proposition that Zs=50+j0.

Owen

Owen, I don't understand where you found a 5% drop in power when you
say it should have been 0%. A 5% drop from what value to what value?
Where did you find that data? If you're referring to 100w Pf with the
50 + j0 load and then the 95w Pf the mismatched load, you've got to
understand that the source was now delivering only 75w, and the 20w of
reflected power added to the 75w = 95w. These are two separate
measurements--the first a stand-alone value and the second the sum of
two values. Is this a point you overlooked?

Walt

Where does it go? (mismatched power)
 

walt wrote in
:

Owen, I don't understand where you found a 5% drop in power when you
say it should have been 0%. A 5% drop from what value to what value?

Hello Walt,

I wrote an article with the mathematical development for the fact that Pf
(as indicated by a properly calibrated directional wattmeter such as the
Bird 43) is independent of load impedance if and only if the Thevenin
source impedance is 50+j0 ohms. The article is at http://vk1od.net/blog/?
p=1028 .

This provides the basis for a simple go/nogo test for the hypothesis that
a particular transmitter, under particular conditions, exibits Zs=50+j0.

Where did you find that data? If you're referring to 100w Pf with the
50 + j0 load and then the 95w Pf the mismatched load, you've got to
understand that the source was now delivering only 75w, and the 20w of
reflected power added to the 75w = 95w. These are two separate
measurements--the first a stand-alone value and the second the sum of
two values. Is this a point you overlooked?

The source is Ch 19a where you report Pf=100, Pr=0 in Step 1; and Pf=95
and Pr=20 in Step 8. I am only interested in the relative values of Pf in
Step 1 (VSWR(50)=1) and Step 8 (VSWR(50)=3).

I conceded in another post that within reasonable expectations of error,
and considering the load case (VSWR(50)=3), that Zs under those
conditions and subject to only one load test is probably very close to 50
+j0, but probably not exactly.

In one test that I conducted with a range of loads, the drop in Pf was up
to 21% with VSWR(50)=1.5, but it varied with different VSWR(50)=1.5 load
angles. Whilst some values supported the proposition that Zs=50+j0,
others indicated that it was significantly different. Zs did not appear
to be constant, it appeared to be load dependent to some extent.

Going back to your case, if the source was 50+j0, we should expect Pf=
100, and if we trust your readings from the Bird 43 as characterising the
load to have rho^2=21%, we would expect the power delivered to the load
to be 78.9W rather than 75W, and error of 0.2dB. Not a large error, but
hard to account for entirely as instrument error.

In the measurements of an IC7000 that I made, the measured output power
on one VSWR(50)=1.5 load was 82.5W when it would have been 104.6W had the
source been 50+j0, an error of 0.8dB. I opined that this test did not
support the proposition that Zs was not 50+j0

Owen

Where does it go? (mismatched power)
 

Owen Duffy wrote in
:

....
In the measurements of an IC7000 that I made, the measured output
power on one VSWR(50)=1.5 load was 82.5W when it would have been
104.6W had the source been 50+j0, an error of 0.8dB. I opined that
this test did not support the proposition that Zs was not 50+j0

Too many "nots", isn't there?

It should read:

I opined that this test did not support the proposition that Zs was 50+j0.

Apologies, Owen.

Where does it go? (mismatched power)
 

On Mon, 14 Jun 2010 02:04:46 GMT, Owen Duffy wrote:

Anyone who accepts that a competent measurement technician, competently
using the Bird 43, can reasonably discern a 5% reduction in Pf when it
should be 0% if Zs=50+j0, might question whether that observation
supports the proposition that Zs=50+j0.

On Mon, 14 Jun 2010 04:18:02 GMT, Owen Duffy wrote:
if we trust your readings ...
Not a large error, but hard to account for entirely as instrument error.

Hi Owen,

Questions, emphatic competency, and reasonableness all in the space of
one sentence. I can see the well where you drew the literary "wax
on."

I will leave you to arguing wether Walt's observed 5% reduction is of any
value.

The follow-on discussion between you and Walt obviously reveals
sufficient value.

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

On 13 jun, 21:27, Owen Duffy wrote:
lu6etj wrote :

...

I should tell owen he was inadvertently moving towards a logical
bifurcation's fallacy because is not absolutely true that Thevenin
theorem can not be used to calculate -for example- circuit efficience.
There are an exception to this limitation = It can be used to
efficience calculations when disconnecting the load in the original
circuit the dissipated power is null.

Miguel, when I said "cannot" I was of course meaning in the general case,
as most readers would understand.

Obviously there are cases where the efficiency calculations will be
correct, the obvious one being where the Thevnin equivalent is identical
to the original network.

If you want to make inferences about a source solely on the basis of its
Thevenin equivalent, you are on dangerous ground because you will not
always be correct.

People making the inference for example, that Zeq of a certain source
cannot be 50+j0 because the conversion efficiency of that source with a
50+j0 load is greater than 50% are wrong.

Owen

Hello Owen, OK at your comment I understand, here my apologies =

In earlier post I write:
"The final Thevenin circuit is an idealization built with an ideal
voltage source in series with an ideal resistor. This new idealized
circuit It is a new born entity whose properties are now fully
described for these only two idealized circuit elements. These are
the
virtues and the defects of reductionist models :("

And in the first post I was thinking about Thevenin resulting scheme
ceases to represent fully the original circuit and place the problem
in a more elementary and different situation that a real world
transmitter, enabling a clear answer to much simpler scheme for the
behavior of standing waves in such elementary context. And I thought
it was pretty clear beacuse was clear enough in my mind... :)
When one write in his own language, normally think is communicating
very well the idea in his mind, until notice his partner take it in
another way! Difficult is multiplied when you are trying communicate
to a mind that also thinks in a different language...!
Later I realized that having formulated in terms of a simple generator
in series with a resistor would not have generated a false
interpretation and not difficult the flow of ideas.

I sincerely believe (correct me if I am wrong) this newsgroup does not
seems make (do?) efforts enough (sufficient?) to totally (completely?)
solve the more simpler cases before skip to more advanced and
sofisticated considerations on more complex systems.
For example: do you (*) recognize Roy Lewallen late example in "Food
for tought" assuming (or conceding that) as not representing a real
rig but a simple constant voltage source in series with a resistor, at
least? to give some credit to his ideas Until now, I could not
know...
(*) I do not know how clearly denote plural in "do you"

73 Miguel - LU6ETJ.
Here 04:46 LT - I am not responsible for anything written by me :)

Where does it go? (mismatched power)
 

lu6etj wrote in
:

....
For example: do you (*) recognize Roy Lewallen late example in "Food
for tought" assuming (or conceding that) as not representing a real
rig but a simple constant voltage source in series with a resistor, at
least? to give some credit to his ideas Until now, I could not
know...

In that article, Roy says "My commercial amateur HF transceiver is
probably typical of modern rigs in that it produces a constant forward
average power into varying load impedances—provided the impedance isn’t
extreme enough to cause the rig to severely cut back its power output."

Assuming that "constant forward average power" means 'as would be
indicated on a directional wattmeter calibrated for Z=50+j0', if that is
true for all load impedances that Pf is constant (within limits), then it
is evidence that Zs=50+j0 (within those limits).

He goes on to say "It turns out that a linear model of my transmitter
(without a transmatch) over its non-shutdown range is very simple—it’s
just a voltage source in series with a resistance." Subject to the
conditions I stated in the previous paragraph, that is correct, but that
whilst that model can be used to determine behaviour of the external
load, there are limits to the inferences that can be drawn about the
internals of the transmitter, including as mentioned in earlier posts,
internal dissipation and efficiency. Roy acknowledges that in the next
paragraph. Only a mischief maker would represent Roy as meaning
otherwise.

From my own experience, I don't agree that HF ham rigs typically produce
constant Pf into varying loads. Walt's transmitter measurements that we
are discussing do not show constant Pf, though the change is fairly
small. But this is a practical measurement project for yourself, don't be
put off by the attempts to discredit measurements with anything but
traceable calibration.

(*) I do not know how clearly denote plural in "do you"

I am not the expert that others are on English language, and we speak a
version of English closer to the English here... but "you" is plural and
singular (but if followed by a verb, it is treated as plural eg "you are
correct"), and you could say to a group "do you agree", though some
people may say "do you all agree" or "do y'all agree", though those might
be seen as asking each member of the group rather than collectively.

Owen

Where does it go? (mismatched power)
 

On Jun 13, 11:19*am, Keith Dysart wrote:
Was your disingenuity deliberate?

No, it was non-existent. The source I referred to was Roy's *specified
source* in his food-for-thought article on forward and reflected
power. Roy used his *specified source* to do his power calculations.
Contrary to what you asserted, Roy certainly did "make an inference
about the internal dissipation of the source". The fact that Roy's
*specified source* doesn't resemble a ham transmitter is irrelevant.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 13, 11:26*am, K1TTT wrote:
because they use the newer 'fields and waves in communication
electronics' by ramo, whinnery, and van duzer that uses reflection
coefficients and calculates voltages, currents, electric and magnetic
fields.

I've got a copy of the 3rd edition of that very book. S-parameters are
covered on page 540 so my question remains. Why don't RF engineers
read the book?
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 13, 11:27*am, K1TTT wrote:
On Jun 13, 3:02*pm, Cecil Moore wrote:

On Jun 13, 9:34*am, K1TTT wrote:

why are they (voltages) indeterminate? *i can calculate them, why can't you?

Please tell me how you can calculate an absolute voltage when Z0
is an unknown variable?

the same way you calculate the power to be zero watts. *no need to
know the z0 if the voltage is zero.

Ah, but you said you could calculate "them", meaning at least two
voltages. You have calculated one voltage to be zero. So what is the
value of one of the other voltages when the Z0 is unknown?
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

K1TTT wrote:
On Jun 9, 10:38 pm, Jim Lux wrote:
In fact, the ration between that stored energy and the amount flowing
"through" (i.e. radiated away) is related to the directivity of the
antenna: high directivity antennas have high stored energy (large
magnetic and electric fields): the ratio of stored to radiated energy
is "antenna Q" (analogous to the stored energy in a LC circuit leading
to resonant rise).

So, high directivity = high stored energy = high circulating energy =
high I2R losses.

this is a relationship i haven't heard of before... and would be very
wary of stating so simply.

I should have used arrows rather than equals signs.
But it's basically a manifestation of Chu's idea combined with practical
materials.

Chu proposed the concept relating directivity and stored energy and
physical size. A passively excited multi element array (like a Yagi) has
to transfer energy from element to element to work, and it follows the
characteristics outlined by Chu. And anything with circulating energy
that gets carried by a conductor is going to have high(er) I2R losses
than something that doesn't.


it may be true for a specific type of
antenna, MAYBE Yagi's, MAYBE rhombics or or close coupled wire arrays,
but some of the most directive antennas are parabolic dishes which i
would expect to have very low Q and extremely low losses.

Interesting case there. Loss isn't all that low (typical parabolic
antennas with their feed have an efficiency of 50-70%), although it IS
low compared to the directivity. And, in fact, there's not much stored
energy (so the Q is low). On the other hand a parabolic antenna is
physically very large compared to a wavelength, so the Chu relationship
holds. I'd have to think about whether one can count the energy in the
wave propagating from feed to reflector surface as "stored", but I think
not. Probably only the E and H fields at the reflector surface.


you could
also have an antenna with very high Q, very high i^2r losses, but very
low directivity, so i would be careful about drawing a direct link
between the two.

Yes.. you're right.. the relations set an upper bound on what's
possible.. That is, for a given directivity, you can get either small
size and large stored energy (the Yagi-Uda or W8JK), or large size and
small stored energy (the parabolic reflector and feed). As you note, a
dummy load has very low directivity.

Where does it go? (mismatched power)
 

On Jun 13, 11:42*pm, Owen Duffy wrote:
Owen Duffy wrote :

...

In the measurements of an IC7000 that I made, the measured output
power on one VSWR(50)=1.5 load was 82.5W when it would have been
104.6W had the source been 50+j0, an error of 0.8dB. I opined that
this test did not support the proposition that Zs was not 50+j0

Too many "nots", isn't there?

It should read:

I opined that this test did not support the proposition that Zs was 50+j0..

Apologies, Owen.

I suppose this will be buried where nobody will read it...

I realized that with the nice instrument-grade directional couplers
that came with a new 100W RF power amplifier, and with the other
equipment on my bench, I can measure RF amplifier/transmitter source
impedance relatively easily and with good accuracy. I strongly
suspect the accuracy will be limited first by how well the setup of
the transmitter/amplifier can be duplicated, and not by the
measurement instruments.

I won't go through the whole test setup, but just say that
substituting an open or short for the connection to the transmitter
yields the expected amplitude return signal, and terminating the line
in a precision 50 ohm calibration standard yields a 47dB return loss,
for the frequency I was measuring (nominally 7MHz, for this first
measurement). The measurement involves sending a signal offset from
the nominal transmitter frequency by a few Hertz at about -20dBm
toward the transmitter, and looking at what comes back.

Measuring a Kenwood TS520S, set up for about 70 watts output, ALC
disabled, operating as a linear amplifier somewhat (about 30 watts)
below its maximum output: result is 56+j16 ohms at the output UHF
connector on the TS520S. That's about 1.4:1 SWR, and at some point
along a lossless line, that's equivalent to about 70+j0 ohms: not
terribly close to 50 ohms. I'm not going to bother with a detailed
error analysis presentation, but I'm confident that the amplitude of
the return loss is accurate within 0.1dB, and the phase angle within
10 degrees, to better than 99% probability.

I may make some more measurements with different amplifier setups and
at different frequencies, but for now, that's it...

Cheers,
Tom

Where does it go? (mismatched power)
 

Making some assumptions about a sensible implementation for the test, an
interesting test Tom, thanks for the writeup.

One of the interesting propostions emerging in the discussion is that Zs
may be dependent on drive level. Your test setup would be an interesting
one to explore that effect (changing drive level alone, no other adjustment
or change).

In my own tests looking for Pf constant (independent of load changes), I
have noted that I get different results from the IC7000 at different drive
levels. Not surprising, as the gain of the active devices are likely to
change significantly in the upper half of the rated power range, and the
effects of either current or voltage saturation are likely to be manifest
at near rated output, and power control loops at maximum output.

If Zs is dependent on drive level in a significant way, then one must ask
what is the application of Zs in the case of an SSB telephony transmitter.

Owen

Where does it go? (mismatched power)
 

On Mon, 14 Jun 2010 11:56:43 -0700 (PDT), K7ITM wrote:

Measuring a Kenwood TS520S, set up for about 70 watts output, ALC
disabled, operating as a linear amplifier somewhat (about 30 watts)
below its maximum output: result is 56+j16 ohms at the output UHF
connector on the TS520S.

Comparing to my table of results for my own TS430S, with similar
initical conditions, that is well within the range of measurements I
have taken.

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

Richard Clark wrote in
:


Comparing to my table of results for my own TS430S, with similar
initical conditions, that is well within the range of measurements I
have taken.

After all the goading about toleranced figures and "vacant adjectives",
this is what you contibute. I dismiss all of your previous comment as a
windup.

Owen

Where does it go? (mismatched power)
 

On Jun 14, 2:56*pm, K7ITM wrote:
On Jun 13, 11:42*pm, Owen Duffy wrote:



Owen Duffy wrote :

...

In the measurements of an IC7000 that I made, the measured output
power on one VSWR(50)=1.5 load was 82.5W when it would have been
104.6W had the source been 50+j0, an error of 0.8dB. I opined that
this test did not support the proposition that Zs was not 50+j0

Too many "nots", isn't there?

It should read:

I opined that this test did not support the proposition that Zs was 50+j0.

Apologies, Owen.

I suppose this will be buried where nobody will read it...

I realized that with the nice instrument-grade directional couplers
that came with a new 100W RF power amplifier, and with the other
equipment on my bench, I can measure RF amplifier/transmitter source
impedance relatively easily and with good accuracy. *I strongly
suspect the accuracy will be limited first by how well the setup of
the transmitter/amplifier can be duplicated, and not by the
measurement instruments.

I won't go through the whole test setup, but just say that
substituting an open or short for the connection to the transmitter
yields the expected amplitude return signal, and terminating the line
in a precision 50 ohm calibration standard yields a 47dB return loss,
for the frequency I was measuring (nominally 7MHz, for this first
measurement). *The measurement involves sending a signal offset from
the nominal transmitter frequency by a few Hertz at about -20dBm
toward the transmitter, and looking at what comes back.

Measuring a Kenwood TS520S, set up for about 70 watts output, ALC
disabled, operating as a linear amplifier somewhat (about 30 watts)
below its maximum output: *result is 56+j16 ohms at the output UHF
connector on the TS520S. *That's about 1.4:1 SWR, and at some point
along a lossless line, that's equivalent to about 70+j0 ohms: *not
terribly close to 50 ohms. *I'm not going to bother with a detailed
error analysis presentation, but I'm confident that the amplitude of
the return loss is accurate within 0.1dB, and the phase angle within
10 degrees, to better than 99% probability.

I may make some more measurements with different amplifier setups and
at different frequencies, but for now, that's it...

Cheers,
Tom

Tom, you stated earlier that you measured the source impedance of a
TS520S transceiver by inserting a somewhat off-resonance signal into
the output terminals when the rig was delivering 70 watts, and the
source impedance was measured as 56+j16 ohms. However, you chose not
to describe the setup or the procedure for obtaining this data.

I'm hungering to learn of the setup and procedure you used, because
I'd like to know what reflection mechanism gave a return signal that
could be discriminated from the 70w output signal from the
transceiver.

In his Nov 1991 QST article Warren Bruene, W5OLY, used what I believe
is a similar procedure, in which he claims he measured the Rs that he
called the 'source impedance' of the RF amp. He used his measurements
in asserting that because his Rs didn't equal RL there could be no
conjugate match when the source is an RF power amp. I have never
believed his procedure and measurements were valid, and I still don't.
So if your setup in any way resembles what Bruene presented in his QST
article I would like to know how you can justify a procedure that
involves inserting an off-set frequency signal rearward into an
operating RF power amp to determine the source impedance.

Walt, W2DU

Where does it go? (mismatched power)
 

On Jun 14, 7:00*pm, walt wrote:
On Jun 14, 2:56*pm, K7ITM wrote:



On Jun 13, 11:42*pm, Owen Duffy wrote:

Owen Duffy wrote :

...

In the measurements of an IC7000 that I made, the measured output
power on one VSWR(50)=1.5 load was 82.5W when it would have been
104.6W had the source been 50+j0, an error of 0.8dB. I opined that
this test did not support the proposition that Zs was not 50+j0

Too many "nots", isn't there?

It should read:

I opined that this test did not support the proposition that Zs was 50+j0.

Apologies, Owen.

I suppose this will be buried where nobody will read it...

I realized that with the nice instrument-grade directional couplers
that came with a new 100W RF power amplifier, and with the other
equipment on my bench, I can measure RF amplifier/transmitter source
impedance relatively easily and with good accuracy. *I strongly
suspect the accuracy will be limited first by how well the setup of
the transmitter/amplifier can be duplicated, and not by the
measurement instruments.

I won't go through the whole test setup, but just say that
substituting an open or short for the connection to the transmitter
yields the expected amplitude return signal, and terminating the line
in a precision 50 ohm calibration standard yields a 47dB return loss,
for the frequency I was measuring (nominally 7MHz, for this first
measurement). *The measurement involves sending a signal offset from
the nominal transmitter frequency by a few Hertz at about -20dBm
toward the transmitter, and looking at what comes back.

Measuring a Kenwood TS520S, set up for about 70 watts output, ALC
disabled, operating as a linear amplifier somewhat (about 30 watts)
below its maximum output: *result is 56+j16 ohms at the output UHF
connector on the TS520S. *That's about 1.4:1 SWR, and at some point
along a lossless line, that's equivalent to about 70+j0 ohms: *not
terribly close to 50 ohms. *I'm not going to bother with a detailed
error analysis presentation, but I'm confident that the amplitude of
the return loss is accurate within 0.1dB, and the phase angle within
10 degrees, to better than 99% probability.

I may make some more measurements with different amplifier setups and
at different frequencies, but for now, that's it...

Cheers,
Tom

Tom, you stated earlier that you measured the source impedance of a
TS520S transceiver by inserting a somewhat off-resonance signal into
the output terminals when the rig was delivering 70 watts, and the
source impedance was measured as 56+j16 ohms. However, you chose not
to describe the setup or the procedure for obtaining this data.

I'm hungering to learn of the setup and procedure you used, because
I'd like to know what reflection mechanism gave a return signal that
could be discriminated from the 70w output signal from the
transceiver.

In his Nov 1991 QST article Warren Bruene, W5OLY, used what I believe
is a similar procedure, in which he claims he measured the Rs that he
called the 'source impedance' of the RF amp. *He used his measurements
in asserting that because his Rs didn't equal RL there could be no
conjugate match when the source is an RF power amp. I have never
believed his procedure and measurements were valid, and I still don't.
So if your setup in any way resembles what Bruene presented in his QST
article I would like to know how you can justify a procedure that
involves inserting an off-set frequency signal rearward into an
operating RF power amp *to determine the source impedance.

Walt, W2DU

OK...

So let's consider making a load-pull measurement of source impedance.
Since we're trying to resolve both resistance and reactance, we need
to change the load in at least two directions that have a degree of
orthogonality. But we could also change the load over a range of
values. For example, we could connect a 51+j0 load directly to the
output port we're trying to measure, and then connect it through
varying lengths of 50.0 ohm lossless coax. 45 electrical degrees of
line would shift the phase of the 51 ohm load so it looks instead like
49.99-j0.99 ohms. 90 electrical degrees shifts the 51 ohm load to
49.02+j0 ohms, and so forth. Measurements of the varying amplitude
output with those loads will give us enough information to resolve the
source resistance and reactance and open-circuit voltage.

For a 51 ohm load on a 50 ohm line, the reflection coefficient
magnitude is 1/100, so if the transmitter is putting out 100Vrms
forward, the reverse is 1Vrms.

Now consider a method to change the line length that doesn't use
individual sections that have to be patched in and out, but rather
uses a "trombone" section that, in theory anyway, could range from
zero length to essentially infinite length. Picture that trombone
section getting longer at a fixed rate, so now the load is rotating
around a circle on the linear reflection coefficient plane (which is,
by the way, exactly the same plane the Smith chart is plotted on); the
circle is centered at zero and is a constant 1/100 amplitude, with
linearly varying phase. So the 1Vrms reverse wave on the line of the
100Vrms forward example arrives back at the amplifier at continuously
varying phase. Imagine that the phase shift is 360 degrees in 1/100
of a second. Now note that the reverse wave corresponds _exactly_ to
a wave offset in frequency from the forward wave by 100Hz. If the
line is continuously lengthening, the offset is negative; if the line
is shortening instead, the offset is positive.

Now, from the point of view of the amplifier, can that scenario be
distinguished from one in which I have a perfect 50 ohm load that
absorbs all the transmitter's output, and a method to introduce a
"reverse" 1.00Vrms wave into the line at a frequency that's offset
from the transmitter's output by 100Hz?

If you believe that the amplifier can distinguish between those two
scenarios, I fear we have nothing more to discuss.

Cheers,
Tom

Where does it go? (mismatched power)
 

On 14 jun, 06:01, Owen Duffy wrote:
lu6etj wrote :

...

For example: do you (*) recognize Roy Lewallen late example in "Food
for tought" assuming (or conceding that) as not representing a real
rig but a simple constant voltage source in series with a resistor, at
least? to give some credit to his ideas Until now, I could not
know...

In that article, Roy says "My commercial amateur HF transceiver is
probably typical of modern rigs in that it produces a constant forward
average power into varying load impedances provided the impedance isn t
extreme enough to *cause the rig to severely cut back its power output."

Assuming that "constant forward average power" means 'as would be
indicated on a directional wattmeter calibrated for Z=50+j0', if that is
true for all load impedances that Pf is constant (within limits), then it
is evidence that Zs=50+j0 (within those limits).

He goes on to say "It turns out that a linear model of my transmitter *
(without a transmatch) over its non-shutdown range is very simple it s
just a voltage source in series with a resistance." Subject to the
conditions I stated in the previous paragraph, that is correct, but that
whilst that model can be used to determine behaviour of the external
load, there are limits to the inferences that can be drawn about the
internals of the transmitter, including as mentioned in earlier posts,
internal dissipation and efficiency. Roy acknowledges that in the next
paragraph. Only a mischief maker would represent Roy as meaning
otherwise.

From my own experience, I don't agree that HF ham rigs typically produce
constant Pf into varying loads. Walt's transmitter measurements that we
are discussing do not show constant Pf, though the change is fairly
small. But this is a practical measurement project for yourself, don't be
put off by the attempts to discredit measurements with anything but
traceable calibration.

(*) I do not know how clearly denote plural in "do you"

I am not the expert that others are on English language, and we speak a
version of English closer to the English here... but "you" is plural and
singular (but if followed by a verb, it is treated as plural eg "you are
correct"), and you could say to a group "do you agree", though some
people may say "do you all agree" or "do y'all agree", though those might
be seen as asking each member of the group rather than collectively.

Owen

Thanks Owen. I believe I quite understand your kind explanation and
reasons, but I am not sure if I can ask my question well enough
indeed!. I beg your patient.
I am not interested yet to make questions about real rigs, but reduce
at first only one specific problem at the most simple theorical model
I can think: an ideal constant voltage source in series with an ideal
resistence loaded at first with simple resistive loads connected
directly and late via ideal lossless TL of differents simple
wavelenghts 1/2, 1/4, 1/8 (I have formed idea about this, but I am not
interested in my concept but your concept about it).
For example, I want to know if you (all) would predict identical Rs
and Rl dissipation in that reduced and theorical context with direct
and remotely connected loads (vinculated via TL).

To this point K1TTT -seem to me- tell me you all would agree to settle
the problem with Telegrapher's equations to obtain TL input Z and then
apply simple circuit theory solution to calculate Rs-Rl dissipation
(before begin Thevenin misleading issue).
I do not want to advance more from here for fear to complicate the
question with my translation.

Thanks again.
Miguel - LU6ETJ

Where does it go? (mismatched power)
 

On 12 jun, 10:10, Cecil Moore wrote:
On Jun 11, 11:24*pm, lu6etj wrote:

As a courtesy to me, a foreigner tourist ham, would you mind stop for
a brief moment your more general differences and tell me if you agree
on the behavior of a Thevenin generator with a series resistance of 50
ohms in relation to changes in impedance of a lossless TL predicted by
the Telegrapher's equations solutions in terms of the power dissipated
on the load resistance and series resistence of Thevenin source?
I am pretty serious about this: until today I could not know if you
agree in that!! :)

Miguel, I don't think there is much disagreement about things that are
easily measured, like voltage and current. One solution to the
telegrapher's equations involves forward and reflected waves of
voltage and current. The conventional way of handling power (energy/
unit-time) is to use the voltages and currents to calculate the power
at certain points of interest. The telegrapher's equations do not tell
us *why* the power is what it is and the energy is where it is. To
obtain the why, one must study the behavior of electromagnetic waves.
How does the energy in electromagnetic waves behave? The telegrapher's
equations and Thevenin source do not answer that question.

For instance: Most readers here seem to think that the only phenomenon
that can cause a reversal of direction of energy flow in a
transmission line is a simple EM wave reflection based on the
reflection model. When they cannot explain what is happening using
that model, they throw up their hands and utter crap like, "Reflected
wave energy doesn't slosh back and forth between the load and the
source". But not only does it "slosh back and forth", it sloshes back
and forth at the speed of light in the medium because nothing else is
possible.

These are the people who have allowed their math models to become
their religion. They will not change their minds even when accepted
technical facts are presented. One response was, "Gobblydegook". (sic)

There is another phenomenon, besides a simple reflection, that causes
reflected energy to be redistributed back toward the load and that is
wave cancellation involving two wavefronts. If the two wavefronts are
equal in magnitude and opposite in phase, total wave cancellation is
the result which, in a transmission line, redistributes the wave
energy in the only other direction possible which is, surprise, the
opposite direction. This is a well known, well understood,
mathematically predictable phenomenon that happens all the time in the
field of optics, e.g. at the surface of non-reflective glass. It also
happens all the time in RF transmission lines when a Z0-match is
achieved.

Using the s-parameter equations (phasor math) at a Z0-match point in a
transmission line:

b1 = s11*a1 + s12*a2 = 0 = reflected voltage toward the source
Square this equation to get the reflected power toward the source.

These are the two wavefronts that undergo total wave cancellation,
i.e. total destructive interference.

b2 = s21*a1 + s22*a2 = forward voltage toward the load
s22*a2 is the re-reflection. Square this equation to get the forward
power toward the load.

If one squares both of those equations, one can observe the
interference terms which indicate why and where the energy goes. All
of the energy in s11*a1 and s12*a2 reverses direction at the Z0-match
and flows back toward the load. All the things that Roy is confused
about in his food-for-thought article on forward and reflected power
are easily explained by the power density equation (or by squaring the
s-parameter equations).

Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)
--
73, Cecil, w5dxp.com

Sorry and thanks Cecil I do not see this kind answer (I still using
normal Google groups reader and loss tracking of your message).
Tomorrow I will analize it with care, now is late here but I do not
want delay my aknowledge.

Miguel

Where does it go? (mismatched power)
 

On Tue, 15 Jun 2010 01:40:16 GMT, Owen Duffy wrote:

Richard Clark wrote in
:


Comparing to my table of results for my own TS430S, with similar
initical conditions, that is well within the range of measurements I
have taken.

After all the goading about toleranced figures and "vacant adjectives",
this is what you contibute.

C'mon Owen,

I have posted my results several times before. I don't think either
of us expected they made a ripple of notice then, but now this one
does when it is confirmed at another bench? I'm flattered.

I dismiss all of your previous comment as a windup.

Golly.

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

lu6etj wrote in
:

....
I am not interested yet to make questions about real rigs, but reduce
at first only one specific problem at the most simple theorical model
I can think: an ideal constant voltage source in series with an ideal
resistence loaded at first with simple resistive loads connected
directly and late via ideal lossless TL of differents simple
wavelenghts 1/2, 1/4, 1/8 (I have formed idea about this, but I am not
interested in my concept but your concept about it).
For example, I want to know if you (all) would predict identical Rs
and Rl dissipation in that reduced and theorical context with direct
and remotely connected loads (vinculated via TL).

To this point K1TTT -seem to me- tell me you all would agree to settle
the problem with Telegrapher's equations to obtain TL input Z and then
apply simple circuit theory solution to calculate Rs-Rl dissipation
(before begin Thevenin misleading issue).
I do not want to advance more from here for fear to complicate the
question with my translation.

Yes, if you are looking for a steady state solution, that is a valid
solution, and almost the easiest solution.

If you are using lossless lines, then the solution can be simpler than the
Telegrapher's Equation which uses hyperbolic trig functions. Simple trig
functions suffice if the line is lossless... but of course, the hyperbolic
functions will also work.

The reason I said "almost the easiest solution" above is that a trig based
simplification of the Telegrapher's Equation is simpler than the hyperbolic
Telegrapher's equation.

Implicit in this is that for the steady state, it doesn't matter how an
external network dictates V/I (Z) at the source/load terminals, whether or
not transmission lines are involved, or their length, the power delivered
by the source to its immediate load is determined by Veq, Zeq of the
source, and Zl of the load.

If an experiment indicates othewise, the experiment is flawed (eg something
is wrong with the assumed values for circuit components, measurement error
etc).

Taking Walt's example, he connected three 50 ohms loads in parallel, then
attached a nominal 50 ohm line of 13.5° electrical length. If you use TLLC
(http://www.vk1od.net/calc/tl/tllc.php) to solve this problem at 4MHz using
RG213 (though I don't know what he used), Zin is 17.78+j10.58. A lossless
solution will have very slightly lower R. Walt measured 17.98 + j8.77 which
suggests that the load at the end of the 50 ohm line was not exactly
16.6667+j0, or the line a little shorter, Zo off spec etc.

Whilst the above focuses on steady state analysis, understand that steady
state analysis is not appropriate for some types of transmitters, but for
systems where the transmission line propagation delay is small compared the
the highest modulating frequency, steady state analysis should be adequate.

Owen

Where does it go? (mismatched power)
 

On Mon, 14 Jun 2010 19:00:47 -0700 (PDT), walt wrote:

I'm hungering to learn of the setup and procedure you used, because
I'd like to know what reflection mechanism gave a return signal that
could be discriminated from the 70w output signal from the
transceiver.

Hi Walt,

For every hundred pundits there is at least one bench worker that is
more productive than them all. Of course I am being facetious, it is
probably closer to a thousand.

In the SARL Forums at:
http://www.sarl.org.za
search for the 19/04/2009 posting by ZS6BIM with the thread name:
Measuring the Output Impedance of a 100W Class AB HF Linear Amplifier
Plenty of data and charted and screen shots of O'scopes.

A very intriguing use of directional couplers (much as I would expect
Tom to have done). It seems to me I've directed you to similar work
(the rat-race-like method seems familiar to our email discussion some
dozen years ago).

The writer had the cojones to use his vector voltmeter at the
transmitter's antenna terminal. Back in 1997 I used my GR bridge to
look down the throat of the fire breathing dragon (TS430S) while it
was asleep and got similar results:
10 MHz 74 Ohms
14 MHz 74 Ohms
18 MHz 60 Ohms
21 MHz 37 Ohms
25 MHz 35 Ohms
28 MHz 48 Ohms
29 MHz 70 Ohms

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

On Jun 14, 9:00*pm, walt wrote:
In his Nov 1991 QST article Warren Bruene, W5OLY, used what I believe
is a similar procedure, in which he claims he measured the Rs that he
called the 'source impedance' of the RF amp. *He used his measurements
in asserting that because his Rs didn't equal RL there could be no
conjugate match when the source is an RF power amp. I have never
believed his procedure and measurements were valid, and I still don't.

IMO, the error that Bruene made is similar in concept to the some of
the errors being made here in this thread. Many people assume that a
simple reflection is the only way to change the direction and momentum
of an EM wave in a transmission line. In "Reflections", Sec 4.3,
Reflection Mechanics of Stub Matching, you describe the role of wave
cancellation associated with interference as a mechanism for
redistributing reflected energy back toward the load. You said, "Wave
interference between these two complimentary waves at the stub point
causes a cancellation of energy flow in the direction toward the
generator." It seems obvious to me that your "virtual short"
reflection at a Z0-match point is a combination of ordinary wave
reflection and interference patterns adding up to zero energy flowing
toward the source. Obviously, interference can also happen inside a
source.

Redistribution of energy can occur associated with destructive/
constructive interference inside the source and has an effect on the V/
I ratios, i.e. on the impedances. Bruene's pinging method completely
ignores the role of interference at the source. The Z = (Vfor+Vref)/
(Ifor+Iref) impedance for coherent waves may be a completely different
value than for the non-coherent pinging waves.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 15, 1:30*am, lu6etj wrote:
Sorry and thanks Cecil I do not see this kind answer (I still using
normal Google groups reader and loss tracking of your message).
Tomorrow I will analize it with care, now is late here but I do not
want delay my aknowledge.

Miguel, some posters here don't seem to realize that your questions
involve a source like the one specified by w7el in his foot-for-
thought article. That source may bear little resemblance to an actual
ham transmitter but it is the one w7el specified and it can be easily
analyzed for source resistor dissipation and tracking of the reflected
energy.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On 15 jun, 05:49, K7ITM wrote:
On Jun 14, 7:00*pm, walt wrote:



On Jun 14, 2:56*pm, K7ITM wrote:

On Jun 13, 11:42*pm, Owen Duffy wrote:

Owen Duffy wrote :

...

In the measurements of an IC7000 that I made, the measured output
power on one VSWR(50)=1.5 load was 82.5W when it would have been
104.6W had the source been 50+j0, an error of 0.8dB. I opined that
this test did not support the proposition that Zs was not 50+j0

Too many "nots", isn't there?

It should read:

I opined that this test did not support the proposition that Zs was 50+j0.

Apologies, Owen.

I suppose this will be buried where nobody will read it...

I realized that with the nice instrument-grade directional couplers
that came with a new 100W RF power amplifier, and with the other
equipment on my bench, I can measure RF amplifier/transmitter source
impedance relatively easily and with good accuracy. *I strongly
suspect the accuracy will be limited first by how well the setup of
the transmitter/amplifier can be duplicated, and not by the
measurement instruments.

I won't go through the whole test setup, but just say that
substituting an open or short for the connection to the transmitter
yields the expected amplitude return signal, and terminating the line
in a precision 50 ohm calibration standard yields a 47dB return loss,
for the frequency I was measuring (nominally 7MHz, for this first
measurement). *The measurement involves sending a signal offset from
the nominal transmitter frequency by a few Hertz at about -20dBm
toward the transmitter, and looking at what comes back.

Measuring a Kenwood TS520S, set up for about 70 watts output, ALC
disabled, operating as a linear amplifier somewhat (about 30 watts)
below its maximum output: *result is 56+j16 ohms at the output UHF
connector on the TS520S. *That's about 1.4:1 SWR, and at some point
along a lossless line, that's equivalent to about 70+j0 ohms: *not
terribly close to 50 ohms. *I'm not going to bother with a detailed
error analysis presentation, but I'm confident that the amplitude of
the return loss is accurate within 0.1dB, and the phase angle within
10 degrees, to better than 99% probability.

I may make some more measurements with different amplifier setups and
at different frequencies, but for now, that's it...

Cheers,
Tom

Tom, you stated earlier that you measured the source impedance of a
TS520S transceiver by inserting a somewhat off-resonance signal into
the output terminals when the rig was delivering 70 watts, and the
source impedance was measured as 56+j16 ohms. However, you chose not
to describe the setup or the procedure for obtaining this data.

I'm hungering to learn of the setup and procedure you used, because
I'd like to know what reflection mechanism gave a return signal that
could be discriminated from the 70w output signal from the
transceiver.

In his Nov 1991 QST article Warren Bruene, W5OLY, used what I believe
is a similar procedure, in which he claims he measured the Rs that he
called the 'source impedance' of the RF amp. *He used his measurements
in asserting that because his Rs didn't equal RL there could be no
conjugate match when the source is an RF power amp. I have never
believed his procedure and measurements were valid, and I still don't.
So if your setup in any way resembles what Bruene presented in his QST
article I would like to know how you can justify a procedure that
involves inserting an off-set frequency signal rearward into an
operating RF power amp *to determine the source impedance.

Walt, W2DU

OK...

So let's consider making a load-pull measurement of source impedance.
Since we're trying to resolve both resistance and reactance, we need
to change the load in at least two directions that have a degree of
orthogonality. *But we could also change the load over a range of
values. *For example, we could connect a 51+j0 load directly to the
output port we're trying to measure, and then connect it through
varying lengths of 50.0 ohm lossless coax. *45 electrical degrees of
line would shift the phase of the 51 ohm load so it looks instead like
49.99-j0.99 ohms. *90 electrical degrees shifts the 51 ohm load to
49.02+j0 ohms, and so forth. *Measurements of the varying amplitude
output with those loads will give us enough information to resolve the
source resistance and reactance and open-circuit voltage.

For a 51 ohm load on a 50 ohm line, the reflection coefficient
magnitude is 1/100, so if the transmitter is putting out 100Vrms
forward, the reverse is 1Vrms.

Now consider a method to change the line length that doesn't use
individual sections that have to be patched in and out, but rather
uses a "trombone" section that, in theory anyway, could range from
zero length to essentially infinite length. *Picture that trombone
section getting longer at a fixed rate, so now the load is rotating
around a circle on the linear reflection coefficient plane (which is,
by the way, exactly the same plane the Smith chart is plotted on); the
circle is centered at zero and is a constant 1/100 amplitude, with
linearly varying phase. *So the 1Vrms reverse wave on the line of the
100Vrms forward example arrives back at the amplifier at continuously
varying phase. *Imagine that the phase shift is 360 degrees in 1/100
of a second. *Now note that the reverse wave corresponds _exactly_ to
a wave offset in frequency from the forward wave by 100Hz. *If the
line is continuously lengthening, the offset is negative; if the line
is shortening instead, the offset is positive.

Now, from the point of view of the amplifier, can that scenario be
distinguished from one in which I have a perfect 50 ohm load that
absorbs all the transmitter's output, and a method to introduce a
"reverse" 1.00Vrms wave into the line at a frequency that's offset
from the transmitter's output by 100Hz?

If you believe that the amplifier can distinguish between those two
scenarios, I fear we have nothing more to discuss.

Cheers,
Tom

Hello Tom,

The method of constant varying phase of a mismatch to determine output
impedance I use in my simulation also (because it saves me time). I
use a second source with some frequency offset instead of the trombone
as you described (the trombone I cannot implement easily in my pspice
package).

When you change the phase very slowly, a soft power supply may change
voltage during the slow variation (as a load change may result in a DC
supply current change). When the difference frequency is sufficiently
high (for example 200 Hz or more), the electrolytics will keep the
supply voltage constant during all phase "steps". Therefore there may
be slight difference between measurement with a set of loads with
increasing phase and the RF injection method (for example with vector
analyzer).

I checked the injection method in simulation for linear circuits (with
both real and complex output impedance), linear active circuits, and
power circuits. I couldn’t find any flaw in the method. So I think
your reasoning is valid.

Some results applicable to PAs, inclusive the implementation in
simulation are in: http://www.tetech.nl/divers/PA_impedance.pdf. Note
that I determine the reflected voltage by means of interference with
the amplifier's output signal (envelope detection) as this saves me
from making a narrow band filter that results in very long run times.

I am now simulating a circuit with 6146 valve and pi-filter output (50
Ohms) and I will add it to the document.

Best regards,

Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me

Where does it go? (mismatched power)
 

Richard Fry wrote:

If reflected power is fictitious, and the number wavelengths of
transmission line of any random impedance compared to the load
connected to it makes no difference in the load seen by the
transmitter, the output power produced by the transmitter, and the
power dissipated in the far-end termination, then what is the reason
you chose a 1/2 wavelength of transmission line in your quoted post?

RF

I chose that length so that the transmission line would have no effect
on the impedance seen by the transmitter. As I've said many times, and
you've continually disagreed with, increased dissipation and/or damage
at the transmitter is due to the impedance it sees, and not by
"reflected power". The example I gave keeps the transmitter load
impedance constant while changing the "reflected power". And it shows
that neither the transmitter nor the load see any changes in dissipation.

If I had chosen a different length line, then changing its Z0 would have
changed the impedance seen by the transmitter, which would have changed
its efficiency by an amount which could have been determined only with
additional knowledge about its characteristics. But replacing the
transmission line with a lumped impedance transforming network would
have exactly the same effect on the transmitter, again illustrating that
the only important factor is the impedance the transmitter sees and not
the "reflected power" in the line it's connected to.

This posting is being made, though, for the benefit of other readers.
I've explained this many times to you before with no noticeable effect
on your understanding.

Roy Lewallen, W7EL

Where does it go? (mismatched power)
 

Richard Fry wrote:
On Jun 11, 12:29 pm, Richard Fry wrote:

If reflected power is fictitious, etc

Followup: Those denying the existence of reflected signals within an
antenna system may wish to view the measurement of such signals, at
the link below.

http://i62.photobucket.com/albums/h8...easurement.gif

RF

Thanks, but I've designed TDR systems and prepared and given classes on
the topic.

Roy Lewallen, W7EL

Where does it go? (mismatched power)
 

Owen Duffy wrote:
lu6etj wrote in news:da3e5147-cad8-47f9-9784-
:

...
OK. Thank you very much. This clarify so much the issue to me. Please,
another question: On the same system-example, who does not agree with
the notion that the reflected power is never dissipated in Thevenin
Rs? (I am referring to habitual posters in these threads, of course)


Thevenin's theorem says nothing of what happens inside the source (eg
dissipation), or how the source may be implemented.
. . .

Cecil has used this fact as a convenient way of avoiding confrontation
with the illustrations given in my "food for thought" essays. However,
those models aren't claimed to be Thevenin equivalents of anything. They
are just simple models consisting of an ideal source and a perfect
resistance, as used in may circuit analysis textbooks to illustrate
basic electrical circuit operation. The dissipation in the resistance is
clearly not related to "reflected power", and the reflected power
"theories" being promoted here fail to explain the relationship between
the dissipation in the resistor and "reflected power". I contend that if
an analytical method fails to correctly predict the dissipation in such
a simple case, it can't be trusted to predict the dissipation in other
cases, and has underlying logical flaws. For all the fluff about
photons, optics, non-dissipative sources, and the like, I have yet to
see an equation that relates the dissipation in the resistance in one of
those painfully simple circuits to the "reflected power" in the
transmission line it's connected to.

Roy Lewallen, W7EL

Where does it go? (mismatched power)
 

lu6etj wrote:

R. R. Well... I forget to say the more important = For the sake of
the advance of the topic please do replace "Thevenin circuit" in my
original question for "an ideal constant voltage source in series with
an ideal resistance" equivalent only to itself :)

Thank you, that's exactly what I've tried to do. But calling all such
simple circuits "Thevenin equivalents" is a convenient way to avoid
having to explain the phenomena they illustrate. So that has been the
tactic of some of the "reflected power" proponents.

Roy Lewallen, W7EL