Transmitter Output Impedance
 

On May 2, 5:33*am, Wimpie wrote:
Are you familiar with the concept of S-parameters where you determine
impedance by measuring of reflection coefficient?

Exactly how do you determine the s-parameters for a single-port black
box? It is my understanding that an s-parameter analysis requires an
input port and an output port to be able to measure the parameters.
Where is the input port on an RF source?

What you seem to be measuring is the effect of one or more physical
impedance discontinuities existing in an environment of interference.
Is what you are measuring the actual dynamic source impedance? If I
understand correctly what Walter Maxwell is saying is that whatever
combinations of physical impedance discontinuities from which you guys
are reflecting your test signals, it/they are *not the source
impedance* which is a V/I ratio that originates in the source. A V/I
ratio and a physical impedance discontinuity do not yield the same
reflection coefficients for a 2-port device.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Transmitter Output Impedance
 

Hello Cecil,

On 2 mayo, 14:37, Cecil Moore wrote:
On May 2, 5:33*am, Wimpie wrote:

Are you familiar with the concept of S-parameters where you determine
impedance by measuring of reflection coefficient?

Exactly how do you determine the s-parameters for a single-port black
box? It is my understanding that an s-parameter analysis requires an
input port and an output port to be able to measure the parameters.
Where is the input port on an RF source?

I used a 2 port VNA frequently for antenna measurements, with the
difference that a single port calibration takes less time than a full
two port calibration.


What you seem to be measuring is the effect of one or more physical
impedance discontinuities existing in an environment of interference.
Is what you are measuring the actual dynamic source impedance? If I
understand correctly what Walter Maxwell is saying is that whatever
combinations of physical impedance discontinuities from which you guys
are reflecting your test signals, it/they are *not the source
impedance* which is a V/I ratio that originates in the source. A V/I
ratio and a physical impedance discontinuity do not yield the same
reflection coefficients for a 2-port device.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

I would recommend you to measure something yourselves, or put it into
a simulation. You will see that it doesn't matter whether you use a
deltaV/deltaI setup (complex values, not magnitudes) or reflection
(time varying phase of VSWR) measurement. Just try to explore other
paths and discover other insights.

It doesn't matter when you are measuring a single port device that
contains a generator in it (as long as your VNA setup is able to
distinguish between the output generated by the single port device and
the reflection towards the VNA).

There is similarity with measuring antenna impedance (single port
measurement) when close to (broadcast) stations. Your antenna is a
generator in that case. You can't use the non-coherent type of VSWR
analyzers as the detector detects the signal from the broadcast
station also. However when using a device with a coherent
(multiplying) detector you can, as the detector doesn't respond to the
output because of the (broadcast) station.

With Tom's HP89410 setup, the injected signal for S11 can be well
within the modulation bandwidth of SSB (that means well within 1 kHz
of the transmitter's carrier frequency).

When you have good understanding of diode detectors, you can even do
it without a VNA by using heterodyning and put your focus on the phase
and amplitude of the beat frequency.

With kind regards,


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

Transmitter Output Impedance
 

On May 2, 8:18*am, Wimpie wrote:
I would recommend you to measure something yourselves, ...

Please take a look at the numerous measurements performed by Walter
Maxwell.

Let's take a simple example of the single-port s-parameter results.
There exists a black box with one exposed port. The impedance is
measured to be 50+j0 and s11 is assumed to be zero when driven by a 50
ohm source. There could be a 50 ohm dummy load in the box but there is
not.

Actually, inside the black box is a 1/4WL Z0=100 ohm transmission line
routed to a 200 ohm resistor. s11 is certainly not zero and is
measured, using the 2-port procedure, to be 0.3333.

Which s11 is correct, 0.0000 or 0.3333?
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Transmitter Output Impedance
 

On 2 mayo, 16:17, Cecil Moore wrote:
On May 2, 8:18*am, Wimpie wrote:

I would recommend you to measure something yourselves, ...

Please take a look at the numerous measurements performed by Walter
Maxwell.

Let's take a simple example of the single-port s-parameter results.
There exists a black box with one exposed port. The impedance is
measured to be 50+j0 and s11 is assumed to be zero when driven by a 50
ohm source. There could be a 50 ohm dummy load in the box but there is
not.

Actually, inside the black box is a 1/4WL Z0=100 ohm transmission line
routed to a 200 ohm resistor. s11 is certainly not zero and is
measured, using the 2-port procedure, to be 0.3333.

Which s11 is correct, 0.0000 or 0.3333?
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Hello Cecil,

Assuming that your black box is a 100 Ohms quarter-wave line with 200
Ohms termination (all inside the black box), S11 (50 Ohms based) = 0,
as the input impedance of your black box is 50 ohms. If you like, you
may see your room temperature black box as a source with about –174dBm/
Hz output and an output impedance of 50 Ohms.

Wim

Transmitter Output Impedance
 

On May 2, 9:57*am, Wimpie wrote:
Assuming that your black box is a 100 Ohms quarter-wave line with 200
Ohms termination (all inside the black box), S11 (50 Ohms based) = 0,....

This is what happens when one changes math models in mid-stream. s11
is NOT zero at the mouth of a stub. The impedance looking into a stub
is IEEE definition (1)(B) and cannot have an s11 of zero unless the Z0
of the stub is infinite, which it is not.

Thanks for proving my point. The single-port s11 is completely
different from the dual-port s11 and that is most likely what is
happening with your attempts to measure the source impedance of an RF
amplifier. How can you possibly promote an experimental approach where
s11 changes by an infinite percentage depending on whether one
measures it as a single-port parameter vs a dual-port parameter???
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Transmitter Output Impedance
 

On 2 mayo, 19:38, Cecil Moore wrote:
On May 2, 9:57*am, Wimpie wrote:

Assuming that your black box is a 100 Ohms quarter-wave line with 200
Ohms termination (all inside the black box), S11 (50 Ohms based) = 0,....

This is what happens when one changes math models in mid-stream. s11
is NOT zero at the mouth of a stub. The impedance looking into a stub
is IEEE definition (1)(B) and cannot have an s11 of zero unless the Z0
of the stub is infinite, which it is not.

From your text I didn't conlude it is a stub (but just a series line
section that transforms 50 OHms into 200 Ohms).

Thanks for proving my point. The single-port s11 is completely
different from the dual-port s11 and that is most likely what is
happening with your attempts to measure the source impedance of an RF
amplifier. How can you possibly promote an experimental approach where
s11 changes by an infinite percentage depending on whether one
measures it as a single-port parameter vs a dual-port parameter???

Please explain (or someone else), as I don't understand anything of
the above with regards to a PA. You described a single-port device and
now starts talking about a two-port device.

If you want to prove that for a two port device the impedance
corresponding to S11 may not be equal to the input or output impedance
of port 1, you are right. When S12*S21 isn't zero, the input
impedance depends on the termination of port 2. This is just S-
parameter math, nothing magic.

For the PA case, the active device is the termination for the two-port
matching network, so in that case S11 measurement equals the input/
output impedance of the PA (just a single port measurement). See it as
the output impedance of the active device is connected to port 2 (of
the matching network), and the VNA (or load) is connected to port 1
(that is the output side of the matching network).

The PA reduces to just a single-port network with a source in it
(compare it with the antenna example where a transmitter is in the
vicinity). As long as the injected signal (or slight mismatch
connected to transmission line with increasing length) is small, you
can apply the small signal approach.

As suggested earlier, you may dive into active load pulling:
http://www.focus-microwaves.com/template.php?unique=232

It doesn't matter whether you show some mismatch to a PA, or a perfect
matched load with a small source in it representing the signal that is
reflected towards the amplifier.

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

Kind regards,


Wim
PA3DJS
www.tetech.nl

Transmitter Output Impedance
 

On May 1, 6:01*pm, walt wrote:
On Apr 27, 2:13*pm, Cecil Moore wrote:



On Apr 27, 10:30*am, Wimpie wrote:

Depending *on the frequency resolution of your VSA, the frequency of
the injected signal can be well within 1 kHz of the carrier, so LC
filters in the PA will not distort the measurement. *In case of a 100W
PA and injection of about 100 mW, the difference in wanted signal and
signal to be rejected is 30 dB (not that large).

Would any competent optical physicist suggest that it is valid to
study the conditions associated with interfering coherent light waves
inside an interferometer by introducing an incoherent light source
into the system? Why would any competent RF engineer suggest that the
system source conditions associated with interfering coherent RF waves
can be studied by introducing an incoherent test signal?
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Cecil suggested reading Chapter 19A in Reflections to view the results
of my extensive measurements of the output resistance (impedance) of
RF power amps, but except for Jim and Richard, it appears that the
others have not. Actually, *Chapter19A is an addition to Chapter 19,
which when taken completely will *provide some information that *will
hopefully change some minds concerning the maximum power delivered. It
should be understood that 'maximum' power delivered is that power
delivered with a specified level of drive.

For example, if the drive level is set to deliver a maximum of 100w,
and the pi-network is adjusted to deliver that maximum power into its
load, the source resistance (impedance) will be the (complex)
conjugate of the load impedance. We're not *talking here about the
very maximum power that the amp can deliver, with max drive, max plate
current, etc.

If you review the 19A portion of you will see beyond a doubt that the
conjugate match exists between the output of the pi-network and its
complex load impedance, and that the maximum power delivered at the
drive level that allows only 100w to be delivered as the maximum.

Further review of all the data presented there will also show that the
output resistance of the amp is non-dissipative, while the dissipative
resistance is that between the cathode and plate. The reason the
efficiency of the amps can exceed 50 percent is because the cathode to-
plate resistance is less than the non-dissipative output resistance,
where that R = E/I appearing at the output of the pi-network.

The earlier portion of Chapter 19, that appears in Reflections 2, can
be downloaded from my web page atwww.w2du.com, click on 'Read
Chapters from Reflections 2', and select Chapter 19.

I hope the review of my measured data will clear up some of the
confusion concerning the output resistance (impedance) of the RF power
amp.

Walt, W2DU

When a source is tuned (e.g. through a pi network or any other
matching network between a PA and its load) such that maximum power is
delivered to the load, it's axiomatic that the source impedance is the
complex conjugate of the load. Is there really a need for a whole
chapter for that?

You say, "We're not talking here about the very maximum power that the
amp can deliver, with max drive, max plate current, etc." I beg to
differ. That is EXACTLY what we are talking about. We're talking
about modern amplifiers that would be destroyed if not for protection
circuits, if they were loaded with a load that resulted in maximum
dissipation in the load. We're talking about even old amplifiers with
enough grid drive that they COULD be loaded to higher power output,
but for reasons of wanting the active devices to survive for a
reasonable length of time (or possibly other reasons), are not loaded
so heavily. If you want to exclude such amplifiers from
consideration, then I would hope none would disagree about the
relationship of the load impedance and the source impedance. You
needn't have made any measurements to convince me of that.

Cheers,
Tom

Transmitter Output Impedance
 

On May 2, 1:41*pm, Wimpie wrote:
Please explain (or someone else), as I don't understand anything of
the above with regards to a PA. You described a single-port device and
now starts talking about a two-port device.

If you recognize the example as a two-port device, you correctly
measure an s11 of 0.3333, (100-50)/(100+50). If you happen to overlook
the second port to which the 200 ohm resistor is attached and treat
the example as a single-port device, you measure an s11 of 0.0000,
(50-50)/(50+50). The example has not changed between those two
measurements so which s11 is correct? Doesn't that fact give you pause
to wonder if you are making essentially the same mistake with the PA
measurements?
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Transmitter Output Impedance
 

Hello Cecil,

On 2 mayo, 23:17, Cecil Moore wrote:
On May 2, 1:41*pm, Wimpie wrote:

Please explain (or someone else), as I don't understand anything of
the above with regards to a PA. You described a single-port device and
now starts talking about a two-port device.

If you recognize the example as a two-port device, you correctly
measure an s11 of 0.3333, (100-50)/(100+50). If you happen to overlook
the second port to which the 200 ohm resistor is attached and treat
the example as a single-port device, you measure an s11 of 0.0000,
(50-50)/(50+50). The example has not changed between those two
measurements so which s11 is correct?

You want to know the output/input impedance of your black box, this
means you should look into the 100 Ohms line with 200 Ohms
termination. We will see 50 Ohms, no matter the reference impedance of
the VNA. If it was a 75 Ohms VNA, it would read S11 = -0.2. What the
200 Ohms resistor sees is not important if you just want to know the
behavior of the single-port black box.

Doesn't that fact give you pause to wonder if you are making essentially the same mistake with the PA
measurements?

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

The only thing I could think of is that you have in mind a setup where
the input of the PA is port number 1 (and the output is port number
2 ) and you carry out a full-port measurement. For a full class-A,
AB, non-saturated PA, this may give useful results. When S12*S21
1, S22 will equal the output impedance, otherwise you have to do the
match.

In real world many amplifiers do not behave as a linear system (food
for discussion) and then the two-port setup will fail, as during S22
measurement the input port is terminated with 50 Ohms (so there is no
source that provides output).

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. As
mentioned before, doing a (slow) manual load pull measurement may give
different results because of bias and supply voltage variations.

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


With kind regards,


Wim
PA3DJS
www.tetech.nl

Transmitter Output Impedance
 

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.

http://micro.magnet.fsu.edu/primer/j...ons/index.html

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."

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

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

Transmitter Output Impedance
 

On 4/25/2011 7:35 PM, Sal M. Onella wrote:
This group has presented members with valuable lessons in antennas and
transmission lines, like how to measure, how to match, etc.

Something I haven't seen is a discussion of the source impedance of
the transmitter. My curiosity was piqued today as I took some baby
steps into EZNEC. A particular antenna had such-and-such VSWR if fed
with a 50-ohm cable and a different value if fed with a 75-ohm cable.

No, the antenna still has the same characteristics. You changed the
impedance at a point remote from the antenna.

While this is hardly news, it got me wondering whether a 75-ohm cable
will load the transmitter the same. Doesn't seem like it.

It depends on a number of factors. For example, use a .5 wavelength 75
ohm cable to feed a 50 ohm resistance. Your transmitter happily thinks
it is 50 ohms. Because it is.

My point: Using 75-ohm cable to improve the match at the antenna
won't help me ... IF ... I suffer a corresponding loss due to
mismatch at the back of the radio. My HF radios, all solid state,
specify a 50 ohm load. As necessary, I routinely use an internal
autotuner and either of two external manual tuners. (I'm aware of the
published 1/12 wavelength matching method.)

Wisdom in any form would be appreciated. Thanks.

"Sal"
(KD6VKW)


Transmitters don't have a clearly defined output impedance and, whatever
output impedance is there, doesn't necessarily mean anything.

Your main concern is to provide a 50 ohm load for the transmitter to see.

Actually, it the feed line and load are both 75 ohms, you will only see
a 1.5:1 SWR. Don't worry about it.

Cheers,
John

Transmitter Output Impedance
 

On 5/3/2011 7:27 PM, Cecil Moore wrote:
On May 3, 6:29 pm, John 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?

No. The experiment is designed as a one-port experiment. You have broken
the law and must go to jail without passing GO.

"Having cancer of the colon is better than no colon at all."

Transmitter Output Impedance
 

On 4 mayo, 01:11, dave wrote:
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.

Hello Dave and John,

When you look to my first posting to this thread, you may conclude
that I agree with you.

We had such a discussion about a year ago where I stated that most RF
amplifiers do not have 50 Ohms output impedance. That statement was
heavily disputed by some persons. I tried to support that statement
with simulations, but without any success.

Regarding the "academic discussion" I also agree. In my professional
career where I designed several RF PA's, only 2 times the output
impedance of the amplifier was of importance. In one of these cases I
couldn't meet the specs and had to insert attenuation (some waste of
power…).

The thing I don't like is that some people criticize methods used by
some of the group members without a solid foundation.

Regarding the two-port single-port issue. One can setup a reasoning
based on a two-port setup, but that significantly complicates the
matter without giving any additional insight. I tried to keep it
simple by referencing to a VSWR measurement (with an antenna analyzer)
of an antenna when a strong transmitter is nearby, but it seems I
wasn't clear enough for all people following this thread.

With kind regards,


Wim
PA3DJS
www.tetech.nl
Remove abc first before setting free the pigeon.

Transmitter Output Impedance
 

On Tue, 3 May 2011 19:06:38 -0700 (PDT), Wimpie
wrote:

only 2 times the output
impedance of the amplifier was of importance.

What a strange ellipsis.

In other words if this statement bears upon the discussion, then it
deserves a fuller context. Why was it of importance?

In one of these cases I
couldn't meet the specs and had to insert attenuation (some waste of
power…).

Ed McMahon: "How much power?"

Just trying to get some perspective on this power sourced in the PA.

73's
Richard Clark, KB7QHC

Transmitter Output Impedance
 

On May 3, 9:06*pm, Wimpie wrote:
Regarding the two-port single-port issue. One can setup a reasoning
based on a two-port setup, but that significantly complicates the
matter without giving any additional insight. *I tried to keep it
simple ...

"Everything should be made as simple as possible, but not simpler."
Albert Einstein

Here's my earlier example:

Source-----Z0=50----x----1/4WL Z0=100----200 ohm load

s11 is measured at point x equal to 0.3333 and also 0.3333 at the
load. Nowhere is s11 equal to 0.0000. Put everything to the right of
point x into a black box and s11 measures to be 0.0000 under exactly
the same conditions??? And you guys want all of us to trust that
measurement enough to predict the disputed source impedance of an RF
amp when it cannot even predict the load impedance in the above very
simple passive circuit?

There are reflected waves at point x (s11*a1) that are equal in
magnitude and 180 degrees out of phase with the reflected waves
transmitted back from the load (s12*a2). The two waves undergo
destructive interference at point x which creates a V/I ratio of 50 at
point x. But the absence of *net* reflected energy at point x does not
mean that there are no reflections at point x. There are actually two
sets of reflections at point x that mask any attempt to determine the
actual value of the load impedance by measuring s11 when the system is
installed inside a black box. It is foolish to presume that there are
no similar interference patterns inside an RF amp. In fact, the only
condition where there is no interference inside a simple voltage
source is when there are no reflections or the reflections are
orthogonal to the source signal.

There is a good discussion of the role of interference in the creation
of virtual impedances in section 4.3 of "Reflections", by Walter
Maxwell. Even though a lot RF engineers scoff at the laws of EM wave
physics from the field of optics, the best explanation of interference
I have ever read is the chapter by the same name in "Optics", by
Hecht. Another good chapter in "Optics" is "The Superposition of
Waves".
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Transmitter Output Impedance
 

On 4 mayo, 14:23, Cecil Moore wrote:
On May 3, 9:06*pm, Wimpie wrote:

Regarding the two-port single-port issue. One can setup a reasoning
based on a two-port setup, but that significantly complicates the
matter without giving any additional insight. *I tried to keep it
simple ...

"Everything should be made as simple as possible, but not simpler."
Albert Einstein

Here's my earlier example:

Source-----Z0=50----x----1/4WL Z0=100----200 ohm load

s11 is measured at point x equal to 0.3333 and also 0.3333 at the
load. Nowhere is s11 equal to 0.0000. Put everything to the right of
point x into a black box and s11 measures to be 0.0000 under exactly
the same conditions??? And you guys want all of us to trust that
measurement enough to predict the disputed source impedance of an RF
amp when it cannot even predict the load impedance in the above very
simple passive circuit?

There are reflected waves at point x (s11*a1) that are equal in
magnitude and 180 degrees out of phase with the reflected waves
transmitted back from the load (s12*a2). The two waves undergo
destructive interference at point x which creates a V/I ratio of 50 at
point x. But the absence of *net* reflected energy at point x does not
mean that there are no reflections at point x. There are actually two
sets of reflections at point x that mask any attempt to determine the
actual value of the load impedance by measuring s11 when the system is
installed inside a black box. It is foolish to presume that there are
no similar interference patterns inside an RF amp. In fact, the only
condition where there is no interference inside a simple voltage
source is when there are no reflections or the reflections are
orthogonal to the source signal.

There is a good discussion of the role of interference in the creation
of virtual impedances in section 4.3 of "Reflections", by Walter
Maxwell. Even though a lot RF engineers scoff at the laws of EM wave
physics from the field of optics, the best explanation of interference
I have ever read is the chapter by the same name in "Optics", by
Hecht. Another good chapter in "Optics" is "The Superposition of
Waves".
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Hello Cecil,

Basically, it doesn't matter what is inside the box. It can be fully
characterized by its impedance versus frequency (small signal
approach).

I am fully aware of that there are (back and forth) reflections in the
quarter wave line inside the black box. However in case of a black
box, you don't know that (black box principle).

Same thing happens in most narrow band antennas. The antenna wire
itself may be subjected to (for example) VSWR = 10, though the
impedance can be 50 Ohm (and therefore doesn’t introduce reflection in
a 50 Ohms feed line). I know that this 50 Ohms is the result of
interfering waves/signals, but an MFJ 259B antenna analyzer, or my
FT7B doesn't (and doesn't need to know it).

One could only guess what is a black box based on its impedance versus
frequency curve, S11 curve or Time domain response (I had to do this
in school).

So if we decide to open the black box and we put the reference plane
inside the box, we get a new thread, something like: "why has a PA
certain output impedance?", or "what is the large signal output
impedance of valves, BJT, FET, etc?".


With kind regards,

Wim
PA3DJS
www.tetech.nl

Transmitter Output Impedance
 

John KD5YI wrote:


Transmitters don't have a clearly defined output impedance and, whatever
output impedance is there, doesn't necessarily mean anything.

Your main concern is to provide a 50 ohm load for the transmitter to see.

actually, to provide an *acceptable* load for the transmitter that
maximizes radiated RF power. That might not be 50 ohms.. and I don't
know that we actually care what it is, unless we're designing amplifiers.

50 ohms is basically a "standard test condition" so that you can compare
amplifiers, and it happens to be convenient that everything is made with
the same standard impedance. That way, your test load (which will
dissipate a fair amount of power in some tests) can be located somewhere
different, and connected by a length of transmission line with the same
impedance, so the test at the amplifier output reference plane is still
valid.

Transmitter Output Impedance
 

On May 4, 10:47*am, Wimpie wrote:
Basically, it doesn't matter what is inside the box. It can be fully
characterized by its impedance versus frequency (small signal
approach).

If what is inside the box doesn't matter, why waste time trying to
measure what is inside the RF amp box? What can be "fully
characterized" is the box's relationship to everything outside of the
box. Exactly what is inside the box cannot be ascertained at all while
we handicap ourselves with a blinders. When we actually look inside an
RF amp, we find active non-linear devices upon which we all have been
warned about trying to use linear assumptions.

It doesn't matter what is inside the box when *conditions outside of
the box* are being considered. EE201. What is inside the box matters
considerably for *conditions inside the box*. For instance, replacing
a 50 ohm dipole with a 50 ohm lumped circuit doesn't change things
between the load and the source. But it causes an almost infinite
change from the load out to the rest of reality.

Anything else I could say on this subject would be repetition.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Transmitter Output Impedance
 

On May 2, 8:09*pm, 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.

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."

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



During 1991 Warren Bruene used the RPG method in which he believes he
measured the source resistance of an RF power amp, which he calls
'Rs'. I have never agreed that his method measures the source
impedance, or that his data has any relevance to anything.

Consequently, I am not impressed with the discussion going on here
concerning applying a signal back into an operating RF power amp to
determine the source impedance. Please define 'source impedance'--
where is it located in the amp? At the plate? At the output of the pi-
network?. And how do you know the data obtained using this method is
correct? Have you verified it by comparing it with data obtained using
another method?

I made a statement in an earlier post that when measuring the output
impedance using the 'load pull' method we're not concerned with the
absolute maximum power that can be delivered, but instead, limiting
the 'maximum' power delivered to that which can be delivered with a
specific level of grid drive, one which allows the power to be
limited to that of a normal operating level. Tom disagrees with this
position, that it is really the ABSOLUTE MAXIMUM power delivery that
should be considered. As you can see, I don't agree with Tom.

I don't know how many on this thread have actually reviewed the
portion of my Chapter 19 that presents the step-by-step procedure I
used in determining the output impedance of the Kenwood TS-830S tx,
which shows precisely the output impedance appearing at the output of
the pi-network.

To summarize the procedure that I maintain will provide an accurate
measurement of the output impedance appearing at the output terminals
of the pi-network is as follows:

1) Adjust the loading and tuning controls of the amp to deliver all
the available power to a complex load in the amount normally used in
operation with the setting of the grid-drive level required to obtain
that output power.

2).Measure the impedance of the complex load.

3) The output impedance, or 'source' impedance of the amp appearing at
the output terminals of the pi-network is the complex conjugate of the
load impedance.

Now, when you measure the source impedance using the externally-
injected signal, does the data from that measurement agree with that
of the load-measuring method? If it does, then I'll agree that the RPG
method is valid. If it doesn't I'll continue to have considerable
doubt as to its validity. But I'd still like to know where the
resistance measured by this method is located in the amp.

Walt

Transmitter Output Impedance
 

On 5/4/2011 4:11 PM, walt wrote:
On May 2, 8:09 pm, Cecil wrote:
On May 2, 5:23 pm, 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.

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."

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



During 1991 Warren Bruene used the RPG method in which he believes he
measured the source resistance of an RF power amp, which he calls
'Rs'. I have never agreed that his method measures the source
impedance, or that his data has any relevance to anything.

Consequently, I am not impressed with the discussion going on here
concerning applying a signal back into an operating RF power amp to
determine the source impedance. Please define 'source impedance'--
where is it located in the amp? At the plate? At the output of the pi-
network?. And how do you know the data obtained using this method is
correct? Have you verified it by comparing it with data obtained using
another method?

I made a statement in an earlier post that when measuring the output
impedance using the 'load pull' method we're not concerned with the
absolute maximum power that can be delivered, but instead, limiting
the 'maximum' power delivered to that which can be delivered with a
specific level of grid drive, one which allows the power to be
limited to that of a normal operating level. Tom disagrees with this
position, that it is really the ABSOLUTE MAXIMUM power delivery that
should be considered. As you can see, I don't agree with Tom.

I don't know how many on this thread have actually reviewed the
portion of my Chapter 19 that presents the step-by-step procedure I
used in determining the output impedance of the Kenwood TS-830S tx,
which shows precisely the output impedance appearing at the output of
the pi-network.

To summarize the procedure that I maintain will provide an accurate
measurement of the output impedance appearing at the output terminals
of the pi-network is as follows:

1) Adjust the loading and tuning controls of the amp to deliver all
the available power to a complex load in the amount normally used in
operation with the setting of the grid-drive level required to obtain
that output power.

2).Measure the impedance of the complex load.

3) The output impedance, or 'source' impedance of the amp appearing at
the output terminals of the pi-network is the complex conjugate of the
load impedance.

Now, when you measure the source impedance using the externally-
injected signal, does the data from that measurement agree with that
of the load-measuring method? If it does, then I'll agree that the RPG
method is valid. If it doesn't I'll continue to have considerable
doubt as to its validity. But I'd still like to know where the
resistance measured by this method is located in the amp.

Walt

It seems to me that we maybe should not be talking about source
impedance but maybe about 'regulation' or some other equivalent word.

For an open-loop source made from real components, I think one will
always find that there is a dE/dI number for output loading that we can
call source resistance.

It does not always mean that there is a physical resistance in the
circuit. It is simply a measure of the ability of the device to provide
an unvarying voltage under conditions of varying load.

Does this make any sense?

John

Transmitter Output Impedance
 

On 4 mayo, 23:11, walt wrote:
On May 2, 8:09*pm, 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.

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."

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

During 1991 Warren Bruene used the RPG method in which he believes he
measured the source resistance of an RF power amp, which he calls
'Rs'. I have never agreed that his method measures the source
impedance, or that his data has any relevance to anything.

Consequently, I am not impressed with the discussion going on here
concerning applying a signal back into an operating RF power amp to
determine the source impedance. Please define 'source impedance'--
where is it located in the amp? At the plate? At the output of the pi-
network?. And how do you know the data obtained using this method is
correct? Have you verified it by comparing it with data obtained using
another method?

I made a statement in an earlier post that when measuring the output
impedance using the 'load pull' method we're not concerned with the
absolute maximum power that can be delivered, but instead, limiting
the 'maximum' power delivered to that which can be delivered with a
specific level of grid drive, one which allows the power *to be
limited to that of a normal operating level. Tom disagrees with this
position, that it is really the ABSOLUTE MAXIMUM power delivery that
should be considered. As you can see, I don't agree with Tom.

*I don't know how many on this thread have actually reviewed the
portion of my Chapter 19 that presents the step-by-step procedure I
used in determining the output impedance of the Kenwood TS-830S tx,
which shows precisely the output impedance appearing at the output of
the pi-network.

To summarize the procedure that I maintain will provide an accurate
measurement of the output impedance appearing at the output terminals
of the pi-network is as follows:

1) Adjust the loading and tuning controls of the amp to deliver all
the available power to a complex load in the amount normally used in
operation with the setting of the grid-drive level required to obtain
that output power.

2).Measure the impedance of the complex load.

3) The output impedance, or 'source' impedance of the amp appearing at
the output terminals of the pi-network is the complex conjugate of the
load impedance.

Now, when you measure the source impedance using the externally-
injected signal, does the data from that measurement agree with that
of the load-measuring method? If it does, then I'll agree that the RPG
method is valid. If it doesn't I'll continue to have considerable
doubt as to its validity. But I'd still like to know where the
resistance measured by this method is located in the amp.

Walt

Hello Walt,

Except for bias or supply voltage change due to load change, load
pulling does give similar results as off-carrier signal injection. I
did this in simulation for various circuits (linear and non-linear).
You may remember that I put something in a document (discussion of
last year, it is still on my website).

You can get different results in case of soft power supply or bias
supplies. In case of manual load pulling, bias/supply voltage may
change (think of change in grid current due to change in RF plate
voltage). There is sufficient time for all voltages and currents to
settle. I added a section on the envelope response due to bias
current/voltage change.

When you use the injection method (for example with 130 Hz offset), it
is like load pulling where you switch the load 200 times/s (more
specifically you rotate the phase of the reflection coefficient as
seen by the PA). In such a situation bias and supply voltages will
settle to an average value resulting in (slightly) different results.

While not relevant for here, but nice to mention, stiffness of bias
supplies has influence on IMD also.


With kind regards,

Wim
PA3DJS
www.tetech.nl

Transmitter Output Impedance
 

On 5 mayo, 00:44, John KD5YI wrote:
On 5/4/2011 4:11 PM, walt wrote:



On May 2, 8:09 pm, Cecil *wrote:
On May 2, 5:23 pm, *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.

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."

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

During 1991 Warren Bruene used the RPG method in which he believes he
measured the source resistance of an RF power amp, which he calls
'Rs'. I have never agreed that his method measures the source
impedance, or that his data has any relevance to anything.

Consequently, I am not impressed with the discussion going on here
concerning applying a signal back into an operating RF power amp to
determine the source impedance. Please define 'source impedance'--
where is it located in the amp? At the plate? At the output of the pi-
network?. And how do you know the data obtained using this method is
correct? Have you verified it by comparing it with data obtained using
another method?

I made a statement in an earlier post that when measuring the output
impedance using the 'load pull' method we're not concerned with the
absolute maximum power that can be delivered, but instead, limiting
the 'maximum' power delivered to that which can be delivered with a
specific level of grid drive, one which allows the power *to be
limited to that of a normal operating level. Tom disagrees with this
position, that it is really the ABSOLUTE MAXIMUM power delivery that
should be considered. As you can see, I don't agree with Tom.

* I don't know how many on this thread have actually reviewed the
portion of my Chapter 19 that presents the step-by-step procedure I
used in determining the output impedance of the Kenwood TS-830S tx,
which shows precisely the output impedance appearing at the output of
the pi-network.

To summarize the procedure that I maintain will provide an accurate
measurement of the output impedance appearing at the output terminals
of the pi-network is as follows:

1) Adjust the loading and tuning controls of the amp to deliver all
the available power to a complex load in the amount normally used in
operation with the setting of the grid-drive level required to obtain
that output power.

2).Measure the impedance of the complex load.

3) The output impedance, or 'source' impedance of the amp appearing at
the output terminals of the pi-network is the complex conjugate of the
load impedance.

Now, when you measure the source impedance using the externally-
injected signal, does the data from that measurement agree with that
of the load-measuring method? If it does, then I'll agree that the RPG
method is valid. If it doesn't I'll continue to have considerable
doubt as to its validity. But I'd still like to know where the
resistance measured by this method is located in the amp.

Walt

It seems to me that we maybe should not be talking about source
impedance but maybe about 'regulation' or some other equivalent word.

For an open-loop source made from real components, I think one will
always find that there is a dE/dI number for output loading that we can
call source resistance.

It does not always mean that there is a physical resistance in the
circuit. It is simply a measure of the ability of the device to provide
an unvarying voltage under conditions of varying load.

Does this make any sense?

John

Hello John,

You are correct with respect to the physical resistor. If you tune a
real class-C amplifier for maximum output power (conjugated match),
you may not reach the highest efficiency, but at least significantly
above 60%. So this shows that there is no "physical" 50 Ohms in the PA
(assuming 50 Ohms load). If so, the efficiency would never exceed
50%.

When you change the load (after tuning for maximum output) from 50
Ohms to (for example) 30 Ohms and would carry out an impedance
measurement around that 30 Ohms load resistance, you will measure
something completely different. With "around 30 Ohms" I mean using 32
and 28 Ohms (for example).

With kind regards,


Wim
PA3DJS
www.tetech.nl

Transmitter Output Impedance
 

On 5/4/2011 11:34 AM, Jim Lux wrote:
John KD5YI wrote:


Transmitters don't have a clearly defined output impedance and,
whatever output impedance is there, doesn't necessarily mean anything.

Your main concern is to provide a 50 ohm load for the transmitter to see.

actually, to provide an *acceptable* load for the transmitter that
maximizes radiated RF power. That might not be 50 ohms.. and I don't
know that we actually care what it is, unless we're designing amplifiers.

Acceptable is what the manufacturer recommends for his gear. What does
this have to do with the device's output impedance?

John

Transmitter Output Impedance
 

John KD5YI wrote:

It seems to me that we maybe should not be talking about source
impedance but maybe about 'regulation' or some other equivalent word.

For an open-loop source made from real components, I think one will
always find that there is a dE/dI number for output loading that we can
call source resistance.

It does not always mean that there is a physical resistance in the
circuit. It is simply a measure of the ability of the device to provide
an unvarying voltage under conditions of varying load.

This is comparable to the concept of "dynamic resistance" and, as well,
to "negative resistance" (as exhibited by gas discharges for instance).

The local slope of the V/I curve is negative, but that doesn't mean that
it has negative power.



Does this make any sense?

John

Transmitter Output Impedance
 

John KD5YI wrote:
On 5/4/2011 11:34 AM, Jim Lux wrote:
John KD5YI wrote:


Transmitters don't have a clearly defined output impedance and,
whatever output impedance is there, doesn't necessarily mean anything.

Your main concern is to provide a 50 ohm load for the transmitter to
see.

actually, to provide an *acceptable* load for the transmitter that
maximizes radiated RF power. That might not be 50 ohms.. and I don't
know that we actually care what it is, unless we're designing amplifiers.

Acceptable is what the manufacturer recommends for his gear. What does
this have to do with the device's output impedance?


Absolutely nothing.. which is the point. Manufacturers have to make
their device work into some sort of loads, but can't test every possible
load, nor necessarily optimize the design for every possible load that
someone might use.

So we use, as a convention, that we'll measure the output power or DC to
RF conversion efficiency or Power Added Efficiency or whatever, with a
50 ohm resistive load hooked up to it.

That doesn't mean that we actually want to use a 50 ohm load in real
life. Maybe I have an amplifier designed to drive a loop antenna with a
resistive 10 ohm impedance. I might optimize it for that, but, because
people want to see a spec sheet, I'm going to have to measure the output
into 50 ohms and put that on the data sheet.

Or, perhaps, the active devices I use happen to have a "natural" output
impedance of, say, 8-10 ohms, so I put a 4:1 transformer on the output.
My box, if I were to measure the output Z, would show about 40 ohms.
I'd test it with a 50 ohm load, make sure I could put out 100W into that
load so I can sell it as a "100 W transmitter" and be done with it.

It might happen that you could hook up a 40 ohm load, or a 20 ohm load,
and it will work just fine, and may even put out more than 100W.

The fact of the matter is that ham manufacturers don't actually specify
the output Z: they just give you a range of outputs into which they'll
guarantee the transmitter won't fail. (e.g. say, 25 to 100 ohms..
perhaps specified a bit sloppily as VSWR 2:1 if driven from a 50 ohm source)

Interestingly, my IC7000 manual doesn't even give a output impedance, or
even a range of legal load impedances (page 150 of the manual).

Page 11 does say "Antenna Connector: Accepts a 50 ohm antenna with a
PL-259 connector"

And Page 15 does say "Use a well matched 50ohm antenna" and "An SWR of
1.5:1 or lower is recommended" doesn't say it won't work, doesn't even
say it only develops rated output power (on page 150) into that
particular load. There is a boxed note that says that if the SWR is
higher than approx 2.0:1, the transceiver's output power is reduced.


And, let's look at the service manual.. Spec page is identical to the
one in the user manual. The procedure calls out a whole bunch of
adjustments to set the "% power" to appropriate levels using a 50 ohm
load, but doesn't say what the power into some other load Z would be.

There is a test that when you hook up 100 ohms, the internal SWR meter
reads 2:1, and 50 ohms reads 1:1. But that's not a test of the
transmitter's output Z, more a calibration of the transmitter as RF
ohmmeter for an external component.


The ALC system actually measures the output VOLTAGE (not power) and
matches that up against the setpoint for the power (implying that if you
hook up a 40 ohm load, and you set the radio for 50W, you'll get a bit
more out).

the Automatic Power Limiting is a current sensor: if it exceeds 22A, it
reduces the drive. In fact, it kind of looks like the reflected HF power
(from the sampler at the filter output) isn't used to control the drive
at all.. it's just used to drive the SWR indicator. Kind of tough to
tell, the prose description is a bit unclear, and I haven't followed all
the signals through on the schematic.



but the summary is,

Transmitter Output Impedance
 

On 5/4/2011 7:54 PM, Jim Lux wrote:
John KD5YI wrote:

Acceptable is what the manufacturer recommends for his gear. What does
this have to do with the device's output impedance?


Absolutely nothing.. which is the point.


Are we arguing the same point?


but the summary is,


That it is a bag of worms? I'm waiting with baited breath...

Transmitter Output Impedance
 

On Wed, 04 May 2011 17:54:53 -0700, Jim Lux
wrote:

That doesn't mean that we actually want to use a 50 ohm load in real
life. Maybe I have an amplifier designed to drive a loop antenna with a
resistive 10 ohm impedance.

Interesting. For a community that is so tight-fisted with cash, and
so brag-hearty with power claims, absolutely no one has ever tossed
their hat into the ring that with lower than 50Ohm termination on
their rig (and I'm not talking about 45Ohms) or with a more than 50Ohm
termination on their rig (and this is certainly achievable with a
72Ohm Dipole and 70 Ohm coax) that they have then proclaimed they
substantially exceeded 100W radiated (or lost to heat for that matter)
by the same degree of offset from 50.

Or, perhaps, the active devices I use happen to have a "natural" output
impedance of, say, 8-10 ohms,

That must date to the 1950s vintage of solid state. Your own IC7000
certainly doesn't suffer that abysmal "natural" output impedance. The
RD70HHF1 for 97W into 50Ohms exhibits 0.77-j0.22 Ohms @ 30MHz, a far
cry from your supposition. Off by 1000%?

It might happen that you could hook up a 40 ohm load, or a 20 ohm load,
and it will work just fine, and may even put out more than 100W.

"May" is lazy, "Does" is more authoritative. Any reports of "Does?"

And Page 15 does say "Use a well matched 50ohm antenna" and "An SWR of
1.5:1 or lower is recommended" doesn't say it won't work, doesn't even
say it only develops rated output power (on page 150) into that
particular load. There is a boxed note that says that if the SWR is
higher than approx 2.0:1, the transceiver's output power is reduced.

Which is curious when the manufacturer of the RD70HHF1 power
transistors gives them a "No destroy" rating into a mismatch of 20:1.
Such is the inertia of 1950s design-think with modern components.

73's
Richard Clark, KB7QHC

Transmitter Output Impedance
 

Dear Wimpie: The content of the paragraph below may well rise above the
noise of this thread.
I expect to learn something of value from your comments about why twice the
(apparent) output Z of an amplifier was of importance and what you did to
have the amplifiers conform.

73, Mac N8TT
-------------------
"Wimpie" wrote in message
...

Hello Dave and John,

snip

Regarding the "academic discussion" I also agree. In my professional
career where I designed several RF PA's, only 2 times the output
impedance of the amplifier was of importance. In one of these cases I
couldn't meet the specs and had to insert attenuation (some waste of
power…).

snip

With kind regards,

Wim
PA3DJS
www.tetech.nl
Remove abc first before setting free the pigeon.
--------------

J. C. Mc Laughlin
Michigan U.S.A.
Home:

Transmitter Output Impedance
 

John KD5YI wrote:
On 5/4/2011 7:54 PM, Jim Lux wrote:
John KD5YI wrote:

Acceptable is what the manufacturer recommends for his gear. What does
this have to do with the device's output impedance?


Absolutely nothing.. which is the point.


Are we arguing the same point?


but the summary is,


That it is a bag of worms? I'm waiting with baited breath...

Exactly..


In fact, as interesting as it would be to measure the output impedance
of my radio, I started to think about what it would buy me, and came to
the conclusion, almost nothing (other than satisfying curiosity).

It *might* be interesting to look at (and write an article for QST/QEX
or something) "optimizing radiated power". Answering the question: do
you really want a 50 ohm match on your antenna analyzer, or do you want
maximum net power at the antenna feedpoint, and what that might mean for
typical 100W solid state rigs, antennas, etc.

(as a practical matter, this is what automatic antenna tuners actually
adjust for: either minimum reflected power, or maximum fwd-ref)

but it's possible that deliberately running a mismatch (as shown on your
rig's SWR meter) might actually result in more radiated power. e.g. if
at 1:1 you have 100W fwd and 0 rev, but at 2:1 you have 150W fwd and 15W
reflected, so you're actually net 135W vs 100W; assuming your rig
doesn't otherwise have any problems.

Transmitter Output Impedance
 

Richard Clark wrote:
On Wed, 04 May 2011 17:54:53 -0700, Jim Lux
wrote:

That doesn't mean that we actually want to use a 50 ohm load in real
life. Maybe I have an amplifier designed to drive a loop antenna with a
resistive 10 ohm impedance.

Interesting. For a community that is so tight-fisted with cash, and
so brag-hearty with power claims, absolutely no one has ever tossed
their hat into the ring that with lower than 50Ohm termination on
their rig (and I'm not talking about 45Ohms) or with a more than 50Ohm
termination on their rig (and this is certainly achievable with a
72Ohm Dipole and 70 Ohm coax) that they have then proclaimed they
substantially exceeded 100W radiated (or lost to heat for that matter)
by the same degree of offset from 50.


Maybe that's because of tradition.. it's not only hams who do that. In
the professional world, there's a strong tendency to make the
"interfaces" between subsystems 50 ohms, even if overall system
efficiency might be improved by, say, a 25 ohm impedance. I've worked
on systems where a DAC produced a balanced signal with a 200 ohm Z, and
the vector modulator at the other end had a balanced input with a 200
ohm Z, but everyone insisted we put a 4:1 transformer on each end, so we
could run 50 ohm coax for a 10cm run.

If you're a amp maker, you're probably going to sell a lot more amps if
you specify and design for 50 ohms (the 10 ohm PA for loop antennas
would be a niche market). Likewise, if you're an antenna builder,
designing a product for 50 ohms will sell more than, say, 35 ohms.


In the cellphone world, because the PA and antenna tend to be developed
as an integrated assembly, they're probably a bit more flexible on this.

There's also the issue of wanting to put a power meter in line (but I
think that if you're using a 10 ohm system, you're probably capable of
figuring out how to measure power to make sure that you're not busting
the FCC limits)



Or, perhaps, the active devices I use happen to have a "natural" output
impedance of, say, 8-10 ohms,

That must date to the 1950s vintage of solid state. Your own IC7000
certainly doesn't suffer that abysmal "natural" output impedance. The
RD70HHF1 for 97W into 50Ohms exhibits 0.77-j0.22 Ohms @ 30MHz, a far
cry from your supposition. Off by 1000%?

Just an example. I know that modern parts are MUCH lower. The point is
that if one knew you had a 3 ohm load, you could design an amplifier
that would directly drive it without a transformer or network, and
potentially get better overall efficiency.




It might happen that you could hook up a 40 ohm load, or a 20 ohm load,
and it will work just fine, and may even put out more than 100W.

"May" is lazy, "Does" is more authoritative. Any reports of "Does?"

The link from several days ago for the gentleman who measured the output
Z being around 40 ohms would be a "does"..



And Page 15 does say "Use a well matched 50ohm antenna" and "An SWR of
1.5:1 or lower is recommended" doesn't say it won't work, doesn't even
say it only develops rated output power (on page 150) into that
particular load. There is a boxed note that says that if the SWR is
higher than approx 2.0:1, the transceiver's output power is reduced.

Which is curious when the manufacturer of the RD70HHF1 power
transistors gives them a "No destroy" rating into a mismatch of 20:1.
Such is the inertia of 1950s design-think with modern components.


Yes, but the design limit might not be the transistor mismatch. It
might be a thermal dissipation limit, or some other component that's the
limiting factor.

(Or even, a desire to not have to answer 1000 customer service inquiries
to explain why it really doesn't matter what you hook up, or to explain,
that, yes, you DO need to actually connect an antenna of some sort)

Amateur radio, particularly in the mass market, is pretty slow to move.