On May 13, 6:51*pm, Wimpie wrote:
On 13 mayo, 22:09, Cecil Moore wrote:
On May 11, 3:09*pm, walt wrote:
Cecil, you are correct!!!
Thanks Walt, it would be interesting to see what else about which I am
correct. W7EL has a food-for-thought publication with one section
about reflections from the source. His source has a 50 ohm source
impedance, by design, so any dynamic load pulling experiment should
result in zero reflections from the source. Question is, does it?
http://eznec.com/misc/Food_for_thought.pdf
The "forward and reverse power section" is at the bottom of the page.
Using the dynamic load pulling method, do the results always indicate
50 ohms for the source impedance no matter what is the load and/or the
transmission line length? If not, the method yields invalid results.
IMO, it is a gross error to presume that all redistribution of energy
is a result of reflections. In W7EL's example, there are no
reflections from the source yet it is obvious that energy is being
redistributed from the source back toward the load in some cases but
not in other cases. IMO, destructive/constructive interference must be
taken into account in order to explain the results. Yet, no one except
yours truly has even mentioned interference effects as a method of
redistributing energy.
Anyone interested in understanding the role of interference at
impedance discontinuities in transmission lines is welcome to read my
article at:
http://www.w5dxp.com/energy.htm
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK
Hello Cecil,
When a source behaves like a 50 Ohms source, it will not rereflect the
reflected power back to the load. In other words the forward power
generated by the source will not change when changing the load. The
reflected power that is absorbed by the source may result in increase
or reduction of power dissipation of the active device. It all depends
on the change of the integral of V*I (for the active device).
With regards to load pulling, output impedance of an amplifier may
change depending on the rate of change of the load. As I mentioned
earlier, manual load pulling may result in change of supply and bias
voltage and current. *Average Voltage and current have time to settle
to the new load condition.
In case of load pulling where the load changes relatively fast, but
well within the matching section's bandwidth (as is the case for the
off-carrier signal injection method) supply and bias voltage change
may be less due to decoupling capacitance.
In simulation this can easily be seen by applying a load step change
and observe the envelope and phase response versus time. Of course
this requires you to model your power supply and bias circuit
correctly. You may see a slow step response that cannot be explained
by the bandwidth of the matching network.
To walt:
I didn’t see your announced comment to my last post (regarding the
forward/reverse power mathematics), did I missed it?
With kind regards,
Wim
PA3DJSwww.tetech.nl
Hello Wim,
I hadn't yet given you my final understanding of it. I understand the
voltage divider action, but I don't yet understand the following quote
from you:
"When terminated according to the numbers above the socket ("100W
into
50 Ohms"), this results in 100W ( Vout = 212.1*50/(50+100) = 70.7V,
just a voltage divider consisting of 100 Ohms and 50 Ohms )."
The portion I don't understand is the mathematical basis for:
"( Vout = 212.1*50/(50+100) = 70.7V,"
And Wim, on your last post above I agree that if the source is 50 ohms
with a real resistor, as in the classical generator used in text
books. However, are you not aware that the output resistance of the RF
power amp is R = E/R that appears at the output. Since this R is non-
dissipative it re-reflects all the reflected power. I have proved this
to be true in the experiment I presented with the Kenwood TS-830S
presentation that I'm sure you have a copy of.
Consequently, the RF power amp is not going to absorb the reflected
power.
At this point my email server is down, but I'll forward a copy of my
QEX article ASAP.
Walt