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#1
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On Jun 12, 7:30*pm, Richard Clark wrote:
Deja Vu all over again. *I suppose you posted this to inspire me to, once again, remind you from Mendenhall's own notes about Class C amplifier construction - and so I will: ... If you believe by your understanding of the clips you quoted from Mendenhall that a conventional, single vacuum tube Class C r-f amplifier provides a functional termination of 50+j0 ohms for energy applied to the output connector of the transmitter, then please explain why that termination allows such signals to reach the plate of the PA tube -- whose non-linear characteristics created the _measured_ r-f intermodulation products and other performance data given in the Mendenhall paper I quoted. RF |
#2
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On Sun, 13 Jun 2010 05:17:54 -0700 (PDT), Richard Fry
wrote: the PA tube -- whose non-linear characteristics From Mendenhall: "VHF amplifiers often exhibit a somewhat unusual characteristic when tuning for maximum efficiency. ... If the amplifier is tuned exactly to resonance, the plate load impedance will be purely resistive and teh load line will be linear." 73's Richard Clark, KB7QHC |
#3
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On Jun 13, 12:12*pm, Richard Clark wrote:
On Sun, 13 Jun 2010 05:17:54 -0700 (PDT), Richard Fry wrote: the PA tube -- whose non-linear characteristics From Mendenhall: "VHF amplifiers often exhibit a somewhat unusual characteristic when tuning for maximum efficiency. ... If the amplifier is tuned exactly to resonance, the plate load impedance will be purely resistive and teh load line will be linear." 'Unusual'? Certainly not unexpected once one thinks about it. For a given controlled power, the minimum dissipation in the controlling device will occur when the minimum voltage occurs at the time of maximum current, and vice versa, i.e. the voltage and current in the load are in phase, or equivalently, the load impedance is resistive. I would expect this to be a fundamental characteristic, and not just for VHF amplifiers. The 60Hz folk are trying to achieve the same result as they strive for a power factor of 1. ....Keith |
#4
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On Sun, 13 Jun 2010 10:03:40 -0700 (PDT), Keith Dysart
wrote: 'Unusual'? Certainly not unexpected once one thinks about it. Hi Keith, The "unusual" was lost in the ellipsis that I will reveal: "... The highest efficiency operating point does not exactly coincide with the lowest plate current because the power output continues to rise for a while on the inductive side of resonance coming out of the dip in the plate current. ..." Continuing, Mendenhall presents the problems of trade-offs between what would seem to be maximum power for worsening characteristics in performance - the goal is what he calls minimizing synchronous AM versus Efficiency. Synchronous AM is a problem that can be introduced by measurement equipment, as Mendenhall relates: "The input impedance of the envelope detector must provide a nearly perfect match ... 30dB return loss ... to the sampling line." This sidebar relates to what Owen characterized as "Usability" where I have recited the objective technical specification to reduce the subjectivity of the term. There's more that could be said, but I am waiting to see if Richard is willing to subscribe to his own reference's writings. If not, and Mendenhall doesn't mince words on the topic, then as Perry Mason would observe "The D.A. is impeaching his own witness!" I've always loved Perry Mason. 73's Richard Clark, KB7QHC |
#5
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On Jun 13, 12:35*pm, Richard Clark wrote:
I am waiting to see if Richard is willing to subscribe to his own reference's writings I responded to you some 20 minutes before you posted, and now await your response. Also please comment on whether or not a Class C amplifier operating on a linear portion of its transfer curve will function as a linear amplifier. RF |
#6
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On Sun, 13 Jun 2010 11:04:50 -0700 (PDT), Richard Fry
wrote: I am waiting to see if Richard is willing to subscribe to his own reference's writings I responded to you some 20 minutes before you posted, and now await your response. So your subscription to your own authority is contingent upon me? I'm flattered. Also please comment on whether or not a Class C amplifier operating on a linear portion of its transfer curve will function as a linear amplifier. "If the amplifier is tuned exactly to resonance, the plate load impedance will be purely resistive and the load line will be linear." I would note that the bulk of his paper concerns frequency products - something the bears quite intimately upon linear operation. Your question is rather opaque in that the appearance of subtext is undoubtedly bound up in the term "linear" which Mendenhall has already employed within the norms of amplifier design. How linear is linear? Give me a technical criteria (subjectivity abounds and is not very informative except for statements of fashion). Express the least departure from a line response in percent that you accept as being linear. If it exceeds Mendenhall's design or commercial product, then I am afraid you won't be satisfied with any answer. 73's Richard Clark, KB7QHC |
#7
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On Jun 13, 1:20*pm, Richard Clark wrote:
How linear is linear? *Give me a technical criteria... A truly linear device does not produce r-f intermodulation products when signals of different frequencies are applied to it at the same time. For many practical uses a linear device does need to be perfectly linear. Let's say that the IM products need to be only - 20 dB w.r.t the lowest amplitude value of the various input frequencies. The Mendenhall paper I quoted conclusively shows that even such nominally linear performance is not an attribute of Class C vacuum tube r-f amplifiers -- especially at, and near the center frequency of the PA output tuning/matching network. That Mendenhall paper also shows that such r-f amplifiers do not present a functional, 50+j0 termination to r-f energy entering via the tx output connector. That is the reason why such energy is present in the PA plate circuit, giving rise to r-f intermodulation and/or changing the power dissipation there. RF |
#8
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On Jun 13, 11:12*am, Richard Clark wrote:
From Mendenhall: ... Would you please respond in your own words directly to my previous statement (repeated below)? If you believe by your understanding of the clips you quoted from Mendenhall that a conventional, single vacuum tube Class C r-f amplifier provides a functional termination of 50+j0 ohms for energy applied to the output connector of the transmitter, THEN PLEASE EXPLAIN WHY THAT TERMINATION ALLOWS SUCH SIGNALS TO REACH THE PLATE OF THE PA TUBE -- whose non-linear characteristics created the _measured_ r-f intermodulation products and other performance data given in the Mendenhall paper I quoted. RF |
#9
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On Sun, 13 Jun 2010 10:14:21 -0700 (PDT), Richard Fry
wrote: Would you please respond in your own words Hi Richard, I conform to Mendenhall to the specific statement I responded to. If you believe What I believe has been succinctly laid out in my subscribing to Mendenhall's explicit statement. THEN PLEASE EXPLAIN WHY I don't see that elaboration is going to improve what has been presented. Yes, it is a difficult concept that many struggle with and few have had experience in making a sufficiently accurate determination of. Consult Walt's 333 line posting and examine how experience comes to bear and through my recitations reveal the dovetail fit to theory. The Only Explanation Possible: Here's a modest proposal, Mendenhall constructed a power amplifier that is within the technical grasp of many here to achieve at a modest workbench. The design is quite spartan. The design is quite understressed (there is nothing "forcing" a conclusion). The design conforms to all engineering standards. Build your own. [It feels strange to have to offer that option to a group of Hams.] Having this amplifier before you, observe all the variables, play with them. Measure their impact on NOhms. Account for the heat with direct measurement and note what does not conform to convention. When that is finished the real work begins. Calibrate your tools and repeat this for accuracy. 73's Richard Clark, KB7QHC |
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