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



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

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



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

"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



-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----