|
Don't have a copy of said dictionary at hand.
"Cecil Moore" wrote in message ... Gene Fuller wrote: There is no "proof" that non-dissipative resistance exists. This term is a "definition", not something that can be proven. Gene, seems to me that the necessity of two "non-equivalent" definitions of "resistance" in the IEEE dictionary is proof of something that needs differentiating in the language. Did trees exist before the word "tree" was invented? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
The phrase "None-dissipative resistance" is utter nonsence. It serves no practical purpose whatsover. It has no educational value such as imaginary jX. Its very mention serves only to confuse, particularly to apprentices in the electrical engineering trade. As a practicing engineer for nigh on 60 years I had never heard of it until the old wives and authors on this newsgroup began wittering about it for the sole but futile purpose of demonstrating how knowledgeable they are and for making money. I have managed very well without it thank you. They are from the same class as they who consider radiation from a dipole is mainly from the middle third. But I suppose they must be tolerated. Live and let live, eh! And, to the particular individual concerned, I managed to write this in plain English without swallowing Websters Morocco-bound Dictionary. Having got that off my chest I will finish this nice glass of Californian Red (yes, American. I don't believe in trade sanctions) and retire to bed. Good night everybody. ---- Reg. |
Reg Edwards wrote:
SNIPPED Having got that off my chest I will finish this nice glass of Californian Red (yes, American. I don't believe in trade sanctions) and retire to bed. Good night everybody. ---- Reg. Good morning Reg, Oh Reg: I've spent more than a year building a mental image of the gentleman englishman only to have it destroyed by "Californian Red". A nice continental Sherry would preserve my image! I use only a gram or two of altar wine once a day in the morning! To each his own, but you do have to rebuild my image of you. :-) |
Reg,
Well said! Now go treat yourself to a Wyndham Estates Bin 555 Shiraz, Tyrell's Long Flat, or perhaps a Cesari Amarone Valpolicella. (what particular Californian Red were you enjoying? There's a Corbett Canyon that's not too bad, but I have found very few that are worthy of mention.) B. |
"Reg Edwards" wrote:
The phrase "None-dissipative resistance" is utter nonsence. It is my understanding that if the reactance of the characteristic impedance of a transmission line is zero, the characteristic impedance is non-dissipative and a pure resistance. The IEEE's "resistance = real part of impedance" is indeed often non-dissipative. If a voltage is in phase with a current, the V/I ratio is a resistance but not necessarily dissipative. That's why the IEEE has two definitions of "resistance". -- 73, Cecil, W5DXP |
Cute...
"Richard Clark" wrote in message ... On Sat, 06 Mar 2004 18:19:20 -0600, Cecil Moore wrote: Richard, how much do you pay for your blinders? Trying to sell yours? |
I wonder what happened to my last post with questions to Richard Harrison.
I don't see it. Comments inserted below. -- Steve N, K,9;d, c. i My email has no u's. "Walter Maxwell" wrote in message ... On Fri, 05 Mar 2004 21:50:55 -0600, Cecil Moore wrote: wrote: Non-dissipative resistance is not well accepted or understood by many otherwise well informed engineers, because it has had little or no (or even incorrect) treatment in EE courses. On Sat. 06 Mar 2004 Cecil Moore wrote: Yet the IEEE recognizes those two types of resistances with different definitions. Definition (A) talks about "dissipation or other permanent loss". Definition (B) simply says "The real part of impedance." Then a note: "Definitions (A) and (B) are *NOT* equivalent ..." (emphasis mine) The resistance in a resistor satisfies definition (A). The characteristic impedance of a transmission line satisfies definition (B). What you said above is true, Cecil, but one more statement applies to Definition (B). Although Definitions (A) and (B) are not equivalent, Definition (B) does include the real part of the impedance of a dissipative resistor. The only way to tell which is which is to determine which develops heat. I still maintain that many otherwise well qualified engineers not aware of Definition (B), and therefore reject the concept of a resistance that doesn't dissipate power. And this applies to much more than the Zo of transmission lines. I consider myself very well qualified and have no problem with both of these conditions and believe I understand them fully. (my post of questions to Richard Harrison would show, if I could see it--wonder if I didn't push the send button) I can only say that I have never heard the term "non-disipative resistance" either professionally or in the hobby. It is not taught. If I had to say why, it is because it is a self-conflicting term. In my nomenclature, all "resistance" dissipates power as heat. In other words, resistance is what resistors have. Resistance can produce the real part of an impedance, but the real part can come from other things. In other situations such as the T-Line, there is a "real part" to the impedance. The Engineer understands that this is due to a combinatin of physical things, not only a resistor, so the term in question refers to something that the Engineer does not need to discuss. When there is only a real part (no reactance), then both are indistinguishable to the source -- so from the source's point of view, they are the same. In the bigger picture, however, there is a difference -- heat vs. power going elsewhere. However, I do not believe it is a good term (loss-less resistance) to use in this case. I would call it an unnecessary complication to add a new term of "loss-less resistance" when you understand all of this. If you were to have a term "loss-less real part", then I'd say you have a more accurate technical term. Although, the Engineer (by this, I mean both me and those I have discussed circuit concepts in school and professionally) knows what is underdiscussion and what the impedance is and where it comes from, so a new term is unnecessary. I don't know if this helps, but it really seems to be adding a term or name when it is not necessary. So while the term may "work" for some people, I believe it is a slight mis-use of the term "resistance". When you first start talking about "loss-less resistance" to a schooled Engineer, he'she gets the wrong idea as to just what you are talking about since it is a conflict in terms. -- Steve N, K,9;d, c. i My email has no u's. And because there still remains many who believe the RF power amplifier absorbs and dissipates reflected power, I chose to try again to dispel that notion in my post in the 'max power theorem' thread. Walt Maxwell, W2DU |
"Steve Nosko" wrote in message ... From: Richard Harrison ) Subject: Lossless Resistance? ate: 2004-03-05 20:06:31 PST Steve Nosko wrote: "Power is calculated from "RMS" values of voltage and current." Steve is correct! I apologize. If you have a square wave of minimum peak to peak value of zero volts, its maximum peak to peak value must be 1.414 x the d-c value. Then, 1.414 Vd-c x 1.414 Id-c = 2 Pd-c. 2 Pd-c x 1/2 t = Pd-c for an average. Best regards, Richard Harrison, KB5WZI Whew! Richard H. For some reason, the above post does not show on my reader. I had to go to Google to see what I was missing. After seeing your answer to my teaser about voltage vs. power dB, I was really getting worried (about U) on this one. You DO know. I think the thing here is that BOTH power and Current are averaged by the chopper. I am, however, surprised that you did not have a "gut feel" for that. I just posted another short explanation of MY take on the term "loss-less resistance". I wonder why I am not seeing some posts.. There is another thread about the maximum power transfer theorm. This thread was about the term "loss less resistance". -- Steve N, K,9;d, c. i My email has no u's. |
I wonder what happened to my last post with questions to Richard Harrison.
I don't see it. Boy! I'm not doing well at all on this Usenet thing today..... Comments inserted below. -- Steve N, K,9;d, c. i My email has no u's. "Walter Maxwell" wrote in message ... On Fri, 05 Mar 2004 21:50:55 -0600, Cecil Moore wrote: wrote: Non-dissipative resistance is not well accepted or understood by many otherwise well informed engineers...no ...treatment in EE courses. On Sat. 06 Mar 2004 Cecil Moore wrote: Yet the IEEE recognizes those two types of resistances with different definitions. Definition (A) talks about "dissipation or other permanent loss". Definition (B) simply says "The real part of impedance." Then a note: "Definitions (A) and (B) are *NOT* equivalent ..." (emphasis mine) The resistance in a resistor satisfies definition (A). The characteristic impedance of a transmission line satisfies definition (B). What you said above is true, Cecil, but one more statement applies to Definition B). Although Definitions (A) and (B) are not equivalent, Definition (B) does include the real part of the impedance of a dissipative resistor. The only way to tell which is which is to determine which develops heat. I still maintain that many otherwise well qualified engineers not aware of Definition (B), and therefore reject the concept of a resistance that doesn't dissipate power. And this applies to much more than the Zo of transmission lines. I consider myself very well qualified and have no problem with both of these conditions and believe I understand them fully. (my post of questions to Richard Harrison would show, if I could see it--wonder if I didn't push the send button) I can only say that I have never heard the term "non-disipative resistance" either professionally or in the hobby. It is not taught. If I had to say why, it is because it is a self-conflicting term. In my nomenclature, all "resistance" dissipates power as heat. In other words, resistance is what resistors have. Resistance can produce the real part of an impedance, but the real part can come from other things. In other situations such as the T-Line, there is a "real part" to the impedance. The Engineer understands that this is due to a combinatin of physical things, not only a resistor, so the term in question refers to something that the Engineer does not need to discuss. When there is only a real part (no reactance), then both are indistinguishable to the source -- so from the source's point of view, they are the same. In the bigger picture, however, there is a difference -- heat vs. power going elsewhere. However, I do not believe it is a good term (loss-less resistance) to use in this case. I would call it an unnecessary complication to add a new term of "loss-less resistance" when you understand all of this. If you were to have a term "loss-less real part", then I'd say you have a more accurate technical term. Although, the Engineer (by this, I mean both me and those I have discussed circuit concepts in school and professionally) knows what is underdiscussion and what the impedance is and where it comes from, so a new term is unnecessary. I don't know if this helps, but it really seems to be adding a term or name when it is not necessary. So while the term may "work" for some people, I believe it is a slight mis-use of the term "resistance". When you first start talking about "loss-less resistance" to a schooled Engineer, he'she gets the wrong idea as to just what you are talking about since it is a conflict in terms. -- Steve N, K,9;d, c. i My email has no u's. And because there still remains many who believe the RF power amplifier absorbs and dissipates reflected power, I chose to try again to dispel that notion in my post in the 'max power theorem' thread. Walt Maxwell, W2DU |
"Steve Nosko" wrote in message ... I wonder what happened to my last post with questions to Richard Harrison. I don't see it. Boy! I'm not doing well at all on this Usenet thing today..... BTW Richard H. No need to apologize. You made me go back and look closely to make sure I had it correct. You saw the error AND you really do understand -- that is the most important. I frequently have the fear that I will speed through a calculation and get a response wrong. Excel is my friend! Besides, I just finished pages and pages of these same calculations spurred by Bob Schraders (error prone) Watt meter article in QST. I hope I always sounded as though I was a discussion of concepts rather than emotion. I was beginning to ask friends how to end the thread politely with someone who has it wrong and just doesn't get it. I won't get insulting (nudge, nudge, wink, wink), but I have this bone in my head which makes me WANT to say "I know I'm right, I tried to help you understand so you go prove it to yourself." without sounding insulting... 73, -- Steve N, K,9;d, c. i My email has no u's. |
On Mon, 8 Mar 2004 15:35:46 -0600, "Steve Nosko"
wrote: I wonder what happened to my last post with questions to Richard Harrison. I don't see it. Boy! I'm not doing well at all on this Usenet thing today..... Comments inserted below. Hi Steve, Newsgroup material moves across the internet in a bucket brigade style of relay using NNTP as the control language. It passes from the source back towards you (such that you can see you own posting) with delay that is variable. For me, that delay is on the order of 1 second to three or four minutes. On bad days it can take as many hours. On Google, the delay is more pronounced. 73's Richard Clark, KB7QHC |
Steve Nosko wrote:
In my nomenclature, all "resistance" dissipates power as heat. Even SQRT(L/C)???? The Z0 of transmission line is a resistance, e.g. 50 ohms, i.e. the "real part of impedance". Seems the two "non-equivalent" IEEE definitions resolve the contradictions in your posting. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Steve Nosko wrote:
For some reason, the above post does not show on my reader. I had to go to Google to see what I was missing. You must be using a computer with Motorola chips. :-) -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Steve Nosko wrote:
I won't get insulting (nudge, nudge, wink, wink), but I have this bone in my head which makes me WANT to say "I know I'm right, I tried to help you understand so you go prove it to yourself." without sounding insulting... Well, Steve, maybe you can tell us exactly what happens at the Z0-match point in the following system. What changes the direction and momentum of the reflected power wave from the load? 100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
"Richard Clark" wrote in message ... On Mon, 8 Mar 2004 15:35:46 -0600, "Steve Nosko" wrote: I wonder what happened to my last post Newsgroup material moves across the internet ... delay that is variable. For me, that delay is on the order of 1 second to three or four I posted on Friday and couldn'r see it on Monday. I went to Googls and at least saw Richard H.s comment but not my own. I may have goofed. For the last few weeks now, i ALWAYS get an error message when posting though they have always shown up in a few seconds (if I look). 73, Steve |
"Cecil Moore" wrote in message ... Steve Nosko wrote: I won't get insulting (nudge, nudge, wink, wink), but I have this bone in my head which makes me WANT to say "I know I'm right, I tried to help you understand so you go prove it to yourself." without sounding insulting... Well, Steve, maybe you can tell us exactly what happens at the Z0-match point in the following system. What changes the direction and momentum of the reflected power wave from the load? 100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load Well, Cecil... First, I don't know what point you refer to as "Z0-match point". If it is the "---x---", I proceed... For the way you pose the problem (with no other constraints), the Z @ "x" seen toward the load is 50 ohms, so all is happy in the universe. The 100W enters the ladder line and exits at the "50 ohm load". I'm obviously assuming I don't have balanced-to-unbalanced-problems and I know the good 'ole G5RV does this. So I'd have to ask you just what point are you examining? Is it the use of balanced T-Line as a matching transformer in an unbalanced system, or something else? I think this setup is a sub-optimal way to do it...but Hey! if it works, use it. Manu suboptimal things are used all the time. I don't know from "momentum" of power. If you asking what is going on within the section of ladder line... The easy explanation is that the waves act such that the two ends look line 50 ohms (OH YES, this DOES assume the TX has a 50 ohm Zout... I think). I'd have to study the detail for a while to come up with my mental model and be able to put it into words. I know that the subject of "which way is power flowing" and "what happens at the source" and "how many reflections really occur" are big topics here. I don't get into these discussions because 1) it's been a long time since I studied it and 2) the path and nitty-gritty detail has faded into the far recesses of my mind and I would surely stumble around for a proper answer. I am currently teaching a class in communications which will get into T-lines later. If I need to get this deeply into the subject, perhaps I'll come back and post my take on it. -- Steve N, K,9;d, c. i My email has no u's. |
"Cecil Moore" wrote in message ... Steve Nosko wrote: In my nomenclature, all "resistance" dissipates power as heat. You need to include the rest of my post to to it justice. I discuss the terminology distinction. I go on the say that the term "real part (of an impedance)" is better suited (as a name) to what some like to call "loss-less resistance". In the chopper case that Richard Harrison posed, there is no parallel to the T-line situation. There, he was comparing a chopper with 50% duty cycle to an equal valued resistor and calling the chopper a "loss-less resistance" Even SQRT(L/C)???? The Z0 of transmission line is a resistance, I don't consider that term (resistance) suitable for this situation. "real part of Z" is better. I think to some this is the same thing, but obviously it is not. I believe this is what is causing all the confusion. Remember, the Z0 is properly called "Characteristic Impedance" or "Surge Impedance". I think this distinction makes the subject easier to understand since it eliminates the confusing term "loss-less resistance". Seems the two "non-equivalent" IEEE definitions resolve the contradictions in your posting. I'd have to read the full this before having an opinion! -- Steve N, K,9;d, c. i My email has no u's. |
I see this paper as a variation on a line of reasoning that goes like
this: 1) A conjugate match results in maximum power delivered to a load, so it is good. 2) A connection where the load has much higher resistance than the Thevenin equivalent source resistance results in high efficiency, so it is also good. 3) Since (1) and (2) are both good, they must be equivalent to each other. Therefore a conjugate match is what it is not. This is an apparent contradiction. 4) The contradiction is resolved by postulating a special kind of resistance that adds to the source resistance. However, it has no physical effect and exists only to resolve the contradiction in (3). All good matches are now conjugate matches and everyone is happy! 73--Nick, WA5BDU "Dave" wrote in message ... have you guys read this one yet? www.qsl.net/w9dmk/MPTT.pdf |
"Nick Kennedy" wrote in message om... I see this paper as a variation on a line of reasoning that goes like this: 1) A conjugate match results in maximum power delivered to a load, so it is good. 2) A connection where the load has much higher resistance than the Thevenin equivalent source resistance results in high efficiency, so it is also good. 3) Since (1) and (2) are both good, they must be equivalent to each other. Therefore a conjugate match is what it is not. This is an apparent contradiction. 4) The contradiction is resolved by postulating a special kind of resistance that adds to the source resistance. However, it has no physical effect and exists only to resolve the contradiction in (3). All good matches are now conjugate matches and everyone is happy! 73--Nick, WA5BDU nick, I like your summary. You have captured the essence of my original efficiency puzzle...except for your #4. My puzzle was not intended to be an equivlent to maximum power transfer. Only an "efficiency enhancement" technique/concept.idea/proposal. I did not intend to imply that my solution was transferring "maximum power" only a higher efficiency than the alternate case that I described. Your #3 is clearly only for the discussion. The "they are both good, therefore, they are equivalent" concept clearly can't be a serious conclusion, just an argument tool. -- Steve N, K,9;d, c. i My email has no u's.. |
Steve Nosko wrote:
"Cecil Moore" wrote: 100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load I don't know from "momentum" of power. If you asking what is going on within the section of ladder line... The easy explanation is that the waves act such that the two ends look line 50 ohms It's a pretty simple question. The energy wave reflected from the load possesses energy and momentum, both of which must be conserved. There is no reflected energy on the 50 ohm coax and therefore no momentum in the reflected waves on the 50 ohm coax. So what changes the direction and momentum of the energy wave reflected from the load? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Steve Nosko wrote:
"Cecil Moore" wrote: Even SQRT(L/C)???? The Z0 of transmission line is a resistance, I don't consider that term (resistance) suitable for this situation. "real part of Z" is better. Steve, the "real part of Z" is the IEEE definition of "resistance", i.e. they are exactly the same thing. Z = R + jX Guess what the R stands for? Methinks the problem is that people confuse "resistor" and "resistance". A resistance is not necessarily a resistor. More often than not, it is simply a V/I ratio. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Steve Nosko wrote:
"Remember, the Zo is properly called "Characteristic impedance and Surge impedance." No doubt impedance is used because some are uncomfortable calling a resistance a resistance when it wastes no energy. I think this is unfortunate because it weakens the important idea that resistance is first and foremost a ratio of voltage to current in which the voltage and current are in-phase. Those who insist that every resistance wastes electrical energy by converting it to heat are still firmly stuck in the d-c era. Terman writes on page 88 of his 1955 edition: "The characteristic impedance Zo is the ratio of voltage to current in an individual wave--- it is also the impedance of a line that is infinitely long or the impedance of a finite length of line when ZL=Zo. It will be noted that at radio frequencies the characteristic impedance is a resistance that is independant of frequency." Best regards, Richard Harrison, KB5WZI |
"Cecil Moore" wrote in message ... Steve Nosko wrote: "Cecil Moore" wrote: 100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load I don't know from "momentum" of power. If you asking what is going on within the section of ladder line... The easy explanation is that the waves act such that the two ends look line 50 ohms It's a pretty simple question. The energy wave reflected from the load possesses energy and momentum, both of which must be conserved. There is no reflected energy on the 50 ohm coax and therefore no momentum in the reflected waves on the 50 ohm coax. So what changes the direction and momentum of the energy wave reflected from the load? oh wow, all the way to momentum calculations now. is it time to point you all at a design for a pure electromagnetic rocket engine that uses the momentum of traveling waves to push itself along?? |
Nick, WA5BDU wrote:
"3) Since (1) and (2) are both good, they must be equivalent to each other." (1) is the maximum power transfer theorem. (2) is an efficiency statement. (1) and (2) are dissimilar. (1) says that the only way to get all the power you can from a source is to adjust the load resistance to equal that of the source and to eliminate any reactance which opposes power transfer. This is demonstrable; no need to argue. (2) Resistance of the type which converts electrical energy to heat and that is in the path of energy flow extracts energy and reduces efficiency. It is indubitably minimized to maximize efficiency. The argument in this thread is on the existence of lossless resistance. Proof of the existence of lossless resistance is demonstrable. We have power sources delivering maximum power without dissipating as much power within as is being delivered to the load. That is conclusive. Best regards, Richard Harrison, KB5WZI |
"Richard Harrison" wrote in message
... [...] The argument in this thread is on the existence of lossless resistance. Proof of the existence of lossless resistance is demonstrable. We have power sources delivering maximum power without dissipating as much power within as is being delivered to the load. That is conclusive. Best regards, Richard Harrison, KB5WZI Ahhh! Thanks, Richard. Now I see the essence of the discussion. I believe there is an inherent conflict in the terms you have, namely; "delivering maximum power" and "without dissipating as much power...as ...the load". This is an interesting tack, but I don't see the demonstration you refer to. I am missing just what it is that gets to the "Loss-less resistance" conclusion. We know that the MPT Theorem says the source and load powers must be equal for THAT situation...so. I think (I hope) you agree that in this case, the limiting factor for the delivered power is the source's "Internal" resistance. Now lets look at a case where the source Z is LESS than the load Z. I will take for a first example (at the risk of being accused of being "stuck" in DC) a standard series regulated power supply. (this is the same example I described with my "change the load Z to a higher value" example earlier.) The supply can be designed to supply many watts, right! Yet they have very low output resistances, right! Lets look he 1- Is it putting out its "maximum power" when loaded to its rating? 2- Is it dissipating an amount equal to the load? 3- What is the limiting factor for its output current (what is it that allows us to only draw its rated current?)? SO, are you saying that we have some "Loss-Less Resistance", in this case, which equals the load resistance? If we were to put a load on it which equaled the 0.05 ohm output resistance, then what situation do we have? Maximum power transfer? In the above case, the series pass transistor power dissipation ability is the limiting factor on the output power of the supply, not the internal resistance. I can make the same case for a switching supply and my "raise the supply voltage" example is the same thing. SO...? -- Steve N, K,9;d, c. i My email has no u's. |
|
Dave wrote: oh wow, all the way to momentum calculations now. I don't think we should expect to see any momentum "calculations" from Cecil. He can't even explain how the energy magically reverses direction and heads back toward the load. That's why he always phrases that as a question. The fact is, in a matched system, energy obviously isn't reflected from the load. 73, Jim AC6XG |
Yikes! Long wind alert!
"Cecil Moore" wrote in message ... Steve Nosko wrote: "Cecil Moore" wrote: 100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load I don't know from "momentum" of power. If you asking what is going on within the section of ladder line... The easy explanation is that the waves act such that the two ends look line 50 ohms It's a pretty simple question. The energy wave reflected from the load First I must understand where we are looking. I think is is best to take small steps in the descriptions. (or is this a troll, Cecil) You want to look at a point on the 450 ohm line, is that correct? Assuming yes... I think you are saying that we have a 450 ohm line and a 50 ohm load and therefore there is a reflection. Is that where you are? ....pause... I would rather take one step at a time and stick to the above as the main discussion, but will comment as follows anyway, so the following may be a digression. possesses energy and momentum, both of which must be conserved. Again, I do not know what momentum is for an EM wave on coax and don't believe it is necessary for this discussion. Waves move along the line and we are trying to come to a common understanding of what happens at the junctions, I think... As far as conservation is concerned, I assume a loss-less line. There is no reflected energy on the 50 ohm coax and therefore no momentum in the reflected waves on the 50 ohm coax. I think you contradicted yourself here. Are you saying that there IS or IS NOT reflected energy/waves on the 50 ohm section (to the left of the "--x--"?? I say there is none. It is loaded by the 1/2 wave of 450 ohm line which presents a 50 ohm appearance. To be honist, I don't think I can remember what happens at this junction. I think there is a jump in voltage since 450 is greater than 50 - for the forward wave. In any case, the sum result of the forward wave and the reflected wave is such that E/I=50 at this junction. This is just to the right of the "--x--". So what changes the direction and momentum of the energy wave reflected from the load? You lost me here. Perhaps you are asking; "If there is a wave reflected at the end with the 50 ohm, back toward the left, how does this power then get absorbed IN that load?" Is this the question? The wave, in the 450 section, reflected from the load is NOT the power being delivered TO the load, so it does not have to be "changed' to be delivered there. The wave traveling to the right on the 450 ohm section is *significantly* diffeernt that what is absorbed in the load. I'd be hard pressed to give a value since school was some odd...mumble mummble years ago... Perhaps this is a stumbling block. The wave traveling to the right on the 450 section can be much larger than the power delivered to the load. Weird, but true. It has to be if there is some bounced back to the left and still deliver 100W to the load. I'm sure there is some math to show this...reflection coef and all... 73, -- Steve N, K,9;d, c. i My email has no u's. |
On Tue, 9 Mar 2004 18:01:12 -0600, "Steve Nosko" Ahhh! Thanks,
Richard. Now I see the essence of the discussion. I believe there is an inherent conflict in the terms you have, namely; "delivering maximum power" and "without dissipating as much power...as ...the load". This is an efficiency statement. This is an interesting tack, but I don't see the demonstration you refer to. I am missing just what it is that gets to the "Loss-less resistance" conclusion. We know that the MPT Theorem says the source and load powers must be equal It says nothing of the sort. The comparisons are Z based. for THAT situation...so. I think (I hope) you agree that in this case, the limiting factor for the delivered power is the source's "Internal" resistance. Now lets look at a case where the source Z is LESS than the load Z. I will take for a first example (at the risk of being accused of being "stuck" in DC) a standard series regulated power supply. (this is the same example I described with my "change the load Z to a higher value" example earlier.) The supply can be designed to supply many watts, right! Yet they have very low output resistances, right! You started with a Z and ended with an R. As you are confined to a DC argument that is allowable, but the logic does not extend to an RF source. Lets look he 1- Is it putting out its "maximum power" when loaded to its rating? Rating is a marketing issue. 2- Is it dissipating an amount equal to the load? Only if the load R equals the internal Ohmic loss. 3- What is the limiting factor for its output current (what is it that allows us to only draw its rated current?)? Thevenin explicitly describes the source Z as the ratio of open circuit voltage divided by short circuit current. Substitute R for Z in this special case. The internal Ohmic loss defines the total current available. This is not necessarily the rated current as the system components may not tolerate the heat even though the current is "available." This is the distinction of the "marketing." SO, are you saying that we have some "Loss-Less Resistance", in this case, which equals the load resistance? You have crossed the line from DC to other less restricted examples. Your logic will fail at some point. The "Loss-Less Resistance" cannot avoid Ohmic losses; but through clever engineering, DC supplies (switchers) can reduce the amount of current through them. If you put as large of a conventional heat sinking on switchers, their capacity would rise (that marketing term of "rated"), but then it intrudes on the quality of switchers for being small (and is a facile response that belies more complex issues). If a switcher is 99% efficient, it still dissipates 1% power in heat from Ohmic loss. If we were to put a load on it which equaled the 0.05 ohm output resistance, then what situation do we have? Maximum power transfer? You've switched back (no pun) to the DC argument? I presume so. Can you envision any other outcome? If we are to allow an infinitely adaptable source, then this becomes a tediously futile exercise. However, that is often the argument offered (hence the violation of Thevenin's requirement for linearity). In the above case, the series pass transistor power dissipation ability is the limiting factor on the output power of the supply, not the internal resistance. It would be difficult to separate the two wouldn't it? This delves back into the marketing term of "rated." You can certainly design a finals deck without a heat sink to power limit it. Anyone can design for failure to prove that outcome. The actual, real answer is that the power out is current density limited at the emitter-collector junction. Nearly all failures of solid state power designs devolve to melt-down, a catastrophic thermal failure borne of resistance. The current density is an abstract expression of the inability of the device to shed heat through another resistance: thermal resistance which is expressed in degrees per Watt. If the junction temperature rises to melting through the concurrent availability of so many Watts - BINGO! Failure. In the parlance of the wire trade, this is called the fusing current. I can make the same case for a switching supply and my "raise the supply voltage" example is the same thing. Only if you neglect heat, sinking, thermal resistance, melt-down and the host of real concerns. SO...? Hi Steve, If you care to look at any power transistor operating curves, absolutely NONE run to "rated" capacity. Across the board, they are "de-rated" at the upper limit (observe the dashed lines that clip the maximum region). Like the specification of GainBandWidth, power ratings are mutable against a myriad of considerations, most of which do not allow full blown power. 73's Richard Clark, KB7QHC |
On Tue, 9 Mar 2004 18:34:43 -0600, "Steve Nosko"
wrote: Yikes! Long wind alert! 100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load I don't know from "momentum" of power. If you asking what is going on within the section of ladder line... .... take small steps in the descriptions. (or is this a troll, Cecil) .... Is that where you are? ...pause... .... Again, I do not know what momentum is for an EM wave on coax and don't believe it is necessary for this discussion. .... I think you contradicted yourself here. Are you saying that there IS or IS NOT .... You lost me here. Perhaps you are asking .... Is this the question? This is going to take a loooooooooong time, Steve. Follow your instincts and accept it only for its entertainment value. 73's Richard Clark, KB7QHC |
Jim Kelley wrote:
I don't think we should expect to see any momentum "calculations" from Cecil. He can't even explain how the energy magically reverses direction and heads back toward the load. It's not magic and is explained on the Melles-Groit web page at: http://www.mellesgriot.com/products/optics/oc_2_1.htm Here's a quote: "Clearly, if the wavelength of the incident light and the thickness of the film are such that a phase difference exists between reflections of p, then REFLECTED WAVEFRONTS INTERFERE DESTRUCTIVELY, and overall reflected intensity is a minimum. If the two reflections are of equal amplitude, then this amplitude (and hence intensity) minimum will be ZERO." Destructive interference stops the energy flow and momentum of the two reflected waves toward the source (at the Z0-match point 'x' below. Continuing: "In the absence of absorption or scatter, the principle of conservation of energy indicates all "lost" reflected intensity will appear as ENHANCED INTENSITY (constructive interference) in the transmitted beam" (traveling toward the load). The energy involved in the destructive interference is not lost and appears as constructive interference in the opposite direction. That energy has changed direction and had its momentum reversed. The fact is, in a matched system, energy obviously isn't reflected from the load. No reflections from the load? Neglecting losses, I calculate 178 watts reflected from the load and 278 watts of forward power on the 450 ohm feedline in the following matched system. 100w XMTR---50 ohm feedline---x---1/2WL 450 ohm feedline---50 ohm load -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Steve Nosko wrote:
"I am missing just what it is that gets to the "Loss-less resistance" conclusion." If part of the source resistance were not lossless, efficiency would be limited to 50%. Fact is, Class-C amplification frequently is 2/3 or 66.6% efficient. Twice the power is delivered to the load as is lost in the source. If we deliver 1000 watts into 50 ohms, we have 224 volts and 4.47 amps as a load. Our source is identical,, 224 volts and 4.47 amps. Its volts to amps is the same ratio, but its loss is not the same power as the power delivered to the load because part of the source resistance is lossless because it is the product of interrupted energy delivery, not energy conversion into heat. With 2/3 efficiency, when we have 1000 watts into the load, we have 500 watts lost in the source. 500 watts lost means only 1/2 the dissipative resistance as we have resistance in the load, and thus the source resistance consists of 25 ohms of dissipative resistance and 25 ohms of non-dissipative resistance. The total source resistance is 50 ohms which matches the load resistance. A match allows maximum power transfer. The less than 360 degrees of energy supplied to the load makes 25 ohms of dissipationless resistance as a part of our 50-ohm source in our example. Keep working on the idea and eventually you may get the model. Best regards, Richard Harrison, KB5WZI |
Steve Nosko wrote:
100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load Assuming yes... I think you are saying that we have a 450 ohm line and a 50 ohm load and therefore there is a reflection. Is that where you are? Yes, there are reflections on the 450 ohm ladder line and none on the 50 ohm coax. There is no reflected energy on the 50 ohm coax and therefore no momentum in the reflected waves on the 50 ohm coax. I think you contradicted yourself here. Are you saying that there IS or IS NOT reflected energy/waves on the 50 ohm section (to the left of the "--x--"?? No contradiction. There's no reflected energy on the 50 ohm coax because those two reflections are cancelled at the Z0-match point 'x'. They are equal in magnitude and opposite in phase. So what changes the direction and momentum of the energy wave reflected from the load? You lost me here. Perhaps you are asking; "If there is a wave reflected at the end with the 50 ohm, back toward the left, how does this power then get absorbed IN that load?" Is this the question? No, there is 178 joules/sec rejected by the load on the 450 ohm line. That energy possesses direction and momentum toward the source. What reverses the direction and momentum of that reflected energy? The wave, in the 450 section, reflected from the load is NOT the power being delivered TO the load, so it does not have to be "changed' to be delivered there. There is 100 joules/sec traveling to the right and none to the left on the 50 ohm coax. There is 278 joules/sec traveling to the right and 178 joules/sec to the left on the 450 ohm ladder-line. How does the rearward traveling 178 joules/sec, rejected by the load, get turned around and join the forward traveling 100 source joules/sec in order to add up to 278 joules/sec of forward power on the ladder line that is incident upon the load? It's a simple question: How does energy rejected by the load become energy incident upon the load? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil, W5DXP wrote:
"It`s not magic and is explained on the Melles-Groit web page---. (how the energy magically reverses direction and heads back toward the load.) I agree. It isn`t magic. Optical examples are good because we can see reflections. I think we have impedances which are inherent from the physical and electrical characteristics that describe the path the electrical energy takes. The path enforces the voltage to current ratio called impedance. First example is Zo. Second example is 377 ohms of free-space. Third example is resistor type resistance. Resistance is a property based on configuration, dimensions, material, and temperature which make a certain voltage to current ratio, which is measured in ohms. One ohm is the resistance at zero degrees C of a uniform column of mercury 106.300 cm long and weighing 14.451 grams. One volt across a resistor of one ohm causes a current of one ampere. A wave travels down a transmission line and it consists of a traveling electric field and an associated traveling magnetic field. Relative strengths of the electric and magnetic fields on a practical uniform transmission line are forced into a ratio Zo which is the square root of the ratio of the inductance per unit length divided by the capacitance per unit length. Zo is measured in ohms. If the line is terminated in an impedance other than Zo, one of the two fields is in limited supply as compared with its travel partner. As the termination only accepts energy in its fixed ohms ratio of voltage to current, surplus energy in the ample field is all rejected and reflected as it has nowhere else to go other than to reverse its course. The reflected wave must conform to Zo the same as the incident must. Seems to me there`s no magic. The waves just do what they must. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
Seems to me there`s no magic. The waves just do what they must. Reckon why some people believe that wave energy is allowed to change directions at the load but not allowed to change directions at the match point? One of the problems in the field of RF (that the optics people don't have) is the effective reflection coefficient Vs the physical reflection coefficient. I have never heard an optics engineer say, "Since reflections are eliminated at the thin-film surface, the effective index of refraction of the thin-film is 1.0." The optics reflection coefficient doesn't change with the magnitude of reflected energy. It is always (n2-n1)/(n2+n1) where 'n' is the index of refraction. Yet RF engineers will say, "Since reflections are eliminated at the 50 ohm to 450 ohm impedance discontinuity, the reflection coefficient is zero." Why isn't the reflection coefficient always (Z2-Z1)/(Z2+Z1) as it is in the field of optics? Note: [(Z2-Z1)/(Z2+Z1)]^2 = [(n2-n1)/(n2+n1)]^2 The term on the left side of the equation is the RF power reflection coefficient. The term on the right side of the equation is the Reflectance (the irradiance optical reflection coefficient). Irradiance is energy per unit time per unit area and is equal to power per unit area. Thus the irradiance of a confined laser beam is equivalent to power in a confined transmission line. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Interesting question and actually BEFORE I read R. Clark's post I decided to say that I will begg off at this point. I may be overcome with curiosity on this question in the near future (like I did for the QST WATTMETER article calculations), but for now I prefer not to take the challenge. Joules are much more than I want to consider. Job, Wife, Mother, Mother-in-law, sister-in-law all needing care and a job that is full of crap. I use this for enjoyable discussions that I can contribute to and this a way more brain power that I want to devote. I suspect you, Cecil, can explain it. I know that a 1/2 wave repeats the load Z, smith Chart and all that and can work with that. you win. 73 -- Steve N, K,9;d, c. i My email has no u's. "Cecil Moore" wrote in message ... Steve Nosko wrote: 100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load Assuming yes... I think you are saying that we have a 450 ohm line and a 50 ohm load and therefore there is a reflection. Is that where you are? Yes, there are reflections on the 450 ohm ladder line and none on the 50 ohm coax. There is no reflected energy on the 50 ohm coax and therefore no momentum in the reflected waves on the 50 ohm coax. I think you contradicted yourself here. Are you saying that there IS or IS NOT reflected energy/waves on the 50 ohm section (to the left of the "--x--"?? No contradiction. There's no reflected energy on the 50 ohm coax because those two reflections are cancelled at the Z0-match point 'x'. They are equal in magnitude and opposite in phase. So what changes the direction and momentum of the energy wave reflected from the load? You lost me here. Perhaps you are asking; "If there is a wave reflected at the end with the 50 ohm, back toward the left, how does this power then get absorbed IN that load?" Is this the question? No, there is 178 joules/sec rejected by the load on the 450 ohm line. That energy possesses direction and momentum toward the source. What reverses the direction and momentum of that reflected energy? The wave, in the 450 section, reflected from the load is NOT the power being delivered TO the load, so it does not have to be "changed' to be delivered there. There is 100 joules/sec traveling to the right and none to the left on the 50 ohm coax. There is 278 joules/sec traveling to the right and 178 joules/sec to the left on the 450 ohm ladder-line. How does the rearward traveling 178 joules/sec, rejected by the load, get turned around and join the forward traveling 100 source joules/sec in order to add up to 278 joules/sec of forward power on the ladder line that is incident upon the load? It's a simple question: How does energy rejected by the load become energy incident upon the load? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
High Wind alert!
"Richard Harrison" wrote in message ... Steve Nosko wrote: "I am missing just what it is that gets to the "Loss-less resistance" conclusion." If part of the source resistance were not lossless, efficiency would be limited to 50%. I disagree. DC or RF: Real source = XX volts Real source resistance = 10 ohms Real load 50 ohms Efficiency is 50% DONE. (and I have no "loss-less resistance" anywhere) I can't figure out where you are going nor what the hole is in the technology that needs this extra stuff... are there formulas for it? Fact is, Class-C amplification frequently is 2/3 or 66.6% efficient. Twice the power is delivered to the load as is lost in the source. Yea. We both know all this. I'm trying to get to the WHY part. The issue is WHERE or WHAT is this loss-less resistance? Where is it? Why is is even needed? If we deliver 1000 watts into 50 ohms, we have 224 volts and 4.47 amps as a load. Our source is identical,, 224 volts and 4.47 amps. Its volts to amps is the same ratio, but its loss is not the same power as the power delivered to the load because part of the source resistance is lossless I stop right here and say: It is because the source resistanace is *less than* the load resistance. Simple as that! I can not come to any other conclusion. because it is the product of interrupted energy delivery, not energy conversion into heat. My eyes glaze over here. "interrupted energy delivery" -- can't get a grip on this. With 2/3 efficiency, when we have 1000 watts into the load, we have 500 watts lost in the source. Yep. I believe the true issue is WHY does this happen? What is the cause of this effect? "Loss-less resistance" or Rs RL? I say the latter, I think you say the former. 500 watts lost means only 1/2 the dissipative resistance as we have resistance in the load, and thus the source resistance consists of 25 ohms of dissipative resistance and 25 ohms of non-dissipative Why is this needed? I think I see. You are saying that because we have a "match" We are at the "conjugate match" condition and therefore *MUST* be at Rs=RL. Here's where I disagree. I believe this is absolutely NOT the case as I say above. It's as simple as the fact that the amplifire IS NOT internal resistance limited, but either dissipation limited or perhaps breakdown voltage limited or cathode emission limited. NOW I THINK I GET IT (your tack)!. You believe that the source resistance MUST equal the load resistance.. Well I say Nope! DC or RF: Real source = XX volts Real source resistance = 10 ohms Real load 50 ohms Efficiency is 50% DONE. What was that guy's name Ocam? of the Ocam's razor fame:"Take the simple answer", or something to that effect. resistance. The total source resistance is 50 ohms which matches the load resistance. I beileve this is where your error of reasoning is. I can not accept that there is some other resistance which is not quantifiable/observable/measureable. The source resistance must be less than the load R for Eff 50%. I believe you are making this much more complicated that is really is. A match allows maximum power transfer. I believe this is what is leading you astray. I believe you are assuming that the maximum power of the maximum power theorem, and the maximum power out of the real transmitter, are the same maximum power. This is where I believe the error in reasoning is that requires the loss-less resistance. They are not the same maximum power. The typical DC power supply is operated with Rs RL. The power output is limited by factors other than the maximum power transfer theorem (its output resistance) suggest. The power supply is not limited by its internal resistanse, but usually its ability to dissipate what little amount of heat (relative to the load power) that is can. I'll stick my neck out (because I believe I have a firm understanding of what physical laws can not be violated) and say that my conclusion is that the tube amplifier we commonly think of, when "tuned up" absolutely can not be operating in an Rs=RL configuration. This is for the very reasons you have stated. Because the power dissipated in the tube is less than in the load, Rs does not = RL. This is is, of course, looking at the load R presented to the tube by its output tank circuit. I believe that this (the RL.-RS stuff) is the natural law which can not be violated. There is indeed SOME load which the tube likes to see in order to get whatever power it can provide to come out. And it AIN'T Rs. The less than 360 degrees of energy supplied to the load makes 25 ohms of dissipationless resistance as a part of our 50-ohm source in our example. Keep working on the idea and eventually you may get the model. Best regards, Richard Harrison, KB5WZI I 'spose it's briefly is an interesting idea for discussion, but sorry Richard, I find it hard to beieve it actually has any merit. My (I think generally accepted) model works fine, it can explain all the phenomona(sp) so far observed and... I work in the field and have never met anyone with this idea and I have yet to see anything in the (uh, oh, here it comes. nose in the air Engineer snob talk) profession which suggests this lossless stuff is there, nor is there anything that remains unexplained which needs some other effect. I guess we close disagreeing. If it works for you... Interesting journey into some serious examination of principles, however. -- Steve N, K,9;d, c. i My email has no u's. |
|
Steve Nosko wrote:
you win. 73 It's not a contest, Steve, it is a search for the truth. I thought maybe you could offer something compelling. The following two questions are very similar: 1. How does RF power (joules/sec) rejected by the load wind up being incident upon that same load some time later in a Z0-matched system? 2. How does irradiance (joules/sec) rejected by a pane of glass wind up being incident upon that same glass pane some time later in a non- glare air/thin-film/glass system? Question #2 was answered decades ago by optics physicists. Question #1 still remains unanswered by RF physicists and engineers even though it has virtually the same answer as Question #2. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Cecil,
With reference to question #2: Who says such a silly thing? Not Melles-Griot, who appear to be your favorite optical reference source. Not Born and Wolf in "Principles of Optics", which is the ultimate optics reference book. Not any professional optics experts I have ever encountered. If you go ahead and solve the antireflective glass problem using standard Maxwell's equations (sorry, Reg and Peter) with standard boundary conditions for E and H fields you will find there is not the slightest bit of confusion. This analysis is shown in many optics and E&M textbooks. In the perfectly antireflective case all of the waves keep moving in the same direction, from air to thin film to glass. If one postulates the existence of a wave in the reverse direction it will be discovered that the amplitude of that wave is zero, meaning it does not exist. There is no "bouncing back and forth" of confused energy not knowing where and how to turn around. The interference model is useful and intuitive for what it is meant to explain. However, don't expect this sort of simple handwaving model to be extendable to all sorts of silliness about energy and momentum transfer. 73, Gene W4SZ Cecil Moore wrote: Steve Nosko wrote: you win. 73 It's not a contest, Steve, it is a search for the truth. I thought maybe you could offer something compelling. The following two questions are very similar: 1. How does RF power (joules/sec) rejected by the load wind up being incident upon that same load some time later in a Z0-matched system? 2. How does irradiance (joules/sec) rejected by a pane of glass wind up being incident upon that same glass pane some time later in a non- glare air/thin-film/glass system? Question #2 was answered decades ago by optics physicists. Question #1 still remains unanswered by RF physicists and engineers even though it has virtually the same answer as Question #2. |
| All times are GMT +1. The time now is 09:08 PM. |
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
RadioBanter.com