[Elecraft] Can I measure antenna impedance with K2?

David Woolley forums at david-woolley.me.uk
Fri Aug 1 05:17:37 EDT 2014


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 <w3fpr at embarqmail.com> 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%




More information about the Elecraft mailing list