I don't really agree with this statement at high power levels.
An high level amplifier is not designed ONLY at a stated output power, but
it's designed to be at the highest gain possible and for the best possible
efficiency at such output. (Imagine a 2 MW Amp that wouldn't be designed
under this concept....)
Designing low level RF amplifiers, it's often required that a single stage
provide an exact and specific amount of gain in spite of device's individual
characteristcs. In such case a simultaneus conjugate match when using a well
chosen (in gain) tube or transistor would probably give different and too
much gain in most of the cases depending by tubes or transistors.
Mismatching the output to a specific gain (the constant gain circle on the
Smith chart gives a lot of impedances) the problem is solved
Although selective mismatching in RF amplifiers is not an heresy, but a
practical, logical and controlled way of decreasing (fixing) stage gain,
efficiency is something unavoidable at high power level in order not to
waste energy and not to stress components without any reason.
Big power amplifiers must be instead maximally efficient, and if the output
exceed the designed level the way to reduce it is not found, i.e.,
mismatching the output but instead obtained reducing the driving power.
This is often done automatically if Amp owner connects the ALC cables to the
sockets ad performs the regulations instead to "turn clockwise" (counter
clockwise in the UK) potentiometers to squeeze the last watt from the Amp.
73,
Mauri I4JMY
-----Messaggio Originale-----
Da: Roger D. Johnson <n1rj@pivot.net>
A: Jon Ogden <jono@enteract.com>
Cc: Amps Reflector <amps@contesting.com>
Data invio: venerdì 1 ottobre 1999 5.51
Oggetto: [AMPS] Conjugate matching and amplifiers
>
> Poor Jon is going to think I'm picking on him but I must disagree on his
> statements about
> conjugate matching of amplifiers. In reality the PA tube is only matched
> to the extent that
> it produces the desired power output. We can see this by the design
> steps required. First,
> the output power is selected. Then we pick a tube and a reasonable anode
> voltage to
> produce that power. Then we compute the necessary current and multiply
> that by a fudge
> factor for the class of operation. The anode voltage is divided by the
> previous value and
> that gives the impedance the tube must work into to produce the desired
> power output. A
> matching network is then designed to transform that impedance to 50 ohms
> (usually).
> Notice that at no time was the actual output impedance or load
> resistance of the tube used!
> The fact that the tube is not "matched" to the load also explains why
> most of the reflected
> power is again reflected at the amplifier and goes back towards the
> antenna.
>
> 73, Roger
>
>
>
> --
> FAQ on WWW: http://www.contesting.com/ampsfaq.html
> Submissions: amps@contesting.com
> Administrative requests: amps-REQUEST@contesting.com
> Problems: owner-amps@contesting.com
> Search: http://www.contesting.com/km9p/search.htm
>
--
FAQ on WWW: http://www.contesting.com/ampsfaq.html
Submissions: amps@contesting.com
Administrative requests: amps-REQUEST@contesting.com
Problems: owner-amps@contesting.com
Search: http://www.contesting.com/km9p/search.htm
|