On Jun 6, 3:21*pm, Wimpie wrote:

Measuring method used: change in resistive load, from voltage change
you can calculate the current change, hence the output impedance. One
note, except two, all where solid state.

Of course, since the source impedance may be anywhere in the complex
plane, you need to change more than just the resistive part of the
load, I believe, to get an accurate picture...

All high efficiency designs (class E, D) that I did have output
impedance far from the expected load impedance. With "far" I mean
factor 2 or factor 0.5. I did not measure that (as it is not
important in virtually all cases), but know it from the overload
simulation/measurement and I did the design myself.

What I've seen in similar situations is that the source impedance is
likely to be strongly reactive, depending on the filter network you
use to get sinusoidal output. In any event, the source impedance is
likely to have a reflection coefficient magnitude that is quite close
to unity. That is exactly what you should expect: there's nothing to
absorb reflections. You could (theoretically at least) use feedback
to make the output look like 50 ohms, but just as you say, Wim ...
why?? There's really almost never any point in doing so.

....


Cheers,
Tom

On Jun 6, 10:06*pm, K7ITM wrote:
On Jun 6, 3:21*pm, Wimpie wrote:

Measuring method used: change in resistive load, from voltage change
you can calculate the current change, hence the output impedance. One
note, except two, all where solid state.

Of course, since the source impedance may be anywhere in the complex
plane, you need to change more than just the resistive part of the
load, I believe, to get an accurate picture...



All high efficiency designs (class E, D) that I did have output
impedance far from the expected load impedance. With "far" I mean
factor 2 or factor 0.5. I did not measure that (as it is not
important in virtually all cases), but know it from the overload
simulation/measurement and I did the design myself.

What I've seen in similar situations is that the source impedance is
likely to be strongly reactive, depending on the filter network you
use to get sinusoidal output. *In any event, the source impedance is
likely to have a reflection coefficient magnitude that is quite close
to unity. *That is exactly what you should expect: *there's nothing to
absorb reflections. *You could (theoretically at least) use feedback
to make the output look like 50 ohms, but just as you say, Wim ...
why?? *There's really almost never any point in doing so.

...

Cheers,
Tom

Again Wim, we're not on the same page, so let me quote from your
eariler post:

"Virtually all high efficient amplifiers work in voltage saturated
mode
and are therefore not operated at maximum available power, therefore
their output impedance doens't match the expected load."

I have not been talking about MAXIMUM available power, only the power
available with some reasonable value of grid drive. In the real world
of amateur radio operations, of which I'm talking, when we adjust the
pi-network for that given drive level, we adjust for delivering all
the AVAILABLE power at that drive level. When all the available power
is thus delivered, the output source resistance equals the load
resistance by definition. We're now not talking about changing the
load, phase or SWR--we're talking about the single condition arrived
at after the loading adjustments have been made.

And Tom, why would the source be reactive when the pi-network is tuned
to resonance? And because the source resistance of the (tube) power
amp is non-dissipative, its reflection coefficient is 1.0 by
definition, and so it cannot absorb any reflected energy, and
therefore re-reflects it.

Walt, W2DU

On 7 jun, 04:29, walt wrote:
On Jun 6, 10:06*pm, K7ITM wrote:



On Jun 6, 3:21*pm, Wimpie wrote:

Measuring method used: change in resistive load, from voltage change
you can calculate the current change, hence the output impedance. One
note, except two, all where solid state.

Of course, since the source impedance may be anywhere in the complex
plane, you need to change more than just the resistive part of the
load, I believe, to get an accurate picture...

All high efficiency designs (class E, D) that I did have output
impedance far from the expected load impedance. With "far" I mean
factor 2 or factor 0.5. I did not measure that (as it is not
important in virtually all cases), but know it from the overload
simulation/measurement and I did the design myself.

What I've seen in similar situations is that the source impedance is
likely to be strongly reactive, depending on the filter network you
use to get sinusoidal output. *In any event, the source impedance is
likely to have a reflection coefficient magnitude that is quite close
to unity. *That is exactly what you should expect: *there's nothing to
absorb reflections. *You could (theoretically at least) use feedback
to make the output look like 50 ohms, but just as you say, Wim ...
why?? *There's really almost never any point in doing so.

...

Cheers,
Tom

Again Wim, we're not on the same page, so let me quote from your
eariler post:

"Virtually all high efficient amplifiers work in voltage saturated
mode
and are therefore not operated at maximum available power, therefore
their output impedance doens't match the expected load."

I have not been talking about MAXIMUM available power, only the power
available with some reasonable value of grid drive. In the real world
of amateur radio operations, of which I'm talking, when we adjust the
pi-network for that given drive level, we adjust for delivering all
the AVAILABLE power at that drive level. When all the available power
is thus delivered, the output source resistance equals the load
resistance by definition. We're now not talking about changing the
load, phase or SWR--we're talking about the single condition arrived
at after the loading adjustments have been made.

Walt, this restriction was not given in the original posting, it is
added by you and you exclude many other practical amplifier solutions
(like push pull 3...30 MHz no-tune amplifiers, fixed tuned single band
amplifiers, emerging high efficiency designs for constant envelope
modulation schemes and/or AM with supply voltage modulation).

I fully agree with your statement on optimum matching given certain
drive and output impedance in case of (pi-filter) matching, no doubt
about this. However there are more flavors as I tried to explain.

Even during SSB modulation with an optimally tuned amplifier without
power supply modulation, the amplifier is most of the time operated
into current saturation mode (instead of optimally tuned output, given
a certain drive). For tetrode/pentode, assuming no voltage saturation,
the RF plate impedance is seldom equal to the conjugated load
impedance (load impedance will be lower). I think for a triode in
common grid operation, this will apply also because of the cathode is
not grounded for AC, hence plate impedance increases. I have to be
careful now to avoid that I have to do a lot of work to fulfill
Richard's requests. I dismantled my PL519 common grid amplifier (22
years ago), so a cannot measure it anymore.

All amateurs that do not have a tuner inside their PA, have to live
with non-optimum VSWR (hence amplifier not operating at optimal
tuning, even under CW) or have to insert a tuner. For the latter case,
the problem now reduces to a cable and tuner loss problem, as after
successful tuning, there will be now power reflected to the PA, hence
the PA's output impedance doesn't matter.

Maybe the JC could provide some background behind his question.


And Tom, why would the source be reactive when the pi-network is tuned
to resonance? And because the source resistance of the (tube) power
amp is non-dissipative, its reflection coefficient is 1.0 by
definition, and so it cannot absorb any reflected energy, and
therefore re-reflects it.

Walt, W2DU

Best regards,

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

On Jun 6, 7:29*pm, walt wrote:
....
And Tom, why would the source be reactive when the pi-network is tuned
to resonance? And because the source resistance of the (tube) power
amp is non-dissipative, its reflection coefficient is 1.0 by
definition, and so it cannot absorb any reflected energy, and
therefore re-reflects it.

Walt, W2DU

If you allow that the thing driving the pi network has an effective
source impedance different from the load that the pi network presents
to it, then clearly the output impedance seen at the other end (the
"50 ohm" end) of the pi network won't be 50 ohms. Try it with some
numbers; for example, assume a pi network that transforms your 50 ohm
load to a 4k ohm load to the amplifier output, and assume an amplifier
output stage that looks like a 20k ohm source. Design the pi network
for a loaded Q of 10. I believe you'll find that the source impedance
seen by the 50 ohm load is about 11+j18 ohms.

As Wim has pointed out, requiring an amplifier to be loaded and driven
in a very particular way unnecessarily dismisses some very important
classes of amplifier. What do you do, for example, with a linear
amplifier? What do you do with an amplifier that drives a voltage
very hard (and for which a simple pi network is inappropriate for
matching to a load)? Perhaps an even more basic question is: why
exactly do we tune a pi network to present a particular load to an RF
amplifier stage? Why should we operate a 6146 with, say, a 3000 ohm
plate load? Why not 1000, or 6000?

And what if I set up a tube and pi network for operation such that the
apparent output source impedance is 50 ohms (while driving a 50 ohm
load), and then I add feedback to the amplifier in such a way that the
operating conditions are not changed, but the impedance looking back
into the plate is changed?

How did we get to the source resistance of "the (tube) power amp"
being non-dissipative? I know there are some of us who don't buy into
that...

Cheers,
Tom

On Mon, 7 Jun 2010 09:19:59 -0700 (PDT), K7ITM wrote:

As Wim has pointed out, requiring an amplifier to be loaded and driven
in a very particular way unnecessarily dismisses some very important
classes of amplifier.

This is the standard rebuff with "the rest of the world works
differently" distractions to Walt having stated a premise, described
initial conditions, taken measurements, and having shown the data
supports his hypothesis. The hypothesis is dismissed through shifting
initial conditions to suit an avoidance of committing an honest answer
to Walt's specific and very explicit observation.

Walt, if this community were pressed for an "up or down" vote:

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

then you would be out in the weeds with your only supporting vote from
me (and maybe from others who would do this by email).

I doubt this will set off a stampede to the ballot box, but what few
votes are stuffed in, I bet they will have the "up or down" stapled to
a dissertation of "however...."

To this last, if I sinned in that regard, I used only Walt's data, his
equipment references, and his citation sources. As no one else seems
to tread that narrow path, much less commit beyond grandiose
statements, I don't feet too bad.

73's
Richard Clark, KB7QHC

On 7 jun, 19:46, Richard Clark wrote:
On Mon, 7 Jun 2010 09:19:59 -0700 (PDT), K7ITM wrote:
As Wim has pointed out, requiring an amplifier to be loaded and driven
in a very particular way unnecessarily dismisses some very important
classes of amplifier. *

This is the standard rebuff with "the rest of the world works
differently" distractions to Walt having stated a premise, described
initial conditions, taken measurements, and having shown the data
supports his hypothesis. *The hypothesis is dismissed through shifting
initial conditions to suit an avoidance of committing an honest answer
to Walt's specific and very explicit observation.

Walt, if this community were pressed for an "up or down" vote:

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

then you would be out in the weeds with your only supporting vote from
me (and maybe from others who would do this by email). *

I doubt this will set off a stampede to the ballot box, but what few
votes are stuffed in, I bet they will have the "up or down" stapled to
a dissertation of "however...."

To this last, if I sinned in that regard, I used only Walt's data, his
equipment references, and his citation sources. *As no one else seems
to tread that narrow path, much less commit beyond grandiose
statements, I don't feet too bad.

73's
Richard Clark, KB7QHC

Hello Richard,

I think, most people that are willing to read will support Walt's
statements on output impedance of an amplifier under matched output
power (given a certain drive condition). However there are many
practical circumstances where Walt's conditions are not met.

As soon as you change the drive (for example during an AM or SSB
transmission), your matching is no longer guaranteed, especially when
using circuits with current behavior (tetrode/pentode, common grid,
etc). See my EL34 posting. Note that under practical circumstances,
this is no problem.

As soon as you change the load (without changing the matching), you
may run into current or voltage saturation, of course depending on
phase of VSWR. As you know many people use non-tune solid-state
amplifiers, so they don't have the possibility to tune/match for
maximum output given certain drive. That means, live with it, or use
an external tuner where you don't tune for maximum power, but for
minimum VSWR presented to the PA.

Amplifiers for constant envelope modulation use saturation to increase
efficiency (and accept the loss in gain). These are deliberately used
under mismatch, therefore the gain is less with respect to a non-
saturating approach. You can also see this from the Pout versus Pin
curve for FM transistors. I know that severak CB owners retuned (or
even modified) the output stage to get 1 dB more power, but they did
forget that the final transistor had dissipate 2dB more. Result: some
japanese transistors became very popular (2SC1306, 1307, 1969, etc).

High efficiency circuits are the extreme case and are entering the
amateur world. Active devices behave like switches, output impedance
can have every value as long as it is close to the edge of the passive
Smith Chart. Amplitude modulation with these topologies can only be
done via supply voltage modulation. Tuning for maximum power with an
external tuner will surely destroy the amplifier if no protection is
present.

For me it was a surprise how this thread developed.

Best regards,


Wim
PA3DJS
www.tetech.nl

On Jun 7, 2:09*pm, Wimpie wrote:
As soon as you change the drive (for example during an AM or SSB
transmission), your matching is no longer guaranteed, ...

AM should be the easiest mode to analyze since it requires linear
finals.
--
73, Cecil, w5dxp.com

I haven't followed this thread since it's been beaten to death so many
times here before. But there's an interesting fact that might have been
overlooked, and might (or might not) be relevant:

If you put a directional coupler such as a Bruene circuit at the output
of a transmitter, and use its "forward" output to control the
transmitter power to keep the "forward" directional coupler output
constant, you'll find that the power output vs load resistance
characteristic is exactly the same as if the transmitter had 50 ohm
output impedance. This is assuming that the directional coupler is
designed for a 50 ohm system, and that the load is purely resistive. It
also assumes that any load is left in place long enough for the feedback
circuit to stabilize. The effective source impedance to a very rapidly
varying load (that is, one changing so fast that the ALC feedback system
doesn't have time to respond) would be the open-loop output impedance
which could be quite different. I haven't taken the time to analyze how
it behaves with a complex load.

I stumbled across this some time ago when designing a rig using this ALC
method and found it interesting. I believe many if not most modern
solid-state transmitters use this ALC method.

Roy Lewallen, W7EL

On 7 jun, 22:36, Roy Lewallen wrote:
I haven't followed this thread since it's been beaten to death so many
times here before. But there's an interesting fact that might have been
overlooked, and might (or might not) be relevant:

If you put a directional coupler such as a Bruene circuit at the output
of a transmitter, and use its "forward" output to control the
transmitter power to keep the "forward" directional coupler output
constant, you'll find that the power output vs load resistance
characteristic is exactly the same as if the transmitter had 50 ohm
output impedance. This is assuming that the directional coupler is
designed for a 50 ohm system, and that the load is purely resistive. It
also assumes that any load is left in place long enough for the feedback
circuit to stabilize. The effective source impedance to a very rapidly
varying load (that is, one changing so fast that the ALC feedback system
doesn't have time to respond) would be the open-loop output impedance
which could be quite different. I haven't taken the time to analyze how
it behaves with a complex load.

I stumbled across this some time ago when designing a rig using this ALC
method and found it interesting. I believe many if not most modern
solid-state transmitters use this ALC method.

Roy Lewallen, W7EL

Hello Roy,

If you have sufficient headroom and under the conditions you
mentioned, you mimic a 50 Ohms source. I think it also works for any
(complex) loads (I couldn’t find why not).

The difference between a real 50 Ohms circuit may be that the phase of
the belonging EMF may change, in many amplifiers phase shift is
somewhat excitation dependent, but who bothers?

I expect such scheme in combination with VSWR measurement also, as
several PAs have a reverse power indicator present (the other half of
a the Bruene circuit). You need the reverse power indication to avoid
destroying active devices and/or intermodulation distortion due to
voltage saturation.

Imagine that you have full reflection |RC| = 1 and it appears at your
active device as RC=-1. You want to maintain the original forward
power. Your active device has to deliver in that case double the
current at zero collector/plate voltage to maintain same forward power
as under matched condition. The actual power delivered to the load is
zero (as the active device supplies current, but no voltage, RC=-1
means a short circuit). This will result in massive dissipation in the
active device.

In case of RC=+1, it has to provide double the voltage with no
current. In other circumstances you will have a significant phase
shift between current and voltage resulting also in increased device
dissipation and inconvenient combinations of instantaneous voltage and
current. So above some value for VSWR, you may have to reduce the
forward power

I had a discussion recently about the power control scheme for TETRA
terminals, but we couldn't find the answer to what is happening under
high VSWR (so we have measure it). It only states VSWR2.

Best regards,


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

On Mon, 07 Jun 2010 13:36:41 -0700, Roy Lewallen
wrote:

I haven't followed this thread since it's been beaten to death so many
times here before. But there's an interesting fact that might have been
overlooked, and might (or might not) be relevant:

If you put a directional coupler such as a Bruene circuit at the output
of a transmitter, and use its "forward" output to control the
transmitter power to keep the "forward" directional coupler output
constant, you'll find that the power output vs load resistance
characteristic is exactly the same as if the transmitter had 50 ohm
output impedance. This is assuming that the directional coupler is
designed for a 50 ohm system, and that the load is purely resistive. It
also assumes that any load is left in place long enough for the feedback
circuit to stabilize. The effective source impedance to a very rapidly
varying load (that is, one changing so fast that the ALC feedback system
doesn't have time to respond) would be the open-loop output impedance
which could be quite different. I haven't taken the time to analyze how
it behaves with a complex load.

I stumbled across this some time ago when designing a rig using this ALC
method and found it interesting. I believe many if not most modern
solid-state transmitters use this ALC method.

Roy Lewallen, W7EL

Hi Roy,

This application that you describe was written up in exactly the same
terms within the recent HP Journals I have posted extracts here. HP
used Directional Couplers (the Bruene circuit, also called a bridge,
qualifies too but uses a non-wave design) to separate out the forward
from the reverse power reflected from the mismatch to create a
reference power. Later, HP and others strapped the signals back into
the source in much the manner you describe.

The rudimentary version can be found in HP Journal v.6 n.1-2. HP
Journal v.12 n.4 strengthens the concept with hard copy sweeps of the
reflection coefficient of a load. By HP Journal v.16 n.6, we have the
description of automatic level control. For the 45 years beyond that
last article, more refinements.

73's
Richard Clark, KB7QHC