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W2DU's Reflections III is now available from CQ Communications,Inc.
On May 24, 10:55*am, walt wrote:
Keith, would you please elaborate on why you believe my analysis of transmitter output impedance is flawed? And what is the basis for your belief that my explanations in Reflections require large chunks of linear circuit theory to be discarded. Could it be because you consider the source resistance in the transmitter to be dissipative, as in the classical generator? If so, you must be made to realize that the source resistance of the transmitter is non-dissipative, which is the reason that its efficiency can exceed 50%. No problems there. There has been much confusion in this area and anything that reduces this confusion is beneficial. Or are you considering the output characteristic of the transmitter to be non-linear? This is not the case, because the effect of energy storage in the tank circuit isolates the non-linear input from the output circuit, which is linear as evidenced by the almost perfect sine wave appearing at the output of the tank. This may be the root of my disagreement. Certainly the output can be an arbitrarily perfect sine wave, but this simply depends on the characteristics of the filter and not on whether the system is linear. But the way the filter transforms the impedances is the crux of the issue. It is my understanding that the input impedance to a filter can be computed by starting with the load impedance applied to the filter and then, using the rules for series and parallel connected components, compute the way through the filter until reaching the input and the result is the input impedance to the filter. Similarly, the output impedance of the filter can be computed by starting with source impedance driving the filter, series and paralleling the components until reaching the output and the result is the output impedance of the filter. The desired impedance for the input to the filter is that impedance which produces the desired load on the tube. And the component values are computed to produce this load on the tube when the correct load is attached to the output. For the output impedance of the filter, the question then becomes: What is the source impedance driving the filter? If the source is constructed as a Class A amplifier, then it depends on the controlling device, and for the simplest of circuits would be Rp of the tube. (Just for clarity, in this discussion Rp is the slope of the plate E/I curve with constant grid voltage. In an ideal tube, these lines are equidistant apart and the slopes are the same. Real tubes, of course, are not so well behaved, but this should not affect the basic discussion.) Since the component values for the filter were chosen to provide the optimum load to the tube, and the optimum load value has no relation to Rp, there is no reason to expect the filter will transform Rp to be the conjugate of the load impedance. For amplifiers where conduction is not for 360 degrees (Class AB, B, C), the controlling device is no longer time-invariant so the rules for linear circuit analysis no longer apply. None-the-less, for example, consider a Class AB amplifier where the tube is only cut off for 1 degree. This short cut-off would not have much affect so the analysis for Class A would apply. As the cut-off period increases the behaviour will diverge more and more from that of the Class A amplifier. Simulations produce some interesting results: Another way of measuring the source impedance is to observe the effect on a reflected wave entering the amplifier from the load. With a Class C amplifier, simulation reveals that the effect on the reflected wave depends on the phase of that wave with respect to the drive signal applied to the tube. As the phase of the reflected wave is changed, the reflection co-efficient experienced by the wave changes. Truly a non-linear behaviour. Intriguingly, when the conduction angle is exactly 180 degrees, this effect largely disappears, and the result is much as if the source impedance of the tube was 2 times Rp, which seems to make some sense since the tube is only conducting one-half of the time. One last question: Are you basing your dissatisfaction of Reflections from reviewing the 2nd or 3rd edition? Chapter 19 has been expanded in the 3rd edition, in which I presented additional proof of my position on the subject that you should be aware of. If you haven't yet seen the addition that appears in the 3rd ed, please let me know so that I can send you a copy of the addition. I have been reading the .pdfs at w2du.com along with correspondence and other writings in QST, QEX and newsgroups. The expanded Chapter 19 at w2du.com offers more experimental evidence that seems to support the hypothesis that the transmitter is conjugate matched to the load after tuning, But given, from circuit analysis, that the output impedance can not be well defined for any but a Class A amplifier, the fascinating question is why is there experimental evidence that agrees with the premise that the output impedance of a tuned transmitter is the conjugate match of the load? One simple example to consider which has similar behaviour is a bench power supply that also has a constant current limiter. Set such a power supply to produce a voltage of 100V (more precisely a maximum voltage) and a current limit of 2A. Apply a variable load. Maximum power will be drawn when the load resistance is 50 ohms. Varying the resistance on either side of 50 ohms will reduce the power which might be misconstrued to suggest that the power supply has an output impedance of 50 ohms, when, in fact, it has a infinite output impedance when the load is below 50 ohms and a zero output impedance when the load is above. I have looked for such a simple explanation in the circuits of the transmitters used in the experiments but was not able to find one. So I am still puzzled by the observations. Also include your email address so I can send it. Keith.dot.dysart.at.gmail.com .dot. = . .at. = @ …Keith |
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