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Hi Richard
Please tell me more about melting the finals and a bit more explanation of what was happening? tnx Hank "Richard Clark" wrote in message ... On Sun, 05 Jun 2005 21:58:06 GMT, "Henry Kolesnik" wrote: In TV broadcasting reflections from the antenna back to the transmitter will be reflected by the transmitter to the antenna and the signal will be rebroadcast albeit at somewhat less power. Hi Hank, That would pretty much reveal the SWR if we knew, wouldn't it? If "somewhat less power" was in 1.2:1 ratio, we wouldn't care so much, but how would the viewer feel about such service? Then depending on the length of transmission line the viewer may see ghosting. I think we, or another correspondent and I have dealt with that at one time. At the time I believe it was called "fringing," not "ghosts." The difference being that what were called ghosts at the dawn of the TV era were separated by fractions of an inch rather than fractions of a mm. As such, ghosts couldn't have been originated by anything shorter than mile length transmission lines that were poorly terminated at both ends. Instead, ghosts were actually transmission path length differentials in a multipath situation. In audio I don't know why and I have run my Collins 30S-1 into ladder line with a 14 to SWR with no one except me knowing! Well, if this is meant to be analogous to fringing/ghosting, I suppose its because a microsecond blur at AF is entirely inaudible. Or are we speaking of SSTV? However, this begs the question, How did you know? All the Collins equipment I taught at school didn't come with a SWR meter. It was wholly unnecessary if you performed the standard tune-up. Matter of fact, back then the only SWR meter I saw was for Ham gear. The finals' tank performed every function of matching as any tuner. However, with the KWT-6, we did use an external tuner, 180-V1 (although I may have this mixed up with another model), for coax feedlines. This was more for its automatic feature where the transmitter could be tuned up with a 50 Ohm load, and the automatic tuner simply did the job of presenting it with the transformed load. However, returning to the point of a transmitter rereflecting a reflection; I know the bare KWT-6 into ladder line employs its tank to protect its final tubes. Without that safeguard, I have seen plates melt - something no one here wants to call dissipation lest it be evidence of an internal resistance. The bare tubes with their native very hi Z would rereflect like nothing else - and this begs the observation - how could you get original any power out of them, past the tremendous mismatch? The tuner/final tank comes back into the equation, and rereflection goes out the window as a property of the transmitter and returns to the domain of matching. If anyone wants to constrain the entire crusade of the rereflecting transmitter to the tube set feeding ladder line - then feel free to do so. However, I don't think I've ever seen a mobile tube rig feeding ladder line - no doubt one day I will. We will probably talk about efficiency. :-) 73's Richard Clark, KB7QHC |
Roy Lewallen wrote:
Anyone who's interested can find more interesting cases in "Food for thought - Forward and Reverse Power.txt" at http://eznec.com/misc/food_for_thought/. And those who aren't interested, well, you're welcome to believe what you choose. Just don't look too closely at the evidence. (*) Anybody fond of the notion that reverse power "goes" somewhere or gets dissipated in the source or re-reflected back needs to come to grips with this problem before building further on the flawed model of bouncing waves of flowing power. Roy, I have had an article for review into QEX for more than two months that explains what is missing from your analysis. Unfortunately, I have not heard a word from QEX since I submitted the article. So I will introduce a concept new to the field of RF but completely understood in the field of optics. I actually introduced this concept three years ago in discussions on r.r.a.a with Dr. Best but I was apparently unable to convey the concept. If I ask you what things can cause 100% reflection, I assume you would list three things: 1. A short-circuit, 2. An open-circuit, and 3. A pure reactance. And that is indeed true for loads upon which a single wave is incident. But the field of optics recognizes an additional thing that can cause 100% reflection and that's wave cancellation. If two coherent EM waves are traveling the same path in the same direction in a transmission line and they are 180 degrees out of phase, the waves will cancel and the energy components in the two waves, which must be conserved, will be 100% reflected in the opposite direction. The following two optics web pages verify that fact for EM waves: (near the bottom of the pages) http://www.mellesgriot.com/products/optics/oc_2_1.htm http://micro.magnet.fsu.edu/primer/j...ons/index.html What is happening in your "food for thought" assertions is that you are neglecting the ability of the phenomenon of wave cancellation to cause 100% reflection of the energy components in the two canceled waves, something that is well understood in the field of optics. Dr. Best also neglected to take interference energy into account in his QEX article on transmission lines. "Optics", by Hecht asserts that for every incidence of constructive interference there must be an equal magnitude of destructive interference to satisfy the conservation of energy principle. (That is unless the source itself is capable of delivering extra power.) By the same token, if the source doesn't absorb constructive interference energy, it must go somewhere else. In a transmission line with only two directions, there is only one other way it can possibly go and it becomes a reflection or a re-reflection. What is happening in your "food for thought" examples is that destructive interference/wave cancellation between the forward wave and reflected wave is occurring at your source. That wave cancellation event is feeding constructive interference energy back into the feedline which joins the forward wave energy. Your discussions ignore the effect of interference energy which must necessarily be conserved. Your argument goes something like this: I am ignoring the constructive/destructive interference energy involved in wave cancellation. Therefore, it never existed in the first place. That's a petitio principii type of argument. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Cecil Moore wrote: Roy, none of my textbook authors think the reflection model is flawed. Walter Johnson goes so far as to assert that there is a Poynting (Power Flow Vector) for forward power and a separate Poynting Vector for reflected power. The sum of those two Power Flow Vectors is the net Poynting Vector. Here's my earlier thought example again. 100w----one second long lossless feedline----load, rho=0.707 SWR = (1+rho)/(1-rho) = 5.828:1 Source is delivering 100 watts (joules/sec) Forward power is 200 watts (joules/sec) Reflected power is 100 watts (joules/sec) Load is absorbing 100 watts (joules/sec) It can easily be shown that 300 joules of energy have been generated that have not been delivered to the load, i.e. those 300 joules of energy are stored in the feedline. Not easy if t 2 sec. :-) The 300 joules of energy are stored in RF waves which cannot stand still and necessarily travel at the speed of light. It's ironic that the first paramater cited in the problem starts with an 'S'. :-) TV ghosting can be used to prove that the reflected energy actually makes a round trip to the load and back. A TDR will indicate the same thing. If either source were monochromatic, I bet I could come up with an example where the surfaces reflect no energy. :-) Choosing to use a net energy shortcut doesn't negate the laws of physics. Particular when characterized as a matter of opinion, it can be like having a religious discussion. 73 ac6xg |
Cecil,
You can't be serious! This is basic stuff found in virtually any intermediate level E&M textbook. The treatment is generally the same; start with the field equations describing the waves, add the material conditions and the boundary conditions, plug and crank. The answers pop right out. No need to invoke any magic incantations about interfering waves or wave cancellation. The interference is the result of the analysis, not the cause. In the classical case, there is absolutely no difference in behavior between "RF" and "optical". The material properties for every situation can vary, but the physical principles do not. Sooo, rather than introducing a new concept, you are perhaps the last person to finally understand the old one. 73, Gene W4SZ Cecil Moore wrote: [snip} So I will introduce a concept new to the field of RF but completely understood in the field of optics. I actually introduced this concept three years ago in discussions on r.r.a.a with Dr. Best but I was apparently unable to convey the concept. If I ask you what things can cause 100% reflection, I assume you would list three things: 1. A short-circuit, 2. An open-circuit, and 3. A pure reactance. And that is indeed true for loads upon which a single wave is incident. But the field of optics recognizes an additional thing that can cause 100% reflection and that's wave cancellation. If two coherent EM waves are traveling the same path in the same direction in a transmission line and they are 180 degrees out of phase, the waves will cancel and the energy components in the two waves, which must be conserved, will be 100% reflected in the opposite direction. The following two optics web pages verify that fact for EM waves: (near the bottom of the pages) http://www.mellesgriot.com/products/optics/oc_2_1.htm http://micro.magnet.fsu.edu/primer/j...ons/index.html What is happening in your "food for thought" assertions is that you are neglecting the ability of the phenomenon of wave cancellation to cause 100% reflection of the energy components in the two canceled waves, something that is well understood in the field of optics. Dr. Best also neglected to take interference energy into account in his QEX article on transmission lines. "Optics", by Hecht asserts that for every incidence of constructive interference there must be an equal magnitude of destructive interference to satisfy the conservation of energy principle. (That is unless the source itself is capable of delivering extra power.) By the same token, if the source doesn't absorb constructive interference energy, it must go somewhere else. In a transmission line with only two directions, there is only one other way it can possibly go and it becomes a reflection or a re-reflection. What is happening in your "food for thought" examples is that destructive interference/wave cancellation between the forward wave and reflected wave is occurring at your source. That wave cancellation event is feeding constructive interference energy back into the feedline which joins the forward wave energy. Your discussions ignore the effect of interference energy which must necessarily be conserved. Your argument goes something like this: I am ignoring the constructive/destructive interference energy involved in wave cancellation. Therefore, it never existed in the first place. That's a petitio principii type of argument. |
On Mon, 06 Jun 2005 15:05:43 GMT, "Henry Kolesnik"
wrote: Please tell me more about melting the finals and a bit more explanation of what was happening? Hi Hank, Direct observation offered a glowing plasma between the filament and the plate. It was football shaped rather than beam-like or an amorphous cloud. One point of the football touched the cherry red plate. Following a quick power-down, that point on the plate did not exist anymore as there was a hole. Couldn't really tell, but no doubt the grid suffered just as much in its own way. I suppose plates have become more robust over the years since that amazing demonstration. I helped fix one friend's Amp when it failed along with his antenna (or t'other way round as the chain of causality would suggest). His Amp simply quit working suddenly during bad weather. Fuses checked - OK. No interlocks were open - OK. The tube looked good at a glance - OK. HV Supply looked good - OK. The filaments failed to light up - odd, but consistent. Time to crack open the case. Pulled the tube and measured its filament continuity - OK. Measured filament supply - OK. Things were really getting strange. Time to bust the chassis open and really look. There on the baseplate were several small pools of solder in a circular pattern - how odd. Close scrutiny of connections revealed bright and soldered wires to everything - OK. Time to look at the tube again. Every pin was solder free - that was on the chassis base plate. [dirge played here] The filaments' wires were making enough contact to measure continuity, but no where enough to support real power. In other tubes I've seen the heat become so extreme that the glass envelope slumped into the vacuum and enclosed the plate structure like taffy. This didn't even crack the glass (or it had simply re-fused). Tube still worked afterwards though (so I would suppose the glass never cracked). One occasion was actually due to a bias problem created when the cathode load shorted. Lack of bias protection sent the circuit into massive conduction. Of course the short came about because of an initial excessive conduction (surprising in its own right because the common failure mode is to open). I've also seen stressed thyratrons so mismatched that they filled the workspace with their purple glow like a floodlight. Thankfully fuses work as I did not want to be near that final testimony. Now, that was the short list of Bottle failures. I have another list of melted state failures too, but their evidence is usually better hidden and less dramatic - heat sinking generally spreads the risk, so to speak. And speaking of heat sinks, I've drawn a number of blisters from those that normally only warmed my hand. Note there should be emphasis on normally warm and the obvious contradiction with blisters. Experience with failure has strongly correlated with heat and mismatch. Heat was born by resistance. Resistance is part of life and amplifiers. Heat comes in two forms. Slow-like, which is generally current based; and sudden, which is generally voltage based. I've felt along heatsinks immediately following failure that were as cool/tepid as usual, or ominously cooler! The sudden heat of arc-over in silicon can destroy just as effectively as the long slow broil of a plate turning to slag. Every mismatch in the Amateur experience is a probablistic spin of the wheel of misfortune. Sometimes the wheel stops on the slow bake that aborbs into heatsinks and you notice unusual smell, or your fan running on too long - Quick! do something, and you survive. Other times there's the snap of finality. Both of these examples are for those with keen senses, and often failure comes as a whimper. Most suffers usually discover what matching is for, even if they don't know how it works. 73's Richard Clark, KB7QHC |
Gene Fuller wrote:
Cecil, You can't be serious! This is basic stuff found in virtually any intermediate level E&M textbook. If you can provide me with a reference that says, wave cancellation can cause reflection of the canceled waves, I will be eternally grateful. I have been able to find references that imply such for light waves, but I have not found one that comes right out and says it for either light waves or RF waves. I agree with you 100%. I had this "basic stuff" taught to me at Texas A&M half a century ago. But Roy's "food for thought" stuff completely ignores exactly that "basic stuff" concerning constructive/destructive interference. Nowhere in his arguments is "interference" even mentioned. I expect him to respond that interference is irrelevant. Dr. Best went so far as to deny that interference is necessary for a Z0-match to occur in a system with reflections. That was around May/June 2001 on this very newsgroup for anyone who wants to Google it. I have been fighting this battle for three years on this newsgroup. Now you say it's "basic stuff". I've agreed for three years, but where have you been all this time? The treatment is generally the same; start with the field equations describing the waves, add the material conditions and the boundary conditions, plug and crank. The answers pop right out. No need to invoke any magic incantations about interfering waves or wave cancellation. The interference is the result of the analysis, not the cause. Some people have forgotten what they learned in college. Their net/ steady-state shortcuts have become reality and scrambled their brains. You are obviously not one of the people at whom I aimed my remarks. I am glad to see that not everyone has been seduced into thinking that interference is irrelevant. In the classical case, there is absolutely no difference in behavior between "RF" and "optical". The material properties for every situation can vary, but the physical principles do not. I know that. You know that. We are on the same side. Now convince the RF gurus of that. Roy calculated the net power at the source and assumed from that figure that there was not enough energy available to support the energy in reflected waves. Sooo, rather than introducing a new concept, you are perhaps the last person to finally understand the old one. No, not the last one. What you say is exactly what is wrong with Roy's arguments that reflected energy doesn't flow from the load back toward the source. I am NOT introducing a new concept. I am introducing a new (or forgotten) concept to some of the RF gurus on this newsgroup. I am (re)introducing destructive/constructive interference concepts to Roy, Dr. Best, and others. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Richard Clark wrote:
Every mismatch in the Amateur experience is a probablistic spin of the wheel of misfortune. Sometimes the wheel stops on the slow bake that aborbs into heatsinks ... Current maximum, Ifor+Iref in phase. Vtot = |Vfor|-|Vref| Other times there's the snap of finality. Voltage maximum, Vfor+Vref in phase. Itot = |Ifor|-|Iref| -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
In my third example, where does the other 10 watts of reflected power
go? If it goes to the load and back, why does it reflect off the source resistor? Roy Lewallen, W7EL Cecil Moore wrote: Roy Lewallen wrote: (*) Anybody fond of the notion that reverse power "goes" somewhere or gets dissipated in the source or re-reflected back needs to come to grips with this problem before building further on the flawed model of bouncing waves of flowing power. Roy, none of my textbook authors think the reflection model is flawed. Walter Johnson goes so far as to assert that there is a Poynting (Power Flow Vector) for forward power and a separate Poynting Vector for reflected power. The sum of those two Power Flow Vectors is the net Poynting Vector. . . . |
I don't see anything on those web pages that's outside the concept of
the ordinary interference of waves, i.e., that they add and subtract where they occupy the same space. I don't at all see the concept of a wave of flowing average energy being bounced back or about by another wave, which is what you're proposing as you have many times in the past. But then, many people find miracles when I see only coincidence, so I'm a bit deficient in that regard. This topic has been previously discussed beyond a tedious degree in this newsgroup; anyone interested can find it via groups.google.com. I don't have anything to add to it, with the exception of this question which is relevant to the topic: Does your analysis produce the result of 2.3 dB loss claimed by H. for a 1.7:1 SWR? Best luck with your QEX article. I look forward to reading it. Roy Lewallen, W7EL Cecil Moore wrote: Roy, I have had an article for review into QEX for more than two months that explains what is missing from your analysis. Unfortunately, I have not heard a word from QEX since I submitted the article. So I will introduce a concept new to the field of RF but completely understood in the field of optics. I actually introduced this concept three years ago in discussions on r.r.a.a with Dr. Best but I was apparently unable to convey the concept. If I ask you what things can cause 100% reflection, I assume you would list three things: 1. A short-circuit, 2. An open-circuit, and 3. A pure reactance. And that is indeed true for loads upon which a single wave is incident. But the field of optics recognizes an additional thing that can cause 100% reflection and that's wave cancellation. If two coherent EM waves are traveling the same path in the same direction in a transmission line and they are 180 degrees out of phase, the waves will cancel and the energy components in the two waves, which must be conserved, will be 100% reflected in the opposite direction. The following two optics web pages verify that fact for EM waves: (near the bottom of the pages) http://www.mellesgriot.com/products/optics/oc_2_1.htm http://micro.magnet.fsu.edu/primer/j...ons/index.html . . . |
Cecil Moore wrote:
Gene Fuller wrote: Cecil, You can't be serious! This is basic stuff found in virtually any intermediate level E&M textbook. If you can provide me with a reference that says, wave cancellation can cause reflection of the canceled waves, I will be eternally grateful. I have been able to find references that imply such for light waves, but I have not found one that comes right out and says it for either light waves or RF waves. . . . I'm afraid that your difficulty in finding a reference is simply due to its not being so. If it is indeed so, it appears that your forthcoming QEX article will be a seminal work, as the first published work to explicitly state that this phenomenon indeed happens (outside of countless newsgroup postings to that effect). Assuming you understand the physics which causes it to happen, I'd think that a professional publication would be a much more appropriate forum than QEX for such an important work. Have you tried any of the IEEE publications? Roy Lewallen, W7EL |
Roy Lewallen wrote:
Does your analysis produce the result of 2.3 dB loss claimed by H. for a 1.7:1 SWR? Cheap friggin' damn shot, Roy, after my posting where I disagreed with H. and agreed with your calculations. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Cecil Moore wrote:
. . . What is happening in your "food for thought" assertions is that you are neglecting the ability of the phenomenon of wave cancellation to cause 100% reflection of the energy components in the two canceled waves, something that is well understood in the field of optics. Dr. Best also neglected to take interference energy into account in his QEX article on transmission lines. . . . If I'm neglecting an important phenomenon, then surely some of my numerical results showing voltages, currents, forward, reverse, and total powers, and power dissipation must be incorrect. And an experiment can be set up to demonstrate their incorrectness and the validity of the alleged phenomenon. I'd appreciate it very much if you or anyone else who finds any incorrect results in that series of essays, or anything else I've written or posted, bring it to my attention so it can be corrected. Roy Lewallen, W7EL |
Roy Lewallen wrote:
Cecil Moore wrote: Gene Fuller wrote: Cecil, You can't be serious! This is basic stuff found in virtually any intermediate level E&M textbook. If you can provide me with a reference that says, wave cancellation can cause reflection of the canceled waves, I will be eternally grateful. I have been able to find references that imply such for light waves, but I have not found one that comes right out and says it for either light waves or RF waves. I'm afraid that your difficulty in finding a reference is simply due to its not being so. Well, Gene says it is really basic stuff (not worthy of a second thought). Which is it? - Not worthy of a second thought or seminal work? If it is indeed so, it appears that your forthcoming QEX article will be a seminal work, as the first published work to explicitly state that this phenomenon indeed happens (outside of countless newsgroup postings to that effect). Assuming you understand the physics which causes it to happen, I'd think that a professional publication would be a much more appropriate forum than QEX for such an important work. Have you tried any of the IEEE publications? Nope, I haven't. I've retired from being a pro. Now I am just an amateur. When two coherent waves traveling in the same path and direction are 180 degrees out of phase, they disappear from existence in that original direction of travel, i.e. they undergo wave cancellation. When they are confined to a transmission line with only two directions, the flow of energy in the original direction ceases. There is no other choice but for the energy in the two cancelled waves to be conserved and to reverse direction and start flowing in the opposite direction. That, my friend, is a reflection. How can you possibly believe that the energy in cancelled waves is not conserved? So to your list of shorts, opens, and pure reactances being able to cause 100% reflection, you can add wave cancellation. Note that wave cancellation cannot happen at a single load with a single incident wave. It can only happen at points where there are waves flowing in opposite directions, e.g. match points on transmission lines with reflections and at sources subjected to reflections. Please don't argue that you have never seen such. Anyone who has looked at an oil film on water has witnessed reflections caused by interference and wave cancellation. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Jim Kelley wrote:
Cecil Moore wrote: It can easily be shown that 300 joules of energy have been generated that have not been delivered to the load, i.e. those 300 joules of energy are stored in the feedline. Not easy if t 2 sec. :-) Of course, my statement is related to steady-state. I don't see anything worth responding to, Jim. Where's the beef? -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Roy Lewallen wrote:
If I'm neglecting an important phenomenon, then surely some of my numerical results showing voltages, currents, forward, reverse, and total powers, and power dissipation must be incorrect. No, they are not incorrect. They are net values and, as such, are simply incomplete. Dr. Best had the same problem in his article which was accurate but incomplete. The problem with incomplete information is that erroneous conclusions are likely to occur. In addition to a closed mind, that's your present problem and Dr. Best's problem with his QEX article. Valid data being incomplete can sometimes be just as bad or worse than data being wrong. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Cecil,
I do not expect that the reference you seek can be found. There is no need to invoke interference or wave cancellation to explain anything, and it is unlikely there is any mathematical formulation that uses interference as one of the input variables. It is totally unnecessary. Maxwell's equations contain everything needed to accurately describe electromagnetic interactions, including wave reflections, cancellations, and interference. Any serious treatment of the subject of electromagnetic interactions begins with the field equations, not with the resulting interference. The sort of description found on the Melles-Griot web site and on their CD-ROM is a handwaving, but comforting, description meant for general understanding, not for detailed analysis. 73, Gene W4SZ Cecil Moore wrote: Gene Fuller wrote: Cecil, You can't be serious! This is basic stuff found in virtually any intermediate level E&M textbook. If you can provide me with a reference that says, wave cancellation can cause reflection of the canceled waves, I will be eternally grateful. I have been able to find references that imply such for light waves, but I have not found one that comes right out and says it for either light waves or RF waves. I agree with you 100%. I had this "basic stuff" taught to me at Texas A&M half a century ago. But Roy's "food for thought" stuff completely ignores exactly that "basic stuff" concerning constructive/destructive interference. Nowhere in his arguments is "interference" even mentioned. I expect him to respond that interference is irrelevant. Dr. Best went so far as to deny that interference is necessary for a Z0-match to occur in a system with reflections. That was around May/June 2001 on this very newsgroup for anyone who wants to Google it. I have been fighting this battle for three years on this newsgroup. Now you say it's "basic stuff". I've agreed for three years, but where have you been all this time? The treatment is generally the same; start with the field equations describing the waves, add the material conditions and the boundary conditions, plug and crank. The answers pop right out. No need to invoke any magic incantations about interfering waves or wave cancellation. The interference is the result of the analysis, not the cause. Some people have forgotten what they learned in college. Their net/ steady-state shortcuts have become reality and scrambled their brains. You are obviously not one of the people at whom I aimed my remarks. I am glad to see that not everyone has been seduced into thinking that interference is irrelevant. In the classical case, there is absolutely no difference in behavior between "RF" and "optical". The material properties for every situation can vary, but the physical principles do not. I know that. You know that. We are on the same side. Now convince the RF gurus of that. Roy calculated the net power at the source and assumed from that figure that there was not enough energy available to support the energy in reflected waves. Sooo, rather than introducing a new concept, you are perhaps the last person to finally understand the old one. No, not the last one. What you say is exactly what is wrong with Roy's arguments that reflected energy doesn't flow from the load back toward the source. I am NOT introducing a new concept. I am introducing a new (or forgotten) concept to some of the RF gurus on this newsgroup. I am (re)introducing destructive/constructive interference concepts to Roy, Dr. Best, and others. |
Roy Lewallen wrote:
In my third example, where does the other 10 watts of reflected power go? If it goes to the load and back, why does it reflect off the source resistor? I have read the third example, which is NOT steady-state, and I don't understand the question. Give me some steady- state values and I will discuss it. ************************************************** ************* However, a reflected wave approaching a source resistor doesn't encounter the source resistor as an isolated load. The source voltage superposes with the reflected voltage at that point which may, in reality, actually turn out to be an impedance discontinuity. ************************************************** ************* Quoting Ramo and Whinnery: "It must be emphasized, as in any Thevenin equivalent cirsuit, that the equivalent circuit was derived to tell what happens in the ***LOAD*** under different load conditions, and significance cannot be automatically attached to a calculation of power loss in the internal impedance of the equivalent circuit." -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Gene Fuller wrote:
There is no need to invoke interference or wave cancellation to explain anything, ... But there is, Gene. It's the only way to correct the present misinformation and old wives' tales being promoted on this newsgroup. It is obvious that the r.r.a.a poster who understands the role that interference plays in the conservation of RF energy is very rare. There is a conspiracy to keep this information from surfacing - "Nothing new", "no need", "irrelevant", "who cares?" Why are you guys afraid to discuss the technical details? This should be an easy question to answer. If two coherent waves of 50 joules/sec each, are traveling in the same path in the same direction and are 180 degrees out of phase, they cancel in that direction of travel. What happens to their 50+50 joules/sec? Hint: energy doesn't cancel and there are only two possible directions. Can you spell R-E-F-L-E-C-T-I-O-N? -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Cecil Moore wrote:
Roy Lewallen wrote: Does your analysis produce the result of 2.3 dB loss claimed by H. for a 1.7:1 SWR? Cheap friggin' damn shot, Roy, after my posting where I disagreed with H. and agreed with your calculations. Sorry, it was an honest question. I saw your posting re the SWR calculations, but guess I missed the one you mention. H's calculations remain a mystery to me, as apparently they do to you. Roy Lewallen, W7EL |
Cecil Moore wrote:
Roy Lewallen wrote: In my third example, where does the other 10 watts of reflected power go? If it goes to the load and back, why does it reflect off the source resistor? I have read the third example, which is NOT steady-state, and I don't understand the question. Give me some steady- state values and I will discuss it. It most certainly is steady state, as are the other two examples. Perhaps I wasn't clear in saying I was referring to my recent posting. Here it is again: [The setup is a 100 volt zero impedance voltage source in series with a 50 ohm resistor driving a half wavelength of 50 ohm transmission line.] --------- Well, shoot, maybe the source resistor dissipates all the reverse power *plus* some more power that comes from somewhere else. So let's try a 200 ohm load. Now the current is 0.4 amp, the power in the 200 ohm load resistor is 32 watts, and the power in the 50 ohm source resistor is 8 watts. The SWR is 4:1, the forward power is 50 watts, and the reverse power is 18 watts. Oops, the source resistor is only dissipating 8 watts but the reverse power is 18 watts. Not only isn't it dissipating all the reverse power, but it isn't even dissipating that extra power that came from somewhere else when we connected the 16.67 ohm resistor. Wonder where the other 10 watts of reverse power went? --------- Must have bounced off the forward wave, then? If so, did it bounce only when it reached the source, and not all along the line? Why? How much of it bounced back at the load and why? Can you show me how you calculate the various source resistor dissipations for this and the other two examples using your bouncing average power model? I'd also welcome H.'s comments on this. Roy Lewallen, W7EL |
Richard
I've seen 3-500Z where the solder had melted out of the filament pins. I have some 811 from my 30L-1 that have holes in the plates, they still work quite well. It's been 15 years since I recall finding the holes in the plate sides and replacing. In additon I can't recall the reason for the holes but it had to be either 1. overdriving (which I don't think I did, 100 watt Icom) 2. high swr which I know has been over 10 to 1, but I can't see how that could do it.. 3. Light loading But in any event for something to melt we need dissipation and only resistance can do that! -- 73 Hank WD5JFR Still learning, un-learning and re-learning |
Roy Lewallen wrote:
Cecil Moore wrote: Roy Lewallen wrote: Does your analysis produce the result of 2.3 dB loss claimed by H. for a 1.7:1 SWR? Cheap friggin' damn shot, Roy, after my posting where I disagreed with H. and agreed with your calculations. Sorry, it was an honest question. I saw your posting re the SWR calculations, but guess I missed the one you mention. H's calculations remain a mystery to me, as apparently they do to you. Roy Lewallen, W7EL Really strange response since you already responded to the posting where I agreed with you. Are you as senile as I? :-) -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Roy Lewallen wrote:
... and the power in the 50 ohm source resistor is 8 watts. Sorry, Roy, the Thevenin equivalent source model expressly forbids you from making that assumption. You are going to have to do better than that. What is the DC power input to the source electronics? -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
On Mon, 06 Jun 2005 20:52:52 GMT, "Henry Kolesnik"
wrote: But in any event for something to melt we need dissipation and only resistance can do that! Hi Hank, Unfortunately as much as you and I agree on that bedrock principle, others with Simpson Ohmmeter in hand would glare goggle eyed at us and say that plate has no resistance to speak of and that no amount of current through its Ohmic resistance could ever bring about enough heat to produce the effects so obviously witnessed. One of the most enraging questions I've asked "If it is not the value I've offered, what value is it?" Well, I've never been given a quantitative answer, however I've seen enough carefully crafted mathematical proofs in this group to replace substantive results so easily seen. There is some irrefutable logic in circulation that clearly reveals that what we've experienced just couldn't be. Glasses will be need to be readjusted for such extreme myopic aberrations. There are two principles involved in what is called Plate Resistance, and the first and foremost is not even related to the plate at all. It is called the work function of the cathode emissivity. So, in fact it is more proper to refer to this usual loss as Cathode Resistance, not Plate Resistance. The cathode is the fundamental limit on power generated. What Plate Resistance is, is the ill termed substitution for Plate dissipation. If folks want to work their Simpson, they would blow an aneurysm trying to measure the resistance from cathode (filaments have the same work function issue too) to plate. In fact, the hobby horse argument of it is not resistance at all, but some figurative charting artifice called a "load line" usually appears in the last gasp. Plate Dissipation is resistance clear and simple in spite of the failure of conventional tools to measure a common physical property. Newton would have recognized it, it is called inertia. Once the work function is overcome (the job of the grid), then Plate voltage dominates through the acceleration of charge beyond the grid, toward the plate. That stream of electrons (and there is no doubt about actual current flow in easily counted, significant populations of electrons) is elevated to 90% the speed of light. This current flow is entirely different from what current flows in the remainder of the Plate load. That is also known as displacement current and electrons are shuffling along at a placid meter per second rate. Plate current and displacement current are equal in amplitude and phase, but not in motion nor kinetics. NOW. When that same stream encounters the Plate - WHAM! If anyone here has walked into the wall, and NOT encountered resistance, then we will call you Casper. Inertia reveals that to slow a mass in a distance results in acceleration (negative in this instance) and that property is called Force. Force over time expends calories and is expressed in any number of systems and units - Watts is one, Degrees is another. We could abstract to Horsepower and Candelas (the plate glows too). We know the speed, not many here would give it much though, but none would know the length interval of going at that speed to going zero (0). It is roughly two atoms distance into the metal of the plate. I will leave those calculations of Force to the student to compute or I can provide it from notes of correspondence with Walt Maxwell and Richard Harrison from a round robin discussion several years ago. Hank, does this fulfill your earlier question as to "what" is happening? I first gave you many examples, I hope this segue into real physics fills in their actuality. Too many correspondents demand that I open the source and point at a 50 Ohm carbon composition resistor that is the "source resistance." 73's Richard Clark, KB7QHC |
Cecil Moore wrote:
Roy Lewallen wrote: ... and the power in the 50 ohm source resistor is 8 watts. Sorry, Roy, the Thevenin equivalent source model expressly forbids you from making that assumption. You are going to have to do better than that. What is the DC power input to the source electronics? Who said anything about a Thevenin equivalent? The circuit I proposed isn't intended to be an equivalent of anything. It's a very simple circuit made with components whose characteristics are well defined and well known. The source is an AC steady state source (which I naively assumed was obvious). There is no DC power input to it. There are no "source electronics" -- there are no electronics at all. All elements in my example are simple electrical circuit components, as described in any elementary electrical circuits textbook. There are 18 watts of "reverse power" in the transmission line, as calculated by those who embrace this concept. The source resistor matches the Z0 of the line. The source resistor dissipates 8 watts. Nothing "forbids" calculation of the dissipation of the resistor -- it's V^2/R, I^2*R, or V*I, take your choice. Any EE freshman, and I'd like to think most amateurs, including you, should be able to calculate it. The total power from the source equals the sum of the dissipation in the two resistors. The power dissipated by the load is the difference between the "forward power" and "reverse power", as you can easily see with a small amount of simple arithmetic. There is nothing "forbidden" about this simple circuit, except perhaps explanation by means of your theory. Where's the "reverse power" going? Any theory that doesn't work in a circuit composed of simple elemental electrical circuit elements is suspect to say the least. Are you saying yours doesn't? Roy Lewallen, W7EL |
Cecil,
Nice try. You first. Describe how you set up this coherent wave/anti-wave pair that happily travel together for some indeterminate distance. Then I will describe what happens when at some arbitrary point and time they decide to annihilate. I will repeat one more time, since you did not seem to understand previously. * Maxwell's equations are all that is needed. * Interference is *derived* from the correct application of Maxwell's equations. It is not an independent physical entity. * Interference may be "intuitive" and it may help your understanding, but it adds precisely nothing to the physics of the problem. Everything you believe is buried in the magic of interference is already built into Maxwell's equations. 73, Gene W4SZ Cecil Moore wrote: Gene Fuller wrote: There is no need to invoke interference or wave cancellation to explain anything, ... But there is, Gene. It's the only way to correct the present misinformation and old wives' tales being promoted on this newsgroup. It is obvious that the r.r.a.a poster who understands the role that interference plays in the conservation of RF energy is very rare. There is a conspiracy to keep this information from surfacing - "Nothing new", "no need", "irrelevant", "who cares?" Why are you guys afraid to discuss the technical details? This should be an easy question to answer. If two coherent waves of 50 joules/sec each, are traveling in the same path in the same direction and are 180 degrees out of phase, they cancel in that direction of travel. What happens to their 50+50 joules/sec? Hint: energy doesn't cancel and there are only two possible directions. Can you spell R-E-F-L-E-C-T-I-O-N? |
Roy Lewallen wrote:
Where's the "reverse power" going? Since we can calculate 23 watts of constructive interference occuring toward the load, using the conservation of energy principle as explained by Hecht in "Optics", we can deduce that the reflected power is engaged in destructive interference inside the black box source. Using Dr. Best's conventions: V1 = 32v, V2 = 18v, Vfor = 50v, Vref = 30v Any theory that doesn't work in a circuit composed of simple elemental electrical circuit elements is suspect to say the least. Are you saying yours doesn't? It works just fine. Pfor = P1 + P2 + constructive interference. But that still looks like a Thevenin equivalent to me, you know, the one we cannot trust for internal power calculations. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Gene Fuller wrote:
Cecil, Nice try. You first. Describe how you set up this coherent wave/anti-wave pair that happily travel together for some indeterminate distance. Then I will describe what happens when at some arbitrary point and time they decide to annihilate. Sure, here's the two coherent reflected waves that cancel at a Z0-matched impedance discontinuity in a transmission line. b1 = s11*a1 + s12*a2 = 0 I'm sure you recognize the S-parameter equation for the reflected voltage flowing toward the source which is the phasor sum of two other reflected voltages. They don't travel together for some indeterminate distance. They are cancelled within the first dl and dt. And they don't annihilate. They simply cancel in the rearward direction. Incidentally, if you square both sides of the equation you get b1^2 = s11^2*a1^2 + s12^2*a2^2 + 2*s11*a1*s12*a2 Pref1 = rho^2*Pfor1 + (1-rho^2)*Pref2 + interference The forward voltage equation toward the load is b2 = s21*a1 + s22*a2 -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Your use of "Dr. Best's conventions" only muddles the matter -- neither
I nor probably any of the other readers have any idea what this is. In any event, the numbers you produced are volts. I and many others know how to calculate forward and reflected voltages and currents, and their sums. What's at issue here is where the imaginary waves of average power are going, what they're bouncing off, and why. Correct me if I'm wrong, but power is generally expressed in watts, BTUs per hour, or other more arcane units, but not volts. Let's try again. The source is providing 40 watts, 32 watts of which is delivered to the transmission line. The transmission line is transferring this 32 watts of power to the load. In the transmission line, we can calculate that there's 50 watts of "forward power", and 18 watts of "reverse power". How much of that 18 watts of reverse power is going through the source resistor to reach the source to "engage in destructive interference"? What does it interfere with? How does whatever it interferes with get there? Does any of the power going either way, forward or reverse, get dissipated in the source resistor? If so, how much and why? If not, why not? In your "explanation", I don't see a single figure in watts, except that "we have 23 watts of constructive interference occuring toward the load". (Where is this interference occurring, that is, just where is the point "toward the load" located? Where did the 23 watt figure come from? How much of it is "forward power" and how much "reverse power"?) It's not an explanation at all, but hand-waving. And why do you insist that every combination of voltage source and resistor be a "Thevenin equivalent"? I suggest you go back and read your basic circuits texts, where you'll find that a Thevenin equivalent is a circuit which is used to substitute for a more complex linear circuit to simplify analysis. The electrical circuit components used here are not a substitute for anything. And there's no rule, except something apparently stuck firmly in your mind(*), which prohibits calculating the power dissipated in a resistor. No matter what it's connected to. I make no claim that the power dissipated by the resistor represents anything but the power dissipated in a resistor, that the resistor represents anything but a resistor and the voltage source anything but a voltage source. It is not a Thevenin equivalent, it's a painfully simple electrical circuit (alas, so simple it's difficult to obfuscate). It doesn't matter if you can trust a Thevenin equivalent for internal power calculations. There is no "internal power" here -- it's all out in the open where we can easily measure and calculate it. There's nothing in that "black box source" except an ideal voltage source. You can find a description of this fundamental electrical circuit component, including its complete terminal characteristics, in the circuit analysis textbook of your choice. People skilled in the art are able to calculate the power it produces by multiplying v across it times i flowing from it, and the average power by applying the mathematical definition of average to the calculated power. In this case, it produces an average of 40 watts (100 volts times 0.4 amp). So please tell us how many watts are going where, what they do when they get there, and why. If you can't, you don't have a theory at all. (*)Forgive me, I just can't shake the image of a certain memorable scene from the movie "The Long Kiss Goodbye" as I write this. Roy Lewallen, W7EL Cecil Moore wrote: Roy Lewallen wrote: Where's the "reverse power" going? Since we can calculate 23 watts of constructive interference occuring toward the load, using the conservation of energy principle as explained by Hecht in "Optics", we can deduce that the reflected power is engaged in destructive interference inside the black box source. Using Dr. Best's conventions: V1 = 32v, V2 = 18v, Vfor = 50v, Vref = 30v Any theory that doesn't work in a circuit composed of simple elemental electrical circuit elements is suspect to say the least. Are you saying yours doesn't? It works just fine. Pfor = P1 + P2 + constructive interference. But that still looks like a Thevenin equivalent to me, you know, the one we cannot trust for internal power calculations. |
Roy Lewallen wrote:
We are not going to get anywhere until you admit there is 68 joules/sec in the feedline that haven't yet made it to the load. Once you admit that fact, everything else will be moot. Let's try again. The source is providing 40 watts, 32 watts of which is delivered to the transmission line. The transmission line is transferring this 32 watts of power to the load. In the transmission line, we can calculate that there's 50 watts of "forward power", and 18 watts of "reverse power". And it is easy to prove that the source has generated 50+18=68 watts that have not been delivered to the load. So I ask you: Where are those 68 joules/sec located during steady-state if not in the forward and reflected power waves? Why will 68 joules/sec be dissipated in the system *after* the source power is turned off? If those 68 joules/sec that have been generated by the source but not delivered to the load are not in the forward and reflected power waves, exactly where are they located? There's really no sense in continuing this discussion until you answer that question. Everything else is just a side argument. The answer to that question will expose the errors in your premises. You are apparently assuming there is not enough energy in the system during steady-state to support the forward and reflected power waves. But that exact amount of energy was supplied during the power-on transient state and will be dissipated during the power-off transient state. If it's not in the forward and reflected power waves, you are going to have to store it somewhere else. Where is that somewhere else? The source has supplied 68 joules/sec that has not reached the load. The forward and reflected power waves require 68 joules/sec. That you don't see the logical connection between those two equal energy values is amazing. But I will get you started on an understanding of the component powers using an S-parameter analysis. How much of that 18 watts of reverse power is going through the source resistor to reach the source to "engage in destructive interference"? reference the S-parameter equation: b1 = s11*a1 + s12*a2 I calculate 11.52 watts. (s12*a2)^2 = 11.52 watts The other 6.48 watts are in (s22*a2)^2 where |a2|^2 = 18 watts What does it interfere with? From the S-parameter equation above, it obviously interferes with s11*a1 . Please reference HP App Note 95-1, available on the web. It should answer most of your questions, in particular pages 16 & 17. |a1|^2 = Power incident on the input of the network |a2|^2 = Power incident on the output of the network |b1|^2 = Power reflected from the input port of the network |b2|^2 = Power reflected from the output port of the network -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Cecil,
You completely ducked the question. How did those waves get there in the first place? Hint: there are no laws for conservation of waves or continuity of waves. It is easy to set up a problem with physically unrealizable inputs. It is pointless to try to solve such a problem, however. We've been around this track a couple of times before. Neither of us has changed. Bye. 73, Gene W4SZ Cecil Moore wrote: Gene Fuller wrote: Cecil, Nice try. You first. Describe how you set up this coherent wave/anti-wave pair that happily travel together for some indeterminate distance. Then I will describe what happens when at some arbitrary point and time they decide to annihilate. Sure, here's the two coherent reflected waves that cancel at a Z0-matched impedance discontinuity in a transmission line. b1 = s11*a1 + s12*a2 = 0 I'm sure you recognize the S-parameter equation for the reflected voltage flowing toward the source which is the phasor sum of two other reflected voltages. They don't travel together for some indeterminate distance. They are cancelled within the first dl and dt. And they don't annihilate. They simply cancel in the rearward direction. Incidentally, if you square both sides of the equation you get b1^2 = s11^2*a1^2 + s12^2*a2^2 + 2*s11*a1*s12*a2 Pref1 = rho^2*Pfor1 + (1-rho^2)*Pref2 + interference The forward voltage equation toward the load is b2 = s21*a1 + s22*a2 |
Richard
Thanks for your explanation, I'm still thinking it over since its been 40 years since I've studied any tube theory. .. I recall many years ago watching a guy trying to fix an old radio. IIRC it had an 80 rectifier and right after the radio was turned on and started to warm up a blue cloud could be seen in the tube as the plates started to turn redder and redder. He would turn it off, change a part and check it again. I don't think he fixed it before the 80 went south. Later I figured out that he probably had a shorted filter cap. In this case the diode's plate dissipation rating was exceeded and it melted. The electrons slamming into the plate have a lot of kinetic energy to transfer. and heat up the plate. This results in lower efficiency as this power isn't delivered to the load. Now the diodes resistance can be easily calculated but I'm not sure how to visualize it. Is this where your term cathode resistance enters the picture? tnx -- 73 Hank WD5JFR "Richard Clark" wrote in message ... On Mon, 06 Jun 2005 20:52:52 GMT, "Henry Kolesnik" wrote: But in any event for something to melt we need dissipation and only resistance can do that! Hi Hank, Unfortunately as much as you and I agree on that bedrock principle, others with Simpson Ohmmeter in hand would glare goggle eyed at us and say that plate has no resistance to speak of and that no amount of current through its Ohmic resistance could ever bring about enough heat to produce the effects so obviously witnessed. One of the most enraging questions I've asked "If it is not the value I've offered, what value is it?" Well, I've never been given a quantitative answer, however I've seen enough carefully crafted mathematical proofs in this group to replace substantive results so easily seen. There is some irrefutable logic in circulation that clearly reveals that what we've experienced just couldn't be. Glasses will be need to be readjusted for such extreme myopic aberrations. There are two principles involved in what is called Plate Resistance, and the first and foremost is not even related to the plate at all. It is called the work function of the cathode emissivity. So, in fact it is more proper to refer to this usual loss as Cathode Resistance, not Plate Resistance. The cathode is the fundamental limit on power generated. What Plate Resistance is, is the ill termed substitution for Plate dissipation. If folks want to work their Simpson, they would blow an aneurysm trying to measure the resistance from cathode (filaments have the same work function issue too) to plate. In fact, the hobby horse argument of it is not resistance at all, but some figurative charting artifice called a "load line" usually appears in the last gasp. Plate Dissipation is resistance clear and simple in spite of the failure of conventional tools to measure a common physical property. Newton would have recognized it, it is called inertia. Once the work function is overcome (the job of the grid), then Plate voltage dominates through the acceleration of charge beyond the grid, toward the plate. That stream of electrons (and there is no doubt about actual current flow in easily counted, significant populations of electrons) is elevated to 90% the speed of light. This current flow is entirely different from what current flows in the remainder of the Plate load. That is also known as displacement current and electrons are shuffling along at a placid meter per second rate. Plate current and displacement current are equal in amplitude and phase, but not in motion nor kinetics. NOW. When that same stream encounters the Plate - WHAM! If anyone here has walked into the wall, and NOT encountered resistance, then we will call you Casper. Inertia reveals that to slow a mass in a distance results in acceleration (negative in this instance) and that property is called Force. Force over time expends calories and is expressed in any number of systems and units - Watts is one, Degrees is another. We could abstract to Horsepower and Candelas (the plate glows too). We know the speed, not many here would give it much though, but none would know the length interval of going at that speed to going zero (0). It is roughly two atoms distance into the metal of the plate. I will leave those calculations of Force to the student to compute or I can provide it from notes of correspondence with Walt Maxwell and Richard Harrison from a round robin discussion several years ago. Hank, does this fulfill your earlier question as to "what" is happening? I first gave you many examples, I hope this segue into real physics fills in their actuality. Too many correspondents demand that I open the source and point at a 50 Ohm carbon composition resistor that is the "source resistance." 73's Richard Clark, KB7QHC |
Gene Fuller wrote:
You completely ducked the question. How did those waves get there in the first place? Hint: there are no laws for conservation of waves or continuity of waves. I answered the question in another posting. The waves got there during the power-on transient state. Conservation of energy is assumed. Hope you don't disagree with that principle. If we make Roy's lossless 50 ohm feedline one second long (an integer number of wavelengths), during steady-state, the source will have supplied 68 joules of energy that has not reached the load. That will continue throughout steady state. The 68 joules of energy will be dissipated by the system during the power-off transient state. What you guys are trying to do is hide 68 joules of energy that cannot be destroyed. Where can you hide it in a transmission line to prove that it is not there in the forward and reflected waves? What is your agenda in trying to deny/hide/disguise/ignore that 68 joules of energy? In the one second example, the forward and reflected waves require 68 joules of energy for their existence in the feedline. The source has supplied 68 joules of energy that has not yet reached the load so it must necessarily still be in the feedline. Wonder where the energy in the forward and reflected waves came from? Shirley, you jest! Incidentally, QEX wants to publish my article. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Please reference HP App Note 95-1, available on the web.
There's only one in every half a million of radio amateurs who have a copy of this document, or have ever heard of its existence, let alone having any chance of finding a readable copy of it within the next dozen years. By which time they will have changed their hobby to keeping tropical fish or making Newtonian telescopes. Or just died. In all likelihood they won't be able to make any sense out of it anyway. Cec, why do you bother to mention it? (smiley) ---- Reg. |
Reg Edwards wrote:
W5DXP wrote: Please reference HP App Note 95-1, available on the web. There's only one in every half a million of radio amateurs who have a copy of this document, or have ever heard of its existence, let alone having any chance of finding a readable copy of it within the next dozen years. Give me a break, Reg. Download it for free and enjoy from: http://www.sss-mag.com/pdf/hpan95-1.pdf In all likelihood they won't be able to make any sense out of it anyway. My dog isn't able to make sense of it either - poor dog. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =---- |
Cecil Moore wrote:
Roy Lewallen wrote: [No, I didn't. Cecil wrote the following paragraph.] We are not going to get anywhere until you admit there is 68 joules/sec in the feedline that haven't yet made it to the load. Once you admit that fact, everything else will be moot. I never said a word about how much energy is stored in the feedline; it's irrelevant. It's stored there during the initial charging to the steady state condition, and the same amount remains there until the steady state condition no longer exists. It's exactly like the DC charge on a capacitor or a DC current through an inductor (which, in fact, is exactly what the feedline stored energy consists of) -- it doesn't have any effect on an AC analysis. [I did write this one.] Let's try again. The source is providing 40 watts, 32 watts of which is delivered to the transmission line. The transmission line is transferring this 32 watts of power to the load. In the transmission line, we can calculate that there's 50 watts of "forward power", and 18 watts of "reverse power". And it is easy to prove that the source has generated 50+18=68 watts that have not been delivered to the load. Surely even you can do the basic circuit analysis which shows that the source is continuously generating 40 watts, not 60. 32 of those are delivered to the load and 8 to the source resistor. I guess you mean 68 joules -- but as I said, it's irrelevant. So I ask you: Where are those 68 joules/sec located during steady-state if not in the forward and reflected power waves? Why will 68 joules/sec be dissipated in the system *after* the source power is turned off? The line's stored energy will be dissipated in either the source or load resistor or both when the source power is turned off. But we're doing a steady state analysis here. If those 68 joules/sec that have been generated by the source but not delivered to the load are not in the forward and reflected power waves, exactly where are they located? There's really no sense in continuing this discussion until you answer that question. Everything else is just a side argument. You tell me -- they can be anywhere you'd like. Just answer the simple questions about the power "waves". The answer to that question will expose the errors in your premises. What exactly is my premise, please? All I've done is to give the currents and powers at significant points in the circuit. Are any of the values incorrect? It's you who has the premise, not me. You are apparently assuming there is not enough energy in the system during steady-state to support the forward and reflected power waves. I'm making no such assumption. I'm questioning the existence of traveling waves of average power, and so far you've failed to give any evidence to convince me otherwise. But that exact amount of energy was supplied during the power-on transient state and will be dissipated during the power-off transient state. If it's not in the forward and reflected power waves, you are going to have to store it somewhere else. Where is that somewhere else? You tell me. My analysis doesn't need to consider the stored energy at all. Apparently yours does, so have at it. The source has supplied 68 joules/sec that has not reached the load. I think you mean 68 joules. The forward and reflected power waves require 68 joules/sec. The "forward power" is 50 watts. The "reverse power" is 18 watts. It requires 32 watts to sustain this. That's the amount of power flowing through the transmission line, from source to load. That's 32 joules/second, not 68. If the line were open circuited, the forward and reverse powers would be equal, and it would take no power to sustain them. That you don't see the logical connection between those two equal energy values is amazing. But I will get you started on an understanding of the component powers using an S-parameter analysis. Where will you find to put those all-important 68 joules in an s-parameter analysis? That's a steady state analysis. How much of that 18 watts of reverse power is going through the source resistor to reach the source to "engage in destructive interference"? reference the S-parameter equation: b1 = s11*a1 + s12*a2 I calculate 11.52 watts. (s12*a2)^2 = 11.52 watts Finally, an actual answer. So of the 18 watts of "reverse power", 11.52 watts is making it to the source to "engage in constructive interference". Does any of it get dissipated in the source resistor, or does it just slide through unscathed? The other 6.48 watts are in (s22*a2)^2 where |a2|^2 = 18 watts What's (s22*a2)^2? The forward power wave? The reverse power wave? If it's something else, does it have a name? Where does it go? Is it getting dissipated in the source resistor, reflected at the transmission line/resistor interface, reflected at the source/resistor interface, get radiated, or what? What does it interfere with? From the S-parameter equation above, it obviously interferes with s11*a1 . What's s11*a1? It must be something inside the source. The source is just that, a source. It has (AC) voltage and current, 100 volts of voltage and 0.4 amps of current. Does s11*a1 reside inside every source, or only some special ones? Apparently 11.52 watts of this s11*a1 gets cancelled by the reverse power wave. How much of it is left over? Please reference HP App Note 95-1, available on the web. It should answer most of your questions, in particular pages 16 & 17. |a1|^2 = Power incident on the input of the network |a2|^2 = Power incident on the output of the network |b1|^2 = Power reflected from the input port of the network |b2|^2 = Power reflected from the output port of the network Nope. Enough hand waving and evasion, we've been here before. [Of all the questions, the sole quantitative answer was "(s12*a2)^2 = 11.52 watts".] I'll leave you to the folks who regard this kind of gobbledegook as convincing evidence. Have fun -- I've got actual work to do while you take care of the visionary leadership part. Roy Lewallen, W7EL |
On Tue, 07 Jun 2005 15:01:26 GMT, "Henry Kolesnik"
wrote: Now the diodes resistance can be easily calculated but I'm not sure how to visualize it. Is this where your term cathode resistance enters the picture? Hi Hank, Getting electrons to "boil" off the cathode (or filament, same thing) is not a simple task otherwise there would be no filaments needed. Even with filaments, Edison current is not very considerable unless you add a monomolecular layer of metal to the surface of either the filament, or the cathode. You may note that some tubes are described as having "Thoriated" filaments. This is that monomolecular addition. Its purpose is to lower the W, the work function of the interface. It is far from odd how physics demonstrates at every stage and in every discipline that interfaces where there is mismatch, there is difficulty in transfering power. When an electron from the interior of the metal crystal approaches the surface, it is repelled by that interface due to the potential of the work function, and is attracted by the bulk material behind it. The Thoriated surface offers a matching mechanism between the bulk metal and the free space beyond the surface. In classic Optics this is known as Index Matching. To give examples as to how well a monomolecular addition performs: A Tungsten filament (no treatment) offers a current density of ½A/cM² This makes for a baseline. A Thoriated Tungsten filament offers a current density of 4A/cM² Then we step back to a cylindrical cathode employing Barium. Such a cathode offers a current density of ½A/cM² That seems rather regressive to use a cathode, but temperatures are telling. The simple Tungsten filament is operating at 2500° K and the cathode needs only to simmer along at 1000° K. This gives considerably longer life and more efficiency (most of the power for heating is lost through radiation). Needless to say, cathodes find more application in low power circuits, or their surfaces are treated with other low work function metals for greater emission. Now, when you add a potential gradient, you also lower the work function of the surface (but it is always an advantage to have it lowered going into this game). This is called the Schottky effect. One might be tempted to simply ask, why don't we up the voltage and discard the filament? This device would be called a Cold Cathode but the potential gradient then rises to the level where you run the risk of secondary emission. When that electron stream strikes the plate and raises the temperature, it is just short enough power to present this secondary emission. But if we were to run at 17KV or so, then the electron stream would be so aggressive as to produce high energy effects such as X-Ray emission. So, to return to the resistance of this all, we have physical impositions of cathode surface area (probably offering the prospect of being greater than filament surface area), current density, and potential difference. Discarding all the extraneous surface units (cM²) and employing the proper division (E/I) we have a resistor that glows in the dark under extremes of operation. How this fails to be source resistance is strictly handwaving and the schematic symbolic mysticism of demanding a carbon composition resistor. 73's Richard Clark, KB7QHC |
On Tue, 07 Jun 2005 11:44:29 -0700, Richard Clark
wrote: When that electron stream strikes the plate and raises the temperature, it is just short enough power to present this secondary emission. But if we were to run at 17KV or so, then the electron stream would be so aggressive as to produce high energy effects such as X-Ray emission. Something felt wrong here. I should have said: ...just short enough energy to present this secondary emission. |
Roy Lewallen wrote:
Cecil Moore wrote: We are not going to get anywhere until you admit there is 68 joules/sec in the feedline that haven't yet made it to the load. Once you admit that fact, everything else will be moot. I never said a word about how much energy is stored in the feedline; it's irrelevant. Do you think it is random coincidence that the amount of energy stored in the feedline is *EXACTLY* the amount of energy required by the forward and reflected waves???? The fact that you think it is irrelevant is simply a flight into a wet dream fantasy. It's stored there during the initial charging to the steady state condition, and the same amount remains there until the steady state condition no longer exists. It's exactly like the DC charge on a capacitor ... What a coincidence! It's *EXACTLY* the amount of energy required by the forward and reflected waves that you say don't exist - and it's RF photons, not DC, so it must travel at the speed of light! You cannot store photons in a capacitor. Surely even you can do the basic circuit analysis which shows that the source is continuously generating 40 watts, ... Of course, but during the power-on transient period, 68 watts is NOT delivered to the load. Please don't make me waste my time calculating the forward and reflected power during the power-on transient period. When you actually do those calculations, you will agree with me that, during the power-on transient phase, 68 watts of power has been stored in the feedline and remains there until the power-off transient phase. It is *EXACTLY* the amount of power required by the forward and reflected waves. The line's stored energy will be dissipated in either the source or load resistor or both when the source power is turned off. But we're doing a steady state analysis here. Which must necessarily include the energy stored in the feedline during the power-on transient condition because it is *still there during steady-state*. Ignoring the energy stored in the feedline during the power-on transient phase is both irrational and illogical. It could even be bad for your mental health. What exactly is my premise, please? from my earlier posting: You are apparently assuming there is not enough energy in the system during steady-state to support the forward and reflected power waves. I'm making no such assumption. I'm sorry, Roy, but that is just BS! Either you admit there is enough energy to support the steady-state forward and reflected waves or you don't. My analysis doesn't need to consider the stored energy at all. And that is exactly why your analysis is wrong. You have, once again, been seduced by the steady-state model and are spreading old wives' tales as a result. Hopefully, you don't really want to do that. The source has supplied 68 joules/sec that has not reached the load. I think you mean 68 joules. 68 joules/sec in your original example. 68 joules in my one-second- long feedline example. Where will you find to put those all-important 68 joules in an s-parameter analysis? That's a steady state analysis. The 68 joules were stored in the feedline during the power-on transient phase. They are still there during steady-state. An S-parameter analysis yields the correct results because it includes the reflected power, |a2|^2, as an energy source, something you deny. Wonder what the S-parameter analysis folk know that you don't choose to admit? Finally, an actual answer. So of the 18 watts of "reverse power", 11.52 watts is making it to the source to "engage in constructive interference". Does any of it get dissipated in the source resistor, or does it just slide through unscathed? It never encounters the source resistor as it is re-reflected by wave cancellation, not by an impedance discontinuity. I have a QEX article coming soon that will explain the details. Stand by. With an unprejudiced open mind, you might actually learn something. What's (s22*a2)^2? Same as (1-rho^2)*Pref2 in ham terms. Reflected power from the load that is re-reflected back toward the load by an impedance discontinuity. What's s11*a1? Same as Vfor1*rho in ham terms. Forward voltage that is reflected back toward the source by an impedance discontinuity. I'll leave you to the folks who regard this kind of gobbledegook as convincing evidence. Do you realize what intellectual shape you would be in if you had adopted that attitude when you were one year old? :-) -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
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