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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-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

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