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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>>>
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