"Bill Turner" wrote in message
...
On Fri, 3 Jun 2005 15:53:31 -0400, "Fred W4JLE" wrote:

First, I would say that 1.7 to 1 is fine, leave it alone. It is what is
expected.

__________________________________________________ ____________

Fred is correct and you can prove it to your own satisfaction if you
like:

Place a field strength meter nearby, close enough so you can read the
meter, and sweep your transmitter across the band. You will find your
power output is remarkably constant whether the SWR is 1:1 or 1.7:1.

There will be some variation of course, but when you find the bandwidth
where it drops no more than about 90% or so, you can operate confidently
anywhere in that region without worrying about SWR.

Works for me.

--
Bill, W6WRT

Just a quick back-of-the-envelope calculation.

SWR' = 1.7

or SWR" = 1.0

Let's assume worst case; all the reflected power is absorbed in the source.
This is not necesssarily the case, but gives us the least signal strength in
the high SWR case, SWR'.

So then, comparing the two cases, the change in power to the load in db is
10*log(SWR'/SWR").

SWR'/SWR" = 1.7

2.3 db, barely detectable, worst case.

So it's a question of how much reflected power can the rig tolerate as well.

73
H.
NQ5H

H. Adam Stevens, NQ5H wrote:

Just a quick back-of-the-envelope calculation.

SWR' = 1.7

or SWR" = 1.0

Let's assume worst case; all the reflected power is absorbed in the source.
This is not necesssarily the case, but gives us the least signal strength in
the high SWR case, SWR'.

So then, comparing the two cases, the change in power to the load in db is
10*log(SWR'/SWR").

SWR'/SWR" = 1.7

2.3 db, barely detectable, worst case.

So it's a question of how much reflected power can the rig tolerate as well.

That calculation completely baffles me.

The ALC in my Icom rig keeps the forward power constant up to the point
where it reduces power, at around 3:1 SWR. This is typical for
commercial rigs. The rig delivers 100 watts to a 1:1 SWR load.
(Techically, this really means a load which, if terminating a 50 ohm
line, will produce 1:1 SWR on that line.) 1.7:1 SWR is a reverse/forward
power ratio of 0.067. The ALC keeps the forward power at 100 watts, so
with 1.7:1 SWR, the reverse power is 6.7 watts. The net power delivered
to the load is 100 - 6.7 = 93.3 watts, or 0.3 dB, not 2.3 dB, less than
the power delivered to a 1:1 load.

Oh, and the "reverse power" isn't "absorbed in the source". Anyone
interested in learning more about this might take a look at "Food for
thought - Forward and Reverse Power.txt" at
http://eznec.com/misc/food_for_thought/.

Roy Lewallen, W7EL

I would disagree with your statement about SWR being absorbed in the source.
The reflected wave is rereflected and aside from losses caused by the 2 way
trip is reradiated.

While your math is correct, your application in my opinion is incorrect.


"H. Adam Stevens, NQ5H" wrote in message
...

"Bill Turner" wrote in message
...
On Fri, 3 Jun 2005 15:53:31 -0400, "Fred W4JLE" wrote:

First, I would say that 1.7 to 1 is fine, leave it alone. It is what is
expected.

__________________________________________________ ____________

Fred is correct and you can prove it to your own satisfaction if you
like:

Place a field strength meter nearby, close enough so you can read the
meter, and sweep your transmitter across the band. You will find your
power output is remarkably constant whether the SWR is 1:1 or 1.7:1.

There will be some variation of course, but when you find the bandwidth
where it drops no more than about 90% or so, you can operate confidently
anywhere in that region without worrying about SWR.

Works for me.

--
Bill, W6WRT

Just a quick back-of-the-envelope calculation.

SWR' = 1.7

or SWR" = 1.0

Let's assume worst case; all the reflected power is absorbed in the
source.
This is not necesssarily the case, but gives us the least signal strength
in
the high SWR case, SWR'.

So then, comparing the two cases, the change in power to the load in db is
10*log(SWR'/SWR").

SWR'/SWR" = 1.7

2.3 db, barely detectable, worst case.

So it's a question of how much reflected power can the rig tolerate as
well.

73
H.
NQ5H

Fred W4JLE wrote:
I would disagree with your statement about SWR being absorbed in the source.
The reflected wave is rereflected and aside from losses caused by the 2 way
trip is reradiated.

Reflected current flowing into the final amp can superpose in
phase with the forward current and, without protection circuitry,
cause the final amp to exceed its dissipation rating.

Another possibility is when the reflected voltage superposes
in phase with the forward voltage, and without protection
circuitry, exceeds the voltage rating of the final.

If all reflected power was always re-reflected, there would
be no need for protection circuitry. The generated power is
*defined* as the forward power minus the reflected power. That
does NOT mean that the reflected power is 100% re-reflected.
--
73, Cecil http://www.qsl.net/w5dxp

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On Sat, 4 Jun 2005 22:02:48 -0400, "Fred W4JLE"
wrote:

I would disagree with your statement about SWR being absorbed in the source.
The reflected wave is rereflected

Hi Fred,

I've seen this thesis offered before. I generally ask, since this is
exactly the same thesis offered for a conjugate match offered by a
tuner, and a tuner is used for this very purpose (rereflecting the
mismatch), if the Transmitter already does it - What is the Tuner for?

Why do you use a tuner? What is a match? Why would anyone seek to
match a Transmitter to its load?

73's
Richard Clark, KB7QHC

Cecil, for every question, there are any number of answers if you refashion
the question. In this case the SWR was defined as !.7:1


"Cecil Moore" wrote in message
...
Fred W4JLE wrote:
I would disagree with your statement about SWR being absorbed in the
source.
The reflected wave is rereflected and aside from losses caused by the 2
way
trip is reradiated.

Reflected current flowing into the final amp can superpose in
phase with the forward current and, without protection circuitry,
cause the final amp to exceed its dissipation rating.

Another possibility is when the reflected voltage superposes
in phase with the forward voltage, and without protection
circuitry, exceeds the voltage rating of the final.

If all reflected power was always re-reflected, there would
be no need for protection circuitry. The generated power is
*defined* as the forward power minus the reflected power. That
does NOT mean that the reflected power is 100% re-reflected.
--
73, Cecil http://www.qsl.net/w5dxp

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You have created an all encompassing case from the original SWR of 1.7:1 No
tuner needed. Under these conditions, the amount of reflected energy
radiated approaches Ivory Soap.


"Richard Clark" wrote in message
...
On Sat, 4 Jun 2005 22:02:48 -0400, "Fred W4JLE"
wrote:

I would disagree with your statement about SWR being absorbed in the
source.
The reflected wave is rereflected

Hi Fred,

I've seen this thesis offered before. I generally ask, since this is
exactly the same thesis offered for a conjugate match offered by a
tuner, and a tuner is used for this very purpose (rereflecting the
mismatch), if the Transmitter already does it - What is the Tuner for?

Why do you use a tuner? What is a match? Why would anyone seek to
match a Transmitter to its load?

73's
Richard Clark, KB7QHC

On Sun, 5 Jun 2005 10:02:17 -0400, "Fred W4JLE"
wrote:

You have created an all encompassing case from the original SWR of 1.7:1

Thanx Fred,

No one has ever been able to answer when the Transmitter rereflects
energy, why they need a tuner.

73's
Richard Clark, KB7QHC

Roy
Worst case.
Get it?
H.

"Roy Lewallen" wrote in message
...
H. Adam Stevens, NQ5H wrote:

Just a quick back-of-the-envelope calculation.

SWR' = 1.7

or SWR" = 1.0

Let's assume worst case; all the reflected power is absorbed in the
source.
This is not necesssarily the case, but gives us the least signal
strength in
the high SWR case, SWR'.

So then, comparing the two cases, the change in power to the load in db
is
10*log(SWR'/SWR").

SWR'/SWR" = 1.7

2.3 db, barely detectable, worst case.

So it's a question of how much reflected power can the rig tolerate as
well.

That calculation completely baffles me.

The ALC in my Icom rig keeps the forward power constant up to the point
where it reduces power, at around 3:1 SWR. This is typical for
commercial rigs. The rig delivers 100 watts to a 1:1 SWR load.
(Techically, this really means a load which, if terminating a 50 ohm
line, will produce 1:1 SWR on that line.) 1.7:1 SWR is a reverse/forward
power ratio of 0.067. The ALC keeps the forward power at 100 watts, so
with 1.7:1 SWR, the reverse power is 6.7 watts. The net power delivered
to the load is 100 - 6.7 = 93.3 watts, or 0.3 dB, not 2.3 dB, less than
the power delivered to a 1:1 load.

Oh, and the "reverse power" isn't "absorbed in the source". Anyone
interested in learning more about this might take a look at "Food for
thought - Forward and Reverse Power.txt" at
http://eznec.com/misc/food_for_thought/.

Roy Lewallen, W7EL

Like I said to Roy.
"Worst case."
Get it?
OF COURSE it isn't absorbed in the source.
H.


"Fred W4JLE" wrote in message
...
I would disagree with your statement about SWR being absorbed in the
source.
The reflected wave is rereflected and aside from losses caused by the 2
way
trip is reradiated.

While your math is correct, your application in my opinion is incorrect.


"H. Adam Stevens, NQ5H" wrote in message
...

"Bill Turner" wrote in message
...
On Fri, 3 Jun 2005 15:53:31 -0400, "Fred W4JLE"
wrote:

First, I would say that 1.7 to 1 is fine, leave it alone. It is what
is
expected.

__________________________________________________ ____________

Fred is correct and you can prove it to your own satisfaction if you
like:

Place a field strength meter nearby, close enough so you can read the
meter, and sweep your transmitter across the band. You will find your
power output is remarkably constant whether the SWR is 1:1 or 1.7:1.

There will be some variation of course, but when you find the
bandwidth
where it drops no more than about 90% or so, you can operate
confidently
anywhere in that region without worrying about SWR.

Works for me.

--
Bill, W6WRT

Just a quick back-of-the-envelope calculation.

SWR' = 1.7

or SWR" = 1.0

Let's assume worst case; all the reflected power is absorbed in the
source.
This is not necesssarily the case, but gives us the least signal
strength
in
the high SWR case, SWR'.

So then, comparing the two cases, the change in power to the load in db
is
10*log(SWR'/SWR").

SWR'/SWR" = 1.7

2.3 db, barely detectable, worst case.

So it's a question of how much reflected power can the rig tolerate as
well.

73
H.
NQ5H