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