what happens to reflected energy ?
 

On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart
wrote:

The 100W forward and 50W reflected have no relation to actual powers

From HP Journal V.3n7-8

DIRECTIONAL couplers have been used
widely in wave guide applications for
such purposes as monitoring power, meas
uring reflections, mixing, and for isolation
of signal sources. All of these applications
make use of the property that power flowing
in one direction in the main branch of the
coupler induces a power flow in only one
direction in the auxiliary circuit.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart
wrote:

The 100W forward and 50W reflected have no relation to actual powers

From HP Journal V.6n.1-2:

Power from the source flows down
the main arms of the two couplers
(Fig. 2) and impinges on the load.
The power split off by the 20 db
forward coupler is all passed to the
forward detector, since the directiv
ity characteristic of the multi-hole
directional couplers prevents any
but a negligible amount of the splitoff
power from turning back and
being absorbed in the coupler's in
ternal termination. The power split
off the incident wave by the 10 db
reverse coupler, however, is essen
tially all absorbed in that coupler's
internal termination because of the
reversed direction of connection of
that coupler (Fig. 2).
If the magnitude of the reflection
coefficient ~.T. of the load is, say, 0.1,
Çf of the incident voltage will be
reflected back toward the source. As
this reflection passes back through
the main arm of the reverse coupler,
a 10 db split occurs and is applied
to the reverse detector. The remain
der of the reflection will proceed
back toward the generator where it
will be absorbed in the generator
impedance and in the termination
in the forward coupler.

********

Of course, in regard to this last sentence, HP engineers didn't know
jack-**** about power reflections, especially what could be absorbed
by the source.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

Hello Walt,

I did some simulations with EL34.

Conditions:
Common Grid circuit, Valve connected as triode, 3.6 MHz.
Vsupply, 600V, Vgrid -58V, bias current 40mA (yes I know that it will
run very hot).

Maximum available input power: 9.2W out of 600 Ohms, sinusoidal.

Anode loaded with 20uH, 96pF, 3800 Ohms (parallel circuit), QL= 8.4,
so the anode voltage has sinusoidal shape.

Without changing the drive (9.2W), 3800 Ohms with 96pF gives highest
output power of 64W (Vanode = 698Vp). So power gain is 7.0.

Output impedance:
Change to 4100 Ohms load gives Zout = 1826 (voltage saturation),
Change to 3500 Ohms load gives Zout = 4350 (current saturation),
So for small change in Zload, Zout will be close the load resistor
that gives maximum power. This supports that under well-matched
conditions, given a certain drive level, Zout = ZLoad.

Input power reduced with 3 dB, output becomes 32.6W (Vanode = 497.5Vp,
gain = 7.08). There is a slight increase in gain showing that at 64W
the amplifier is not into deep voltage saturation.

Output impedance at 32.6W output:
Change from 3800 to 3500 Ohms gives Zout = 8100 Ohms.
This shows that reducing the drive level (without changing the
matching) results in Zout not equal to Zload for this triode common
grid circuit.

Also shown that small changes in load (reduction or increase seen at
the anode) results in changes in output impedance. That is why I used
three resistance values for determining Zout at well-matched output.
I used low frequency to avoid endless matching because of parasitic
capacitances and I also skipped the output matching (I have to work
sometimes, so time is limited).

It is not strange to see an RF swing above supply voltage as this is a
common grid circuit with cathode grounded for DC.

If you like, I could do same exercise for modern amplifier topologies,
but be prepared to see unexpected results (As Tom, K7ITM already
noted).

Best regards,


Wim
PA3DJS
www.tetech.nl
remove abc first in case of PM.

what happens to reflected energy ?
 

On Jun 6, 1:01*pm, K1TTT wrote:
as in the other thread, what is the mechanism of that 'interaction'
between waves? *i contend there can be no 'interaction' between
forward and reflected waves if the device is linear.

David, you are preaching to the choir. I explained before that there
is NO interaction between forward and reflected waves because they are
traveling in different directions. The only time that coherent,
collimated waves can interact is when they are traveling in the same
direction in a transmission line.

Keith's argument requires that forward waves and reflected waves
interact. I say they cannot interact in a constant Z0 environment. You
say they cannot interact. We are on the same side of this argument.

The mechanism of the interaction of two coherent, collimated waves
traveling in the same direction is that the superposition process is
irreversible. The source photons and the reflected photons are
indistinguishable.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

On Jun 7, 1:37*am, Richard Clark wrote:
All of these applications
make use of the property that power flowing
in one direction in the main branch of the
coupler induces a power flow in only one
direction in the auxiliary circuit.

Power flow??? Heaven forbid.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

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