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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 |
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