Transmitter Output Impedance
 

On 3 mayo, 02:09, Cecil Moore wrote:
On May 2, 5:23*pm, Wimpie wrote:

I am very sorry Cecil, but I still don't see the point where the
discussed method may go wrong.

Everyone seems to be charging ahead, willy-nilly, without seeing the
point which is that there are other effects present besides
reflections.

Therefore carrying out a single-port measurement with a slightly off-
carrier frequency (to create non-coherence) under required output
conditions, will result in a meaningful output impedance.

Nope, it won't because virtual impedances don't cause reflections.
Only physical impedance discontinuities cause reflections. The rest of
the redistribution of RF energy is caused by the superposed
interaction between forward and reflected waves, i.e. interference
effects. Most hams do not understand the role of interference in the
redistribution of RF energy. Hope this helps.

The signal injection is just a way of emulating a non 50 Ohm
termination where you need to change load and or cable length. By
using slightly off-carrier frequency, you emulate a changing phase of
reflection coefficient. That emulated reflected signal goes to the PA
and interferes with the forward signal (produced by the PA). That
interference is required as this modifies current and voltage at the
PA's active device.


http://micro.magnet.fsu.edu/primer/j...interference/w...

Please pay close attention to the last paragraph. "... when two waves
of equal amplitude and wavelength that are 180-degrees ... out of
phase with each other meet, they are not actually annihilated, ... All
of the photon energy present in these waves must somehow be recovered
or redistributed in a new direction, according to the law of energy
conservation ... Instead, upon meeting, the photons are redistributed
to regions that permit constructive interference, so the effect should
be considered as a redistribution of light waves and photon energy
rather than the spontaneous construction or destruction of light."

To be honest, I, and my ham friends, never had to use photon theory to
solve problems in both amateur and professional RF-Engineering. So I
think this doesn’t contribute to the discussion.

You guys are presuming that reflections are the only thing you are
seeing and that is just not true. You are also seeing interference
effects without realizing it so your conclusions are doomed to failure
unless you can differentiate between constructive/destructive
interference and reflected waves. Since there has been no mention of
interference effects, I am forced to conclude that you guys are
ignorant of such effects.

Worth to try yourself:

1. Set up a simple linear circuit in simulation (for example AC
voltage source some impedance in series). You can use whatever (free)
program you like. Use a program that allows transient (non-linear)
simulation as you are working with two frequencies at the same time.
I used Beige Bag spice A/D professional version 4 (not free)

2. Determine the output impedance via load pulling (for various loads
if you like)

3. Do the same with the "VNA" setup (with the PA producing RF output).
You need to observe the envelope variations or you have to insert a
narrow band filter to suppress the carrier. I prefer observing the
sinusoidally changing envelope (as this shows distortion
immediately (intermod products) and saves me the settling time for the
filters).

In simulation you have ideal current meters, so if you use voltage and
current directly (and drop the directional couplers) you can skip the
conversion from reflection coefficient to impedance.

Here you will see that the output impedance is independent of VSWR
(as we started with a linear circuit) and both methods give same
results.

If you like, you could now implement a real PA in simulation and do
the simulation again. Note that tuning the amplifier in simulation is
(very) time consuming. You will notice that, depending on the
amplifier you implemented, impedance will change with increasing VSWR
presented to the amplifier (valid for both methods).

I did all the above with various circuits to prove the fitness of the
method. For some circuits there was a difference in result. After
evaluation, that difference was caused by changes in (bias) supply
during "manual" load pulling (that initially did not happen during the
slightly off-carrier signal injection). With stiff bias conditions,
the difference disappeared to within the accuracy limits.

--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK


With kind regards,


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

Transmitter Output Impedance
 

On May 3, 9:01*am, Wimpie wrote:
To be honest, I, and my ham friends, never had to use photon theory to
solve problems in both amateur and professional RF-Engineering. So I
think this doesn’t contribute to the discussion.

Knowing that EM RF waves must necessarily obey the laws of photon
physics can save one from all sorts of technical blunders. But feel
free to blunder on.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Transmitter Output Impedance
 

On 5/3/2011 2:55 PM, Cecil Moore wrote:

73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

So, a heart attack is better than no heart at all?

Transmitter Output Impedance
 

On 3 mayo, 21:55, Cecil Moore wrote:
On May 3, 9:01*am, Wimpie wrote:

To be honest, I, and my ham friends, never had to use photon theory to
solve problems in both amateur and professional RF-Engineering. So I
think this doesn’t contribute to the discussion.

Knowing that EM RF waves must necessarily obey the laws of photon
physics can save one from all sorts of technical blunders. But feel
free to blunder on.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Hello Cecil,

Maybe you should change your attitude from "it is wrong" to "it may be
true". This might open your mind to study other insights.

When measuring an RF current in a PA, everybody ignores the
quantisation of charge. Is this lack of knowledge?

The frequency of our waves is sufficiently low to ignore any
quantisation effects due to photons. Is this lack of knowledge? In my
opinion a good Engineer knows what he can ignore and what not to
efficiently solve a certain task.

Starting a discussion about photons, in my opinion, is a way to reduce
the S/N ratio of the discussion.

With kind regards,


Wim
PA3DJS
www.tetech.nl.

Transmitter Output Impedance
 

On May 3, 4:15*pm, Wimpie wrote:
Maybe you should change your attitude from "it is wrong" to "it may be
true". This might open your mind to study other insights.

When an s11 measurement is wrong by an infinite percentage, I cannot
see how it could possibly be "true" - sorry.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Transmitter Output Impedance
 

On May 3, 3:58*pm, John KD5YI wrote:
On 5/3/2011 2:55 PM, Cecil Moore wrote:
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

So, a heart attack is better than no heart at all?

Don Hubbard was my best friend and that was one of his pet sayings. I
assume he meant it was better to have bad breath than to be an SK,
which he is.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Transmitter Output Impedance
 

On May 3, 2:01*pm, Wimpie wrote:
On 3 mayo, 02:09, Cecil Moore wrote:



On May 2, 5:23*pm, Wimpie wrote:

I am very sorry Cecil, but I still don't see the point where the
discussed method may go wrong.

Everyone seems to be charging ahead, willy-nilly, without seeing the
point which is that there are other effects present besides
reflections.

I think most of the discussion is ignoring lots of things... first of
all you must answer the question; Does it really matter? If it does,
then i will assume that you are a designer and know how to apply the
tube characteristics to come up with a design that matches your
selected tube to the expected load... but does that process really
ever describe the 'output impedance'?? Then you must also consider
the tuning parameters employed... sure, you can measure the output
impedance at a given operating point, but answer the question again;
Does it really matter? Or do you need to measure it over a wide range
of operating conditions? Every operator i have seen tune up one of my
amps has done it a little bit different... heck, i don't even do it
exactly the same twice in a row i bet. So when i am running an amp
into a switchable set of 7 different antenna combinations on a given
band, can tune from one end of the band to the other without touching
the settings, and can make an infinite number of small adjustments to
the drive, tune, and load settings, and on some bands can tweak a
tuner after the amp to 'make it happier', do I really care what the
'output impedance' really is? As long as the matching network
provides adequate adjustment so i can get out the desired power into
my various loads while keeping the tubes within their operating
limits, do i really care what the 'output impedance' really is at any
one set of conditions that i may never exactly duplicate again? I
think not. So this boils down to an academic discussion, and as in
many cases where no one has, or can, make an exact statement of the
problem the specific answer remains elusive. So consider this, until
you have a complete statement of the problem you will never be able to
derive a value that any two of you will agree on, let alone actually
try to set up a measurement of.

Transmitter Output Impedance
 

On 5/3/2011 4:15 PM, Wimpie wrote:
On 3 mayo, 21:55, Cecil wrote:
On May 3, 9:01 am, wrote:

To be honest, I, and my ham friends, never had to use photon theory to
solve problems in both amateur and professional RF-Engineering. So I
think this doesn’t contribute to the discussion.

Knowing that EM RF waves must necessarily obey the laws of photon
physics can save one from all sorts of technical blunders. But feel
free to blunder on.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Hello Cecil,

Maybe you should change your attitude from "it is wrong" to "it may be
true". This might open your mind to study other insights.

When measuring an RF current in a PA, everybody ignores the
quantisation of charge. Is this lack of knowledge?

The frequency of our waves is sufficiently low to ignore any
quantisation effects due to photons. Is this lack of knowledge? In my
opinion a good Engineer knows what he can ignore and what not to
efficiently solve a certain task.

Starting a discussion about photons, in my opinion, is a way to reduce
the S/N ratio of the discussion.

With kind regards,


Wim
PA3DJS
www.tetech.nl.

Hi, Wim -

I believe his confusion is the one-port vs two-port problem. I don't
have a problem with your explanation. But, I think he is throwing in a
port where he should not.

Cheers,
John

Transmitter Output Impedance
 

On May 3, 6:11*pm, dave wrote:
So this boils down to an academic discussion, and as in
many cases where no one has, or can, make an exact statement of the
problem the specific answer remains elusive.

When one's basic premises violate the laws of physics, it is very
difficult to come up with a valid answer.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Transmitter Output Impedance
 

On May 3, 6:29*pm, John KD5YI wrote:
I believe his confusion is the one-port vs two-port problem. I don't
have a problem with your explanation. But, I think he is throwing in a
port where he should not.

The second port is actually there on the other side of the black box,
but it has been overlooked. Doesn't it give anyone pause to realize
that the s11 parameter changes by an infinite percentage depending
upon whether it is measured as a one-port system or as a two-port
system? If that can happen with a passive black box, consider how much
more complicated it might be with an RF source included in the black
box.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK