[Elecraft] Can I measure antenna impedance with K2?

[Don Wilhelm]

P-T,

No, a conjugate match will assure 100% power transfer.

Look at it this way - what the conjugate match says is that if you cut a 
transmission line at any point, looking one way at that cut point, you 
will have some impedance - example is 70 + j20.  Now look the other way 
and the impedance will be the conjugate match - 70 - j20.  That is the 
condition that exists.
It also happens to be the condition for maximum power transfer.

Since that cut can be made at any and all points along the transmission 
line - think what would happen if the efficiency at each of those points 
(when connected together) were 50% - nothing would get to the antenna.  
So we know that 50% is *not* the efficiency of any and all junctions of 
any conjugate match.

Mixing the conjugate match concept with the maximum power transfer 
theorem is getting us into confusion - there are 'holes' in that 
combination.  Yes, they work together, but not seamlessly.
If a generator has an internal impedance of 50 ohms, the maximum power 
transfer will be only if the load to that generator is also 50 ohms.  
That says nothing about a conjugate match.

Now to further complicate things, the internal impedance of a generator 
has nothing to do with the efficiency of that generator - especially if 
we are discussing a PA output stage.  Bringing the conjugate match 
concept into the internal design of a PA stage is in error - it just 
does not work that way.  The conjugate match only applies to the output 
of that amplifier stage.

In other words, if we send 10 watts into a feedline (or ATU), all 10 
watts will go to the load (antenna) except for losses in the feedline.

I am not about to embark on the design and efficiency of a PA stage on 
this reflector, so take the conjugate match and maximum power transfer 
theorem only to the terminals at the PA stage and all will make sense.  
They do not apply to the internals of that stage.

73,
Don W3FPR

On 7/31/2014 7:32 PM, Per-Tore Aasestrand wrote:
> Hi Don,
>
> 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%
>
> P-T
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