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#11
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On Sat, 16 Aug 2003 09:38:18 GMT, Richard Clark
wrote: On Sat, 16 Aug 2003 07:07:16 GMT, "George, W5YR" wrote: Richard, It always astounds me how the technically challenged are wont to launch ad hominem attacks instantly Do you accept that yours similarly deflates your credentials? It would at least place us on equal footing - in the gutter I suppose, but I brought my snorkel. ;-) 73/72, George Hi George, If you suffer the heat of taking a stand, so much for a test of faith. I have not seen you respond to the chain of evidence I have supplied to these matters. I will offer that this body of work long preceded my missive. You might or might not find that work interesting/correct/or worthy of your attention, but that does not erase if from the archive nor detract its intrinsic merit in desired measure to bruised ego. Pick any ONE of your cherished notions that I so soiled and put it up for a clear and concise examination. OR Let me head that off with a very simple question that most dodge; and in fact lies at the very heart of your subject line: Given the premise that the amateur's rig output Z is NOT 50 Ohms, what value does it have (cite any assembly of conditions)? You will no doubt get many thumps on the back from well-wishers who spit in my direction. How many will offer a numeric response to that technical enquiry? I can forecast that will stand at the current exchange rate of 0. I will also forecast there will be either total silence, or scattered muttering about why they wouldn't engage such a scurvy fellow as me. And yet the absence of that number from the discussion under this subject line mocks the charter of this group more than my humor did you. Is it lower than 50? Higher than 50? How much? The stunned silence in response to such simple, forced speculation is more a result of intellectual catatonia than moral indignation. Those who have offered numbers (I count among them), who have revealed methods of their derivation (I count among them), who performed actual bench work (I count among them), who offer rationale as to the subject's correlation to other observables (I count among them) is notable in contrast to those who have nothing to show but the shallow rhetoric of impotent denial. 73's Richard Clark, KB7QHC Hi All, I note by the absence of response that at least my mystical powers of clairvoyance are unparalleled. No one dare step up to the bar to answer the question: "What is the Z of a transmitter, if it is not 50 Ohms?" For those who would rather argue the mystic ability than answer the question, I will allow that this same absence may be due in part to the Blaster virus and the power black out. Now that I've braced up your crutches, let's proceed with a telling example of both the academic principle and the practical implication. I will simply choose a value for you. In other words, we will venture where these angels fear to tread. We will start with a deliberately mismatch transmitter, and a deliberately mismatched load. We will then throw in the practical necessity of line loss and ask the question that is my acid test for the wide-eyed inventors: "Does it make more than 1dB difference?" The scenario begins: "A 50-Ohm line is terminated with a load of 200+j0 ohms. The normal attenuation of the line is 2.00 decibels. What is the loss of the line?" Having stated no more, the implication is that the source is matched to the line (source Z = 50+j0 Ohms). This is a half step towards the full blown implementation such that those who are comfortable to this point (and is in fact common experience) will observe their answer and this answer a "A = 1.27 + 2.00 = 3.27dB" "This is the dissipation or heat loss...." we then proceed: "...the generator impedance is 100+0j ohms, and the line is 5.35 wavelengths long." "A = 1.27 + 2.00 + 1.62 = 4.9 decibels" Thus the answer to my question is: Yes. 1.62dB is greater than 1dB. Now, as to the application of this knowledge to the typical user. It becomes: "does my standard of 1dB meet the thresh hold of your caring?" Perhaps not and even 3 dB may be of no concern. For such folk I offer my best wishes and we each proceed happily on in life. [This, of course, presumes they do not in fact have a rig that exhibits a 600 Ohm output Z and hence the danger of nay saying the obvious without expressing a value to replace it.] Now, as to the application of this knowledge to the critical user. By this I mean those here who want to have a complete answer, and being thus informed can make their own choices. Is there anyone corresponding here that want to dispute that this is the charter of this group? I have then twice shown how a transmission line being bound by two reflecting planes introduces a Mismatch Uncertainty. This example has enlarged on that slightly through my advice that this uncertainty can be reduced to zero through the description of all paths. As I have also pointed out in the past, this is a simple truism of wave interference math - very simple. The fact of the matter is that nearly every correspondent to this forum employs a transmitter designed to and exhibiting 50 Ohms source Z. The simple fact of the matter is that none of those same correspondents will typically encounter that additional 1.67dB because of this. Those who choose to operate their transmitter outside of this specification may; but those same operators rarely, if ever, examine the evidence of Mismatch Uncertainty because they never move their load nor their SWR meter (the path never changes). They instead will observe a reading in their complacency and accept the error without being aware (unless they have read this, that is). I will add that even when operating outside of the characteristic source Z, that is not significantly off enough to match the issue portrayed above unless you cut power dramatically - and even then the issue is moot even though the loss is not. So, part and parcel to the subject header above and having shown how ignorance and rejection of the obvious has a concomitant loss; the question, as always, remains: Given the premise that the amateur's rig output Z is NOT 50 Ohms, what value does it have (cite any assembly of conditions)? What constitutes the silence from this point on? My proven clairvoyant skill being elevated by the day, the Blaster virus, the power outage, or that same intellectual catatonia? 73's Richard Clark, KB7QHC |
#12
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Richard Clark wrote:
Richard Clark wrote: Given the premise that the amateur's rig output Z is NOT 50 Ohms, what value does it have (cite any assembly of conditions)? What constitutes the silence from this point on? Nobody knows and nobody cares. All voltage to current ratios are determined by the system outside of the transmitter. The only thing a transmitter need furnish is a voltage (or current or power). No matter what the output Z of a transmitter, it can put out a voltage (or current or power). What happens inside a transmitter doesn't affect anything except transmitter efficiency. Any coherent energy re-reflected inside the transmitter simply superposes with the forward wave and becomes indistinguishable from the generated power. If modulation is added and the feedline is long enough, the re- reflection could be detected. With an unmatched TV generator and about 1000 ft of open-wire line, the TV ghosts would give an indication of how much reflected power is actually re-reflected inside the generator. For those who assert there are no reflections from a generator, this would be an easy experiment to run. -- 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! =----- |
#13
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On Sun, 17 Aug 2003 14:41:58 -0500, W5DXP
wrote: What constitutes the silence from this point on? Nobody knows and nobody cares. Hi Cecil, Then I count you in that group who finds the additional 1.62dB loss as inconsequential. I also note you have no answer to the question, but in that regard I wish you well, and we are both content. 73's Richard Clark, KB7QHC |
#14
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Richard Clark wrote:
W5DXP wrote: What constitutes the silence from this point on? Nobody knows and nobody cares. Then I count you in that group who finds the additional 1.62dB loss as inconsequential. I also note you have no answer to the question, but in that regard I wish you well, and we are both content. I think everyone appreciates the fact that if he/she destroys the efficiency of his/her transmitter by whatever means, then that is not inconsequential. But that's not really what the discussion is about, is it? SWR doesn't depend upon the efficiency of the transmitter, does it? -- 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! =----- |
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On Sun, 17 Aug 2003 15:09:40 -0500, "H. Adam Stevens"
wrote: "Richard Clark" wrote in message news ![]() On Sun, 17 Aug 2003 14:41:58 -0500, W5DXP wrote: What constitutes the silence from this point on? Nobody knows and nobody cares. Hi Cecil, Then I count you in that group who finds the additional 1.62dB loss as inconsequential. I also note you have no answer to the question, but in that regard I wish you well, and we are both content. 73's Richard Clark, KB7QHC Hi Richard Since Cecil's reference to "nobody" implicitly includes me, I beg to be removed from that group. How to measure source impedance. Connect an adjustable load directly to the source, forget the transmission line. When the load gets hottest for the least power into the source, their impedances are matched. 73 H. NQ5H -------1.62 db here, 1.62 db there-------;^) Hi OM, You are stricken from that group then. Your conscription is ended and we still all remain content by universal acclamation. ;-) Myself I can tolerate up to 10dB without care so 1.62dB is hardly as monumental as the vivid reactions to this news would evoke. I only test claims to the level of 1dB to be generous. I would note that most claims fail to achieve even that value. 73's Richard Clark, KB7QHC |
#16
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"H. Adam Stevens" wrote:
How to measure source impedance. Connect an adjustable load directly to the source, forget the transmission line. When the load gets hottest for the least power into the source, their impedances are matched. Unless of course that shuts down, or even destroys, the source device in the process. -- Floyd L. Davidson http://web.newsguy.com/floyd_davidson Ukpeagvik (Barrow, Alaska) |
#17
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"Ralph Mowery" wrote in message ...
proper voltage? When you connect speakers to an amplifier, do you worry about what the source impedance is, so long as it's low enough to not materially affect damping? If not, why would you worry about I would not go there on the audio . The speakers do need to match the design of the amp just as the load on a transmitter needs to match the design impedance. Most power output devices are designed to produce maximum power and /or minimum distortion into a specific load. Just my point, Ralph... you should worry about your speakers being within the range of LOAD impedances that the amplifier is designed for, but you seldom would worry about the SOURCE impedance. You should worry about the antenna-load you present to your transmitter or amplifier being within the range of impedances for which the transmitter or amplifier is designed, but why worry about the transmitter or amplifier source impedance? (Some folk worry about audio amp damping factor, but it tends to be grossly overemphasized...see postings over the years by Dick Pierce in the audio groups for the simple explanation why it doesn't matter all that much.) Cheers, Tom |
#18
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Richard Clark wrote in message . ..
On Fri, 15 Aug 2003 07:45:32 GMT, "George, W5YR" wrote: About all that anyone needs to know about an amateur radio amplifier (transmitter) in order to use it properly is the output power level and the required load resistance. Hi George, And us folk need never worry about what is beneath the hood as long as we don't need a mechanic. Sheesh, haven't you learned to turn the knob until the meter is full scale? All you have to really remember to push the button before you talk! .... Richard, I'm seriously worried about you. Was that posting, and many of the others I've seen from you recently, a plea for help? I'm sure we can come up with appropriate help lines for you to call, if only you will tell us what's at the root of your problem. Regards, Tom |
#19
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#20
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Floyd Davidson wrote:
"Unless of course that shuts down ,or even destroys the source devices in the process (adjustable load gets hottest for least power to the source as proposed by H. Adam Stevens)." H.`s variable load can determine a match because maximum power transfers if the load impedance is the conjugate of the source. But, least power input may not coincide with maximum power output. Low internal source impedance as compared with the load impedance usually is more efficient, but not necessarily so. With impedance produced by switched-off time as in the Class C amplifier, impedance is not only a function of internal loss. A Class A amplifier has constant power input so that the more power delivered to a load, the less power is dissipated in the amplifier. Simple determination of source impedance is to divide the open-circuit voltage by the short-circuit current out of the device. That requires a source impedance that is the same at matched loading as with a shorted load. As Floyd notes, the extremes, open and shorts, can shut-down or destroy from too much voltage or current. Near the region of a conjugate match, less severe voltage and current are likely. Maximum power transfer to the load occurs at the conjugate match point. Heat rise would be the likely cause of damage to a transmitter at maximum output. Power output measurements can be made quickly to limit transmitter and load temperature rise in and around the load impedance which gives maximum power output. A versatile load can be used to determine the conjugate match and thus the source impedance. The transmitter may not be designed to deliver maximum power output even momentarily and may have built-in protection to prevent it. Feedback may not increase maximum power output without regard to distortion. In an audio amplifier, the output impedance may be made to look like a short to the speaker and thereby put the brakes on its movement as it tends to ring on after excitation. The same feedback lowers distortion at a given power level, or said another way increases power output for a given level of distortion, but maximum power output without regard to distortion is unaffected by the feedback. R-F amplifier negative feedback likewise has no effect on the all-out maximum power. Best regards, Richard Harrison, KB5WZI |
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