Ian wrote -
I don't believe transmitter output impedance is
fundamentally
un-knowable... only that it is extremely difficult to
measure in
practice. All the measurement techniques I've heard
of seem to be
subject to either theoretical weaknesses or problems
of practical
accuracy.

====================================

You lot will drive me to despair.

What's the problem?

You have just designed the PA.

You are familiar with its very simple internal
circuitry.

You have its component values, including the active
ones.

Hasn't it yet occurred to you engineering Ph.D's all
you have to do is calculate it.

Sewer kids in Rio have better arithmetical aptitudes.
---
Reg

"Dr. Slick" wrote in message
om...
"Tarmo Tammaru" wrote in message
...
"Dr. Slick" wrote in message
om...

But they never explain WHY a lossy line can INCREASE the
reflected power! The lossless line would not attenuate the reflected
wave at all!

I don't trust their claims on this.

If you get more power reflected than you send into a passive
network, you are getting energy from nowhere, and are thus violating
conservation of energy.


But the reflection coefficient is for Voltage. I think the clew lies in "The
main point of interest lies in the fact that we cannot, in general,
superpose the average powers carried by incident and reflected waves on a
dissipative line, although we could do so on a lossless line" A/C/F
.

I don't see the problem. 100 /_30 degrees divided by 2/_5 degrees is
50/_15
degrees. Different phase angle. By general case they mean not the
lossless
case.


I believe you mean 50 @ 25 degrees.


Yeah, I started typing this line before I had decided what numbers to use.



Also, they go from equation 5.12 to 5.13 without showing us how
they got there.

They use the identity e**jx = cos x + jsin x



Yes? And? How did they get the Zo=(Zn-1)/(Zn+1) from this?


I think the math is the same as for a lossless line

As I said out front. The book is copyrighted 1960. There is a certain
life
to these things.

Tam


But it seems to be out of print, perhaps with good reason...


Slick
The "print file" for a book used to be stored on hundreds of tin or lead
plates. Two N pages per plate. After printing some number of books, these
plates would have been recycled. I don't know that there was not a newer
edition.

Tam/WB2TT

Let's start all over on this.

1. You have a transmission line of Zo= 1 - j1
2. Zo* = 1 + j1
3 Short the end of the line
4. By the classic formula RC= (0 - (1 - j1))/(0 + (1 - j1))
= -(1 - j1)/(1 - j1) = -1
5. By Besser's formula RC = (0 - (1 + j1))/(0 + (1 - j1))
which is (-1 -j1)/(1 - j1)
6. The above is SQRT(2)/_-135 / SQRT(2) /_-45 = 1/_-90
7. Note that RC has angle -90, not 180.
8. Reflected voltage is now (V+) x (1/_-90). This is in quadrature with V+,
not in opposition with.
9. This won't meet the boundary condition that ( V+) + (V-) = 0
10. You can use a different phase angle for Zo if you like, but the only Zo
phase angle that will give you RC = 1/_180 is 0.

Tam/WB2TT

Dr. Slick wrote:
I meant, how did they derive Reflection Coefficient=(Zn-1)/Zn+1)?
They certainly don't show us! This would be key to answering our questions.

Not taking sides here, just hopefully contributing something.

The s-parameters normalize voltages to the square root of Z0
such that the square of 'a1' is incident power normalized to
Z0=1.0. The standard Smith Chart is normalized to Z0=1.0. Thus
s-parameter quantities can be plotted directly on a standard
Smith Chart. The above equation looks as if Z0=1.0 for
normalization purposes. If 50 ohms is normalized to 1.0,
then Zn is probably Z1 also normalized to 50 ohms, i.e.
(Z1-Z0)/(Z1+Z0) = [(Z1/Z0)-(Z0/Z0)]/[(Z1/Z0)+(Z0/Z0)] =
(Zn-1)/(Zn+1) = s11. The 'n' in Zn may mean Z1 "normalized".
--
73, Cecil http://www.qsl.net/w5dxp



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Reg Edwards wrote:
Ian wrote -
I don't believe transmitter output impedance is
fundamentally
un-knowable... only that it is extremely difficult to
measure in
practice. All the measurement techniques I've heard
of seem to be
subject to either theoretical weaknesses or problems
of practical
accuracy.

====================================

You lot will drive me to despair.

What's the problem?

You have just designed the PA.

You are familiar with its very simple internal
circuitry.

You have its component values, including the active
ones.

Hasn't it yet occurred to you engineering Ph.D's all
you have to do is calculate it.

Sewer kids in Rio have better arithmetical aptitudes.

Sorry, I don't have an engineering PhD... but if *you* think it's so
simple, then lay your money down: show us.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

Reg Edwards wrote:
Hasn't it yet occurred to you engineering Ph.D's all
you have to do is calculate it.

Of course, but can you prove that impedance is what is "seen"
by the incident reflected waves? Dr. Bruene attempted it and
AFAIAC, didn't succeed because his ping frequency was different
from the operating frequency.
--
73, Cecil http://www.qsl.net/w5dxp



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"Tarmo Tammaru" wrote in message ...

But they never explain WHY a lossy line can INCREASE the
reflected power! The lossless line would not attenuate the reflected
wave at all!

I don't trust their claims on this.

If you get more power reflected than you send into a passive
network, you are getting energy from nowhere, and are thus violating
conservation of energy.


But the reflection coefficient is for Voltage. I think the clew lies in "The
main point of interest lies in the fact that we cannot, in general,
superpose the average powers carried by incident and reflected waves on a
dissipative line, although we could do so on a lossless line" A/C/F


But the square of the MAGNITUDE of the Voltage RC is the power
RC.

They never tell us why, and i don't think a lossy line will
increase your chances of getting rho1. In fact, i don't believe this
is possible with a passive network.





Also, they go from equation 5.12 to 5.13 without showing us how
they got there.

They use the identity e**jx = cos x + jsin x



Yes? And? How did they get the Zo=(Zn-1)/(Zn+1) from this?


I think the math is the same as for a lossless line



What is not understood is how one gets from:

Voltage R. C.= (Vr/Vi)e**(2*y*z)

where y=sqrt((R+j*omega*L)(G+j*omega*C))
and z= distance from load

To:

Voltage RC=(Z1-Z0)/(Z1+Z0) for purely real Zo
or Voltage RC=(Z1-Z0*)/(Z1+Z0)

And i have NO problems with the normalized formula,
AS LONG AS Zo IS PURELY REAL.

If Zo is complex, then Zo*/Zo is certainly NOT equal to
one!

I don't really trust this book too much, maybe that's why
it is out of print.



The "print file" for a book used to be stored on hundreds of tin or lead
plates. Two N pages per plate. After printing some number of books, these
plates would have been recycled. I don't know that there was not a newer
edition.

Tam/WB2TT



A book can always be reprinted if there is a demand.
Possibly no one bought the book because it's incorrect?


Slick

"Tarmo Tammaru" wrote
"Reg Edwards" wrote

What's the problem?

You have just designed the PA.

You are familiar with its very simple internal
circuitry.

You have its component values, including the active
ones.

Hasn't it yet occurred to you engineering Ph.D's
all
you have to do is calculate it? [Source
resistance].

Reg,

I am *not* a PhD, but, I can tell you what the
problem is. You don't know
what the transistor's Norton equivalent collector
(drain) resistance is at
DC, let alone RF. It is generally not on data sheets.
They will often
specify an "output impedance". This is a convenience
for people who *think*
they are doing conjugate matching. The real component
of this has absolutely
nothing to do with a particular device, outside of
second order effects
caused by series lead inductances.

Check out Motorola application notes AN282A, 721, and
1033. The following is
a quote from AN1033:

"The output impedance of a microwave power transistor
is usually defined as
the conjugate of the load impedance required to
achieve the desired
performance. A typical output equivalent circuit is
shown in Fig1. { Shows
current source in parallel with Cout and resistor
labeled transformed load
impedance. From these two nodes there are inductor
Lc and Lcom going to the
output terminals}. The capacitor Cout is nearly
equal to the collector -
base capacitance Cob specified for the selected
transistor. Lc is the
inductance of the bond wire used to bridge from the
collector metallization
area to the package output lead, and Lcom represents
the inductive effects
of the common element bond wire"

"For correct operation of the transistor, the
ultimate load impedance must
be transformed to a real impedance across the current
generator. This real
impedance is determined by

RL = (Vcc - VCEsat)^2/(2Pout)

The load impedance presented to the package terminals
will contain the real
impedance at the current generator, transformed to a
lower value by the low
pass section formed by Cout and the parasitic
inductances Lc and Lcom.
Usually the reactive part of the load impedance is
made inductive to tune
out the residual capacitance of the device."

One of the other ap notes mentioned collector
resistance to the extent that
it is "high". This kind of analysis is not limited
to transistors. For
kicks, I calculated the load for an 813 tube, and got
the same value as a
book I was reading, which used a roundabout method.
The other ap note also
made a point of the fact that if you were actually
conjugate matched, the
efficiency could never be more than 50%.
==================================

Tam,

The only problem is that caused by entirely unnecessary
complications introduced by people being too clever.

First design a single-ended, class-A, audio amplifier
like a 6V6 beam-tetrode tube plus a transformer to
drive an 8-ohm speaker.

Then calculate the source resistance seen from the
speaker looking back into the amplifier.

To save time searching for manufacturers data, assume:

DC plate supply is 250 volts.
Peak signal at plate 200 volts.
Internal plate resistance, from mnfr's curves, is 100
K-ohms.
Audio power output = 5watts into 8-ohms load.
Peak volts across load = 8.944 volts peak.
Transformer turns ratio = 22.36

Then ask yourself "Is the source resistance of the
amplifier in the same ball-park as the 8-ohm speaker ?"

Is there anything like a conjugate match ?

Which is what it's all about.

For Class-B and Class-C conditions, just multiply
internal plate resistance by a constant which depends
on plate current operating angle. Curvature of tube
characteristics can be taken into account for a higher
order of accuracy. And all this was sorted out in the
early 1920's.

Modern Western radio engineers' education seems to have
been sadly neglected. The next generation's primary
education is taking place in the rat-ridden sewers of
Rio and in the medicine-less, water-less and
electricity-less ruins of Baghdad.

Cecil, in the meantime I'm looking forward to the day
when Texas vintages appear on the shelves of my local
supermarket. ;o)
---
Reg.

Texas has a long, proud history of producing fine fermented beverages.
Perhaps you can find a sample in a specialty shop. Ask for Pearl or Lone
Star beer.

Roy Lewallen, W7EL

Reg Edwards wrote:
. . .
Cecil, in the meantime I'm looking forward to the day
when Texas vintages appear on the shelves of my local
supermarket. ;o)

Reg Edwards wrote:
Cecil, in the meantime I'm looking forward to the day
when Texas vintages appear on the shelves of my local
supermarket. ;o)

Want me to have the Texas winery send you a bottle, Reg?
--
73, Cecil http://www.qsl.net/w5dxp



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Reg:

[snip]
Amen Brother!

--
Peter K1PO
Indialatic By-the-Sea, FL.

==========================

Amen ?

Peter, does "Amen" imply worship of Terman and the ARRL
Handbook has now been heretically switched to a mere
box of electronics ? Heaven help us!
----
Reg, G4FGQ
[snip]

Not if you work it all out from first principles and write the computer
programs yourself!

:-)

Reg, I seldom use "store bought" or "downloaded" computer programs where I
can know neither
the underlying Physics nor the numerical methods in use...

But when I do use "store bought" or "downloaded" computer programs, like
President Regan, I "trust but verify"!

Like you,I write most of my design programs myself, and I have an extensive
lot of such... some of which have been sold commercially by my former
employees. Unlike you, I however use trusty proven Fortran, not that modern
computer scientists abortion Pascal. :-)

Fortran is proven technology not obsolete junk.... and in the newer
releases, Fortran 90/95 has only the "new" stuff that you really need, and
not a lot of "fashionable" computer science features.

--
Peter K1PO
Indialantic By-the-Sea, FL.

"Tarmo Tammaru" wrote in message ...
Let's start all over on this.

1. You have a transmission line of Zo= 1 - j1
2. Zo* = 1 + j1
3 Short the end of the line
4. By the classic formula RC= (0 - (1 - j1))/(0 + (1 - j1))
= -(1 - j1)/(1 - j1) = -1
5. By Besser's formula RC = (0 - (1 + j1))/(0 + (1 - j1))
which is (-1 -j1)/(1 - j1)
6. The above is SQRT(2)/_-135 / SQRT(2) /_-45 = 1/_-90
7. Note that RC has angle -90, not 180.
8. Reflected voltage is now (V+) x (1/_-90). This is in quadrature with V+,
not in opposition with.
9. This won't meet the boundary condition that ( V+) + (V-) = 0
10. You can use a different phase angle for Zo if you like, but the only Zo
phase angle that will give you RC = 1/_180 is 0.

Tam/WB2TT


You are correct. But please note that the rho is 1 in both cases.

And i believe that the conjugate equation is correct, that the
reflected voltage will have a phase shift of twice the phase of the
Zo, which in this case would be 45 degrees.

And you bring up an excellent point Tam, that the only Zo that
will give you a RC=-1 (or 1 /_ 180 ), with an ideal short as a Zload,
would be a purely REAL Zo, with no reactance.

So again, the "normal" equation fails in this respect. I'll state
once again that the normal equation assumes a purely real Zo.

Besser and Kurokawa and the ARRL are all correct.


Slick

"Dr. Slick" wrote in message
om...
"Tarmo Tammaru" wrote in message
...
Let's start all over on this.

1. You have a transmission line of Zo= 1 - j1
2. Zo* = 1 + j1
3 Short the end of the line
4. By the classic formula RC= (0 - (1 - j1))/(0 + (1 - j1))
= -(1 - j1)/(1 - j1) = -1
5. By Besser's formula RC = (0 - (1 + j1))/(0 + (1 - j1))
which is (-1 -j1)/(1 - j1)
6. The above is SQRT(2)/_-135 / SQRT(2) /_-45 = 1/_-90
7. Note that RC has angle -90, not 180.
8. Reflected voltage is now (V+) x (1/_-90). This is in quadrature with
V+,
not in opposition with.
9. This won't meet the boundary condition that ( V+) + (V-) = 0
10. You can use a different phase angle for Zo if you like, but the only
Zo
phase angle that will give you RC = 1/_180 is 0.

Tam/WB2TT


You are correct. But please note that the rho is 1 in both cases.

And i believe that the conjugate equation is correct, that the
reflected voltage will have a phase shift of twice the phase of the
Zo, which in this case would be 45 degrees.

And you bring up an excellent point Tam, that the only Zo that
will give you a RC=-1 (or 1 /_ 180 ), with an ideal short as a Zload,
would be a purely REAL Zo, with no reactance.

So again, the "normal" equation fails in this respect. I'll state
once again that the normal equation assumes a purely real Zo.

Besser and Kurokawa and the ARRL are all correct.

************************************************** *****

Read this again. The *normal* equation works. It is the ARRL/Besser equation
that does not.

Tam

************************************************** *****


Slick

"Ian White, G3SEK" wrote
Reg Edwards wrote:

The only problem is that caused by entirely
unnecessary complications
introduced by people being too clever.

First design a single-ended, class-A, audio
amplifier like a 6V6
beam-tetrode tube plus a transformer to drive an
8-ohm speaker.

Then calculate the source resistance seen from the
speaker looking back
into the amplifier.

To save time searching for manufacturers data,
assume:

DC plate supply is 250 volts.
Peak signal at plate 200 volts.
Internal plate resistance, from mnfr's curves, is
100 K-ohms. Audio
power output = 5watts into 8-ohms load. Peak volts
across load = 8.944
volts peak. Transformer turns ratio = 22.36

Then ask yourself "Is the source resistance of the
amplifier in the
same ball-park as the 8-ohm speaker ?"

Is there anything like a conjugate match ?

Which is what it's all about.

For Class-B and Class-C conditions, just multiply
internal plate
resistance by a constant which depends on plate
current operating
angle. Curvature of tube characteristics can be
taken into account for
a higher order of accuracy.


Kind of as expected: Reg chops the problem down to
the easy bit,
explains that in detail, and airily dismisses the
hard part in one final
paragraph.

And all this was sorted out in the early 1920's.

And that problem-solving method - make the problem
fit the solution, and
chop the rest off - was invented even longer ago, by
an ancient Greek
bandit.
========================================

Life would have been made less complicated for you if I
had stopped at "Which is what it (the foregoing) is all
about."

After all the years of haggling on this newsgroup are
you not yet convinced the internal resistance of radio
transmitters is not 50 ohms?

Reg,

A truly wonderful example. If you don't mind, I would like to add to this.
See below.

"Reg Edwards" wrote in message
...
"Tarmo Tammaru" wrote
"Reg Edwards" wrote
Tam,

The only problem is that caused by entirely unnecessary
complications introduced by people being too clever.

First design a single-ended, class-A, audio amplifier
like a 6V6 beam-tetrode tube plus a transformer to
drive an 8-ohm speaker.

Then calculate the source resistance seen from the
speaker looking back into the amplifier.

To save time searching for manufacturers data, assume:

DC plate supply is 250 volts.
Peak signal at plate 200 volts.
Internal plate resistance, from mnfr's curves, is 100
K-ohms.
Audio power output = 5watts into 8-ohms load.
Peak volts across load = 8.944 volts peak.
Transformer turns ratio = 22.36

Now, let's see what would happen if Reg had tried to "conjugate Match" the
speaker to the 6V6.

Given that the plate resistance is 100K, and desired Po is 5 W, The voltage
required to generate 5W across 100K can easily be shown to be

VDC = 50 + SQRT (5 x 2 x 100000) = 50 + 1000 = 1050V

This is about 3 times the rated maximum plate voltage.

The efficiency of an amplifier, based on using the Norton equivalence, is
reduced by the amount

(1 + Rs/Rl) = 1 + 100K/100K = 2

Note that in Reg'd design the load impedance was

8 x 22.36 x22.36 = 4000 Ohms, and his efficiency was reduced by only

1 + 4000/100000 = 1.04

So, by matching, the efficiency has neen reduced by 2/1.04 =1.923.

Anybody for going down to the local power company and conjugate matching all
thouse pesky AC generators?

Tam/WB2TT

Tarmo Tammaru wrote:
Anybody for going down to the local power company and conjugate matching all
thouse pesky AC generators?

That's why Edison thought AC would never catch on. I'm looking
for a ham amp with the internal impedance of a 100 MW AC
generator. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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I notice I did not run spell check on my previous! It is a rainy day here;
so, let me add something else that may not be obvious to some:

In general, the addition of negative feedback to change the output impedance
does not change the dynamic range.

Consider the following. An ideal DC op amp operating from a 10V supply with
0 output impedance, and infinite gain.

1. Put a 1K resistor in series with the output.

2. You now have an op amp with 1K output impedance.

3. Connect the inverting input of the op amp to the other end of the 1K
resistor. Call that the output terminal. and connect a load resistor Rl from
there to ground. ( The gain from the non inverting input to Rl =1 in the
linear range)

4. Because of the infinite feedback, the output impedance is now 0 again,
but all of the load current still flows through the 1K series resistor. It
will not, for instance, deliver 5V into a 910 Ohm resistor because the
amplifier will have saturated before that point. For a given RL, the maximum
voltage you can get out is (10 x Rl)/(1000 + Rl) with or without feedback.

Tam/WB2TT
"Cecil Moore" wrote in message
...
Tarmo Tammaru wrote:
Anybody for going down to the local power company and conjugate matching
all
thouse pesky AC generators?

That's why Edison thought AC would never catch on. I'm looking
for a ham amp with the internal impedance of a 100 MW AC
generator. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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On Mon, 1 Sep 2003 10:19:39 +0100, "Ian White, G3SEK"
wrote:

Richard Clark wrote:
On Sun, 31 Aug 2003 19:42:42 +0100, "Ian White, G3SEK"

Motorola publishes the equivalent series (or parallel) resistance for
the MRF-xxx (pick your own that corresponds to the actual device used
for the finals in your own transmitter).

Sorry, that horse won't run. RF power transistor data sheets specify the
load impedance that needs to be presented *to* the transistor from the
outside world, in order for the device to function as specified.

Clearly you did not look at such a sheet, and certainly not from
Motorola.

The MRF421 is used in two of my HF rigs and you have yet to offer what
you have. You didn't look did you? Have you ever looked? Have you
ever repaired a Finals' deck? Ever design one that is comparable?

From that specification sheet(s):
Figure 7 - Series Equivalent Impedance, Zin
shown with both Smith Chart and tabular results over the range of 2
MHz to 30 MHz:
Freq. Zin
30 0.7 - j.5
15 1.39 - j1.1
7.5 2.8 - j1.9
2.0 5.35 - j2.2

Figure 9 - Output Resistance versus Frequency, Zout
slightly less than 2 Ohms at 1.5 MHz to 1.0 Ohms at 30 MHz shown in a
clear and unambiguous chart over the entire range of frequency and
resistance described as Rout, Parallel Equivalent Output Resistance,
(Ohms).

Both specifications are observed for the Transistor powered at 12.5V
with a collector quiescent current of 150mA, and supporting an output
power of 100W PEP.

Sometimes they state the conjugate of the required load impedance, and
sometimes they don't say which.

Sometimes? You don't seem to sure, and you offer nothing specific.


That is not the output impedance *of* the transistor itself (except in
certain special cases). However, the data sheets sometimes do
ambiguously call it the "output impedance". It's a confusing mess.

The data sheets are quite ordinary to the designers however. I've
never sat through a design seminar where any RF design engineer has
made such a statement as yours (much less the outright howler of
circularity below). In fact their queries related to exactly these
specifications. How is it that your experience is different?


However, I'd hoped by now that we were all agreed about this particular
dead horse. It's buried somewhere in the Google archives of this
newsgroup.

The rest of the argument about transforming the impedance from the
device collector/drain anode to the output socket is correct, but it's
being applied to the wrong impedance. The answer you'll find at the
output socket will always be 50 ohms, because all you've done is lead
your horse around in a very tight circle.

Odd that in a chain of signal flow, that you see it being circular.
I've yet to observe a commercial rig with its antenna strapped back
into its driver input. Reading schematics is a fine art, and I have
taught more than 200 students how to do it with far more complex
Transceivers, like Collins equipment KWT-6 (AN/URC-32 HF) or Ship's
Radar AN/SPS-10, VHF/UHF AN/SRC20-21 and so on.

The quality of documentation for those pieces of equipment include
mini-treatises covering the engineering details of design, and
specific to these issues of matching. You would be well advised to
garner such gear, as you could, with attending documentation to
augment your education. One still popular item that I taught also
comes from Collins in the form of the R-390A.

I'd better stop now, before some seriously tasteless horse metaphors
come to mind.

Hi Ian,

Regrettably, you would be outclassed in metaphor usage by an old salt
who could express the word **** as a verb, noun, adjective, adverb,
conjunction....

For those lurkers who get short-shrifted of quality in these
half-debates I offer another Motorola reference: "RF Device Data
Volume II." Observe in AN2821, "Systemizing RF Power Amplifier
Design," in the sub-section "Amplifier Design":
"After selection of a transistor with the required
performance capabilities, the next step in the design
of a power amplifier is to determine the large-signal
input and output impedance of the transistor."
...
"Having determined the large-signal impedances, the
designer selects a suitable network configuration
and proceeds with his network synthesis."

ALL may note this is exactly what has been described by me. ALL may
note nothing of equal scope and depth is offered in rebuttal - if one
were to elevate those responses to such a lofty description.

73's
Richard Clark, KB7QHC

"Tarmo Tammaru" wrote in message ...

Read this again. The *normal* equation works. It is the ARRL/Besser equation
that does not.

Tam


I can admit that i'm wrong, but only if you give me good
reasons, which is something you have failed to do.

Read THIS again.

But please note that the rho is 1 in both cases.

And i believe that the conjugate equation is correct, that the
reflected voltage will have a phase shift of twice the phase of the
Zo, which in this case would be 45 degrees.

And you bring up an excellent point Tam, that the only Zo that
will give you a RC=-1 (or 1 /_ 180 ), with an ideal short as a Zload,
would be a purely REAL Zo, with no reactance.

So again, the "normal" equation fails in this respect. I'll state
once again that the normal equation assumes a purely real Zo.

Besser and Kurokawa and the ARRL are all correct.


Slick

On Mon, 1 Sep 2003 14:12:08 -0400, "Tarmo Tammaru"
wrote:


In general, the addition of negative feedback to change the output impedance
does not change the dynamic range.


Hi Tam,

H.W. Bode, of Bell Labs, developed a body of work that proves you
completely wrong. Negative feedback enhances every facet of amplifier
design and to claim otherwise is wholly misinforming.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote in message
...
On Mon, 1 Sep 2003 14:12:08 -0400, "Tarmo Tammaru"
wrote:


In general, the addition of negative feedback to change the output
impedance
does not change the dynamic range.


Hi Tam,

H.W. Bode, of Bell Labs, developed a body of work that proves you
completely wrong. Negative feedback enhances every facet of amplifier
design and to claim otherwise is wholly misinforming.

73's
Richard Clark, KB7QHC

Hi Richard,

Everything, except the output dynamic range. Please go through my numbers.
The equation I gave is for the point at which the amplifier goes non linear,
and there actually is no feedback. There is a large error voltage at the
input, but the amp can't do anything about it because it is already in
saturation. For a 5V input/output and a 1K load the output terminal of the
op amp is already at VCC. A more practical example might have been a 13.8
volt power supply regulator running off 16 volts, with a current sensing
resistor in series with the output. Anyway, it was a rainy afternoon kind
of thing.

Tam/WB2TT

Richard,

Let me be the lurker on for the day. Actually, you have a good example. I
see the parallel equivalent resistance for the device at 1.5 MHz is about 2
Ohms. At 12.5V with a Vcesat of an optimistic .5V, the power that can be
delivered at conjugate match is 12^2/(2 * 2) = 36W. Not the rated 100.

The 421 is a fairly inefficient transistor. The 50W MRF450 has a *series*
equivalent output impedance at 30 MHz of 174 - j.5. The parallel resistance
will be about 175. Matched power will be 12^2/(2 * 175) = 0.4W. There is
nothing wrong with that.

One of their ap notes states that on newer devices they specify the
conjugate of the desired load, rather than the intrinsic output impedance.
On some devices you simply can't tell which they mean. On the very popular
MRF150, it is footnoted as being "conjugate of optimum load". It may be that
the FET resistance is too high to be meaningful.

Tam/WB2TT
"Richard Clark" wrote in message
...
On Mon, 1 Sep 2003 10:19:39 +0100, "Ian White, G3SEK"
The MRF421 is used in two of my HF rigs and you have yet to offer what
you have. You didn't look did you? Have you ever looked? Have you
ever repaired a Finals' deck? Ever design one that is comparable?

From that specification sheet(s):
Figure 7 - Series Equivalent Impedance, Zin
shown with both Smith Chart and tabular results over the range of 2
MHz to 30 MHz:
Freq. Zin
30 0.7 - j.5
15 1.39 - j1.1
7.5 2.8 - j1.9
2.0 5.35 - j2.2

Richard Clark wrote:
On Mon, 1 Sep 2003 10:19:39 +0100, "Ian White, G3SEK"
wrote:

Richard Clark wrote:
On Sun, 31 Aug 2003 19:42:42 +0100, "Ian White, G3SEK"

Motorola publishes the equivalent series (or parallel) resistance for
the MRF-xxx (pick your own that corresponds to the actual device used
for the finals in your own transmitter).

Sorry, that horse won't run. RF power transistor data sheets specify the
load impedance that needs to be presented *to* the transistor from the
outside world, in order for the device to function as specified.

Clearly you did not look at such a sheet, and certainly not from
Motorola.

The MRF421 is used in two of my HF rigs and you have yet to offer what
you have. You didn't look did you? Have you ever looked?

Of course I have, over several years: looked at data sheets (Motorola
more than any others), at numerous application notes, and at Dye and
Granberg's textbook.

Have you
ever repaired a Finals' deck?

If you mean, have I worked in the mobile radio industry, then the answer
is no.

Ever design one that is comparable?

Oh yes: designed, built, used - and understood.


From that specification sheet(s):
Figure 7 - Series Equivalent Impedance, Zin
[snip]
Yes, Zin is the true input impedance of the device - but "Zout" is not.

Figure 9 - Output Resistance versus Frequency, Zout
slightly less than 2 Ohms at 1.5 MHz to 1.0 Ohms at 30 MHz shown in a
clear and unambiguous chart over the entire range of frequency and
resistance described as Rout, Parallel Equivalent Output Resistance,
(Ohms).

"Described" is the operative word. What these terms really mean is a
different matter - see the quotes from Motorola below.


Sometimes they state the conjugate of the required load impedance, and
sometimes they don't say which.

Sometimes? You don't seem to sure, and you offer nothing specific.

I am very sure. Some manufacturers state the conjugate, others don't.
Even Motorola have changed their terminology over the years: sometimes
it's "output impedance" or "Zout", sometimes it's "Z(subscript OL)",
sometimes it's "Z(superscript *)(subscript OL)".

Motorola's AN1526 was written in the 1990s to clear up this mess. This
is the key paragraph [my comments in square brackets]:

"Almost every RF power device in Motorola's RF Device Data Book has a
section identifying the device's large-signal series equivalent input
and output impedances. Most often, the device output impedance is
referred to as 'the complex conjugate of the optimum load impedance into
which the device operates at a given output power, voltage and
frequency.' That is certainly a statement requiring some careful
thought, especially since the term 'output impedance' is somewhat
misleading [so even Motorola admit that]. ... as described in [AN282]
it is the conjugate of the LOAD impedance at the fundamental operating
frequency which allowed the transistor to 'function properly' [when the
load impedance was varied in a test jig]."

AN1526 goes on to explain this in much more detail. All of it is
consistent with my earlier statement that the parameter sometimes
labelled "output resistance" is actually the LOAD that needs to be
presented *to* the device from the outside world, for optimum
performance.



That is not the output impedance *of* the transistor itself (except in
certain special cases). However, the data sheets sometimes do
ambiguously call it the "output impedance". It's a confusing mess.

The data sheets are quite ordinary to the designers however. I've
never sat through a design seminar where any RF design engineer has
made such a statement as yours (much less the outright howler of
circularity below). In fact their queries related to exactly these
specifications. How is it that your experience is different?

Maybe it's because I haven't swallowed someone else's half-digested
information, but have done my own thinking and not given up till it
really made sense.


However, I'd hoped by now that we were all agreed about this particular
dead horse. It's buried somewhere in the Google archives of this
newsgroup.

The rest of the argument about transforming the impedance from the
device collector/drain anode to the output socket is correct, but it's
being applied to the wrong impedance. The answer you'll find at the
output socket will always be 50 ohms, because all you've done is lead
your horse around in a very tight circle.

Odd that in a chain of signal flow, that you see it being circular.

I never mentioned signal flow - I'm talking about your circular logic!

The output network is designed to transform a 50-ohm load at the output
into the complex load impedance that needs to be presented to the
collector. But you incorrectly claim that the data sheet gives the
"output impedance" of the device itself. You then claim that by
transforming that "device output impedance" through the output network,
you can calculate the output impedance of the transmitter. Since you are
simply back-tracking through the network design calculations, the answer
you get will always be 50 ohms! It's circular logic in the tightest
possible loop.


For those lurkers who get short-shrifted of quality in these
half-debates I offer another Motorola reference: "RF Device Data
Volume II." Observe in AN2821, "Systemizing RF Power Amplifier
Design," in the sub-section "Amplifier Design":

That's actually AN282A.

"After selection of a transistor with the required
performance capabilities, the next step in the design
of a power amplifier is to determine the large-signal
input and output impedance of the transistor."
...
What you cut out here was the key reference to "collector LOAD
resistance", presumably because you missed the significance of "load".

"Having determined the large-signal impedances, the
designer selects a suitable network configuration
and proceeds with his network synthesis."

ALL may note this is exactly what has been described by me. ALL may
note nothing of equal scope and depth is offered in rebuttal

AN282, from which you quote, was first published in 1968. The truth
about load impedance is in there to be seen, but I'd be the first to
agree that it's not stated clearly. Motorola then confused the issue by
continuing to talk ambiguously about "output impedance" for at least
another 20 years.

AN1526, from which I'm quoting above, supersedes AN282. It was written
about 25 years later in an attempt to clear up that inherited mess of
loose definitions... but apparently with limited success.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

On Mon, 1 Sep 2003 17:52:43 -0400, "Tarmo Tammaru"
wrote:
Hi Richard,

Everything, except the output dynamic range. Please go through my numbers.
The equation I gave is for the point at which the amplifier goes non linear,
and there actually is no feedback. There is a large error voltage at the
input, but the amp can't do anything about it because it is already in
saturation. For a 5V input/output

What exactly does this mean? That you expect 5V in will track with 5V
out? Say so. Shortcuts in specification lead to disaster. If this
is not what you mean (and you did not specify an input in the
original) your ambiguity allows for any interpretation.

and a 1K load the output terminal of the
op amp is already at VCC. A more practical example might have been a 13.8
volt power supply regulator running off 16 volts, with a current sensing
resistor in series with the output. Anyway, it was a rainy afternoon kind
of thing.

Tam/WB2TT

Hi Tam,

Let's review as you asked:

Consider the following. An ideal DC op amp operating from a 10V supply with
0 output impedance, and infinite gain.

1. Put a 1K resistor in series with the output.

Hence the "ideal" Op Amp is crippled from the start, any expectations
of superlative performance have already been abandoned. Further, the
"ideal" Op Amp is not rail limited, nor is it convention to discuss Op
Amp designs with single ended supplies. These are more short cuts
that could have been expressed with very little greater length to
conventional usage.


2. You now have an op amp with 1K output impedance.

#2 in fact adds absolutely nothing but repetition.


3. Connect the inverting input of the op amp to the other end of the 1K
resistor. Call that the output terminal. and connect a load resistor Rl from
there to ground. ( The gain from the non inverting input to Rl =1 in the
linear range)

That being the new output of the amplifier, R1 as you call it, now
loads that amp to ground. (By the way, if you are going to have two
resistors, convention would label them R1 R2; not R and R1.)

The gain is NOT 1 by sheer, obvious placement of the resistors you
describe. You elsewhere supply a gain that does not agree with this
#3. What you imply are the mu and beta gains, but you do not really
go into that distinction, nor do you perform the math that bears on
their usage.


4. Because of the infinite feedback,

Infinite feedback? Poor specification where I have to presume you are
in error and meant infinite gain (for the previously "ideal" amplifier
- which it is not now). If the output were strapped back to the
inverting input, that "might" qualify as infinite, but your load is
the gain determinant of the amplifier. If you meant that the Op Amp
output is impressed upon the inverting input, that goes without saying
for all linear applications doesn't it? If by this your statement of
a gain of 1 above was along the same lines, it suffers by similar
degree.

the output impedance is now 0 again,

No, it is not, you have ascribed (by description) a 1000 Ohm output
impedance (resistance). The amplifier is not the Op Amp component, it
is the assembly of components presented to the load and is modified by
the gain that you incorrectly ascribe above.

but all of the load current still flows through the 1K series resistor.

Which confirms my statement and directly follows from the addition you
originally offer.

It
will not, for instance, deliver 5V into a 910 Ohm resistor because the
amplifier will have saturated before that point. For a given RL, the maximum
voltage you can get out is (10 x Rl)/(1000 + Rl) with or without feedback.


With OR without feedback? Which is it? Do you have feedback or don't
you? The additional baggage of your statement both adds nothing, and
is self conflicting. Do I now have the choice to express it has no
feed back?

Dynamic range is not the same thing as rail limited and rail limiting
is certainly not within the canon of "ideal" amplifiers. Dynamic
range is dimensionless and generally described in dB and is a function
of noise. Feed back has a direct correlation to the amount of noise
added by the amplifier and thus impacts Dynamic Range directly.

It would have taken a whole lot less to simply use the conventional
741 Op Amp as an example, warts and all, to express the same issue
which merely points out that a poor design works poorly. You even
anticipate this poor aspect through the modifications after the fact:
A more practical example might have been a 13.8
volt power supply regulator running off 16 volts, with a current sensing
resistor in series with the output.
which illuminates how a design engineer builds from known limitations
toward known loads. In other words, if the engineer faces a 10V rail
limitation, he could have as easily added a DC-DC up converter to
solve it. Anyone can trap another through crafted specifications.
I've got several many squirrels up a tree right now.

73's
Richard Clark, KB7QHC

On Mon, 1 Sep 2003 19:17:40 -0400, "Tarmo Tammaru"
wrote:

Richard,

Let me be the lurker on for the day. Actually, you have a good example. I
see the parallel equivalent resistance for the device at 1.5 MHz is about 2
Ohms. At 12.5V with a Vcesat of an optimistic .5V, the power that can be
delivered at conjugate match is 12^2/(2 * 2) = 36W. Not the rated 100.


Hi Tam,

rated 100 WHAT?

Sheesh.

73's
Richard Clark, KB7QHC

On Tue, 2 Sep 2003 00:28:12 +0100, "Ian White, G3SEK"
wrote:


AN1526, from which I'm quoting above, supersedes AN282. It was written
about 25 years later in an attempt to clear up that inherited mess of
loose definitions... but apparently with limited success.


Hi Ian,

25 years after the 60's, and yet AN282 is still published by Motorola
in 1991 by my copy. Further, it appears to survive through to 91
without any appearance of AN1526 which superseded it. This would
suggest that this application note, if published in the 90's
represents quite a bulk of material (nearly half again the total AN's
by number following 1991) published in 10 Years? 1968 + 25 = 1993
which suggests rather a revolution in thought over two years and
clearly not within the scope of credibility. These publishing date
games are too inspecific with your reference clearly in front of you.
Don't you know to surer accuracy?

I still see nothing of substance, merely suggestion:
That is certainly a statement requiring some careful
thought, especially since the term 'output impedance' is somewhat
misleading [so even Motorola admit that]. ... as described in [AN282]
it is the conjugate of the LOAD impedance at the fundamental operating
frequency which allowed the transistor to 'function properly' [when the
load impedance was varied in a test jig]."
Does not say what it is, but what it was by testing - hardly
revolutionary nor upsetting.

Certainly you could come up with a smoking gun couldn't you? Complete
with an actual, demonstrable specification for the item I offered
(seeing as you still lack any concrete example). What does your new
and updated resource say about the MRF 421. Does it abandon that
discussion entirely to this new-age era of all being unknowable?

This is still nay-saying and cut-and-paste philosophy without any
correlatives to actual equipment in the field. Why is it so simple to
correlate for me, and for you to dispute through t'ain't so with no
further elaboration?

For example you countered my query as to "did you build your own RF
deck" and the paucity of specifics could create a vacuum. What
impedes your own counter examples of actual implementation? What
prevents your discussion of design issues considered that are germane
to this side-bar? Where is the scope and depth? What service do you
offer those mythical "lurkers" or do we agree that they are simply an
egoists device?

If you think for yourself, be prepared to stand and deliver.
Otherwise this commentary is more posture than content.

73's
Richard Clark, KB7QHC

Why of course that's the explanation! But we're sure thankful we've got
you to tell us what the *real* device characteristics are, and how
things really work! Or at least to give us programs that give us the
for-sure right answers when the underlying principles are too hard for
us to understand. Thanks!

Roy Lewallen, W7EL

Reg Edwards wrote:
Evidently data sheets and books are written by
marketing and sales departments.

Hi Richard,

Thanks for your comments. The next time I do anything like this I will make
the thing more real world. Should have done the power supply thing. I have
built enough of those. As I said, this was a rainy afternoon thought. I
sprinked in some commenys below.

Tam
"Richard Clark" wrote in message
...
On Mon, 1 Sep 2003 17:52:43 -0400, "Tarmo Tammaru"
wrote:
Hi Richard,

Everything, except the output dynamic range. Please go through my
numbers.
The equation I gave is for the point at which the amplifier goes non
linear,
and there actually is no feedback. There is a large error voltage at the
input, but the amp can't do anything about it because it is already in
saturation. For a 5V input/output

What exactly does this mean? That you expect 5V in will track with 5V
out? Say so. Shortcuts in specification lead to disaster. If this
is not what you mean (and you did not specify an input in the
original) your ambiguity allows for any interpretation.

and a 1K load the output terminal of the
op amp is already at VCC. A more practical example might have been a 13.8
volt power supply regulator running off 16 volts, with a current sensing
resistor in series with the output. Anyway, it was a rainy afternoon
kind
of thing.

Tam/WB2TT

Hi Tam,

Let's review as you asked:

Consider the following. An ideal DC op amp operating from a 10V supply
with
0 output impedance, and infinite gain.

1. Put a 1K resistor in series with the output.

Hence the "ideal" Op Amp is crippled from the start, any expectations
of superlative performance have already been abandoned. Further, the
"ideal" Op Amp is not rail limited, nor is it convention to discuss Op
Amp designs with single ended supplies. These are more short cuts
that could have been expressed with very little greater length to
conventional usage.
**********************************
I am making a not quite perfect amplifier from a perfect one. I used that so
I would not have to get into details about 80 db gain, .5V dropout voltage,
10 meg input impedance, 20 Ohm output impedance, etc. Single rail
amplifiers, including these that have a common mode range that includes
ground are quite common.
***********************************

2. You now have an op amp with 1K output impedance.

#2 in fact adds absolutely nothing but repetition.
************************************
right, but I want the reader to consider the 1K as part of the amplifier,
and I have placed it at a point where its effect is obvious
*************************************

3. Connect the inverting input of the op amp to the other end of the 1K
resistor. Call that the output terminal. and connect a load resistor Rl
from
there to ground. ( The gain from the non inverting input to Rl =1 in the
linear range)

That being the new output of the amplifier, R1 as you call it, now
loads that amp to ground. (By the way, if you are going to have two
resistors, convention would label them R1 R2; not R and R1.)

The gain is NOT 1 by sheer, obvious placement of the resistors you
describe.
***************************************
The gain is precisely 1 at the point that I call the amplifier output. The
gain from input to the output terminal of the original amplifier is
variable, and depends on the load.
***************************************
You elsewhere supply a gain that does not agree with this
#3. What you imply are the mu and beta gains, but you do not really
go into that distinction, nor do you perform the math that bears on
their usage.

************************
Precisely. People might read this who have no idea what mu and beta are, but
they know Ohm's law.
************************

4. Because of the infinite feedback,

Infinite feedback? Poor specification where I have to presume you are
in error and meant infinite gain (for the previously "ideal" amplifier
- which it is not now). If the output were strapped back to the
inverting input, that "might" qualify as infinite, but your load is
the gain determinant of the amplifier. If you meant that the Op Amp
output is impressed upon the inverting input, that goes without saying
for all linear applications doesn't it? If by this your statement of
a gain of 1 above was along the same lines, it suffers by similar
degree.
***************************
I have reduced the open loop gain A of the amplifier by the ratio
Rload/(Rload + 1K), but it is still infinite. I may be using terms that are
not universally used. When feedback is used to reduce gain by say 20 db, it
is common to refer to this as 20 db of feedback (May be an audio term). I
have a unity gain amplifier, as I have defined it, and reduced the open loop
gain by infinity.
***************************8

the output impedance is now 0 again,

No, it is not, you have ascribed (by description) a 1000 Ohm output
impedance (resistance). The amplifier is not the Op Amp component, it
is the assembly of components presented to the load and is modified by
the gain that you incorrectly ascribe above.
********************************
I am defining the 1K as part of the amplifier, but I placed it at a point
where you could see it. As defined, if you placed +6V on the noninverting
input, and a load of 10K on the output, the output voltage would be 6V. If
you changed the load to 2K, the output voltage would still be 6V. The
impedance is 0. A person might assume he could easily obtain 11 ma from such
a low impedance source. He can not. Not even at .01V.
**********************************
but all of the load current still flows through the 1K series resistor.

Which confirms my statement and directly follows from the addition you
originally offer.

It
will not, for instance, deliver 5V into a 910 Ohm resistor because the
amplifier will have saturated before that point. For a given RL, the
maximum
voltage you can get out is (10 x Rl)/(1000 + Rl) with or without
feedback.


With OR without feedback? Which is it? Do you have feedback or don't
you?
***********************
I realize this sounds unclear. Consider the node that drives the 1K
resistor. It can not go higher that 10V; the short circuit current is 10 ma.
This is true both for
a) loop is closed. That is the voltage across the load ic connected to the
non inverting input.

b) loop is open with inverting input at ground and noninverting input at
+10V
************************

The additional baggage of your statement both adds nothing, and
is self conflicting. Do I now have the choice to express it has no
feed back?

Dynamic range is not the same thing as rail limited and rail limiting
is certainly not within the canon of "ideal" amplifiers. Dynamic
range is dimensionless and generally described in dB and is a function
of noise. Feed back has a direct correlation to the amount of noise
added by the amplifier and thus impacts Dynamic Range directly.

It would have taken a whole lot less to simply use the conventional
741 Op Amp as an example, warts and all, to express the same issue
which merely points out that a poor design works poorly. You even
anticipate this poor aspect through the modifications after the fact:
**************************
I didn't want to do that. Instead of a 741 I would use an LM358.
**************************
A more practical example might have been a 13.8
volt power supply regulator running off 16 volts, with a current sensing
resistor in series with the output.
which illuminates how a design engineer builds from known limitations
toward known loads. In other words, if the engineer faces a 10V rail
limitation, he could have as easily added a DC-DC up converter to
solve it.
***************************************
I didn't want to build a good circuit, I wanted to build a bad circuit, and
show that all the feedback in the world was not going to make it a good
circuit.
***************************************

Anyone can trap another through crafted specifications.
I've got several many squirrels up a tree right now.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote in message
...
On Mon, 1 Sep 2003 19:17:40 -0400, "Tarmo Tammaru"
wrote:

Richard,

Let me be the lurker on for the day. Actually, you have a good example. I
see the parallel equivalent resistance for the device at 1.5 MHz is about
2
Ohms. At 12.5V with a Vcesat of an optimistic .5V, the power that can be
delivered at conjugate match is 12^2/(2 * 2) = 36W. Not the rated 100.


Hi Tam,

rated 100 WHAT?

Not the rated 100 Watts.
This is just like the 6V6 thing we did earlier.

Tam

Sheesh.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
On Tue, 2 Sep 2003 00:28:12 +0100, "Ian White, G3SEK"
wrote:


AN1526, from which I'm quoting above, supersedes AN282. It was written
about 25 years later in an attempt to clear up that inherited mess of
loose definitions... but apparently with limited success.


Hi Ian,

25 years after the 60's, and yet AN282 is still published by Motorola
in 1991 by my copy. Further, it appears to survive through to 91
without any appearance of AN1526 which superseded it. This would
suggest that this application note, if published in the 90's
represents quite a bulk of material (nearly half again the total AN's
by number following 1991) published in 10 Years? 1968 + 25 = 1993
which suggests rather a revolution in thought over two years and
clearly not within the scope of credibility. These publishing date
games are too inspecific with your reference clearly in front of you.
Don't you know to surer accuracy?

I only know what Motorola say in HB215/D, 'RF Application Reports'. As
you know, Motorola don't go back and re-write old application notes -
they only publish newer ones, leaving users to sort out which aspects
have been superseded and which are still valid.

AN282 is still included in HB215/D and older compilations because it
contains valuable information on other topics.

The application notes themselves are not dated. The list of references
in AN1526 states that AN282(A?) was published in 1968. The publication
date for AN1526 can be bracketed to the early 1990s (after the latest
dated reference that it quotes, and before 1995 when it was re-published
in HB215/D).

That is sufficient to establish my point: that the thinking in AN1526 is
based on about 25 years further experience after AN282.



I still see nothing of substance, merely suggestion:
That is certainly a statement requiring some careful
thought, especially since the term 'output impedance' is somewhat
misleading [so even Motorola admit that]. ... as described in [AN282]
it is the conjugate of the LOAD impedance at the fundamental operating
frequency which allowed the transistor to 'function properly' [when the
load impedance was varied in a test jig]."
Does not say what it is, but what it was by testing - hardly
revolutionary nor upsetting.

Certainly you could come up with a smoking gun couldn't you?

If you can't get it from the key paragraph I quoted, then read all 15
pages of AN1526. If you still can't see that your notion about "device
output impedance" is shot clear through, then neither Motorola and I can
help.

Complete
with an actual, demonstrable specification for the item I offered
(seeing as you still lack any concrete example). What does your new
and updated resource say about the MRF 421.

You know perfectly well that AN1526 won't say anything about your
specific pet device, so what was the point of asking that question?

Does it abandon that
discussion entirely to this new-age era of all being unknowable?

And that is an even worse travesty of what Motorola and I are saying.

If you want to measure the *true* output impedance of an MRF 421 - as
distinct from the load impedance given in the data sheet - then go ahead
and do it. After all, you're the one who claims it is an important
design parameter.

I'm the one who says it is (a) not what you think it is; and (b) not
important anyway.

Now it's up to other people to judge the technical truth of the matter.



--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

"Roy Lewallen" wrote

- - - - - right answers when the underlying
principles
are too hard for us to understand.

===========================

Roy, there you go again - blaming the principles.

On Tue, 2 Sep 2003 08:54:45 +0100, "Ian White, G3SEK"
wrote:

Certainly you could come up with a smoking gun couldn't you?

If you can't get it from the key paragraph I quoted, then read all 15
pages of AN1526. If you still can't see that your notion about "device
output impedance" is shot clear through, then neither Motorola and I can
help.

Complete
with an actual, demonstrable specification for the item I offered
(seeing as you still lack any concrete example). What does your new
and updated resource say about the MRF 421.

You know perfectly well that AN1526 won't say anything about your
specific pet device, so what was the point of asking that question?

Quite obvious isn't it? (Speaking of rhetorical questions....) You
fail to offer ANY of your personal knowledge outside of the
cut-and-paste rebuttal that you loath as being inferior to thinking
for oneself. I supplied my experience and thought both with
correlations and supplemental insights that you condemn through
association; and you cannot dredge up any affect your understanding of
AN1526 bears on the literal design application of the MRF 421. In
short, an academic appeal to it being unknowable with the concomitant
ivory tower snub of application (including your own!).

Does it abandon that
discussion entirely to this new-age era of all being unknowable?

And that is an even worse travesty of what Motorola and I are saying.

You AND Motorola? Are you two in a joint partnership? Twice you draw
on this stale illusion where your own original experience offers a
vacuum of discussion.

If you want to measure the *true* output impedance of an MRF 421 - as
distinct from the load impedance given in the data sheet - then go ahead
and do it. After all, you're the one who claims it is an important
design parameter.

This merely underlines you having ignored my having posted both
commentary AND data to that effect, and you remain silent in regard to
your own efforts that could prove insightful to the meaning imparted
by AN1526 to you. To this point, and through my prodding you have yet
to offer any substance of its importance aside from snippets that are
drawn from an unknown context that you challenge me to review. And to
what end if I were to; and offer your understanding lacked in its
regard? More denial and little detail of substantiation? Who does
your thinking for you? You toss that in my face and then abandon the
field when I put it to you to explain how it bore to the application
you built under your own hands. Was this piece-de-resistance a
Heathkit? Why does its detail of implementation remain cloaked from
your discussion?

I'm the one who says it is (a) not what you think it is; and (b) not
important anyway.

Is there any doubt? (a) remains deliberately vague and (b) is,
frankly, contradicted by the hew and cry that attends your
considerably extended fulmination.


Now it's up to other people to judge the technical truth of the matter.

Ah Ian,

The TRUTH. As if there is only one answer and its altar is not to be
approached. Appeals to educating the lurker is vanity. I enjoy that
game as much as the rest of you, especially when you guys, like
wallflowers, come up so stylistically drab and technically
un-prepared. :-)

Too many mix Truth with explicit admission that this discussion is
not important anyway.
The quality of your rebuttal already proves your sentiment.

Now, your offering any further discussion that relates to your
experience; the role of this AN1526 to it; and some, even if plagued,
specification for any source you designed to, if that follows; then we
may actually get around to a dialog over the topic. Anything less
will be repetition and would again belie your closing sentiment.

73's
Richard Clark, KB7QHC

"Richard Clark" wrote in message
Hi Tam,

You missed the point. 100 Watts PEP is not 100 Watts continuous (such
as your computation leads to).
............................................

Hi Richard,

The voltage swing at the collector can only go between roughly ground a
2*VCC. For a continuous wave it does that for every cycle of the RF signal.
For SSB, it only reaches that on voice peaks, but it can get no larger.
Although you will be on the ragged edge, you can tune up a pi network linear
on a constant tone, and operate with speech.

BTW, did you notice the 174 Ohm Zo of the MRF450. Clearly, you can't
conjugate match that. I am convinced that is a real number, and not a
conjugate of load number. They do not give any min or max limits on this. I
looked at some small signal transistors, and found that Ro can vary by more
than an order of magnitude from unit to unit.

I think that virtually all current HF ham transmitters are push pull, and
use feedback. That will affect the output impedance.


We have all been quoting Motorola literature. I am going to look at what
Philips, and the Japanese have to say on this.

Tam/WB2TT

On Tue, 2 Sep 2003 21:18:03 -0400, "Tarmo Tammaru"
wrote:


"Richard Clark" wrote in message
Hi Tam,

You missed the point. 100 Watts PEP is not 100 Watts continuous (such
as your computation leads to).
............................................

Hi Richard,

The voltage swing at the collector can only go between roughly ground a
2*VCC.

You've got the cart before the horse. Your computation converted to a
PEP valuation is on par and excursions of 2*VCC is not suggested by
anyone.

For a continuous wave it does that for every cycle of the RF signal.
For SSB, it only reaches that on voice peaks, but it can get no larger.

Why voice peaks? What confines PEP to voice? What defines PEP as
voice modulated signals characterizing Z? Two-tone tests are
sprinkled through the literature when it comes to modulation - I don't
think they mean duo-tonic renditions of whistling dixie.


Although you will be on the ragged edge, you can tune up a pi network linear
on a constant tone, and operate with speech.

Where in the specification with an equivalent circuit is there a pi
network? You are probably looking at power decoupling.

BTW, did you notice the 174 Ohm Zo of the MRF450. Clearly, you can't
conjugate match that. I am convinced that is a real number, and not a
conjugate of load number. They do not give any min or max limits on this. I
looked at some small signal transistors, and found that Ro can vary by more
than an order of magnitude from unit to unit.

Small signal characterization? Other than being two orders of
magnitude off, you should stick with large signal characteristics.
Seems like this exact discussion has been gone through before.


I think that virtually all current HF ham transmitters are push pull, and
use feedback. That will affect the output impedance.

Making an appeal to "modern" equipment? Look at the schematic for
your own rig and describe the negative feedback path and its
magnitude. What you are looking at is neutralization. That may be
feedback, but it is far from the gain determination characteristic in
the classic sense defined by Bode. If you didn't have it, you would
be in trouble stability-wise.


We have all been quoting Motorola literature. I am going to look at what
Philips, and the Japanese have to say on this.

Tam/WB2TT


Hi Tam,

Look at
http://www.semelab.com/
which offers to allow you to select FETs by specifying Z (how about
that? 10 years after Motorola gave up in confusion - by some accounts)

http://www.polyfet.com/Dsheet%5CSM724.pdf
or
http://www.polyfet.com/dsheet%5CSR341.pdf
chosen at random, another outlet that foolishly treads where Motorola
gave up in confusion - by some accounts

http://www.semiconductors.philips.co...F145_CNV_2.pdf
about the only HF Power transistor in their repertoire (I could be
wrong as they too offer input and output Z where Motorola gave up in
confusion - by some accounts)

Hi All,

You know fellas, this goes far afield from a simple bench test to
perform against 2 resistors and a hank of transmission line. As no
one handles these "advanced" topics of schematic reading and parts
specification, don't you think you might want to prove you can turn on
a transmitter and read a meter?

If you cannot accept a figure clearly displaying the Z characteristics
of your output finals, how do you think you are going to argue
something simpler such as line loss? It requires no advanced degree,
only proof of performing a task suitable to an amateur. Surely you
can manage that little.

Too much breezing on in place of work. The chorus of appeals of
"doing it for the lurker" must have them rolling in the aisles. :-)

73's
Richard Clark, KB7QHC

Richard Clark wrote:
Does it abandon that
discussion entirely to this new-age era of all being unknowable?

And that is an even worse travesty of what Motorola and I are saying.

You AND Motorola? Are you two in a joint partnership?

I and the Motorola technical reference that I quoted.

Recall what happened: I gave my technical point of view. You challenged
it, asking if I'd ever read any Motorola literature. I quoted a
reference from Motorola, supporting my point of view. Now you attempt to
smear me and Motorola both.

Twice you draw on this stale illusion where your own original
experience offers a vacuum of discussion.

When you quote Motorola AN282A, it is a reference. When I quote AN1526,
it is a "stale illusion."

Along with the "new-age... unknowable" stuff, these are cheap smears
that discredit only you.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

Reg Edwards wrote:

Now it's up to other people to judge the technical
truth of the matter.

=================================

Justice from 'peers' on THIS newsgroup ??????


I wasn't feeling in need of justice - it's more about encouraging people
to think for themselves.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

Ian, G3SEK wrote:
"..It`s more about encouraging people to think for themselves."

Why shoulld people worry with reflection coefficients or SWR?

Terman says:
SWR is important because it is easily measured, and SWR directly
indicates reflection in a system.

Reflection coefficient is defined in my dictionary as:
The vector ratio between the electric fields associated with the
reflected and incident waves, at the junction of a uniform transmission
line and a mismatched terminating impedancee.

Other planes and junctions producing an impedance discontinuity are also
cited as producing a reflection and thus a reflection coeficient.

The dictionary quantifies the reflection coefficient as:
(Z2 - Z1) / (Z2 + Z1)
where Z1 = source Z
and Z2 = load Z.

Absolute values are used above for the reflection coefficient when
relating it to SWR.
This is the voltage divider fraction.

Terman says on page 97 of his 1955 opus:
"Reflection coefficient (rho) = (VSWR-1) / (VSWR + 1).
"S" ---sometimes called (VSWR) to distinguish it from the standing-wave
ratio expressed as a power ratio, which is (Emax / Emin) squared.

best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
The dictionary quantifies the reflection coefficient as:
(Z2 - Z1) / (Z2 + Z1)

Terman says on page 97 of his 1955 opus:
"Reflection coefficient (rho) = (VSWR-1) / (VSWR + 1).

Please note that if Z1 and Z2 are characteristic impedances of
transmission lines at an impedance discontinuity point, these
two equations yield different results.
--
73, Cecil http://www.qsl.net/w5dxp



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
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Richard,

Do yourself a favor and try to get a copy of Motorola ap note AN762. It
describes several amplifiers, including a version with the MRF421. It is
clearly stated that the MRF has an output rating of 100W, PEP OR CW. Also,
the amplifier has negative feedback.

Tam/WB2TT

On Thu, 4 Sep 2003 20:25:01 -0400, "Tarmo Tammaru"
wrote:

Richard,

Do yourself a favor and try to get a copy of Motorola ap note AN762. It
describes several amplifiers, including a version with the MRF421. It is
clearly stated that the MRF has an output rating of 100W, PEP OR CW. Also,
the amplifier has negative feedback.

Tam/WB2TT


Hi Atm,

"Continuous collector current could go as high as 21.3 A at
13.6 V operated into any load."

By Ohm's law, and variations, P = 289.68W and with a collector
efficiency 55% (however, only spec'd at 180W) results in 159W in RF
products (not all within band) over a full cycle. This spec is for
the tandem configuration common to most finals' decks found in amateur
equipment, not a single transistor and not at 12V. Further, the
current is shared by each transistor through alternation as the two
are in series feed to the primary of T3. At the drive levels offered,
the transistors each offer ballpark 1.2 Ohms T3 is specified at 1:5
(again, just as I described in past messages) and offers a 25:1
impedance transform would at a first pass evaluation offers exactly
what I said it would.

Motorola (in their confusion) offers:
"For example, in the 180 Watt version the input
transformer is of 16:1 impedance ratio, making
the secondary impedance 3.13 Ohm with a 50 Ohm
interface."
...
"It should be noted that in the lower power versions
[common to the experience and quality of gear found
in amateur application - rwc] the input and output
impedances are higher...."

At the end of AN762 they offer a design for low pass filtering (to
remove some of those out of band RF products) which specifically
includes the source specification of (-gasp!-) 50 Ohms, and a load of
50 Ohms. Of course, this all occurred prior to some accounts of the
great Motorola confusion of the early 90's that rendered all such
advice - um, well, who knows?

As for the negative feedback. Again, this is exactly what I said it
was "The Input Frequency Correction Network." This hardly qualifies
in the classic Bode sense of Z stabilization so commonly found in AF
amplifiers.

Now I did you a favor by reading it to you.

73's
Richard Clark, KB7QHC