Re: Can I measure antenna impedance with K2?

Posted by David Woolley (E.L) on Aug 01, 2014; 9:17am
URL: http://elecraft.85.s1.nabble.com/Can-I-measure-antenna-impedance-with-K2-tp7591798p7591845.html

The efficiency would be even less, in practice, if designed that way.

The power transfer theorem isn't useful in most real world cases.  In
practice, the optimum load for a PA will be very different from its AC
resistance (the reactive part should be corrected for by the fixed part
of the output matching circuitry).  For an AF PA the AC resistance will
generally be much lower than the optimum load line resistance, because
of the heavy feedback used.  For RF PA's there will be lots of parasitic
impedances and relatively little, if any, feedback will be applied, so
the situation is more complex.  Without feedback, I have a feeling the
resistive component will be high, as it would be at AF, without
feedback, but I've not looked into the details.  (The K2 low power PA
does have some negative feedback, so it is possible that it acts as a
voltage source, i.e. has an AC resistance much lower than optimum load.)

Generally, however, the optimum load is determined much more by the
target output power and the available power supply voltage. To a first
approximation, for ideal class B push pull, you want Vcc**2/RL to be
twice(?) the PEP.  You then transform 50 ohms to that optimum value,
which is probably reactive, with the fixed part of the matching network.

It is the 50 ohm design value for the load that the KATn matches to the
antenna, not the actual source impedance of the transmitter.  If you
look towards the transmitter, you will not see the complex conjugate of
the feedline input impedance, however, if you assume 50 ohms and the ATU
L and C values, you will get the complex conjugate of the feedline input
impedance.

--
David Woolley
Owner K2 06123

On 01/08/14 00:32, Per-Tore Aasestrand wrote:

>
> On 1 August 2014 01:16, Don Wilhelm <[hidden email]> wrote:
>
> If one uses the voltage divider example, yes the maximum efficiency is 50%,
>> But the output of a PA stage is not a resistor, and the collector load
>> "resistance" is set by the designer to produce the output power desired.
>>
>
> I fully agree.
>
> But will not a conjugate match also imply a max efficiency of 50%

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