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Sci.Electronics.Basics -> VSWR doesn't matter?
There are 62 messages in this thread.
You are currently looking at messages 20 to 40.
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Author: Jimmie DDate: 13:37 12-03-07
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"billcalley" <billcalley@yahoo.com> wrote in message
news:1173661709.309774.178770@q40g2000cwq.googlegroups.com...
> We are all told that VSWR doesn't matter when using low loss
> transmission lines, 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?
> For the RF to bounce back and forth, wouldn't the transmitter's
> impedance have to be very, very high (or low) when the reflected RF
> energy hit its output stages? I know I'm missing something vital
> here...
>
> Thanks!
>
> -Bill
>
Your mistake is that you assume the output of the tx is 50 ohms, in the case
you stated the transmitter must be matched to the impedance it sees looking
into the transmission line.
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Author: Richard ClarkDate: 13:39 12-03-07
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On Mon, 12 Mar 2007 11:00:04 GMT, Jan Panteltje
<pNaonStpealmtje@yahoo.com> wrote:
>Also normally, there is a pi type filter (to prevent harmonics), between
>amplifier and antenna.
>This filter _WILL_ match the antenna to the output impedance of the
>transmitter, so _even_ if the transmitter output impedance is very
>very low (low voltage high current output stage for example), the reflected
>power will be nicely converted to match the transmitter, and heat up the
>output amp, with its possible destruction as result.
Hi Jan,
Actually, there is a transformer there in the typical Ham transmitter
(and probably in every general purpose power source) that typically
transforms the native Z to the output Z. This is a step up for solid
state, and step down for tubes. In the solid state rigs, it is a
literal transformer feeding the 1-2 Ohms through a 5:1 winding ratio
to a switched bank of low pass filters. This stuff is mud ordinary.
As for the reflected energy, depending upon the phase it will either
combine destructively (heat) or constructively (cool) in the extremes.
There are, of course, 179 degrees of variation between these extremes
before they repeat themselves again. Cooling, of course, is something
of a misnomer as nothing useful is happening (poor power transfer) so
perhaps the terms should be destructive through uselessly benign.
73's
Richard Clark, KB7QHC
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Author: Antonio VernucciDate: 14:30 12-03-07
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> 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?
In my opinion the simplest way to answer your question is that you are assuming
that the transmitter is equivalent to a 50 ohm load, which is not true because
the transmitter is instead equivalent to the series of a 50 ohm load and a
voltage generator.
A simple DC example grossly clarifies thre issue: connect a 12V battery to the
series of a 50-ohm load and another 12V battery. How much current flows through
the load? Naught (assuming the correct polarity). So no power is dissipated in
it.
73
Tony I0JX
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Author: Brian HowieDate: 14:36 12-03-07
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In message <1173661709.309774.178770@q40g2000cwq.googlegroups.com>,
billcalley <billcalley@yahoo.com> writes
> We are all told that VSWR doesn't matter when using low loss
>transmission lines, 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?
>For the RF to bounce back and forth, wouldn't the transmitter's
>impedance have to be very, very high (or low) when the reflected RF
>energy hit its output stages? I know I'm missing something vital
>here...
It matters when it changes suddenly, like mine did recently on my 70MHz
beam, when one of the elements came off in a gale.
Brian GM4DIJ
--
Brian Howie
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Author: Dan BloomquistDate: 14:48 12-03-07
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billcalley wrote:
> We are all told that VSWR doesn't matter when using low loss
> transmission lines, 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.
As pointed out, VSWR does matter. A lot of bouncing means you heat the
transmission line with the power instead of radiating the power.
'Doesn't matter', really means it can be tolerated if need be.
> 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?
> For the RF to bounce back and forth, wouldn't the transmitter's
> impedance have to be very, very high (or low) when the reflected RF
> energy hit its output stages? I know I'm missing something vital
> here...
Here is what you are missing. In the case of the output, (real/resistive
component of the transmitter), seeing the reflected wave, it is _not_
reflecting that power back up the transmission line as you think it is.
It would go back to that real impedance and heat the transmitter. Here
is what is done with a miss match in the real world.
trans-output -> match -> line -> antenna
The 'match' is where the magic happens. All the energy coming down the
line that got reflected from the antenna 'sees' the 'trans-output ->
match' as a perfect reflector and gets bounced back[*]. On the other
side of the match is the trans->output. There the trans->output sees a
perfect impedance, (technically, the conjugate of the trans->output), so
that all the power travels through the match toward the antenna.
The magic is that when the match is tuned, both of the above conditions
are satisfied.
*The reflected wave sees a purely reactive reflector not just because of
the network but also because of the output power of the transmitter.
Without transmitter power the impedance as seen from the load will
dramatically change.
Best, Dan.
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Author: Richard ClarkDate: 15:03 12-03-07
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On Mon, 12 Mar 2007 13:37:00 -0400, "Jimmie D"
<GFENDER@carolina.rr.com> wrote:
>Your mistake is that you assume the output of the tx is 50 ohms,
Hi Jimmie,
At the risk of yet another, non-quantitative reply I will repeat:
>>a question that has NEVER been answered by those who know what the
>>transmitter output Z ISN'T:
>> "What Z is it?"
>in the case
>you stated the transmitter must be matched to the impedance it sees looking
>into the transmission line.
THAT is true, and it brings us to the point of all this energy
sloshing around until the antenna finally dissipates it out into the
Ęther. It is the reflection off the mismatch of the tuner (the
mismatch seen by the antenna as source to the line going back) that
prevents energy from presenting any destructive results to the source
- the whole point of using a tuner in the first place.
73's
Richard Clark, KB7QHC
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Author: Jimmie DDate: 15:12 12-03-07
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"Richard Clark" <kb7qhc@comcast.net> wrote in message
news:ad8bv21rnpi0ij33kssnkscl2k0sqcutt6@4ax.com...
> On Mon, 12 Mar 2007 13:37:00 -0400, "Jimmie D"
> <GFENDER@carolina.rr.com> wrote:
>
>>Your mistake is that you assume the output of the tx is 50 ohms,
>
> Hi Jimmie,
>
> At the risk of yet another, non-quantitative reply I will repeat:
>>>a question that has NEVER been answered by those who know what the
>>>transmitter output Z ISN'T:
>>> "What Z is it?"
>
>>in the case
>>you stated the transmitter must be matched to the impedance it sees
>>looking
>>into the transmission line.
>
> THAT is true, and it brings us to the point of all this energy
> sloshing around until the antenna finally dissipates it out into the
> Ęther. It is the reflection off the mismatch of the tuner (the
> mismatch seen by the antenna as source to the line going back) that
> prevents energy from presenting any destructive results to the source
> - the whole point of using a tuner in the first place.
>
> 73's
> Richard Clark, KB7QHC
Correct but I just want to remember that the purpose of the tuner is to
match the impedance of the transmitter to the impedance of the antenna/
transmission line.The standing waves can be viewed as a reflect voltage, a
reflect current or as a reflected impedance. Besides I thought there had
been enough quanitative analysis of the question and was hoping a simple
answer may be enough to turn on the light bulb for the OP. If he still
wanted to know more I figure he would ask.
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Author: Richard ClarkDate: 16:19 12-03-07
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On Mon, 12 Mar 2007 19:30:59 +0100, "Antonio Vernucci"
<vernucci@tin.it> wrote:
>A simple DC example grossly clarifies thre issue: connect a 12V battery to the
>series of a 50-ohm load and another 12V battery. How much current flows through
>the load? Naught (assuming the correct polarity). So no power is dissipated in
>it.
Hi Tony,
Turn the second battery over. Double the power is dissipated in it.
Phase, you can't live with it, you can't live without it.
73's
Richard Clark, KB7QHC
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Author: Antonio VernucciDate: 16:24 12-03-07
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>>A simple DC example grossly clarifies thre issue: connect a 12V battery to the
>>series of a 50-ohm load and another 12V battery. How much current flows
>>through
>>the load? Naught (assuming the correct polarity). So no power is dissipated in
>>it.
>
> Hi Tony,
>
> Turn the second battery over. Double the power is dissipated in it.
>
> Phase, you can't live with it, you can't live without it.
>
> 73's
> Richard Clark, KB7QHC
Of course. Mine was just a DC example to illustrate things in a simple manner.
When the transmitter is properly tuned, the phase relationship is such that the
reflected wave does not get dissipated at all into the 50 ohm output of the
transmitter, and is then reflected back to the antenna
Tony I0JX
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Author: John Smith IDate: 16:43 12-03-07
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Bob wrote:
> On Mar 12, 1:08 am, "billcalley" <billcal...@yahoo.com> 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
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Author: Jim BackusDate: 18:27 12-03-07
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On Mon, 12 Mar 2007 01:25:00 UTC, Tim Wescott <tim@seemywebsite.com>
wrote:
> * I am _not_ going to start the Big Transmitter Output Impedance Debate.
> sed denizens -- just don't comment on what a transmitter's "actual"
> output impedance may be, lest you start a flame war.
>
OK ;-))
--
Jim Backus running OS/2 Warp 3 & 4, Debian Linux and Win98SE
bona fide replies to j <dot> backus <the circle thingy> jita <dot>
demon <dot> co <dot> uk
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Author: Jimmie DDate: 22:47 12-03-07
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"Dan Bloomquist" <public21@lakeweb.com> wrote in message
news:6ghJh.4761$ya1.2347@news02.roc.ny...
> billcalley wrote:
>
>> We are all told that VSWR doesn't matter when using low loss
>> transmission lines, 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.
>
> As pointed out, VSWR does matter. A lot of bouncing means you heat the
> transmission line with the power instead of radiating the power. 'Doesn't
> matter', really means it can be tolerated if need be.
>
>> 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?
>> For the RF to bounce back and forth, wouldn't the transmitter's
>> impedance have to be very, very high (or low) when the reflected RF
>> energy hit its output stages? I know I'm missing something vital
>> here...
>
> Here is what you are missing. In the case of the output, (real/resistive
> component of the transmitter), seeing the reflected wave, it is _not_
> reflecting that power back up the transmission line as you think it is. It
> would go back to that real impedance and heat the transmitter. Here is
> what is done with a miss match in the real world.
>
> trans-output -> match -> line -> antenna
>
> The 'match' is where the magic happens. All the energy coming down the
> line that got reflected from the antenna 'sees' the 'trans-output ->
> match' as a perfect reflector and gets bounced back[*]. On the other side
> of the match is the trans->output. There the trans->output sees a perfect
> impedance, (technically, the conjugate of the trans->output), so that all
> the power travels through the match toward the antenna.
>
> The magic is that when the match is tuned, both of the above conditions
> are satisfied.
>
> *The reflected wave sees a purely reactive reflector not just because of
> the network but also because of the output power of the transmitter.
> Without transmitter power the impedance as seen from the load will
> dramatically change.
>
> Best, Dan.
>
Saying that SWR doesnt matter is a rather broad statement(like saying never
or always) but I have know of antenna systems having an SWR of 30:1 and his
was normal. The feedline was balanced line made of 1 inch copper. Of course
an SWR lie this on coax could be fatal to coax and equipment. A more common
example of this is the 1/4 wl matching section on a J-pole antenna. It
matches 50 ohms to a few Kohms so an SWR of 60: 1 or so would not be unusal
here.S oas long as the feedline can handle the current and voltage peaks
without much los it doesnt matter much as long as the source impedance is
matched to the impedance at the input to the transmission line.Im sure there
is a practical limit though.
Jimmie
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Author: David G. NagelDate: 23:55 12-03-07
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Tim Wescott wrote:
> billcalley wrote:
>
>> We are all told that VSWR doesn't matter when using low loss
>> transmission lines, 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?
>> For the RF to bounce back and forth, wouldn't the transmitter's
>> impedance have to be very, very high (or low) when the reflected RF
>> energy hit its output stages? I know I'm missing something vital
>> here...
>>
> That's assuming you use an antenna tuner. The tuner will transform the
> transmitter's output impedance* just as it transforms the line. Were
> the transmitter output impedance actually at 50 ohms, on the other side
> of the tuner it would have the same VSWR as the line when everything was
> tuned up.
>
> Having said that, the VSWR _does_ matter somewhat when using low loss
> lines, both because the line loss is low but not zero, and the tuner
> loss will tend to go up as you correct for higher and higher VSWR.
>
> * I am _not_ going to start the Big Transmitter Output Impedance Debate.
> sed denizens -- just don't comment on what a transmitter's "actual"
> output impedance may be, lest you start a flame war.
>
If you want a quick lesson in high vswr find a ham with an old tube
transmitter and see if he will hook it up to a mismatched load. The
cherry red plates are the reflected energy being absorbed. Transistors
will just turn to smoke under the same conditions.
Dave WD9BDZ
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Author: Jerry MartesDate: 00:25 13-03-07
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"David G. Nagel" <nagel@core.com> wrote in message
news:12vc861j53n70d2@corp.supernews.com...
> Tim Wescott wrote:
>> billcalley wrote:
>>
>>> We are all told that VSWR doesn't matter when using low loss
>>> transmission lines, 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?
>>> For the RF to bounce back and forth, wouldn't the transmitter's
>>> impedance have to be very, very high (or low) when the reflected RF
>>> energy hit its output stages? I know I'm missing something vital
>>> here...
>>>
>> That's assuming you use an antenna tuner. The tuner will transform the
>> transmitter's output impedance* just as it transforms the line. Were the
>> transmitter output impedance actually at 50 ohms, on the other side of
>> the tuner it would have the same VSWR as the line when everything was
>> tuned up.
>>
>> Having said that, the VSWR _does_ matter somewhat when using low loss
>> lines, both because the line loss is low but not zero, and the tuner loss
>> will tend to go up as you correct for higher and higher VSWR.
>>
>> * I am _not_ going to start the Big Transmitter Output Impedance Debate.
>> sed denizens -- just don't comment on what a transmitter's "actual"
>> output impedance may be, lest you start a flame war.
>>
> If you want a quick lesson in high vswr find a ham with an old tube
> transmitter and see if he will hook it up to a mismatched load. The cherry
> red plates are the reflected energy being absorbed. Transistors will just
> turn to smoke under the same conditions.
>
> Dave WD9BDZ
Hi david
Wouldnt it be OK to have a high VSWR along the transmission line if the
"tank ckt" can be adjusted to match the load to the transmitter output
impedance? That is, the VSWR along the transmission could concievely be
high, yet, with proper "tank ckt" adjustment that impedance seen by the
output circuit (plate) wouldnt result in a "cherry red plate".
What I am asking is ? is the transmission line VSWR directly related to
"plate reddening"?
I'm more asking than *telling*.
Jerry
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Author: Roy LewallenDate: 00:50 13-03-07
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David G. Nagel wrote:
>>
> If you want a quick lesson in high vswr find a ham with an old tube
> transmitter and see if he will hook it up to a mismatched load. The
> cherry red plates are the reflected energy being absorbed. Transistors
> will just turn to smoke under the same conditions.
Unfortunately, you'd be learning the wrong lesson.
The cherry color is due to the transmitter being loaded with an
impedance it's not designed for, causing the final to run at low
efficiency. You can disconnect the antenna and replace it with a lumped
RC or RL impedance of the same value and get exactly the same result.
Alternatively, you can attach any combination of load and transmission
line which give the same impedance, resulting in a wide variation of
"reflected energy", and get exactly the same result. All that counts is
the impedance seen by the transmitter, not the VSWR on the line or the
"reflected power".
The problem is that the idea of "reflected energy" turning the plates
hot is so easy to understand, that people aren't willing to abandon it
simply because it isn't true.
See http://eznec.com/misc/Food_for_thought.pdf for more.
Roy Lewallen, W7EL
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Author: Roy LewallenDate: 00:53 13-03-07
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Jerry Martes wrote:
>
> Wouldnt it be OK to have a high VSWR along the transmission line if the
> "tank ckt" can be adjusted to match the load to the transmitter output
> impedance? That is, the VSWR along the transmission could concievely be
> high, yet, with proper "tank ckt" adjustment that impedance seen by the
> output circuit (plate) wouldnt result in a "cherry red plate".
Yes! All that matters to the transmitter is the impedance it sees. It
doesn't know or care that you've mathematically separated the delivered
power into "forward" and "reverse" components. It doesn't know or
care
what the SWR is on the transmission line connected to it, or even if a
transmission line is connected at all.
> What I am asking is ? is the transmission line VSWR directly related to
> "plate reddening"?
Absolutely not.
> I'm more asking than *telling*.
That's the first step in learning.
Roy Lewallen, W7EL
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Author: BobDate: 01:32 13-03-07
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> > 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
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Author: David G. NagelDate: 01:34 13-03-07
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Jerry Martes wrote:
> "David G. Nagel" <nagel@core.com> wrote in message
> news:12vc861j53n70d2@corp.supernews.com...
>> Tim Wescott wrote:
>>> billcalley wrote:
>>>
>>>> We are all told that VSWR doesn't matter when using low loss
>>>> transmission lines, 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?
>>>> For the RF to bounce back and forth, wouldn't the transmitter's
>>>> impedance have to be very, very high (or low) when the reflected RF
>>>> energy hit its output stages? I know I'm missing something vital
>>>> here...
>>>>
>>> That's assuming you use an antenna tuner. The tuner will transform the
>>> transmitter's output impedance* just as it transforms the line. Were the
>>> transmitter output impedance actually at 50 ohms, on the other side of
>>> the tuner it would have the same VSWR as the line when everything was
>>> tuned up.
>>>
>>> Having said that, the VSWR _does_ matter somewhat when using low loss
>>> lines, both because the line loss is low but not zero, and the tuner loss
>>> will tend to go up as you correct for higher and higher VSWR.
>>>
>>> * I am _not_ going to start the Big Transmitter Output Impedance Debate.
>>> sed denizens -- just don't comment on what a transmitter's
"actual"
>>> output impedance may be, lest you start a flame war.
>>>
>> If you want a quick lesson in high vswr find a ham with an old tube
>> transmitter and see if he will hook it up to a mismatched load. The cherry
>> red plates are the reflected energy being absorbed. Transistors will just
>> turn to smoke under the same conditions.
>>
>> Dave WD9BDZ
>
> Hi david
>
> Wouldnt it be OK to have a high VSWR along the transmission line if the
> "tank ckt" can be adjusted to match the load to the transmitter output
> impedance? That is, the VSWR along the transmission could concievely be
> high, yet, with proper "tank ckt" adjustment that impedance seen by the
> output circuit (plate) wouldnt result in a "cherry red plate".
> What I am asking is ? is the transmission line VSWR directly related to
> "plate reddening"?
> I'm more asking than *telling*.
>
> Jerry
>
>
Jerry;
The point I was trying to make is that the reflected current is
disapated as heat in the finals if the transmitter isn't matched to the
load.
In a tube radio the tank circuit is the equivilent of an antenna
match/tuner and converts the 2000 or so ohms at the plate to the 50 ohms
of the transmission line and the unknown ohms of the mis matched antenna.
Dave WD9BDZ
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Author: Owen DuffyDate: 01:34 13-03-07
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don@manx.misty.com (Don Klipstein) wrote in
news:slrneva0no.m7j.don@manx.misty.com:
...
> 1) The transmitter may well have output impedance matching the
> characteristic impedance of the transmission line. RF power reflected
> back in this case gets converted to heat in the output stage of the
> transmitter, in addition to whatever heat the output stage already has
> to dissipate.
...
I thought we put one of these fires out just a few days ago!
Owen
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Author: BobDate: 02:00 13-03-07
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On Mar 12, 4:56 am, Richard Clark <kb7...@comcast.net> wrote:
> On 11 Mar 2007 20:39:46 -0700, "Bob" <b...@mailinator.com> 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
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