On Mar 12, 1:08 am, "billcalley" wrote:
We are all told that VSWR doesn't matter when using low loss
transmission lines,

No, we are not all told that.

since the RF energy will travel from the
transmitter up to the mismatched antenna, where a certain amount of
this RF energy will reflect back towards the transmitter; after which
the RF will then reflect back up to the antenna -- where the energy is
eventually radiated after bouncing back and forth between the
transmitter and antenna. I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?

The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage. It's only 50ohm once it becomes a moving
wave in the transmission line.

Bob9

On 11 Mar 2007 20:39:46 -0700, "Bob" wrote:

The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage.

Hi Bob,

Well, aside from the initial misunderstanding of how power gets to the
load (much less back, and then to the load again); I will put to you
a question that has NEVER been answered by those who know what the
transmitter output Z ISN'T:
"What Z is it?"

73's
Richard Clark, KB7QHC

Triode or pentode? ;o)

Tim

--
Deep Fryer: A very philosophical monk.
Website @ http://webpages.charter.net/dawill/tmoranwms

"Richard Clark" wrote in message
...
On 11 Mar 2007 20:39:46 -0700, "Bob" wrote:

The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage.

Hi Bob,

Well, aside from the initial misunderstanding of how power gets to the
load (much less back, and then to the load again); I will put to you
a question that has NEVER been answered by those who know what the
transmitter output Z ISN'T:
"What Z is it?"

73's
Richard Clark, KB7QHC

On Mon, 12 Mar 2007 00:09:26 -0600, "Tim Williams"
wrote:

Triode or pentode? ;o)

For you, cascode, push-pull triodes.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
I will put to you
a question that has NEVER been answered by those who know what the
transmitter output Z ISN'T:
"What Z is it?"

It really doesn't matter except for overall
efficiency. A 10 ohm source outputting 100 volts
into a local load of R +/- jX sources the same
amount of power as a 100 ohm source outputting
100 volts into the same local load.
--
73, Cecil http://www.w5dxp.com

On Mar 12, 4:56 am, Richard Clark wrote:
On 11 Mar 2007 20:39:46 -0700, "Bob" wrote:

The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage.

Hi Bob,

Well, aside from the initial misunderstanding of how power gets to the
load (much less back, and then to the load again); I will put to you
a question that has NEVER been answered by those who know what the
transmitter output Z ISN'T:
"What Z is it?"

73's
Richard Clark, KB7QHC

As Tim Williams alludes, it depends on the transmitter design.
It will often be complex rarther than resistive. Since the active
device changes impedance during a single cycle of the RF
signal it may not even be adequately described by a single
value in ohms for a paticular frequency if you wish to
analyse the case of forward and reflected power.

Consider a class C or class E output stage with an
output transistor that is low impedance during
most of the positive half of a cycle of signal and mostly
somewhere near open circuit for the negative half
of the cycle. It seems to me that the effect of reflected
power is going to be different depending its phase
relative to the forward power.
I think this also applys to a lesser extent to a class
A PA with a nice hi-Q tank circuit.

As usually whan this topic comes up, It don't feel
like we have arrived at a usefull and convincing model
of what happens, possibly because simple
descriptions don't cover everything.

Bob

On 12 Mar 2007 23:00:59 -0700, "Bob" wrote:

As Tim Williams alludes, it depends on the transmitter design.

Hi Bob,

No quantifiable answer I see. It's not unexpected, everyone who knows
what it isn't has never been able to say what it is. It seems like
the stock answer you give the cop who asks if you know the speed
limit.
"No. But I wasn't speeding!"

The dependency here started with a conventional Ham transmitter, one
so ordinary as to be a commodity. The design is not so exotic as to
elude a very simple value - except for those who know it isn't 50
Ohms.

73's
Richard Clark, KB7QHC

Bob wrote:
On Mar 12, 1:08 am, "billcalley" wrote:
We are all told that VSWR doesn't matter when using low loss
transmission lines,

No, we are not all told that.

since the RF energy will travel from the
transmitter up to the mismatched antenna, where a certain amount of
this RF energy will reflect back towards the transmitter; after which
the RF will then reflect back up to the antenna -- where the energy is
eventually radiated after bouncing back and forth between the
transmitter and antenna. I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?

The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage. It's only 50ohm once it becomes a moving
wave in the transmission line.

Bob9



Kewl, then I'll just run a tap directly to the 5000 ohm plates and start
a long chat up ... what the heck is all those pi matching components in
the way of the rf? Probably some loss there! ROFLOL!!!

JS
--
http://assemblywizard.tekcities.com

The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage. It's only 50ohm once it becomes a moving
wave in the transmission line.

Bob9

Kewl, then I'll just run a tap directly to the 5000 ohm plates and start
a long chat up ... what the heck is all those pi matching components in
the way of the rf? Probably some loss there! ROFLOL!!!

JS

It appears that you agree with that part of my post but you are
drawing
an invalid conculsion from it. I never suggested that the passive
matching
network usually found in a transmitter output is unnecessary.

Bob

"Bob" wrote in message
ups.com...
On Mar 12, 1:08 am, "billcalley" wrote:
We are all told that VSWR doesn't matter when using low loss
transmission lines,

No, we are not all told that.
The active part of the transmitter output isn't 50 ohm.
That would cause half the power to be lost as heat in
the output stage. It's only 50ohm once it becomes a moving
wave in the transmission line.

Bob9


In that case...

Half power is only lost when terminated to a 50-ohm load;
i.e. no standing waves. What happens when there's a
mismatch and reflected energy



I'll go stand in a corner...