what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

On Jun 7, 5:00*pm, Cecil Moore wrote:
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

This is what YOU said:

What happens to the 50 joules/second of reflected energy depends upon
the phasing between the source wave and the reflected wave at the
source impedance. What most amateurs don't understand is there are
two mechanisms that can redistribute reflected energy back toward the
antenna. Those mechanisms are a re-reflection based on the physical
reflection coefficient (what RF engineers understand) and wave
interaction resulting in constructive/destructive interference (what
most RF engineers don't seem to understand) because, unlike optical
physicists, have not been forced to follow the energy flow.

There is no interaction between the waves, they may be superimposed,
and maybe their photons are indistinguishable, but there is no
'interaction'. interaction implies that one wave affects the other,
energy is transferred or fields are affected, such things occur in non-
linear media, but not in linear ones.

what happens to reflected energy ?
 

On Jun 7, 12:19*pm, K7ITM wrote:
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

Hello Tom,

Well, the reason I chose to discuss only the tube amp with a pi-
network filter and impedance transformer is that that arrangement is
the only one I've measured, and that I'm not sufficiently acquainted
with other arrangements to discuss them.

As to your question, why not use plate loads of 3000, 6000 or 1000?
Because the transceivers I measured using two 6146s in parallel are
Kenwood TS-830S and Heathkit HW-100. I don't remember the exact plate
load with the HW-100, close to 1400 ohms, while the TS-830S was 1400
ohms. I had no control over those plate loads because the plate and
grid voltages were preset, and 1400 ohms is the RL I measured at the
input of the pi-network when the grid drive and the network were
adjusted to deliver precisely 100w to the load 50-ohm load.

Now concerning the non-dissipative source resistance of the tube-type
power amp. There are two separate resistances in the amp, the cathode-
to-plate resistance, Rpd, that accounts for all the dissipation in the
tube; and the output-source resistance that is non-dissipative. It is
a common myth that an RF power amp cannot have an efficiency greater
than 50% when conjugate matched to the load, because half the RF power
is dissipated in the source resistance. This is not true, because when
the amp is operating properly, resistance Rpd is less than the output
source resistance, thus allowing more power delivered to the load than
that dissipated in the plate-to-cathode resistance. The output source
resistance is derived from the voltage-current ratio E/I that appears
at the output terminals of the pi-network. A ratio, as such, cannot
dissipate energy, but the load it feeds does.

Using an example from Terman's Radio Engineer's Handbook I explain
this phenomenon in great detail in Chapter 19 in Reflections, and
further in Chapter 19A, an addition to Chapter 19. Both are available
on my web page at www.w2du.com. Chapter 19 appears in 'Read Chapters
from Reflections 2', and 19A appears in 'Preview Chapter from
Reflections 3'. The entire Chapter 19 appears in Reflections 3, which
is now available from CQ. I invite you to review these Chapters.

Walt, W2DU

what happens to reflected energy ?
 

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.

what happens to reflected energy ?
 

Wimpie wrote:
. . .
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.
. . .

Yes, I was referring to the operating region at which normal power is
maintained. As you pointed out, rigs employing this method will only
maintain the same "forward" power over some range of SWR beyond which
they'll begin reducing it, otherwise they'd self-destruct in the
attempt. In the reduced power region, the apparent 50 ohm output
impedance of course no longer holds.

Roy Lewallen, W7EL

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

On Mon, 7 Jun 2010 12:09:23 -0700 (PDT), Wimpie
wrote:

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.

So, that is one vote for (with bulky provisos), and one vote against
(with bulky provisos) going into the ballot box.

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

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

It seems writing the phrase "Conjugate Match is shown by Walt's data"
is a fearsome (loathsome) step to take, or can be taken with
qualifiers and other soporifics to deaden the pain.

Walt's data shows evidence of a Conjugate Match.

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 a current NARDA specification:

GENERAL COUPLER OPERATION
A coaxial directional coupler has the general appearance
of a section of coaxial line, with the addition of a second
parallel section of line and with one end terminated (see
Figure 9). These two sections are known as the main and
auxiliary lines. The two lines are internally separated from
each other; the amount of spacing between lines determines
the amount of RF energy that may be transferred
from the main line to the auxiliary line. In operation, assume
that energy is fed into port A of the main line. Most of
this energy will appear at output port B of the main line.
However, a fraction of this energy (determined by coupling
value) will also appear at the coupled port C, of the
auxiliary line.
A dual-directional coaxial coupler, such as the
reflectometer coupler, consists essentially of two single-
ended couplers connected back-to-back. Perhaps
the most important characteristic of the directional coupler
(and the one from which its name originates) is its
directivity.
....
For reflectometry applications,
the dual directional coupler, incorporating two
auxiliary outputs, permits the simultaneous sampling of incident
and reflected power.
....
RF power applied to
the load is reflected to some degree depending on load
characteristics, thereby resulting in a voltage standing
wave ratio (VSWR) which is reflected back to the main line
output port. this reflected power is coupled out of the reflected
output port at a level 10 dB down from the reflected
power level at the load.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On Mon, 07 Jun 2010 15:49:04 -0700, Richard Clark
wrote:

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.

HP Journal Nov. 1970 is dedicated to all system elements going into
one box, 8620A, 8632A (with options for an external Directional
Coupler and Power Meters).

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On Jun 7, 4:21*pm, K1TTT wrote:
There is no interaction between the waves, they may be superimposed,
and maybe their photons are indistinguishable, but there is no
'interaction'. *interaction implies that one wave affects the other,
energy is transferred or fields are affected, such things occur in non-
linear media, but not in linear ones.

Note that I was not talking about forward and reflected waves in a
constant Z0 transmission line. I am talking about the four wavefront
components that are generated at an impedance discontinuity. It has
been proven in experimentally that those waves indeed interact. In
fact, the transfer of destructive interference energy from wave
cancellation to the areas that permit constructive interference is
obviously interaction since the canceled waves disappear in their
original direction of travel. But, as the FSU web page says, they are
not annihilated - their energy components simply change direction.

How can you possibly argue that wave cancellation doesn't require wave
interaction? Those two waves completely disappear in the direction of
destructive interference. Dr. Best argued that those two waves don't
interact and continue propagating (completely devoid of energy)
forever in the direction of original travel. I asked him to prove that
his phantom waves exist but he could not. Is there such a proof
available?

At a 1/4WL thin-film coating on non-reflective glass, when the
internal reflected wave arrives with equal magnitude and 180 degrees
out of phase with the external reflection, wave cancellation occurs.
That is an *obvious* effect that one wave has on the other. Wave
cancellation is an obvious interaction. In the s-parameter equation:

b1 = s11*a1 + s12*a2 = 0

the s11*a1 wavefront has obviously interacted with the s12*a2
wavefront to accomplish wave cancellation.
--
73, Cecil, w5dxp.com