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Old March 2nd 05, 10:22 PM
gwhite
 
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Ken Smith wrote:

In article , gwhite wrote:
[...]
You entirely missed the point. You don't know the output impedance because you
don't have a way of determining it by swinging the output full-scale.


You don't have to swing the output full-scale to measure the impedance.


Playing along with the idea that there is some meaningful fixed Z of the device
for large swings, yes you would have to do so to prove the concept. You would
need to prove that output Z was the same for driving 1 W into the output as for
driving 100 W into the output. I also predict that even the small signal output
Z of the power amp will not be that conjugate impedance you think it is for a
properly designed PA. (I am not making a claim that it would *never* be so for
any PA.)

Any change in the load, no matter how small, will cause a change in the
output voltage and the output current.


Likewise, a change in the output Z would do the same thing. Since you're
presuming linearity, we can include gain linearity. I.e., the gain with "-10
dB" of drive is the same as the gain with "0 dB" drive. I'll define the 0 dB
gain as associated with the 1 db compression point. Since the gain is defined
as linear (really fixed regardless of drive), and the load is fixed, something
must have "caused" the compression. A way to *model* the compression is a
changed output Z as a function of drive. While I realize this is an
unconventional view of output compression modeling, I believe it is fair, since
you are making the linear presumption. I think this is fair also because the
impedance concept is a linear/sinusoid one. Under that presumption, you've given
me license to disregard distortion.


From these you can calculate the
output impedance at the current operating point.

When a transistor is operating under large signal conditions into a tuned
load, there is still an output impedance and this impedance still
discribes what will happen for small changes in the load.


Let's do another example.

Say the device we've selected has an Imax rating of 1 amp and a generator
resistance of 100 ohms. Per standard linear theory, we do our norton model of
Igen in parallel with the 100 ohms. Under standard conjugate matching theory,
we should load it with 100 ohms.
Now with the 100 ohm load, we get a 50 V peak for Imax = 1 amp. But what if
both our DC supply and device breakdown won't allow this? We have a practical
limiting Vmax not at all included in linear theory. Due to breakdown or supply
rail concerns, we'll see our Imax quite short of the 1 amp we expect when the
device is loaded with 100 ohms. We won't be getting all the power out of it we
"expect" because of practical limitations not built into linear conjugate
matching theory.

How do we select the best load, since conjugate loading clearly does not use the
device to its full potential? We seek Ropt, or what is commonly referred to as
the load line match.

Ropt = Vmax/Imax

where Ropt Rgen, if not

(Rgen + Ropt)/(Rgen*Ropt) = Vmax/Imax


So even looking into the PA output in the small signal sense (or tweaking the
impedance as you suggest), we won't likely see Ropt = Rgen, because we are
dealing with some practical design limitations not accounted for in linear
theory.

Perhaps a couple of quotes from Cripps would be nice:

http://www.amazon.com/exec/obidos/tg.../-/0890069891/

"The load-line match is a real-world compromise that is necessary to extract the
maximum power from RF transistors and at the same time keep the RF voltage swing
within specified limits and/or the available DC supply." p13

"A final note here concerns the nebulous and highly questionable concept of
large signal impedance. The reason for the load-line match is to accommodate the
maximum allowable current and voltage swings at the transistor output. That says
nothing about the impedance of the device, which remains the same throughout the
linear range. Once a device starts to operate in a significantly nonlinear
fashion, the apparent value of the impedances will change, but the whole concept
of impedance starts to break down as well, because the wave forms no longer are
sinusoidal." p14