W5DXP wrote:
The answer has been found but most people think they already know
everything there is to know. The answer is contained in the following
example. The sources are signal generators with circulator loads (SGCL).
The (50) or (150) subscript is the ohmic value of the circulator load
resistor. The signal generators are phase-locked and can be turned on
and off independently.

100W SGCL(50)--1WL 50 ohm feedline--+--1/2WL 150 ohm feedline--33.33W SGCL(150)
Pfwd1-- Pfwd2--
--Pref1 --Pref2

Step 1: With the 100W SGCL(50) on and the 33.33W SGCL(150) off, the
following conditions exist:

Pfwd1 = 100W, Pref1 = 25W, Pfwd2 = 75W, Pref2 = 0W

Step 2: Now turn on the 33.33W SGCL(150). At the instant the rearward-
traveling signal reaches the impedance discontinuity, the following
conditions exist at the discontinuity:

Pfwd1 = 100W, Pref1 = 0W, Pfwd2 = 133.33W, Pref2 = 33.33W

All anyone has to do is figure out what happened to Pref1 = 25W in
Step 1 the instant the 33.33W arrived at the impedance discontinuity
in Step 2. This is not a transient buildup condition. If we ignore
any distortion in the 33.33W Pref2 wavefront, this is an immediate
event and Pref1 is immediately canceled by an equal magnitude and
opposite phase Pref2(1-|rho|^2) wavefront at the moment it first
arrives.

You're sourcing and sinking an additional 33.33 watts, and yet the
wattmeter can't discern the difference between this scenario and the 100
watt, single source scenario.

The example illustrates perfectly the shortcomings of the idea of power
flow, as well as some of the faulty conclusions that can be drawn from
measurements made by a directional power meter.

73, AC6XG

Jim Kelley wrote:

W5DXP wrote:

Point is that "maximum possible power" will cause a lot of transmitters
to exceed their maximum power rating and overheat.

How much is the maximum possible power? H-Bomb? Gamma ray burst? Big
Bang? ;-)

Maximum power is when you turn your metal 6L6 upside down in a glass of
water during a contest. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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Jim Kelley wrote:
If the sig gen is putting out 100 watts, with 3 watts reflected and 97
watts going to the load, ...

The forward power is 103 watts and the reflected power is 3 watts in
the given example. The signal generator equipped with a circulator
load is putting out 103 watts.
--
73, Cecil http://www.qsl.net/w5dxp



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Jim Kelley wrote:

W5DXP wrote:

The answer has been found but most people think they already know
everything there is to know. The answer is contained in the following
example. The sources are signal generators with circulator loads (SGCL).
The (50) or (150) subscript is the ohmic value of the circulator load
resistor. The signal generators are phase-locked and can be turned on
and off independently.

100W SGCL(50)--1WL 50 ohm feedline--+--1/2WL 150 ohm feedline--33.33W SGCL(150)
Pfwd1-- Pfwd2--
--Pref1 --Pref2

Step 1: With the 100W SGCL(50) on and the 33.33W SGCL(150) off, the
following conditions exist:

Pfwd1 = 100W, Pref1 = 25W, Pfwd2 = 75W, Pref2 = 0W

Step 2: Now turn on the 33.33W SGCL(150). At the instant the rearward-
traveling signal reaches the impedance discontinuity, the following
conditions exist at the discontinuity:

Pfwd1 = 100W, Pref1 = 0W, Pfwd2 = 133.33W, Pref2 = 33.33W

All anyone has to do is figure out what happened to Pref1 = 25W in
Step 1 the instant the 33.33W arrived at the impedance discontinuity
in Step 2. This is not a transient buildup condition. If we ignore
any distortion in the 33.33W Pref2 wavefront, this is an immediate
event and Pref1 is immediately canceled by an equal magnitude and
opposite phase Pref2(1-|rho|^2) wavefront at the moment it first
arrives.

You're sourcing and sinking an additional 33.33 watts, and yet the
wattmeter can't discern the difference between this scenario and the 100
watt, single source scenario.

But that sourcing and sinking is occurring *INSIDE* the SGCL(150). The net
power inside SGCL(150) is 100W dissipated, *exactly* like the other scenario.

The example illustrates perfectly the shortcomings of the idea of power
flow, as well as some of the faulty conclusions that can be drawn from
measurements made by a directional power meter.

The shortcoming I notice is your sidestepping of the question: What happened
to Pref1=25W? It just seems to have disappeared when we turned on SGCL(150)
and the 33.33W wavefront arrived at the impedance discontinuity. What could
have possibly made Pref1 disappear?
--
73, Cecil http://www.qsl.net/w5dxp



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It's not clear why you'd make that assumption.

A transmission line doesn't radiate; a traveling wave antenna does.
The feedpoint impedance of a traveling wave antenna is dictated by both
the termination resistance and by radiation. The characteristic
impedance of a transmission line isn't a function of either.

You can make a transmission line with an arbitrarily high impedance,
although because of the nearly logarithmic relationship between wire
diameter/spacing and Z0, values of more than a few hundred ohms (for
open wire line, and considerably less for coax) become impractical. Of
course, you'd have to go to very low frequencies to make use of very
high Z0 transmission line, to avoid line radiation or, in the case of
coax, propagation of other than TEM waves.

And, just out of curiosity, how would you construct a traveling wave dipole?

Roy Lewallen, W7EL

W5DXP wrote:
Reg Edwards wrote:

The feedpoint impedance of traveling wave antennas is usually about
600-800 ohms according to The ARRL Antenna Book.


Anyway, if you have correctly quoted it, the ARRL Antenna Book is wrong,


What is the maximum Z0 possible with transmission line? Seems that would
be the feedpoint impedance of a traveling wave dipole.

"Roy Lewallen" wrote in message
...
Tarmo Tammaru wrote:
Roy,

You are cheating. In the steady state there is no load on your source.
Regardless of what the Bird meter reads. Do one of the following:

Why is this cheating? There is reverse power on the line. The source is
not absorbing the reverse power. You and others have said, without
qualification, that it does. I've shown a case where it doesn't.

What you have done is to put a resonant parallel tuned circuit on the output
end of the 50 Ohm resistor. After 1/2 cycle of RF, the voltage on the two
ends of the resistor is equal. Hence no power is delivered from the source,
provided it is lossless coax.

Tell me if I am wrong, but you could put an HP oscillator and a transformer
that delivers 70.7V into an open circuit in place of the 100W amp, and in
the steady state you would reach the same conditions. It would just take
longer. If the Bird read 100W before, it would still read 100W. Assume it is
a perfect WM that does not absorb any power. If you buy the tuned circuit
analogy, the WM is in effect measuring circulating current in the tuned
circuit.

Here is what I think the problem with trying to understand reflected power
in transmitters is. On the one hand, you have forward and reflected power in
and out of a black box. On the other hand, open up the black box and measure
the IMPEDANCE back towards the load. Without knowledge of reflections, we
know from the Smith chart how a misterminated line is going to change the
load on the collector of the transistor, and hence cause it to change its
power output. So, in the impedance analysis there is no concept of absorbing
reflected power.


You don't like talking about pulses, but that is the way I was taught
reflections. We used "Electromagnetic Energy Transmission and Radiation" by
Adler, Chu, and Fano. 1960 (gulp!)

Tam/WB2TT

Roy Lewallen wrote:
A transmission line doesn't radiate; a traveling wave antenna does.

The energy flowing in a flat transmission line goes somewhere.
The energy flowing in a traveling wave antenna goes somewhere.
Both systems are flat so they seem similar to me. A terminated
Rhombic radiates and has approximately the same feedpoint impedance
as wide spaced transmission line.

And, just out of curiosity, how would you construct a traveling wave
dipole?

In my head or in real life? A terminated Sloping-V comes to mind.
--
73, Cecil http://www.qsl.net/w5dxp



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Roy Lewallen wrote:
Why is this cheating? There is reverse power on the line. The source is
not absorbing the reverse power.

If "the source is not absorbing the reverse power", then the reverse
power is being re-reflected to become the forward power. During steady-
state, the forward power is not coming from the source so it has to be
coming from a re-reflection process. It can only come from one of those
two places. In a lossless line, you could disconnect the line at the
source and there would still exist forward energy and reflected energy.
The reflected energy would be reflected from one open end and the forward
energy would be reflected from the other open end. It's essentially a super-
conducting ring that has been cut and straightened out.

Make the stub one second long and you will have 200 joules stored in
the line. The source will have supplied all of that 200 joules with
the voltage and current in phase, i.e. it was supplied as joules/sec.
--
73, Cecil http://www.qsl.net/w5dxp



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Ian White, G3SEK wrote:


* The output impedance of the transistor doesn't come into the story at
all - not when characterizing RF power devices that are not operating in
class A. Even the device manufacturer doesn't know or care what it is.
Neither need we.



Tubes and transistor power amplifiers quite oftem
use negative feedback to improve SSB linearity.
Improvements of 5 to 10 dB are common. The
negative feedback reduces the internal impedance
of the tube and transistor amplifiers. The
tube/transistor data sheets do not consider this
factor.

Again, we usually don't really know or care much
about the values of the internal impedances.

But there is a special case. Voice/music/data tube
transmitters operating at low frequencies have a
problem called "sideband clipping" where the plate
tank selectivity may be too sharp and reduces the
modulation bandwidth. The internal impedance tends
to broaden the response at resonance. When
designing the tank circuit this effect may have to
be included.

Bill W0IYH

On Fri, 18 Jul 2003 22:32:13 -0500, W5DXP
wrote:


Maximum power is when you turn your metal 6L6 upside down in a glass of
water during a contest. :-)

What I have learned so far:

If you have a 100 watt transmitter, the watt meter shows 3 watts
reflected. I deliver 103 watts to the antenna. I now know where the
reflected power go's. But where did it come from? If I could find a
way to have 100 watts reflected I could put 200 watts to the antenna
from a 100 watt transmitter.

If my transmitter has an output impedance of 50 ohms, all the
reflected power will be absorbed by my PA. I need to find a way to
change my output impedance to something other than 50 ohms?

If I could make my SB-401 act like a radar transmitter, and time the
pulses correctly, I could cancel out the reflected power pulses.

For some reason I need a circulator on my SB-401.

To get max power out my 6146's I need to turn them upside down in a
glass of water? :-)