what happens to reflected energy ?
 

On 9 jun, 02:07, Richard Clark wrote:
On Tue, 8 Jun 2010 12:17:37 -0700 (PDT), Wimpie
wrote:

When we go back to my first posting in this thread (the first reply)
where I stated that amplifiers do not show 50 Ohms in general, is not
that strange and doesn't violate Walt's conclusions.

Hi Wimpie,

Time to rewind:

On Thu, 27 May 2010 19:32:40 -0700 (PDT), walt wrote:

The source resistance data reported in Secs 19.8 and 19.9 were
obtained using the load variation method with resistive loads. Note
that of the six measurements of output source resistance reported in
Table 19.1, the average value of the resistance is 50.3 ohms obtained
with the reference load resistance of 51.2 ohms, exhibiting an error
of only 1.8 percent.

Considering these two explicit initial conditions and findings:



1. Using a Kenwood TS-830S transceiver as the RF source, the tuning
and loading of the pi-network are adjusted to deliver all the
available power into a 50 + j0-ohm load with the grid drive adjusted
to deliver the maximum of 100 watts at 4 MHz, thus establishing the
area of the RF power window at the input of the pi-network, resistance
RLP at the plate, and the slope of the load line. The output source
resistance of the amplifier in this condition will later be shown to
be 50 ohms. In this condition the DC plate voltage is 800 v and plate
current is 260 ma. DC input power is therefore 800 v ? 0.26 a = 208 w.
Readings on the Bird 43 wattmeter indicate 100 watts forward and zero
watts reflected. (100 watts is the maximum RF output power available
at this drive level.) From here on the grid drive is left undisturbed,
and the pi-network controls are left undisturbed until Step 10.

2. The amplifier is now powered down and the load resistance RL is
measured across the input terminals of the resonant pi-network tank
circuit (from plate to ground) with an HP-4815 Vector Impedance Meter.
The resistance is found to be approximately 1400 ohms. Because the
amplifier was adjusted to deliver the maximum available power of 100
watts prior to the resistance measurement, the averaged resistance RLP
looking into the plate (upstream from the network terminals) is also
approximately 1400 ohms. Accordingly, a non-reactive 1400-ohm resistor
is now connected across the input terminals of the pi-network tank
circuit and source resistance ROS is measured looking rearward into
the output terminals of the network. Resistance ROS was found to be 50
ohms.

Quite explicit. *Rated power into the rated Z load revealed a
conjugate basis Z match (afterall 50=50) or an image basis Z match
(50=50). *The pi-network simply transformed the source/load. *Anyone
skilled with the concept of the Smith Chart can immediately follow
that logical construct.

We even get the plate resistance which conforms to within published
specifications for the pair of tubes.

We even get the near classic result of a conjugate basis Z match
efficiency (however, only in the first pass approximation).

How asking the same question:

* "Does Walt's data support the evidence of a Conjugate Match?"

migrates into responses to other issues is a strange dance. *I ask "up
or down," and everyone wants to vote sideways.

When you don't touch the plate and load capacitor and change from 50
Ohms to a load with VSWR = 2, you violate Walt's conditions

I have not the vaguest idea where you got the idea that I changed
anything. *The totality of my discussion never budged from steps 1 and
2 and I deliberately and clearly confined all matters there and
repeated them more than occasionally.

In case of doubt use the forward power measurement technique,

I have, and the data I've asked from you was not intended for some
one-time design for a unique application. *This is an Ham radio group,
and the object under consideration is an Ham transmitter. *Feel free
to substitute freely among a similar class (frequency range, power
output) such that others might compare their own. *

when it
changes under varying load, your amplifier doesn’t behave as a 50 Ohms
source.

This is the most curious statement that eventually comes down the
pike. *It is like I am talking to a novice to explain:
* * * * "That is what the tuner is for!"

Maybe I had to be more precise to
mention the exceptions: forward power control loop as mentioned by
Roy,

What Roy is describing is nothing close to Walt's hypothesis that I
have confined to a single point observation. *No one has advanced
claims made that demonstrate:
1. *Constant Z across all loads;
2. *Constant Z across all frequencies;
3. *Constant Power across all loads;
4. *Constant Power across all frequencies.
No competent Ham expects that of any Ham transmitter.

Walt reported a 50 Ohm source Z from a tube rig. *I have measured a 50
Ohm Z from two of my own transistor rigs. *The difference in
technology is not an issue. *I have also been responsible for
measuring RF power and source/load specifications within the chain of
national standards laboratories. *Stretching the term RF, this spanned
from DC to 12GHz for power levels up to 100W. *The sources across the
board exhibited 50 Ohms Source Z.

Your own national lab is represented at:http://www.vsl.nl/
They undoubtedly use the same reference citations as our National
Institutes for Science and Technology.

***********

Putting the entirety of this discussion aside, and returning to your
very special project with an 9 Ohm source of RF at 8 Mhz. *How
efficient was its operation feeding a 50 Ohm line to a 50 Ohm load?
Can you document this with schematics, parts description, and measured
data at key points from drain through to the load? *This was the level
of available resources that came with Walt's discussion.

73's
Richard Clark, KB7QHC

Hello Richard,

We can continue mentioning measurement results, do other claims, but
that may not converge. Your results show 50 Ohms output impedance,
mine show other. Such results cannot be verified easily, so this
remains food for endless discussions.

I decided to use another approach that can be verified by people
having a (spice) simulator. I did simulations for some circuits that
can be expected in the amateur world and aren't exotic. I put them
over he http://www.tetech.nl/divers/PA_impedance.pdf . Now people
can figure out what it IS and get their own educated opinion. I used
low frequencies to avoid discussion about the validity of the
simulations (parasitic inductances, etc).

From these simulations it is clear that small changes in load or drive
level result in large change of output impedance. This knowledge
resulted in my statement: "an RF PA isn't a 50 Ohms source", what was/
is heavily disputed by you.

Of course you can design to have better output VSWR under varying load
or drive, but this is generally not done for systems that just have to
deliver the power (transmitters).

If you want to reproduce some of the results yourself, we can compare
our spice netlists. I recently added a real class C example as this
was heavily discussed in another thread.

Best regards,


Wim
PA3DJS
www.tetech.nl
When you remove abc, PM will reach me.

what happens to reflected energy ?
 

On Jun 13, 5:08*pm, Wimpie wrote:
On 9 jun, 02:07, Richard Clark wrote:



On Tue, 8 Jun 2010 12:17:37 -0700 (PDT), Wimpie
wrote:

When we go back to my first posting in this thread (the first reply)
where I stated that amplifiers do not show 50 Ohms in general, is not
that strange and doesn't violate Walt's conclusions.

Hi Wimpie,

Time to rewind:

On Thu, 27 May 2010 19:32:40 -0700 (PDT), walt wrote:

The source resistance data reported in Secs 19.8 and 19.9 were
obtained using the load variation method with resistive loads. Note
that of the six measurements of output source resistance reported in
Table 19.1, the average value of the resistance is 50.3 ohms obtained
with the reference load resistance of 51.2 ohms, exhibiting an error
of only 1.8 percent.

Considering these two explicit initial conditions and findings:

1. Using a Kenwood TS-830S transceiver as the RF source, the tuning
and loading of the pi-network are adjusted to deliver all the
available power into a 50 + j0-ohm load with the grid drive adjusted
to deliver the maximum of 100 watts at 4 MHz, thus establishing the
area of the RF power window at the input of the pi-network, resistance
RLP at the plate, and the slope of the load line. The output source
resistance of the amplifier in this condition will later be shown to
be 50 ohms. In this condition the DC plate voltage is 800 v and plate
current is 260 ma. DC input power is therefore 800 v ? 0.26 a = 208 w.
Readings on the Bird 43 wattmeter indicate 100 watts forward and zero
watts reflected. (100 watts is the maximum RF output power available
at this drive level.) From here on the grid drive is left undisturbed,
and the pi-network controls are left undisturbed until Step 10.

2. The amplifier is now powered down and the load resistance RL is
measured across the input terminals of the resonant pi-network tank
circuit (from plate to ground) with an HP-4815 Vector Impedance Meter.
The resistance is found to be approximately 1400 ohms. Because the
amplifier was adjusted to deliver the maximum available power of 100
watts prior to the resistance measurement, the averaged resistance RLP
looking into the plate (upstream from the network terminals) is also
approximately 1400 ohms. Accordingly, a non-reactive 1400-ohm resistor
is now connected across the input terminals of the pi-network tank
circuit and source resistance ROS is measured looking rearward into
the output terminals of the network. Resistance ROS was found to be 50
ohms.

Quite explicit. *Rated power into the rated Z load revealed a
conjugate basis Z match (afterall 50=50) or an image basis Z match
(50=50). *The pi-network simply transformed the source/load. *Anyone
skilled with the concept of the Smith Chart can immediately follow
that logical construct.

We even get the plate resistance which conforms to within published
specifications for the pair of tubes.

We even get the near classic result of a conjugate basis Z match
efficiency (however, only in the first pass approximation).

How asking the same question:

* "Does Walt's data support the evidence of a Conjugate Match?"

migrates into responses to other issues is a strange dance. *I ask "up
or down," and everyone wants to vote sideways.

When you don't touch the plate and load capacitor and change from 50
Ohms to a load with VSWR = 2, you violate Walt's conditions

I have not the vaguest idea where you got the idea that I changed
anything. *The totality of my discussion never budged from steps 1 and
2 and I deliberately and clearly confined all matters there and
repeated them more than occasionally.

In case of doubt use the forward power measurement technique,

I have, and the data I've asked from you was not intended for some
one-time design for a unique application. *This is an Ham radio group,
and the object under consideration is an Ham transmitter. *Feel free
to substitute freely among a similar class (frequency range, power
output) such that others might compare their own. *

when it
changes under varying load, your amplifier doesn’t behave as a 50 Ohms
source.

This is the most curious statement that eventually comes down the
pike. *It is like I am talking to a novice to explain:
* * * * "That is what the tuner is for!"

Maybe I had to be more precise to
mention the exceptions: forward power control loop as mentioned by
Roy,

What Roy is describing is nothing close to Walt's hypothesis that I
have confined to a single point observation. *No one has advanced
claims made that demonstrate:
1. *Constant Z across all loads;
2. *Constant Z across all frequencies;
3. *Constant Power across all loads;
4. *Constant Power across all frequencies.
No competent Ham expects that of any Ham transmitter.

Walt reported a 50 Ohm source Z from a tube rig. *I have measured a 50
Ohm Z from two of my own transistor rigs. *The difference in
technology is not an issue. *I have also been responsible for
measuring RF power and source/load specifications within the chain of
national standards laboratories. *Stretching the term RF, this spanned
from DC to 12GHz for power levels up to 100W. *The sources across the
board exhibited 50 Ohms Source Z.

Your own national lab is represented at:http://www.vsl.nl/
They undoubtedly use the same reference citations as our National
Institutes for Science and Technology.

***********

Putting the entirety of this discussion aside, and returning to your
very special project with an 9 Ohm source of RF at 8 Mhz. *How
efficient was its operation feeding a 50 Ohm line to a 50 Ohm load?
Can you document this with schematics, parts description, and measured
data at key points from drain through to the load? *This was the level
of available resources that came with Walt's discussion.

73's
Richard Clark, KB7QHC

Hello Richard,

We can continue mentioning measurement results, do other claims, but
that may not converge. Your results show 50 Ohms output impedance,
mine show other. Such results cannot be verified easily, so this
remains food for endless discussions.

I decided to use another approach that can be verified by people
having a (spice) simulator. *I did simulations for some circuits that
can be expected in the amateur world and aren't exotic. *I put them
over he *http://www.tetech.nl/divers/PA_impedance.pdf. *Now people
can figure out what it IS and get their own educated opinion. I used
low frequencies to avoid discussion about the validity of the
simulations (parasitic inductances, etc).

From these simulations it is clear that small changes in load or drive
level result in large change of output impedance. *This knowledge
resulted in my statement: "an RF PA isn't a 50 Ohms source", what was/
is heavily disputed by you.

Of course you can design to have better output VSWR under varying load
or drive, but this is generally not done for systems that just have to
deliver the power (transmitters).

If you want to reproduce some of the results yourself, we can compare
our spice netlists. I recently added a real class C example as this
was heavily discussed in another thread.

Best regards,

Wim
PA3DJSwww.tetech.nl
When you remove abc, PM will reach me.

Hi Wim,

I quote a statement you made in your above post:

"From these simulations it is clear that small changes in load or
drive
level result in large change of output impedance. "

Now I agree that changing the drive level will result in a change in
output impedance. However, let's say you have just adjusted the amp to
deliver all the available power into a given load at a given drive
level, and now you change the load, but leave the drive level and all
other adjustments of the amp untouched. Are you saying this change in
load changes the output impedance? If so, please explain why.

In addition, which situation has the greatest priority, data obtained
from precise measurements or data obtained solely from simulations?

Walt, W2DU

what happens to reflected energy ?
 

On Sun, 13 Jun 2010 14:08:59 -0700 (PDT), Wimpie
wrote:

This knowledge
resulted in my statement: "an RF PA isn't a 50 Ohms source", what was/
is heavily disputed by you.

Hi Wim,

Your conclusion:
"Under maximum power output matching given certain drive (not
necessarily being the maximum drive), the output impedance equals the
load resistor (the so-called "conjugated match condition")."

It was not the only conclusion, but it certainly disputes what you say
above.

Given the easy access to real amplifiers available to Hams at their
own bench, and further given that they do not present any more
complexity than your own simulation; then I have to wonder why we have
wandered into strange topologies.

I am not interested in trying to follow 23 pages of dense presentation
where you could have simulated Walt's finals, validated or rejected
his data, and THEN formed your own conclusions. If this is a problem
of simulation (you don't have that tube handy) what about topology?

It looks more like a link coupled tank circuit from pre-WWII days. The
Tank Q looks to be non-existent. When I rummage through the paper I
find 4.4! This is certainly beyond the standard for PA design - and
we have swapped out of the tube into a mosfet (would you care to stick
to one objective?). Elsewhere (wandering back in tubes) I find a
value of 10.3 (elsewhere it is 11) which inhabits the lowest margin of
good design. Then an itinerant report of a cathode tank (????) whose
Q is 0.44. Q appears to be of little concern.

Again, typical Ham equipment shows Q as low as 10 - maybe, and that
would be for a dog; but those that I have seen typically fall around
15, sometimes 20. As the Z transformation in plate/drain circuits
varies by the square of Q, this is not inconsequential and it once
again has me wondering why the strange topology?

No, your paper lacks focus by trying to be all things (tubes, mosfets,
Class AB, Class C). You examine the most inconsequential details as
if they were equally important as those details that bear on your
concern. The paper lacks structure. Headings are nothing more than
feature descriptions, not argument development. There is a lot of
discussion of the idiosyncrasies of simulation - I am not interested
and that discussion creates the impression of hidden errors. Really,
this stuff goes into an appendix not as the object of the narrative.
The graphs are pretty diversions, but they don't add much. In the
software industry, this is spaghetti design.

I would prefer a conventional topology. I suppose that has come
through clear. Pick one significant point, re-edit this into 4 pages
and then you might have something interesting. This advice comes from
one of our American writers, Mark Twain:
"If I had more time, I would have written you a shorter letter."

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On Jun 10, 5:52 pm, K1TTT wrote:
On Jun 9, 2:04 am, Keith Dysart wrote:

Exactly. And yet, were you to attach a DC-couple directional
wattmeter to the line, it would indicate 50W forward and 50W
reflected (using your corrected numbers).

only as you monitored the initial transient. once the transient has
passed there is no power flowing in the line.

I agree completely with the second sentence. But the first is in
error. A directional wattmeter will indeed show 50W forward and 50W
reflected on a line with constant DC voltage and 0 current. I’ll
start by providing support for your second sentence and then get
back to reflected power.

Choose an arbitrary point on the line and observe the voltage. The
observed voltage will be a function of time. Let us call it V(t).
At the same point on the line, observe the current and call it I(t).

The energy flowing (i.e. power) at this point on the line will
be P(t)=V(t)I(t). If the voltage is measured in volts and the current
in amps, then the power will be in joules/s or watts.
We can also compute the average energy flow (power) over some
interval by integrating P(t) over the interval and dividing by
the length of the interval. Call this Pavg. For repetitive
signals it is convenient to use one period as the interval.

Note that V(t) and I(t) are arbitrary functions. The analysis
applies regardless of whether V(t) is DC, a sinusoid, a step,
a pulse, etc.; any arbitrary function.

To provide some examples, consider the following simple setup:
1) 50 ohm, Thevenin equivalent circuit provides 100 V in to an
open circuit.
2) The generator is connected to a length of 50 ohm ideal
transmission line. We will stick with ideal lines for the
moment, because, until ideal lines are understood, there
is no hope of understanding more complex models.
3) The ideal line is terminated in a 50 ohm resistor.

Set the generator to DC:
V(t)=50V, wherever it is measured on the line
I(t)=1A, wherever it is measured on the line
P(t)=50W, wherever it is measured on the line
Pavg=50W
No disagreement, I hope.

Set the generator to generate a sinusoid:
V(t)=50sin(wt)
I(t)=1sin(wt)
P(t)=25+25sin(2wt)
Pavg=25W
describes the observations at any point on the line.
Measurements at different points will have a different
phase relationship to each due to the delay in the line.
Note carefully the power function. The power at any point
on the line varies from 0 to 50W with a sinusoidal
pattern at twice the frequency of the voltage sinusoid.
The power is always positive; that is, energy is always
flowing towards the load.

Now let us remove the terminating resistor. The line is
now open circuit.

Again, set the generator to DC. After one round trip
time, the observations at any point on the line will
be:
V(t)=100V
I(t)=0A
P(t)=0W
Pavg=0W

The story for a sinusoid is more interesting...
The observations now depend on the location on the
line where the measurements are performed.
At a current 0:
V(t)=100sin(Wt)
I(t)=0A
P(t)=0W
Pavg=0W
At a voltage 0:
V(t)=0V
I(t)=2sin(wt)
P(t)=0W
Pavg=0W
Halfway between the current 0 at the load and the
preceding voltage 0 (that is, 45 degrees back
from the load:
V(t)=70.7sin(wt)
I(t)=1.414cos(wt)
P(t)=50sin(2wt)
Pavg=0W
This tells us that for the first quarter cycle of the
voltage (V(t)), energy is flowing towards the load,
with a peak of 50W, for the next quarter cycle, energy
flows towards the source (peak –50W), and so on, for
an average of 0, as expected.

For a sinusoid, when the load is other than an open
or a short, energy flows forward for a greater
period than it flows backwards which results in a
net transfer of energy towards the load.

At the location of current or voltage minima on
the line, energy flow is either 0 or flows towards the
load.

All of the above was described without reference to forward
or reflected voltages, currents or power. It is the basic
circuit theory description of what happens between two
networks. It works for DC, 60 HZ, RF, sinusoids, steps,
pulses, you name it. If there is any disagreement with the
above, please do not read further until it is resolved.

Now it is time to introduce forward and reflected waves to
the discussion. Forward an reflected waves are simply a
mathematical transformation of V(t) and I(t) to another
representation which can make solving problems simpler.

The transformation begins with assuming that there is
a forward wave defined by Vf(t)=If(t)*Z0, a reflected
wave defined by Vr(t)=Ir(t)*Z0 and that at any point
on the line V(t)=Vf(t)+Vr(t) and I(t)=If(t)-Ir(t).

Since we are considering an ideal line, Z0 simplifies
to R0.

Simple algebra yields:
Vf(t)=(V(t)+R0*I(t))/2
Vr(t)=(V(t)-R0*I(t))/2
If(t)=Vf(t)/R0
Ir(t)=Vr(t)/R0

Now these algebraically transformed expressions have all
the capabilities of the original so they work for DC,
60 HZ, RF, sinusoids, steps, pulses, you name it.

Some people look at the expression If(t)=Vf(t)/R0 and
think it looks a lot like a travelling wave so they decide
to compute the energy being moved by this wave:
Pf(t)=Vf(t)*Vf(t)/R0
Pr(t)=Vr(t)*Vr(t)/R0

So let us explore the results when applied to the DC examples from
above.

Terminated with 50 ohms:
Vf(t)=50V
If(t)=1A
Vr(t)=0V
Ir(t)=0A

Pf(t)=50W
Pr(t)=0W

This all looks good, no reflected wave and the appropriate
amount of energy is being transported in the forward wave.

Let us try the open ended line:
Vf(t)=50V
If(t)=1A
Vr(t)=50V
Ir(t)=1A

which yield
V(t)=Vf(t)+Vr(t)=100V
I(t)=If(t)-Ir(t)=0A

which is in agreement with the observations. But there is
definitely a forward voltage wave and a reflected voltage
wave. Carrying on to compute powers:
Pf(t)=50W
Pr(t)=50W
and
P(t)=Pf(t)-Pr(t)=0W

So there you have it. The interpretation that the forward
and reflected wave are real and transport energy leads to
the conclusion that on a line charged to a constant DC
voltage, energy is flowing in the forward direction and
energy is flowing in the reflected direction.

Because it is obvious that there is no energy flowing on
the line, this interpretation makes me quite uncomfortable.

Now there are several ways out of this dilemma.

Some say they are only interested in RF (and optics), and
that nothing can be learned from DC, steps, pulses or other
functions. They simply ignore the data and continue with
their beliefs.

Some claim that DC is special, but looking at the equations,
there is no reason to expect the analysis to collapse with
DC.

The best choice is to give up on the interpretation that the
forward and reflected voltage waves necessarily transport
energy. Consider them to be a figment of the analysis technique;
very convenient, but not necessarily representing anything real.

As an added bonus, giving up on this interpretation will remove
the need to search for “where the reflected power goes” and
free the mind for investment in activities that might yield
meaningful results.

....Keith

what happens to reflected energy ?
 

On 13 jun, 23:43, walt wrote:
On Jun 13, 5:08*pm, Wimpie wrote:



On 9 jun, 02:07, Richard Clark wrote:

On Tue, 8 Jun 2010 12:17:37 -0700 (PDT), Wimpie
wrote:

When we go back to my first posting in this thread (the first reply)
where I stated that amplifiers do not show 50 Ohms in general, is not
that strange and doesn't violate Walt's conclusions.

Hi Wimpie,

Time to rewind:

On Thu, 27 May 2010 19:32:40 -0700 (PDT), walt wrote:

The source resistance data reported in Secs 19.8 and 19.9 were
obtained using the load variation method with resistive loads. Note
that of the six measurements of output source resistance reported in
Table 19.1, the average value of the resistance is 50.3 ohms obtained
with the reference load resistance of 51.2 ohms, exhibiting an error
of only 1.8 percent.

Considering these two explicit initial conditions and findings:

1. Using a Kenwood TS-830S transceiver as the RF source, the tuning
and loading of the pi-network are adjusted to deliver all the
available power into a 50 + j0-ohm load with the grid drive adjusted
to deliver the maximum of 100 watts at 4 MHz, thus establishing the
area of the RF power window at the input of the pi-network, resistance
RLP at the plate, and the slope of the load line. The output source
resistance of the amplifier in this condition will later be shown to
be 50 ohms. In this condition the DC plate voltage is 800 v and plate
current is 260 ma. DC input power is therefore 800 v ? 0.26 a = 208 w.
Readings on the Bird 43 wattmeter indicate 100 watts forward and zero
watts reflected. (100 watts is the maximum RF output power available
at this drive level.) From here on the grid drive is left undisturbed,
and the pi-network controls are left undisturbed until Step 10.

2. The amplifier is now powered down and the load resistance RL is
measured across the input terminals of the resonant pi-network tank
circuit (from plate to ground) with an HP-4815 Vector Impedance Meter.

what happens to reflected energy ?
 

On 14 jun, 02:13, Richard Clark wrote:
On Sun, 13 Jun 2010 14:08:59 -0700 (PDT), Wimpie
wrote:

This knowledge
resulted in my statement: "an RF PA isn't a 50 Ohms source", what was/
is heavily disputed by you.

Hi Wim,

Your conclusion:
"Under maximum power output matching given certain drive (not
necessarily being the maximum drive), the output impedance equals the
load resistor (the so-called "conjugated match condition")."

It was not the only conclusion, but it certainly disputes what you say
above.

Given the easy access to real amplifiers available to Hams at their
own bench, and further given that they do not present any more
complexity than your own simulation; then I have to wonder why we have
wandered into strange topologies.

My findings are from practice before I had the opportunity to do PA
simulations.

I am not interested in trying to follow 23 pages of dense presentation
where you could have simulated Walt's finals, validated or rejected
his data, and THEN formed your own conclusions. *If this is a problem
of simulation (you don't have that tube handy) what about topology? *

23 pages is not that much as graphs and circuit diagrams dominate. If
you can provide me a circuit diagram I will figure out whether I can
simulate it or not.


It looks more like a link coupled tank circuit from pre-WWII days. The
Tank Q looks to be non-existent. *When I rummage through the paper I
find 4.4! *This is certainly beyond the standard for PA design - and
we have swapped out of the tube into a mosfet (would you care to stick
to one objective?). *Elsewhere (wandering back in tubes) I find a
value of 10.3 (elsewhere it is 11) which inhabits the lowest margin of
good design. *Then an itinerant report of a cathode tank (????) whose
Q is 0.44. *Q appears to be of little concern.

Fully agree with the low cathode Q, therefore it doesn't affect the
simulations results significantly.


Again, typical Ham equipment shows Q as low as 10 - maybe, and that
would be for a dog; but those that I have seen typically fall around
15, sometimes 20. *As the Z transformation in plate/drain circuits
varies by the square of Q, this is not inconsequential and it once
again has me wondering why the strange topology?

Tell me what Q I should use (evt, give me component values) in the
simulation, and I will change it for you.


No, your paper lacks focus by trying to be all things (tubes, mosfets,
Class AB, Class C). *

This is to support my statement that under real world amateur
conditions many PAs do not obey ZL = Zout*. If I have some time I will
add a push-pull transfomer coupled amplifier also.

You examine the most inconsequential details as
if they were equally important as those details that bear on your
concern. *The paper lacks structure. *Headings are nothing more than
feature descriptions, not argument development. *There is a lot of
discussion of the idiosyncrasies of simulation - I am not interested
and that discussion creates the impression of hidden errors. *Really,
this stuff goes into an appendix not as the object of the narrative.
The graphs are pretty diversions, but they don't add much. *In the
software industry, this is spaghetti design.

I would prefer a conventional topology. *I suppose that has come
through clear. *Pick one significant point, re-edit this into 4 pages
and then you might have something interesting. *This advice comes from
one of our American writers, Mark Twain:
* * * * "If I had more time, I would have written you a shorter letter."

73's
Richard Clark, KB7QHC

simulation is general practice in the design flow, I know it has
limitations. Of course I could use ADS, but most people don't have
that at home, so I decided to use spice as this is used by many.

If you believe I am so wrong, why don't you present some simulations
to show that my original statement is completely wrong?

Best regards,


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

what happens to reflected energy ?
 

On Mon, 14 Jun 2010 06:06:06 -0700 (PDT), Wimpie
wrote:

23 pages is not that much as graphs and circuit diagrams dominate. If
you can provide me a circuit diagram I will figure out whether I can
simulate it or not.

Hi Wim,

I'm glad you wish to examine this more. I will send you Mendenhall's
400W FM design done 42 years ago to this month and nearly day. I will
also send a partial schematic for Walt's TS-830s.

Mendenhall's is not very different from your own model except with a
far more powerful VHF Tube, however, it is a tetrode design too.
Walt's is a pair of pentodes. See what you can do.

Fully agree with the low cathode Q, therefore it doesn't affect the
simulations results significantly.

Why a Common Grid design for a tetrode tube?

Tell me what Q I should use (evt, give me component values) in the
simulation, and I will change it for you.

15 is one that Terman cites, and Mendenhall cites Terman, and Walt's
(if I recall correctly) also find a similar value. Given your
especially low frequency simulation, component values are way out from
those commonly encountered in HF/VHF work.

No, your paper lacks focus by trying to be all things (tubes, mosfets,
Class AB, Class C). *

This is to support my statement that under real world amateur
conditions many PAs do not obey ZL = Zout*. If I have some time I will
add a push-pull transfomer coupled amplifier also.

Then you should do the finals deck for a transistor rig for that
topology. It would bring you into the 21st century (even if the
details still date from the 1970s). You will also find more
simulation options.

simulation is general practice in the design flow, I know it has
limitations. Of course I could use ADS, but most people don't have
that at home, so I decided to use spice as this is used by many.

One of the titans of linear design, Bob Pease of National
Semiconductor had little sympathy for those who discovered the
problems of simulation. However, those who read his work were sure to
avoid pain and trauma. Consult:
http://www.national.com/rap/

Having said that, it is somewhat ironic that we all use simulation for
antennas - so we are all aware of limitation and possibility. I have
used many simulators from a spectrum of disciplines.

If you believe I am so wrong, why don't you present some simulations
to show that my original statement is completely wrong?

You have more time and passion for it. I have already had a career in
testing these issues specifically and proof is at the bench. I was
trained to demanding methods for this and I worked with the most
accurate instruments in the world. However, I don't expect equal
results. If you wish to work further with the two schematics I send
you, then that would interest me far more than the banalities of
photonic explanations and spreadsheet solutions.

I am not saying anything you have done is wrong, I am saying that your
style is getting in the way of communication. It reads like a
detective novel, not a like presentation. You discover conclusions,
you don't seem to test for expectations. Surprise and mystery arrive
haphazardly through the narrative. Inconsequential details are mulled
over like Holmes describing the characteristics of cigar ash to
Watson.

You are trying to do too many things in one place and the topic begins
to blurs and the goal is distracted. For instance, this thread's
subject line of "what happens to reflected energy?" is never
summarized in your final conclusion - that is not good reportage. I
went to that last section and saw that discussion missing. Closer
examination revealed it by parts as seemingly trivial observations.

For style, start with a good grounding. Establish a base line. Do a
Monte Carlo analysis of a known design with known characteristics.
Then introduce a hypothesis tied to one variable. Push the variable
and note how it either conforms or strays from your hypothesis.
Present your work. Have a discussion to shake out problems of
understanding (from anyone). Then move on to either shift the
topology, or further stress the existing design. Build on a solid
foundation.

Schematics will follow as soon as I can make their file sizes suitable
for email servers (about 1-2 MB).

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On 14 jun, 20:00, Richard Clark wrote:
On Mon, 14 Jun 2010 06:06:06 -0700 (PDT), Wimpie
wrote:

23 pages is not that much as graphs and circuit diagrams dominate. *If
you can provide me a circuit diagram I will figure out whether I can
simulate it or not.

Hi Wim,

I'm glad you wish to examine this more. * I will send you Mendenhall's
400W FM design done 42 years ago to this month and nearly day. *I will
also send a partial schematic for Walt's TS-830s.

Mendenhall's is not very different from your own model except with a
far more powerful VHF Tube, however, it is a tetrode design too.
Walt's is a pair of pentodes. *See what you can do.

Fully agree with the low cathode Q, therefore it doesn't affect the
simulations results significantly.

Why a Common Grid design for a tetrode tube?

Because I built such circuit myself based on two PL519 television
tubes and to reduce the number of runs to tune the amplifier in
simulation (I still need some time to work also).
*

Tell me what Q I should use (evt, give me component values) in the
simulation, and I will change it for you.

15 is one that Terman cites, and Mendenhall cites Terman, and Walt's
(if I recall correctly) also find a similar value. *Given your
especially low frequency simulation, component values are way out from
those commonly encountered in HF/VHF work.

Over here 3.6 MHz is an amateur band, we may even go lower than that.
I will change the real class C ciruit to Q = 15. Q = Rload/XL?


No, your paper lacks focus by trying to be all things (tubes, mosfets,
Class AB, Class C).

This is to support my statement that under real world amateur
conditions many PAs do not obey ZL = Zout*. If I have some time I will
add a push-pull transfomer coupled amplifier also.

Then you should do the finals deck for a transistor rig for that
topology. *It would bring you into the 21st century (even if the
details still date from the 1970s). *You will also find more
simulation options.

simulation is general practice in the design flow, I know it has
limitations. Of course I could use ADS, but most people don't have
that at home, so I decided to use spice as this is used by many.

One of the titans of linear design, Bob Pease of National
Semiconductor had little sympathy for those who discovered the
problems of simulation. *However, those who read his work were sure to
avoid pain and trauma. *Consult:http://www.national.com/rap/

Having said that, it is somewhat ironic that we all use simulation for
antennas - so we are all aware of limitation and possibility. *I have
used many simulators from a spectrum of disciplines.

If you believe I am so wrong, why don't you present some simulations
to show that my original statement is completely wrong?

You have more time and passion for it. *

I have very limited time for this, so we have to keep it very
efficient.

I have already had a career in
testing these issues specifically and proof is at the bench. * I was
trained to demanding methods for this and I worked with the most
accurate instruments in the world. *However, I don't expect equal
results. *If you wish to work further with the two schematics I send
you, then that would interest me far more than the banalities of
photonic explanations and spreadsheet solutions.

I am not saying anything you have done is wrong, I am saying that your
style is getting in the way of communication. *It reads like a
detective novel, not a like presentation. *You discover conclusions,
you don't seem to test for expectations. *Surprise and mystery arrive
haphazardly through the narrative. *Inconsequential details are mulled
over like Holmes describing the characteristics of cigar ash to
Watson. *

Except the high Z value for the Common Cathode amplifier, all results
are as expected. Otherwise I did not make my statement on the output
impedance of PAs.


You are trying to do too many things in one place and the topic begins
to blurs and the goal is distracted. *For instance, this thread's
subject line of "what happens to reflected energy?" is never
summarized in your final conclusion - that is not good reportage. *I
went to that last section and saw that discussion missing. *Closer
examination revealed it by parts as seemingly trivial observations.

You disputed my statement on Zout and came up with the "what IS it"
attitude, I only provided information so that other people can get
their own opinion. I prefer to keep the focus on that.


For style, start with a good grounding. *Establish a base line. *Do a
Monte Carlo analysis of a known design with known characteristics.
Then introduce a hypothesis tied to one variable. *Push the variable
and note how it either conforms or strays from your hypothesis.
Present your work. *Have a discussion to shake out problems of
understanding (from anyone). *Then move on to either shift the
topology, or further stress the existing design. *Build on a solid
foundation.

This is not a thesis, I just did some simulation to show that things
can be different.

Schematics will follow as soon as I can make their file sizes suitable
for email servers (about 1-2 MB).

73's
Richard Clark, KB7QHC

Before I am going to spend lots into time this, some points:

1. Do you accept the "measuring" method with injecting a signal with
slight frequency difference as mentioned in the document?

2. Do you accept the way of presentation (envelope change versus time
and converted to output VSWR as used in the document)?

3. Do you accept B^2 spice A/D professional, version for as suitable
platform?

If we cannot agree on this, it is of no use to continue.

I limit the simulation to low HF frequencies as based on a circuit
diagram alone we cannot guess the parasitics for high HF/VHF use and
this may be food for more discussion. Please also mention what
components you want into the simulation as simulating a full TS830
will take to long.

All relevant info will be made available to the group, inclusive
original circuit diagrams and the diagrams I used for simulation. I
hope that I can get a suitable simulation model for the tubes.

Regarding circuit diagrams, can you reduce the color depth (gray scale
or B/W) and convert them to PNG as this is non-lossy coding.

I am looking forward to the results.

Best regards,


Wim
PA3DJS
www.tetech.nl

what happens to reflected energy ?
 

On Jun 14, 12:09*pm, Keith Dysart wrote:
On Jun 10, 5:52 pm, K1TTT wrote:

On Jun 9, 2:04 am, Keith Dysart wrote:

Exactly. And yet, were you to attach a DC-couple directional
wattmeter to the line, it would indicate 50W forward and 50W
reflected (using your corrected numbers).

now remember, this is AFTER you disconnected the load resistor.


only as you monitored the initial transient. *once the transient has
passed there is no power flowing in the line.

I agree completely with the second sentence. But the first is in
error. A directional wattmeter will indeed show 50W forward and 50W
reflected on a line with constant DC voltage and 0 current. I’ll
start by providing support for your second sentence and then get
back to reflected power.


no, it won't... as i will explain below.

Choose an arbitrary point on the line and observe the voltage. The
observed voltage will be a function of time. Let us call it V(t).
At the same point on the line, observe the current and call it I(t).


if you can choose an arbitrary point, i can choose an arbitrary
time... call it a VERY long time after the load is disconnected.

The energy flowing (i.e. power) at this point on the line will
be P(t)=V(t)I(t). If the voltage is measured in volts and the current
in amps, then the power will be in joules/s or watts.
We can also compute the average energy flow (power) over some
interval by integrating P(t) over the interval and dividing by
the length of the interval. Call this Pavg. For repetitive
signals it is convenient to use one period as the interval.

Note that V(t) and I(t) are arbitrary functions. The analysis
applies regardless of whether V(t) is DC, a sinusoid, a step,
a pulse, etc.; any arbitrary function.

To provide some examples, consider the following simple setup:
1) 50 ohm, Thevenin equivalent circuit provides 100 V in to an
* *open circuit.

this is the case we were discussing... the others are irrelevant at
this point


2) The generator is connected to a length of 50 ohm ideal
* *transmission line. We will stick with ideal lines for the
* *moment, because, until ideal lines are understood, there
* *is no hope of understanding more complex models.
3) The ideal line is terminated in a 50 ohm resistor.

Set the generator to DC:
* V(t)=50V, wherever it is measured on the line
* I(t)=1A, wherever it is measured on the line
* P(t)=50W, wherever it is measured on the line
* Pavg=50W
No disagreement, I hope.

only if this is a case WITH a load... this is not what we were
discussing above.



Set the generator to generate a sinusoid:
* V(t)=50sin(wt)
* I(t)=1sin(wt)
* P(t)=25+25sin(2wt)
* Pavg=25W
describes the observations at any point on the line.
Measurements at different points will have a different
phase relationship to each due to the delay in the line.
Note carefully the power function. The power at any point
on the line varies from 0 to 50W with a sinusoidal
pattern at twice the frequency of the voltage sinusoid.
The power is always positive; that is, energy is always
flowing towards the load.

Now let us remove the terminating resistor. The line is
now open circuit.

OK, now we are back to where i had objections above.


Again, set the generator to DC. After one round trip
time, the observations at any point on the line will
be:
* V(t)=100V
* I(t)=0A
* P(t)=0W
* Pavg=0W


ok, p(t)=0w, i=0a, i connect my directional wattmeter at some very
long time after the load is disconnected so the transients can be
ignored and i measure 0w forever. there is no power flowing in the
line.


The story for a sinusoid is more interesting...
The observations now depend on the location on the
line where the measurements are performed.
At a current 0:
* V(t)=100sin(Wt)
* I(t)=0A
* P(t)=0W
* Pavg=0W
At a voltage 0:
* V(t)=0V
* I(t)=2sin(wt)
* P(t)=0W
* Pavg=0W
Halfway between the current 0 at the load and the
preceding voltage 0 (that is, 45 degrees back
from the load:
* V(t)=70.7sin(wt)
* I(t)=1.414cos(wt)
* P(t)=50sin(2wt)
* Pavg=0W
This tells us that for the first quarter cycle of the
voltage (V(t)), energy is flowing towards the load,
with a peak of 50W, for the next quarter cycle, energy
flows towards the source (peak –50W), and so on, for
an average of 0, as expected.

For a sinusoid, when the load is other than an open
or a short, energy flows forward for a greater
period than it flows backwards which results in a
net transfer of energy towards the load.

At the location of current or voltage minima on
the line, energy flow is either 0 or flows towards the
load.

All of the above was described without reference to forward
or reflected voltages, currents or power. It is the basic
circuit theory description of what happens between two
networks. It works for DC, 60 HZ, RF, sinusoids, steps,
pulses, you name it. If there is any disagreement with the
above, please do not read further until it is resolved.

so you have derived voltages and currents at a couple of special
locations where superposition causes maximum or minimum voltages or
currents... this of course ONLY works after sinusoidal steady state
has been achieved, so your cases of connecting and disconnecting the
load must be thrown out. and how do you find a maximum or minimum
when the load is matched to the line impedance? standing waves are
always a trap... step back from the light and come back to the truth
and learn how to properly account for forward and reflected waves with
voltages or currents and all will be revealed.



Now it is time to introduce forward and reflected waves to
the discussion. Forward an reflected waves are simply a
mathematical transformation of V(t) and I(t) to another
representation which can make solving problems simpler.

The transformation begins with assuming that there is
a forward wave defined by Vf(t)=If(t)*Z0, a reflected
wave defined by Vr(t)=Ir(t)*Z0 and that at any point
on the line V(t)=Vf(t)+Vr(t) and I(t)=If(t)-Ir(t).

Since we are considering an ideal line, Z0 simplifies
to R0.

Simple algebra yields:
* Vf(t)=(V(t)+R0*I(t))/2
* Vr(t)=(V(t)-R0*I(t))/2
* If(t)=Vf(t)/R0
* Ir(t)=Vr(t)/R0

Now these algebraically transformed expressions have all
the capabilities of the original so they work for DC,
60 HZ, RF, sinusoids, steps, pulses, you name it.

hmmm, they work for all waveforms you say... even dc?? ok, open
circuit load, z0=50, dc source of 100v with 50ohm thevenin
resistance...

If(t) = 100/50 = 2a
Ir(t) = 100/50 = 2a

so there is a reflected 2a current you say? where does that go when
it gets back to the source? we know that when the 100v source is open
circuited as it must be in this case that it develops 100v across its
terminals... sure doesn't seem like any way that 2a can go back into
the source. it can't be reflected since the source impedance matches
the line impedance. so where does it go? maybe those simple
algebraic expressions have to be reconsidered a bit.

well, maybe i missed something... lets look at the voltages:

Vf(t)=(100+50*2)/2 = 100
Vr(t)=(100-50*2)/2 = 0

hmm, you do seem to get the right constant voltage... but it kind of
messes up your power flow since Vr=0 there can be no power in that
reflected wave, so it really doesn't make it much of a wave.

but wait, the equations must be consistent so

Ir(t)=0/50

So there really is no reflected current... but there is forward
current? so where does that forward current go if there is no load?
it doesn't reflect back if Ir=0... does it just build up at the open
end? going to be lots of trons sitting there after a while.





Some people look at the expression If(t)=Vf(t)/R0 and
think it looks a lot like a travelling wave so they decide
to compute the energy being moved by this wave:
* Pf(t)=Vf(t)*Vf(t)/R0
* Pr(t)=Vr(t)*Vr(t)/R0

So let us explore the results when applied to the DC examples from
above.

Terminated with 50 ohms:
* Vf(t)=50V
* If(t)=1A
* Vr(t)=0V
* Ir(t)=0A

* Pf(t)=50W
* Pr(t)=0W

This all looks good, no reflected wave and the appropriate
amount of energy is being transported in the forward wave.

Let us try the open ended line:
* Vf(t)=50V
* If(t)=1A
* Vr(t)=50V
* Ir(t)=1A

which yield
* V(t)=Vf(t)+Vr(t)=100V
* I(t)=If(t)-Ir(t)=0A

which is in agreement with the observations. But there is
definitely a forward voltage wave and a reflected voltage
wave. Carrying on to compute powers:

sorry, you can't have a 'voltage wave' without a 'current wave'...
just doesn't work i'm afraid. to have one you must have the other.

* Pf(t)=50W
* Pr(t)=50W
and
* P(t)=Pf(t)-Pr(t)=0W

So there you have it. The interpretation that the forward
and reflected wave are real and transport energy leads to
the conclusion that on a line charged to a constant DC
voltage, energy is flowing in the forward direction and
energy is flowing in the reflected direction.

obviously incorrect, by reductio ad absurdum if i remember the latin
correctly.


Because it is obvious that there is no energy flowing on
the line, this interpretation makes me quite uncomfortable.

Now there are several ways out of this dilemma.

Some say they are only interested in RF (and optics), and
that nothing can be learned from DC, steps, pulses or other
functions. They simply ignore the data and continue with
their beliefs.

Some claim that DC is special, but looking at the equations,
there is no reason to expect the analysis to collapse with
DC.

The best choice is to give up on the interpretation that the
forward and reflected voltage waves necessarily transport
energy. Consider them to be a figment of the analysis technique;
very convenient, but not necessarily representing anything real.

As an added bonus, giving up on this interpretation will remove
the need to search for “where the reflected power goes” and
free the mind for investment in activities that might yield
meaningful results.

...Keith

what happens to reflected energy ?
 

On Mon, 14 Jun 2010 12:32:02 -0700 (PDT), Wimpie
wrote:

Before I am going to spend lots into time this, some points:

1. Do you accept the "measuring" method with injecting a signal with
slight frequency difference as mentioned in the document?

Hi Wim,

I got only a hint of what you were trying to do because it was buried
in so much conflicting agendas.

However, my first impression was that it looked worth discussing. My
second impression was that it was not needed. Still, that is worth
discussing too.

2. Do you accept the way of presentation (envelope change versus time
and converted to output VSWR as used in the document)?

I hadn't the slightest idea what that was all about. Like I said, too
many things going on. From your email, it sounds like too much
complexity to accomplish what was done with meter indications.

For others, this might be found in the section called "Output Tuning"
in:
"Fine Tuning FM Final Stages," by Mendenhall (google for a link or
email me for a copy).

3. Do you accept B^2 spice A/D professional, version for as suitable
platform?

I have no problem with what you are comfortable with. I am sure a
suitable simulation will evolve.

If we cannot agree on this, it is of no use to continue.

I limit the simulation to low HF frequencies as based on a circuit
diagram alone we cannot guess the parasitics for high HF/VHF use and
this may be food for more discussion. Please also mention what
components you want into the simulation as simulating a full TS830
will take to long.

All relevant info will be made available to the group, inclusive
original circuit diagrams and the diagrams I used for simulation. I
hope that I can get a suitable simulation model for the tubes.

Regarding circuit diagrams, can you reduce the color depth (gray scale
or B/W) and convert them to PNG as this is non-lossy coding.

I am looking forward to the results.

We will resolve the details offline. I would encourage you to save
effort (we both have things to do) by doing a "quick and dirty" first
pass just to see how little we may have left to do (it could be that
simple). There is more to read than there is to simulate.

For others, the material I have linked Wim to probably tips the
download scale at 50MB. 44 MB is one document alone that others here
might find useful:
"EIMAC Care and Feeding of Power Grid Tubes" 44MB at:
http://www.g8wrb.org/data/Eimac/care...grid_tubes.pdf

It will brush away the cobwebs.

73's
Richard Clark, KB7QHC