Good evening, I am still fiddling and pondering the observations on this
problem. Gary's observation means that no matter the band the input
impedance is the same except for the "intervention" of the coax between the
exciter and the AMP. That is hard to believe, b/c my 8"jumper cable seems
to keep the 160, 80 and 20 meter input "happy" (No SWR) but on 40 it all
goes to "pot."
I can't believe that TT built non-matching rigs in the 1990s. Something
else has to be wrong.
Kris KM2KM
Merschrod
123 Warren Road
Ithaca, NY 14850
www.merschrod.net
----- Original Message -----
From: "CSM(r) Gary Huber" <glhuber@msn.com>
To: "Discussion of Ten-Tec Equipment" <tentec@contesting.com>
Sent: Monday, May 31, 2010 10:16 PM
Subject: Re: [TenTec] Ten-Tec 411 Centaur Amp Problem
In the case of the Centaur 411, the input is simply a 25 Ohm resistor in
parallel with the cathodes of the 4 811s (which are also in parallel).
The input load impedance of the Centaur is always less than 25 Ohms and
thus
the length of input coaxial cable plays a significant role in transforming
the SWR for the driving solid state transmitter.
73,
Gary - AB9M
(NOT an E.E.)
--------------------------------------------------
From: <wow_chf@hotmail.com>
Sent: Monday, May 31, 2010 8:32 PM
To: "Discussion of Ten-Tec Equipment" <tentec@contesting.com>
Subject: Re: [TenTec] Ten-Tec 411 Centaur Amp Problem
Yes indeed. In fact, there were two different lengths in the Collins
30S1
manual, one for their transceiver and one for their transmitter.
In my article the explanation as to why this works is as follows:
"In cases involving RF signals, some time will pass during the 'round
trip
of the reflected energy and the phase of the reflection will also depend
upon this length of time. Imagine that a resistor in a black box is at
the
end of a length of cable. From the outside world this length of cable
will
give the reflection from the resistor a phase shift since the signal must
make a round trip through the length. If a 100 ohm resistor has an SWR of
2,
a cable long enough to invert the signal after the round trip will make
it
look like a 25 ohm resistor, also with an SWR of 2 but with inversion (a
cable with a multiple of 1/4 wavelength would do the trick). Since the
impedance looking into this black box is a function of the SWR and the
cable
length, it can be seen that intentionally mismatched lines can be used to
transform one impedance into another. Notice that the 1/4 wave cable
inverts
the impedance and preserves the SWR. This impedance inversion may be used
to
match two impedances at a particular frequency by connecting them with a
1/4
wave cable with an impedance equal to the geometric mean of the two
impedances. (The geometric mean is the square-root of their product.) A
50
ohm, 1/4 wave cable will match a 25 ohm source to a 100 ohm load :
sqrt(25
x
100) = 50. Of course, it is not always easy to find the desired impedance
cable!
Multiples of 1/2 wavelength will give enough delay that the reflection is
not inverted and the impedance will be the same as the load. Such cables
may
be used to transfer the load impedance to a remote location without
changing
its value (at one frequency).
Other cable lengths will transform an impedance which differs from the
cable's impedance with a reactive component. If the load is a lower
impedance than the cable, a length below 1/4 wave will have an inductive
component and above 1/4 (but below 1/2) wave a capacitive component. If
the
load is a higher impedance than the cable, the reverse is true. Above 1/2
wavelength, the reactance will alternately look capacitive and inductive
in
1/4 wave multiples. This reactance will combine with the load's reactance
and offers the possibility of resonating the reactive component of the
load.
Therefore, a cable with the "right" length and impedance can match a
source
and load with different resistance and reactance values. Obviously, these
calculations can become quite involved and most engineers resort to a
Smith
chart, a computer program or perhaps the most common method, trial and
error
with a SWR meter or return loss bridge!"
Sorry for so much bandwidth.
73 and Happy DXing,
Mike
W2AJI
-------------------------------------------------
From: "James Duffer" <dufferjames@hotmail.com>
Sent: Monday, May 31, 2010 8:15 PM
To: "Ten Tec" <tentec@contesting.com>
Subject: Re: [TenTec] Ten-Tec 411 Centaur Amp Problem
I recall from some Collins manual for their amp 30S1 they specified a
length for the coaxial cable between the KWM-2 and the amp. In that
case
that was their fix.
Jim de wd4air
From: k9yc@audiosystemsgroup.com
To: tentec@contesting.com
Date: Mon, 31 May 2010 13:19:38 -0700
Subject: Re: [TenTec] Ten-Tec 411 Centaur Amp Problem
On Mon, 31 May 2010 15:55:31 -0400, wow_chf@hotmail.com wrote:
>I submitted the article draft to QST, and it was questioned by the
>Technical
>Review folks, and although I have been published in QST before, they
>could
>not see how this would "transform" the apparent input SWR.
Because it's a POOR fix for the fundamental problem, which is something
wrong
in the input circuit of the power amp causing a mismatch. The proper
fix
is
to find and correct the problem in the input circuit. Adding coax is a
band-
aid.
73,
Jim K9YC
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