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Richard Harrison wrote:
Cecil, W5DXP wrote: "The superposed forward voltage and reflected voltage can damage an unprotected transmitter." To do so, they would be in-phase and not out-of-phase. Nope, they don't have to be in-phase. The reflected voltage can arrive with any phase with respect to the forward voltage. You trimmed out the example I gave. I will repeat it so you can study it closer. Just a for instance - assume the transmitter is putting out 70.7v in phase with 1.4a at zero deg. The arriving reflected wave is 50v at 90 deg and 1.0a at -90 deg. The load seen by the transmitter is 86.6v at 35 deg and 1.72a at -35 deg. Over voltage and over current exist at the transmitter output. The forward power is 100w and the reflected power is 50w. The net power being delivered to the reactive "load" seen by the transmitter is 86.6*1.72*cos(70.4) = 50w. In the above example, the designed for output voltage is 70.7v and the designed-for output current is 1.4a. The superposed voltage is 86.6v, higher than the designed-for voltage. The superposed current is 1.72a, higher than the designed-for current. In this example, we have both over-voltage and over-current occurring *at the same time* even when the forward voltage and reflected voltage are 90 degrees out of phase. The superposed voltage is not smaller than the forward voltage until the phase angle between them is in the neighborhood of 120 degrees. Two superposed voltages with a 90 degree phase angle are *always* larger than either voltage component, i.e. SQRT(V1^2+V2^2) is always larger than either V1 or V2. The superposed voltage will be higher than the forward voltage for any phase angle from zero to 90 degrees. At some phase angle higher than 90 degrees, in the neighborhood of 120 degrees, the superposed voltage starts to decrease. If you sit down and draw these phasors on a piece of paper, you will discover that you are mistaken. It appears that you are thinking one-dimensionally instead of two-dimensional phasors. -- 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:
how does a transmitter happen to always be "in-phase" to any reflection? It doesn't. The reflected voltage can obviously be 90 degrees out of phase with the forward voltage in which case, the interference term equals zero, and the superposed voltage is SQRT(Vf^2+Vr^2), i.e. greater than Vf. (The argument reminds me of Gary Coffman's one-dimensional "spitting up the fire hose" argument.) -- 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 Harrison wrote:
Standing waves display interference between incident and reflected waves whiich ideally have in-phase and out-of-phase constituents. In the following, the interference term equals zero. XMTR--------1/8WL 50 ohm lossless coax------291.5 ohm load Forward power = 100w, reflected power = 50w. Vf = 70.7v at zero degrees, If = 1.4a at zero degrees Vr = 50v at 90 degrees, Ir = 1.0a at -90 degrees Superposing: Vtot = SQRT(Vf^2+Vr^2) = 86.6v over-voltage condition Itot = SQRT(If^2+Ir^2) = 1.72a over-current condition The phase angle between Vtot and Itot is about 70.4 degrees, i.e. the source sees a highly reactive impedance. Vtot is NOT in phase with Vf, Itot is NOT in phase with If, Vtot is NOT in phase with Itot. Vf is NOT in phase with Vr. No two voltages are in phase. No two currents are in phase. -- 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:
"Nope, they don`t have to be in-phase." Cecil is right. The third side of a triangle can be longer or shorter than one of the sides. I should have said, it is entirely possible that the superposed forward and reflected voltages can damage an unprotected transmitter, The reflected voltage can add to the forward voltage applied to the transmitter. The phase of the reflected voltage should be out of phase with the transmitter source voltage to maximize the volts across the internal impedance of the transmitter if we want to damage an unprotected transmitter. Best regards, Richard Harrisin, KB5WZI |
Richard Harrison wrote:
Cecil, W5DXP wrote: "Nope, they don`t have to be in-phase." Cecil is right. The third side of a triangle can be longer or shorter than one of the sides. I should have said, it is entirely possible that the superposed forward and reflected voltages can damage an unprotected transmitter, The reflected voltage can add to the forward voltage applied to the transmitter. The phase of the reflected voltage should be out of phase with the transmitter source voltage to maximize the volts across the internal impedance of the transmitter if we want to damage an unprotected transmitter. Now assuming, as Richard H. logically did, that we don't want to damage an unprotected transmitter, let's introduce the concept of an active match Vs a passive match. The match at the transmitter output is passive, not active. Consider the following system: XMTR---1/4WL 300 ohm feedline---1800 ohm load What does it really mean when we say that the XMTR "sees" 50 ohms? It means that an interference pattern is established at the XMTR that re-reflects (re-directs) the reflected energy back toward the load. The transmitter has absolutely nothing to do with that reflection. We can reconfigure the system as follows: XMTR---1WL 50 ohm feedline--+--1/4WL 300 ohm feedline---1800 ohm load Now we can see that the match is passive and the XMTR has nothing to do with the match. Destructive interference on the XMTR side of '+' eliminates the reflections. The resulting equal magnitude of constructive interference on the load side of '+' re-reflects (re-directs) the reflected energy back toward the load. Optics engineers understand interference. Seems RF engineers might not fully understand interference. Interference is the cause of the elimination of reflections in optic systems. Interference is the cause of elimination of reflections in RF systems, including at antenna tuners. -- 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 posed a problem in which a transmitter seeking a 50-ohm
load (I suppose) is attached to 1 WL of 50-ohm cable which is attached to 1/4-wave of 300-ohm feedline which is attached to an 1800-ohm load. Cecil wrote: "Destructive interference on the XMTR side of "+" (where 50-ohm cable meets the 300-ohm Q-matching section) eliminates the reflections." It seems to me, the product of 50 (the cable Zo) and 1800 (the load Z) is 90,000. The sq.rt. of 90,000 is 300 (the Q-section impedance). The numbers are right, so in my opinion, the Q-section converts the 1800-ohms to 50-ohms. This is the 50-ohms likely prescribed for the transmitter`s load. The 50-ohms presented by the Q-section to the cable should result in a match and thus there should be no reflected energy in the 50-ohm cable to cause interference. The reflections should all be in the 300-ohm Q-section. Best regards, Richard Harrison, KB5WZI |
On Sun, 05 Sep 2004 18:06:07 GMT, Richard Clark
wrote: |On Sun, 5 Sep 2004 12:47:45 -0500, (Richard |Harrison) wrote: | |The tank circuit is mostly a harmonic filter providing a very high |impedance to the fundamental frequency and shorting out the harmonics. | |Hi Richard, | |Even here, the Goatman offered in his notes that his finals tank |(actually a series resonant Z match) offered a loaded Q of 2! (If I |read his scribblings correctly.) Yeah and he also calculates (http://www.techatl.com/wrek/docs/gnm69_25.htm) the required plate load resistance as: Eb Rl ~ ------- Idc Which for class C is off by about a factor of 2, but with Eb = 2500 and Idc = .25, he does the division and comes up with Rl = 1000. Hey what the heck, what's a factor of 10 among friends. If the calculation is done more accurately: Eb - Eg2 Rl = --------- K * Idc Where K = 2 for Class C and Eg2 = 300 (screen voltage) Then Rl ~ 4 Kohm Since the minimum output capacitance (Cp) of a 4CX300 is 4.5 pF, the parallel equivalent of Rl and Cp is Rp ~ 3920, Xp ~ -388. Thus the minimum possible Q ~ 10, which to someone who has built a few VHF amplifiers, sounds much more plausible. For example here's one I designed and built not much later that the WREK(ed) transmitter. http://www.qsl.net/n7ws/K7CVT_Amp.html But it gets worse. Try as I might with the component values he specifies, I cannot develop a plate load Z anywhere close to what is necessary. He has a lot more inductance that he thinks, so maybe that helps and I suspect his output lowpass filter (seen in the photos but not on the schematic) is part of the matching network. I'm really surprised that with the construction and documentation presented he could get FCC type acceptance. |
Richard Harrison wrote:
Cecil, W5DXP posed a problem in which a transmitter seeking a 50-ohm load (I suppose) is attached to 1 WL of 50-ohm cable which is attached to 1/4-wave of 300-ohm feedline which is attached to an 1800-ohm load. Cecil wrote: "Destructive interference on the XMTR side of "+" (where 50-ohm cable meets the 300-ohm Q-matching section) eliminates the reflections." It seems to me, the product of 50 (the cable Zo) and 1800 (the load Z) is 90,000. The sq.rt. of 90,000 is 300 (the Q-section impedance). The numbers are right, so in my opinion, the Q-section converts the 1800-ohms to 50-ohms. This is the 50-ohms likely prescribed for the transmitter`s load. The 50-ohms presented by the Q-section to the cable should result in a match and thus there should be no reflected energy in the 50-ohm cable to cause interference. The reflections should all be in the 300-ohm Q-section. Exactly, thanks to destructive interference. In S-parameter terms: b1 = a1*s11 + a2*s12 where b1 is the normalized reflected voltage back toward the source and is equal to zero because complete destructive interference between a1*s11 and a2*s12 causes their phasor sum to equal zero. Those two voltage components cancel to zero toward the source. That destructive interference is necessary and sufficient for a matched system. The result is constructive interference toward 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 Moore wrote:
Walter Maxwell wrote: Thus, no transmission line is necessary. For example, the device can be connected directly to the antenna terminals, or any other device you desire to determine the mismatch, and power it directly from the signal source--no transmission line is needed on either port for the device to indicate the degree of mismatch. Assume a 100+j100 ohm load and a 100-j100 ohm transmitter directly connected with no transmission line. The system is matched. Are there any reflections? Now install a transmission line. Will an SWR meter read the same thing in both cases? What's the Zo of the transmission line? Is it sufficiently long to act as a transmission line? Cecil, methinks you posed an incomplete question :-) |
On Tue, 07 Sep 2004 18:12:35 -0700, Wes Stewart
wrote: He has a lot more inductance that he thinks, so maybe that helps and I suspect his output lowpass filter (seen in the photos but not on the schematic) is part of the matching network. I'm really surprised that with the construction and documentation presented he could get FCC type acceptance. Hi Wes, Still and all, a good story of the exploit. Well given the measurements, it seemed some filtering was necessarily unmentioned. And given the FCC type acceptance (obviously allowed), the measurements (or rather the quality of the gear) were sufficient. I especially find the scrawled notes submitted with the acceptance application a time capsule back to the days before computers (or seemingly the IBM selectric). Still and all, he described where he was going, and offered how he thought he got there. To translate that to today's specifications "missing" the output Z of transmitters (obviously part and parcel to the canon of the design engineer who built them) because of their irrelevance - that is a stretch of imagination right off the showroom floor. And then to notice in the ad copy, they can build to other output Z's... I find the novel modulation techniques interesting though. Seems like an alphabet soup of modes has sprung up over the years. 73's Richard Clark, KB7QHC |
On Wed, 08 Sep 2004 02:42:21 GMT, Dave Shrader
wrote: Cecil, methinks you posed an incomplete question :-) WHAT!!? NEVER! Well hardly ever (apologies to G&S). |
Dave Shrader wrote:
Cecil Moore wrote: Assume a 100+j100 ohm load and a 100-j100 ohm transmitter directly connected with no transmission line. The system is matched. Are there any reflections? Now install a transmission line. Will an SWR meter read the same thing in both cases? What's the Zo of the transmission line? Is it sufficiently long to act as a transmission line? Cecil, methinks you posed an incomplete question :-) Yep, a deliberate act to get you to consider all possibilities. What Z0 did you assume? For what Z0 is your SWR meter calibrated? -- 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! =----- |
On Wed, 08 Sep 2004 07:17:03 -0500, Cecil Moore
wrote: What Z0 did you assume? He hardly ever says a BIG BIG Z! ...well give three cheers and three cheers more... (further apologies to G&S) |
God bless the Cween.
Walt ======================== Really Walt, the introduction of God and the royal family to your conjugal match philosophy is a little far-fetched. I never suspected it but what brand of Bourbon are you on tonight? I recognise the symptoms. (Funny smiley) ---- Reg. |
Reg wrote:
"Really Walt, the introduction of God and the royal family to your conjugal match philosophy is a little far-fetched." They like to be on the right side too, but sometimes even the royals are wrong. I went all the way to London in july to wade in the Diana Ditch. When I arrived, it was surrounded by a wire fence posted with this notice: "Diana, Princess of Wales Memorial Fountain - There have been a number of slippage incidents in a section of the Fountain. The safety of our visitors is very important to us. As a precaution, there will be no acess to the Fountain until a full investigation has been completed. We apologise for this inconvenience. The Royal Parks." Well written explanation. Maybe next year? My 5 year old granddaughter, a student in Queen`s College, enjoyed the Hyde Park Lido, so the outing was rewarding anyway. Best regards, Richard Harrison, KB5WZI |
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On Mon, 6 Sep 2004 23:49:42 +0000 (UTC), "Reg Edwards"
wrote: snip Although I may jokingly profess ignorance of American electrical measuring instruments, for several years I was the Government-approved Head of Laboratory of a measurement standards laboratory of second echelon to the British National Physical Laboratory. I was familiar with the excellent qualities of HP, GR, Fluke and similar instruments. I played an original part in the conversion of the assessment of National worst-case measurement uncertainties to statistical uncertainty assssment. But I don't brag about it. ---- Reg, G4FGQ Well, Reg, your joking concerning the measuring instruments misled me. Sorry about that. But I don't call your leveling with us on your experience and accomplishments bragging. The more we know about our friends on line the greater the enhancement of our communication, even the humor will be more appreciated. I know that I'll have more appreciation of Punchinello than before. Now, from a previous post of yours: Really Walt, the introduction of God and the royal family to your conjugal match philosophy is a little far-fetched. I never suspected it but what brand of Bourbon are you on tonight? I recognise the symptoms. (Funny smiley) ---- Reg. Yeah, Reg, I must have been on something pretty strong when I said that, but I can't remember what it was. Another subject.: I'm wondering if I have your correct email address. During the past two years, or more, I've sent you several emails with no response. Yet you respond to my posts on this rraa in a cordial and friendly manner. In one of my emails I asked if I had said something that offended you so that I could apologize, but no response. During a cruise on the North and Baltic Seas two years ago I was in London at both the beginning and end of the cruise. I emailed you twice while in London and three times from the ship, asking for your phone number so I talk with you to learn how to find you for a visit. I first used the address you show on your posts with the 'zzz', but it came back--wrong address. I then used your address without the 'zzz' with no response. Can you help me understand what went wrong, because I really wanted to meet you while in England. Walt |
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On Thu, 09 Sep 2004 05:47:51 GMT, Richard Clark wrote:
On Wed, 8 Sep 2004 23:17:56 -0500, (Richard Harrison) wrote: Reg wrote: "Really Walt, the introduction of God and the royal family to your conjugal match philosophy is a little far-fetched." They like to be on the right side too, but sometimes even the royals are wrong. Perhaps Reggie is more a fan of the Sex Pistols: "God save the Queen the fascist regime, they made you a moron a potential H-bomb." (my regrets to G&S) Apparently the only G&S the Pistols knew about was 'Pirates'. |
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