"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

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

H. Adam Stevens, NQ5H wrote:
Roy
Worst case.
Get it?
H.

No, sorry, I don't. The result will be as I posted in the best case,
worst case, and typical case.

What conditions would cause it to be different?

Roy Lewallen, W7EL

Good Lord Roy, I thought you knew better.

If the match at the load is not perfect, energy is refleced back to the
source, are you with me so far?

I can easily build a source that absorbs all the reflected power: A zero
impedance source in series with a resistor that matches the transmission
line impedance.

73
H.


"Roy Lewallen" wrote in message
...
H. Adam Stevens, NQ5H wrote:
Roy
Worst case.
Get it?
H.

No, sorry, I don't. The result will be as I posted in the best case,
worst case, and typical case.

What conditions would cause it to be different?

Roy Lewallen, W7EL

H. Adam Stevens, NQ5H wrote:
Good Lord Roy, I thought you knew better.

If the match at the load is not perfect, energy is refleced back to the
source, are you with me so far?

I can easily build a source that absorbs all the reflected power: A zero
impedance source in series with a resistor that matches the transmission
line impedance.


Let's see, I put a 100 volt zero impedance source in series with a 50
ohm resistor, connect that to a half wave transmission line terminated
with 150 ohms. The current will be 100/200 = 0.5 amp, the power in the
150 ohm load is 37.5 watts, the power in the 50 ohm source resistor is
12.5 watts. The SWR is 3:1, the forward power is 50 watts, the reverse
power is 12.5 watts. Sure enough, the power in the source resistor
equals the reverse power. Good job. That sure must be the worst case,
all right.

Just to check, I'll change the load resistor to 16.67 ohms. Now the
current is 1.5 amps, the power in the 16.67 ohm load is 37.5 watts, and
the power in the source resistor is 112.5 watts. The SWR is still 3:1,
the forward power is 50 watts just like before, and the reverse power is
12.5 watts just like before. Hm. The reverse power is 12.5 watts, but
the source resistor is now dissipating 112.5 watts. Must be worse than
the worst case.

Well, shoot, maybe the source resistor dissipates all the reverse power
*plus* some more power that comes from somewhere else. So let's try a
200 ohm load. Now the current is 0.4 amp, the power in the 200 ohm load
resistor is 32 watts, and the power in the 50 ohm source resistor is 8
watts. The SWR is 4:1, the forward power is 50 watts, and the reverse
power is 18 watts. Oops, the source resistor is only dissipating 8 watts
but the reverse power is 18 watts. Not only isn't it dissipating all the
reverse power, but it isn't even dissipating that extra power that came
from somewhere else when we connected the 16.67 ohm resistor. Wonder
where the other 10 watts of reverse power went?(*)

So using your simple criterion of a zero impedance source and resistor
equal to the transmission line impedance, and by only changing the load
resistance, we've got cases whe

-- The source resistor dissipation equals the reverse power
-- The source resistor dissipation is greater than the reverse power
-- The source resistor dissipation is less than the reverse power

And none of these will explain the loss figure you gave earlier.

Guess I don't know better after all.

Anyone who's interested can find more interesting cases in "Food for
thought - Forward and Reverse Power.txt" at
http://eznec.com/misc/food_for_thought/. And those who aren't
interested, well, you're welcome to believe what you choose. Just don't
look too closely at the evidence.

(*) Anybody fond of the notion that reverse power "goes" somewhere or
gets dissipated in the source or re-reflected back needs to come to
grips with this problem before building further on the flawed model of
bouncing waves of flowing power.

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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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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Fred W4JLE wrote:
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

!.7:1???? I must admit, that's a new one on me. :-)

My statements were general, and apply to any SWR.
--
73, Cecil http://www.qsl.net/w5dxp

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