Owen Duffy wrote:
You didn't differentiate between the lossy and lossless solutions. The
difference isn't very much, I figure about 1100-j2000 vs 1000-j2170
which is real hard to resolve on the Smith chart.

Owen, it's pretty obvious that you were manufactured with a
lower tolerance (+/- 0.1%) than I was (+/- 20%). :-) Over on
QRZ.com, I suggested that the average amateur radio operator
has a hard time implementing a choke in a high-impedance
environment. W8JI said he could do it. I responded that, by
no stretch of the imagination, was he an "average amateur
radio operator". :-)
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 30 Nov 2005 17:08:27 GMT, Cecil Moore wrote:


Owen, it's pretty obvious that you were manufactured with a
lower tolerance (+/- 0.1%) than I was (+/- 20%). :-) Over on

Perhaps...

Now, I think you have told us over several articles that you are using
an SGC500 into a 30 ohm load on 7.15MHz. That 30 ohm load is a result
of feeding a dipole with 90' of ladderline, which I estimate has 0.9dB
of loss under those circumstances.

If your transmitter was delivering 500W to the feedline, about 100W is
lost in the feeder.

Do you know how much power your amplifier delivers to the feedline? It
is likely that with a load VSWR of 1.7 it may have reduced output, it
is also possible that it is delivering even more than 500W to the low
Z load.

Owen
--

Owen Duffy wrote:
Now, I think you have told us over several articles that you are using
an SGC500 into a 30 ohm load on 7.15MHz.

Now please be a gentleman and please don't go putting words in my mouth.
Here a quote of my exact words:

"Speaking of indirect measurements - let's say the feedline Z0 is 380
ohms with a VF of 0.9 and a length of 90 ft. The measured resistance
at the current maximum point is 30 ohms on 7.15 MHz."

Clearly, "let's say", is a hypothetical postulate. I freely admit that
I pulled those values out of thin air. Going back to my web page reveals
that the feedpoint impedance on 40m for my 130 ft. dipole was really 38
ohms. Nonetheless, I can still make my point assuming the 30 ohm value
which would have been perfectly acceptable to me.

If your transmitter was delivering 500W to the feedline, about 100W is
lost in the feeder.

With a 1.7:1 SWR???? Maybe you should reprogram your calculator to take
the square root???? The ratio of Pref/Pfor for an SWR of 1.7:1 is 0.067.
Methinks you might be using the voltage reflection coefficient?

500(0.067) is 34 watts, not 100 watts. The SGC-500 laughes at 34 watts
reflected. (I swear that is true. I have heard it laughing to itself in
the wee hours during a contest.) Seriously, that amp is not known as
"The Brick" just because it looks like a brick.

Do you know how much power your amplifier delivers to the feedline? It
is likely that with a load VSWR of 1.7 it may have reduced output, it
is also possible that it is delivering even more than 500W to the low
Z load.

An SWR of 1.7:1 is nothing to worry about unless you think the percentage
power reflected is the same as the percentage voltage reflected. Don't feel
bad, many others have made that same mistake. Most people are programmed not
to think within a power/energy context and it gets them into trouble with
such concepts as "reflected power just sloshes around from side-to-side" and
"gobbledegook" applied to any attempt to track energy in a transmision line.

The SGC-500 is speced to tolerate an SWR of 6:1. That means that it can
dissipate more than half of its output power and keep on ticking. I
don't recommend allowing that to happen but that spec is why I don't worry
at all about reflected power unless the SWR is in excess of 2:1.

If we keep arguing, one of us is bound to make a mistake that the other
catches. I would guess that your above mistake bothers you a lot more
than it bothers me. :-)
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:

Owen Duffy wrote:
If your transmitter was delivering 500W to the feedline, about 100W is
lost in the feeder.

With a 1.7:1 SWR???? Maybe you should reprogram your calculator to take
the square root???? The ratio of Pref/Pfor for an SWR of 1.7:1 is 0.067.
Methinks you might be using the voltage reflection coefficient?

Sorry, I read that as 100W lost in the amplifier. Macular Degeneration
strikes again.
--
73, Cecil http://www.qsl.net/w5dxp

On Wed, 30 Nov 2005 21:35:32 GMT, Cecil Moore wrote:

Owen Duffy wrote:
Now, I think you have told us over several articles that you are using
an SGC500 into a 30 ohm load on 7.15MHz.

Now please be a gentleman and please don't go putting words in my mouth.

Well, there was some uncertainty, and it is why I opened with "I
think...".

Here a quote of my exact words:

"Speaking of indirect measurements - let's say the feedline Z0 is 380
ohms with a VF of 0.9 and a length of 90 ft. The measured resistance
at the current maximum point is 30 ohms on 7.15 MHz."

Clearly, "let's say", is a hypothetical postulate. I freely admit that
I pulled those values out of thin air. Going back to my web page reveals
that the feedpoint impedance on 40m for my 130 ft. dipole was really 38
ohms. Nonetheless, I can still make my point assuming the 30 ohm value
which would have been perfectly acceptable to me.

If your transmitter was delivering 500W to the feedline, about 100W is
lost in the feeder.

With a 1.7:1 SWR???? Maybe you should reprogram your calculator to take
the square root???? The ratio of Pref/Pfor for an SWR of 1.7:1 is 0.067.
Methinks you might be using the voltage reflection coefficient?

Ok, I saw your later post where you have note that you were on the
wrong track here.


500(0.067) is 34 watts, not 100 watts. The SGC-500 laughes at 34 watts
reflected. (I swear that is true. I have heard it laughing to itself in
the wee hours during a contest.) Seriously, that amp is not known as
"The Brick" just because it looks like a brick.

Do you know how much power your amplifier delivers to the feedline? It
is likely that with a load VSWR of 1.7 it may have reduced output, it
is also possible that it is delivering even more than 500W to the low
Z load.

An SWR of 1.7:1 is nothing to worry about unless you think the percentage
power reflected is the same as the percentage voltage reflected. Don't feel
bad, many others have made that same mistake. Most people are programmed not
to think within a power/energy context and it gets them into trouble with
such concepts as "reflected power just sloshes around from side-to-side" and
"gobbledegook" applied to any attempt to track energy in a transmision line.

The SGC-500 is speced to tolerate an SWR of 6:1. That means that it can
dissipate more than half of its output power and keep on ticking. I
don't recommend allowing that to happen but that spec is why I don't worry
at all about reflected power unless the SWR is in excess of 2:1.

Lets leave that issue alone.


If we keep arguing, one of us is bound to make a mistake that the other
catches. I would guess that your above mistake bothers you a lot more
than it bothers me. :-)

Well, I think we are agreed that you made a mistake in identifying a
mistake, if I am not mistaken!

No, back on track, I thought you might have measured forward and
reflected power at the amplifier output on the 30 ohm load, to deduce
the net forward power, then by allowing for the line loss, you would
have the net power at the feedpoint (most of which will be radiated in
some direction or another).

One could then calculate the performance of the feed configuration
compared to what would be delivered to an ideal nominal load with no
feed loss.

The whole excercise goes nowhere, because it seems that the 30 ohms
scenario is "hypothetical".

Owen
--

Owen Duffy wrote:
Well, I think we are agreed that you made a mistake in identifying a
mistake, if I am not mistaken!

Your original mistake was in making a posting that caused me to
misunderstand. :-) Your loss calculator gives 0.762 dB loss for
90 feet of Wireman #445 with an SWR of 12.7:1 on 7.15 MHz.

The whole excercise goes nowhere, because it seems that the 30 ohms
scenario is "hypothetical".

I knew it was thirty-something ohms so I said "30". It was
actually 38 ohms indicating an SWR at the source of 1.3:1.
That's acceptable losses for me. If I used an antenna tuner
to achieve a 1:1 match, it would probably be a wash.

Which is probably a good question. At what SWR should one
install an antenna tuner? My IC-706 seems perfectly
happy at 2:1.
--
73, Cecil http://www.qsl.net/w5dxp

Reg Edwards wrote:

Gerry,

The load is the antenna - about which the SWR meter knows absolutely
nothing. All the the meter has to work with is the input impedance of
the tuner or the transmission line.

Line input Z = R+jX and to aggravate matters the meter discards all
information about X.
----
Reg.



The load that counts is what the transmitter sees. Which is what his
instrument measures.

tom
K0TAR

On Wed, 30 Nov 2005 23:42:33 GMT, Cecil Moore wrote:

Owen Duffy wrote:
Well, I think we are agreed that you made a mistake in identifying a
mistake, if I am not mistaken!

Your original mistake was in making a posting that caused me to
misunderstand. :-) Your loss calculator gives 0.762 dB loss for
90 feet of Wireman #445 with an SWR of 12.7:1 on 7.15 MHz.

Now, you're trying to trick me... did you mean Wireman 554?

It does show 0.76dB for 90' of 554 with a 30+j0 input Z at 7.15MHz.

The 0.9dB stated earlier was (as stated) for 552 which had a Zo closer
to your 380 ohms, whereas 554 is 360 ohms.


The whole excercise goes nowhere, because it seems that the 30 ohms
scenario is "hypothetical".

I knew it was thirty-something ohms so I said "30". It was
actually 38 ohms indicating an SWR at the source of 1.3:1.
That's acceptable losses for me. If I used an antenna tuner
to achieve a 1:1 match, it would probably be a wash.

Which is probably a good question. At what SWR should one
install an antenna tuner? My IC-706 seems perfectly
happy at 2:1.

Given that some radios (including the IC706-IIG) reduce drive power at
high VSWR as a protection mechanism, you may want to install an ATU to
develop full power output. My recollection is that the power output of
the IC706-IIG is significantly down at VSWR=2, but it probably also
depends on the actual load impedance.

A likely scenario could be that the radio only develops 50W of output,
and only 40W makes it to the feedpoint (assuming 1dB feed loss). The
ATU might raise that to 100W of output, less tuner loss and feed loss
giving 75 to 80W at the feedpoint. Mere fraction of an S point... but
if you have a 100W transmitter, might as well use it, and besides, the
technical challenge of achieving that goal and measuring the
achievement might be part of what amateur radio is about.

Owen
--

The load that counts is what the transmitter sees. Which is what
his
instrument measures.

tom
K0TAR

========================================

I fully agree the transmitter load is what the transmitter sees.

But his meter does not measure it.

His meter simply tells him whether or not its resistance is 50 ohms.

Which is all he wants to know anyway.
----
Reg, G4FGQ

On Thu, 1 Dec 2005 06:46:58 +0000 (UTC), "Reg Edwards"
wrote:

But his meter does not measure it.

Jerry has a slotted line - described here in loving detail some time
ago. It and the meter so attached is perfectly capable of measuring
all the factors stated.

Which is all he wants to know anyway.

which is an assumption (or a forced argument, take your pick) as he
has already reconciled any mis-understandings.