Richard
This is in response to your answer of last night. Before going to bed I got
out the book REFLECTIONS II by Walt Maxwell W2DU. I'm typing verbatim from
page 2-2 and 23-1 for those who don't have the book.page 2-2 "Contrary to
what many believe, it is not true that when a transmitter delivers power
into a line with reflections, a returning wave sees an internal generator
resistance as a dissipative load. Nor is the reflected wave converted to
heat and, while at the same time damaging the final amplifier....the
reflected power is entirely conserved...." from page 23-1 "One of the most
serious misconceptions concerned reflected power reaching the tubes in the
RF amplifier of the transmitter. The prevalent, but erroneous thinking was
that the reflected power enters the amplifier, causing tube overheating and
destruction. However, I dispelled this misconception in the above mentioned
publications, using wave-mechanics treatment, discussed here in greater
detail, by showing that when the pi-network tank is tuned to resonance, a
virtual short circuit to rearward traveling waves is created at the input of
the network. Consequently, instead of the reflected power reaching the
tubes of the amplifier, it is totally re-reflected toward the load by the
virtual short circuit appearing only to waves at the network input".
I'm guessing it's a virtual short because the pi-network is resonant but
what happens if it is a bit off. Also what happens in a transistor final
with no pi?

73 Hank WD5JFR


"Richard Clark" wrote in message
...
On Tue, 25 May 2004 02:59:52 GMT, "Henry Kolesnik"
wrote:

Richard

You say my question has been answered but I haven't seen the answer, too
old, too stupid, whatever but I still would like to understand. So,
would
you be kind enough to give me a better understanding of the "mechanism"
in
the transmitter final that can dissipate and transform yet it can't
dissipate a reflection because there's some kind of one way device that
acts
like a checkvalve or diode that reflects.
tnx

Hi Hank,

There is no answer as you phrase the question because the transmitter
does dissipate power reflected to it. I can easily imagine you may
find no understanding as I go hyperbolic in the following lines. :-)

It can also reflect (some or all of) what it does not absorb, but only
if it does not match the line connected to it. The degree to which it
absorbs/reflects is determined by the ratio of its Z to the line/load
at the antenna terminal. In this case (bear with me), the source Z is
NOW found at the load's mismatch, transformed through the line. This
means matching works both ways (often the singlemost ignored aspect of
obvious reciprocity in any of these arguments). Such circuit analysis
is called superposition. This is merely an academic way of saying you
look at the problem from both points of view where you reverse roles
(source becomes load, and load becomes source). This is a common
practice learned in first quarter circuit analysis when Kirchhoff's
laws are developed (Thevenin/Norton equivalent circuits also emerge).

For example (a strained one at that), if you had a 500 Ohm Source
characteristic Z, you will never launch much power into a 50 Ohm
system because it would immediately hit a reflective interface at the
antenna connector. This means that a 500 Ohm source, when confronted
by power going towards it from a 50 Ohm system will reflect most of
that power (but how did we get this power into the system in the first
place - Karma?) This, by design, will never happen in any transistor
ham rig built in the last quarter century.

So, this is why you see such a vacuum of response when you (the
general readership "you") ask:
"What is the source Z if it is not 50 Ohms?"
Silence guarantees they either don't know (a fatal admission for egos)
or if they offered a value, we could all balance the checkbook and
that would end the game being played. Hence we are treated to all
these suppositions of shorts and opens or magic reflectors hidden
beneath the hood. The frequent toss-off comment of "no one knows" is
called projection, a psychological salve for the ego meaning if I
don't know, then certainly no one else does either - or they are
wrong.

The answer is the transmitter source Z is 50 Ohms at rated power. If
a watt of power is chooglin' down the line toward it, that 50 Ohms is
going to dissipate into a watt worth of calories. This can be argued
with wave mechanics, or lumped circuit equivalents - doesn't matter
because it's all the same calories. Modern rigs can tolerate this
watt through limited feedback and level controlling circuitry - if you
paid more than a kilobuck for your rig that is. For the rest of us,
it's a spin of the wheel and you take your chance. 1 watt hardly
amounts to much, but are you that lucky that it is ONLY 1 watt? Are
we to suppose those unfortunate souls who blasted their rig
transmitting into a mismatch lost it only because they lacked enough
magic pixie dust? I will bet no rig was lost to a cold snap with
frosty finals.

I've already answered about where the heat can be found, so we will
conserve bandwidth.

=== WARNING to the logic impaired, the following is a supposition ===

Now, lets simply accept those answers that require the magic pixie
dust of total reflection from the transmitter. Fine, the mismatched
antenna reflects some power, this power returns to the transmitter,
the transmitter simple routes it ALL back to the antenna,
round-and-round until the antenna finally radiates it. If this were
true, what do we need tuners for? Take a survey of everyone who
chants this mantra of the rig reflecting, and ask if they have a tuner
in the line. Is it a paper weight holding down their license? Is it
a line stretcher because they needed another foot extension between
the antenna lead and their rig? Dare I point out the utter failure of
their faith?

I will pause to allow those answers to emerge and enjoy the thrashing
dawn of a new era of metaphysics. Ooooohhhhhhmmmmms Laaaaaaaaaaw!

73's
Richard Clark, KB7QHC

"Henry Kolesnik" wrote in message
.. .
Richard
This is in response to your answer of last night. Before going to bed I
got
out the book REFLECTIONS II by Walt Maxwell W2DU. I'm typing verbatim
from
page 2-2 and 23-1 for those who don't have the book.page 2-2 "Contrary to
what many believe, it is not true that when a transmitter delivers power
into a line with reflections, a returning wave sees an internal generator
resistance as a dissipative load. Nor is the reflected wave converted to
heat and, while at the same time damaging the final amplifier....the
reflected power is entirely conserved...." from page 23-1 "One of the
most
serious misconceptions concerned reflected power reaching the tubes in the
RF amplifier of the transmitter. The prevalent, but erroneous thinking
was
that the reflected power enters the amplifier, causing tube overheating
and
destruction. However, I dispelled this misconception in the above
mentioned
publications, using wave-mechanics treatment, discussed here in greater
detail, by showing that when the pi-network tank is tuned to resonance, a
virtual short circuit to rearward traveling waves is created at the input
of
the network. Consequently, instead of the reflected power reaching the
tubes of the amplifier, it is totally re-reflected toward the load by the
virtual short circuit appearing only to waves at the network input".
I'm guessing it's a virtual short because the pi-network is resonant but
what happens if it is a bit off. Also what happens in a transistor final
with no pi?

73 Hank WD5JFR
Henry,
Here is an example of what you just said. Take a sine wave source, and
connect it to a 1/4 wave section of shorted transmission line through a
series resistor R. The reflected wave will reach this resistor 1/2 cycle
later, and will be in phase with the source. For a lossless transmission
line, there will be *0 Volts across the resistor*. There will be 0 current
through the resistor, and the reflected wave will be re reflected for all
values of R, including R=Z0, because the reflected wave will not "know" what
R is. You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Tam/WB2TT

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance? Hint: it's the net current
that is zero at the input of a 1/4WL stub. The forward current
and reflected current are equal in magnitude and opposite in
phase. Their individual magnitudes can be very large.
--
73, Cecil http://www.qsl.net/w5dxp





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Cecil wrote,

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?


In the case of a lossless, 1/4WL stub, (no such thing, except on this
newsgroup) in the steady state you can disconnect the transmitter and
the currents will still be there. Current doesn't flow, Cecil, charge flows.
Current is just the rate of flow of charge, dQ/dt. In a 1/4WL stub, charge,
and the fields associated with it, are in a state of oscillation. Over time,
their average movement in space is zero. It's too bad you have to think of
current
as like the water in a big river that has to flow from one place to another
in order to exist. (That's wrong, too. Current in a river is the rate of flow
of
water past a point, not the water itself.) Sloppy semantics sure screw up
understanding.
73,
Tom Donaly, KA6RUH

Tdonaly wrote:
In the case of a lossless, 1/4WL stub, (no such thing, except on this
newsgroup) in the steady state you can disconnect the transmitter and
the currents will still be there.

Since that configuration doesn't exist in reality, only God
can cause what you are asserting. Why am I not surprised that
you need a supernatural being to prove your arguments?

Current doesn't flow, Cecil, charge flows.

:-) :-) Having to resort to trivial arguments is the sure sign
of a loser. :-) :-)
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?

The question is a little misleading because the direction of the flow of
current changes every half cycle and is transverse, or orthogonal to the
direction of wave propagation. In a transmission line, the current
flows through Z0, ostensibly, which is essentially the impedance from
one conductor to the other at every point along the transmission line.
Other than that, superposed forward and reflected waves behave just as
you described, naturally.

73, Jim AC6XG

"Jim Kelley" wrote in message
...
Cecil Moore wrote:

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?

The question is a little misleading because the direction of the flow of
current changes every half cycle and is transverse, or orthogonal to the
direction of wave propagation. In a transmission line, the current
flows through Z0, ostensibly, which is essentially the impedance from
one conductor to the other at every point along the transmission line.
Other than that, superposed forward and reflected waves behave just as
you described, naturally.

73, Jim AC6XG
There is no current in the steady state. The steady state voltage is
independent of source impedance, which affect only how long it takes to
reach that. I ran a simulation on this, and you can see that as the voltage
builds up, the current decreases

Tam

Tam/WB2TT wrote:

"Jim Kelley" wrote in message
...
Cecil Moore wrote:

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?

The question is a little misleading because the direction of the flow of
current changes every half cycle and is transverse, or orthogonal to the
direction of wave propagation. In a transmission line, the current
flows through Z0, ostensibly, which is essentially the impedance from
one conductor to the other at every point along the transmission line.
Other than that, superposed forward and reflected waves behave just as
you described, naturally.

73, Jim AC6XG
There is no current in the steady state. The steady state voltage is
independent of source impedance, which affect only how long it takes to
reach that. I ran a simulation on this, and you can see that as the voltage
builds up, the current decreases

Tam

Hi Tam,

The simulation would be fun to play with. What do you use?

73 de jk

SWCAD from Linear. I also built up a model of an ideal SWR meter.

Tam

Tam/WB2TT wrote:
There is no current in the steady state.

There is no *net* current in the steady state at the input of a
shorted 1/4WL stub. What do you think the current at the short
is? How did that large amount of current get there without flowing?
--
73, Cecil http://www.qsl.net/w5dxp



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