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Jimmy wrote:
"Efficiency is based on how much of your signal your antenna turns into heat compared to the amount radiated and nothing more." With some reshuffle of terms, Terman seems to agree. On page 893 of his 1955 edition: "The efficiency of the antenna as a radiator is the ratio: Rradiation / Rradiation + Rloss----." Best regards, Richard Harrison, KB5WZI |
On Sat, 21 Feb 2004 19:04:24 GMT, "Jimmy"
wrote: Gain and efficecey have nothing to do with each other Hi Jimmy, In the classic sense, this is very true. However, when we look at gain integrated over all dimensions, we arrive at the concept of the isotropic reference. As a basis of comparison (skip the sophistry of there being no such physical device) this too reveals how gain and efficiency can be compared. In other words, with gain you can play the numbers to claim efficiency by sweeping the bad news under the carpet if you simply ignore the integration factor. A prime example is the recent glowing reports of the cfa. The "inventor" claims that his "FCC" tests reveal a gain of his antenna over the standard monopole, through substitution with an actual broadcast band, licensed transmitter. Well, perhaps in the direction of the nearby tower it was supposed to replace. When you look in every other direction, and step well away from the source (say, like were the listeners actually live and listen); then it is a different game altogether. Only 20 miles out and you find the cfa 30dB down into the mud compared to FCC standard charts (I won't go deeply into the fact that the comparison BC station had a ****-poor antenna itself 10dB down from those same charts). True, no one did a helicopter flyover to vindicate the cfa's redemption of superb cloud warming capabilities, but that was not where the listening audience lives. Further, given that the cfa's poor performance conformed to modeling along the testing scenario, it was hardly an indictment of the models that they did not reveal the glow in the sky. So, when you see claims based against gain tied to arguments of efficiency (apply any special terminology you wish) ask the hard questions and observe the answers. Do they respond with technical specifications that answer the issue, or are they laden with conflict and personality? 73's Richard Clark, KB7QHC |
Good evening Art, I have a few minutes to put some thoughts on email
regarding efficiency per unit length of an antenna, or antenna elements. I hope I don't glaze your eyes :-) Efficiency by definition is work delivered to a load divided by the total work available. The work [power] delivered to the load is the work available minus losses all divided by work available. Your car engine may have 100 horsepower available but losses [heat] in the transmission, drive shaft, differential, oils, fluids, bearings, tires and wheels may limit horsepower delivered to the road to 30 horsepower. In this model then the efficiency is 30%. Note: bandwidth is not a factor. It is total work[power] delivered divided by total work[power] available. For an antenna the following definition is applicable by similarity: Efficiency = Radiated Power/Total power. By definition an antenna is a linear device and the Principle of Reciprocity is applicable. That is, it has the same efficiency either transmitting or receiving. Now, total power = I^2*Reffective Reffective = Radiation Resistance [Rradiation] + Loss Resistance. Radiated Power = I^2*Rradiation. Where I = Io*cos[wt + b]. Where wt = frequency, b = phase shift along antenna element. By definition radiation resistance is that determined by integrating the total radiated power over the surface of the sphere containing that power. [1] For a half wave dipole that converges to the value at a current maximum. So, radiation resistance for a 1/2 wave infinitely thin dipole is 73 ohms at the current maximum.[1] As one moves away from the current maximum the radiation resistance increases as 1/cosine(angle from the maximum)^2 i.e. 1/cos^2[theta].[2] Now, in a uniform cross section antenna the Loss Resistance per unit length is constant. So, the Radiation Resistance at the ends of the antenna is infinite. [Cosine 90 degrees = 0] That means the efficiency at the ends of a 1/2 wavelength dipole is 100%. Isn't that a surprise?? [It's the same for a dipole or a Yagi!!!!] The Radiation Resistance at the 45 degree point from the current maximum of a thin 1/2 wavelength dipole is 73 ohms/(cos^2(45 degrees)) = 146 Ohms. The conclusion is that the efficiency of an antenna element varies along it's length and can vary between maximum of infinity at the ends and have a minimum value, that depends on the length of the antenna, at a current maximum. The total antenna efficiency is measured in the radiated pattern by integrating the power density per square steradian [or square degrees] over the full surface of a sphere divided by the power into the antenna. So, a Yagi with 8.14 dBi (6 dBd + 2.14 dBi) gain has concentrated it's radiated power into 15.38% of the three dimensional space defined by the surface of a sphere. [See Note 1.] Now if the Yagi is 95 % efficient and a dipole is 95% efficient the 6 dBd value remains constant. _____________ Note 1 [I'm using degrees instead of steradians to simplify understanding. The Science/math is the same] The diameter of the earth is 360 degrees as measured from E-W and N-S. The surface is then proportional to 360^2 = 129600 square degrees. [I will be dividing the constant Pi in a subsequent step so it is deleted here.] A 6 dBd Yagi is also +2.14 dBi giving a net gain of 8.14 dBi. This normalizes the value to a sphere. So, 8.14 dBi = 10*Log(129600/X) X = 19938.4 square degrees. The Yagi has concentrated it's pattern into a piece of the surface of the sphere that represents 15.38% of the total surface [19938.4/129600]. This is Gain NOT efficiency. _____________ Reference [1]: Antennas, Kraus, McGraw-Hill 1950, Chapter 5, section 5-6, pages 143-146. Reference [2]: Antennas, Kraus, McGraw-Hill 1950, Chapter 5, section 5-7, pages 147-148. |
Reg, G4FGQ wrote:
"What allows a class-C amplifier to exceed 50% efficiency is a small operating angle." Exactly, and during the majority of the degrees it`s switched completely off. It draws no current and suffers no "IsquaredR loss" during the amplifier off-time. Impedance is approximately E/I, but I is the average I, which is much less than the bursts of I during the conduction angle. The switched-off time makes the I in the denominator of E/I very small indeed and the solution to Ohm`s law is a high impedance without the dissipation of a resistance that remains in place continuously while agitating the atoms of a poor conductor to limit current. Instead, we have a low-resistanc in high conducton for short spurts. On-time is limited, instead of conduction, to produce a certain effective resistance. An automobile Kettering ignition system may use a dwell-meter to indicate how much of the time the points are closed. An ohmmeter indicates the resistance between its test prods. The two test circuits are almost the same although limitation of the deflection of the dwell-meter is different from limitation of the deflection of the ohmmeter due to the difference between limited conduction angle ignition points, and the continuous conduction through a current-limiting resistor. There`s an analogy between Class C and Class A amplifiers in there somewhere. Best regards, Richard Harrison, KB5WZI |
Dave Shrader wrote:
Lots of good stuff. Reciprocity is applicable to antennas. Antenna gain and directivity are the same transmitting or receiving. Impedance is the same as a source or load. Dave wrote: "That is, it has the same efficiency transmitting or receiving." I hadn`t given that much thought but it seems to me there may be a difference. When an antenna is receiving, it is excited by the received signal, resulting in voltage and current on the antenna. The antenna doesn`t care about the source of the signal. If the antenna is conjugately matched to the receiver, radiation resistance is the source resistance of the signal feeding the receiver. Half the signal power is consumed in the source resistance (radiation resistance) and half is consumed in the receiver. The half consumed in the radiation resistance is re-radiated. The antenna doesn`t know that re-radiation is uncalled for. If the antenna is mismatched to the receiver, more than 50% of all power received is re-radiated, depending upon the severity of the mismatch. If we have a Class C amplifier feeding power to the same antenna and enjoying a conjugate match, we can have a source that takes less than 50% of the available energy. So, the transmitting antenna system can be more efficient than the receiving antenna system, it seems to me. Best regards, Richard Harrison, KB5WZI |
Dave wrote,
Where I = Io*cos[wt + b]. Where wt = frequency, b = phase shift along antenna element. wt is radians per second times seconds which is just radians, an angle. b is time related also. Are you thinking of perhaps kl? By definition radiation resistance is that determined by integrating the total radiated power over the surface of the sphere containing that power. Are you thinking of Rr=2Prad/|Io|^2? [1] For a half wave dipole that converges to the value at a current maximum. So, radiation resistance for a 1/2 wave infinitely thin dipole is 73 ohms at the current maximum.[1] Balanis gives (eta/4pi)Cin(2pi)=73. Cin is .5772+ln(2pi)-Ci(2pi) which is approximately equal to 2.435. Ci is a tabulated function. So, the Radiation Resistance at the ends of the antenna is infinite. [Cosine 90 degrees = 0] That means the efficiency at the ends of a 1/2 wavelength dipole is 100%. Isn't that a surprise?? [It's the same for a dipole or a Yagi!!!!] The Radiation Resistance at the 45 degree point from the current maximum of a thin 1/2 wavelength dipole is 73 ohms/(cos^2(45 degrees)) = 146 Ohms. The conclusion is that the efficiency of an antenna element varies along it's length and can vary between maximum of infinity at the ends and have a minimum value, that depends on the length of the antenna, at a current maximum. The total antenna efficiency is measured in the radiated pattern by integrating the power density per square steradian [or square degrees] over the full surface of a sphere divided by the power into the antenna. So, a Yagi with 8.14 dBi (6 dBd + 2.14 dBi) gain has concentrated it's radiated power into 15.38% of the three dimensional space defined by the surface of a sphere. [See Note 1.] Now if the Yagi is 95 % efficient and a dipole is 95% efficient the 6 dBd value remains constant. Very entertaining. 73, Tom Donaly, KA6RUH |
Know what ya mean Ed. Used to mention Maxwell every now and then
myself but lets face facts, the coffee just isn't that good. Know whut I mean! Ed Price wrote: "Butch" wrote in message ... Time out!! You people are taking all this far to seriously. Just throw an aerial out the window, feed it to your rig via a tuner, and enjoy Amateur radio. Butch Magee KF5DE It just not that simple, Butch. I'm sure you have heard that Ham radio is a hobby that has many facets; construction, public service, contesting, field trips, QRP DX, etc. Some of our members get their kicks merging theory with rag chewing. I don't think there's any structure to this sub-category, other than to require at least one mention of Maxwell in every discussion. Ed WB6WSN |
Richard Harrison wrote:
SNIP Dave wrote: "That is, it has the same efficiency transmitting or receiving." I hadn`t given that much thought but it seems to me there may be a difference. When an antenna is receiving, it is excited by the received signal, resulting in voltage and current on the antenna. SNIP: Agree The antenna doesn`t care about the source of the signal. If the antenna is conjugately matched to the receiver, radiation resistance is the source resistance of the signal feeding the receiver. SNIP: This resistance is the Radiation resistance of the antenna, i.e. approximately 73 ohms in a thin 1/2 wavelength dipole. Half the signal power is consumed in the source resistance (radiation resistance) and half is consumed in the receiver. SNIP: Not quite. Half is RE-RADIATED. [It does not dissipate it radiates!][See your next statement]. The other half is delivered to the transmission line sub-system then to the receiver. The half consumed in the radiation resistance is re-radiated. SNIP: Agree The antenna doesn't know that re-radiation is uncalled for. SNIP: I wonder if this statement is the root of our misunderstanding? My understanding is that the antenna does not have to know anything other than passively allowing the Laws of Nature [Physics] to operate. If the antenna is mismatched to the receiver, more than 50% of all power received is re-radiated, depending upon the severity of the mismatch. SNIP: Have to think about what you are trying to say. If the antenna has received a 10^-12 watt signal and 4*10^13 watts is delivered to the transmission line and 5*10^-13 watts is reradiated then 1*10^13 watts is energizing a standing wave in the antenna. If we have a Class C amplifier feeding power to the same antenna and enjoying a conjugate match, we can have a source that takes less than 50% of the available energy. SNIP: Help me understand what you are trying to say. So, the transmitting antenna system can be more efficient than the receiving antenna system, it seems to me. SNIP: I probably disagree. But, I do not fully understand what you are trying to say in the previous paragraph. Deacon Dave Best regards, Richard Harrison, KB5WZI |
Dave Shrader wrote:
"Help me understand what you are trying to say." I`ll elaborate. Efficiency is output / input. 1/2 or more of the power received by a receiving antenna is re-radiated. Nearly all of the power received by a transmitting antenna is transmitted. Considering the energy available to the antenna, the job done by the transmitting antenna system as compared with the job done by the receiving antenna system, the transmitting system is better. A receiving antenna must be resonant to enable full acceptance of available energy, and it must be matched to avoid re-radiation of more than 50% of the energy it is able to grab. If off-resonance, the receiving antenna has too-high impedance for significant induced current. Of course, we have such good receivers we can do without good efficiency. A transmitting antenna will radiate energy proportional to the current in the antenna. Ronold W.P. King says in "Transmission Lines, Antennas, and Wave Guides": "---the power (Io squared)(Ro) supplied to a highly conducting antenna (of Copper), with Ro taken from the curves of Sec. 10, is for practical purposes all radiated to the more or less closely coupled universe outside the antenna, while that used in heating the antenna itself is negligible." This information is on page 113. Inefficiency is to be found elsewhere from the transmitting antenna itself. We may use inefficient transmission lines and our wave generator, the transmitter, may be inefficient. We usually try to keep their losses low. It is not uncommon to produce RF in a Class C amplifier with an efficiency of 70%. With reasonable lines and antennas, nearly 100% of this power output can be radiated, producing appropriate millivolts per meter at one mile from the antenna. This is not completely reversible due to re-radiation of 1/2 or more of all the power a receiving antenna can grab. The hope for point to point wireless power transmission is in using antennas like large dishes, for example, which concentrate power within such a small angle that the receiving antenna captures all the transmitted beam. Similarly, all re-radiated power is beamed back to the transmitting antenna for another trip to the receiver. Best regards, Richard Harrison, KB5WZI |
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Richard Harrison wrote:
1/2 or more of the power received by a receiving antenna is re-radiated. Nearly all of the power received by a transmitting antenna is transmitted. Expanding a bit to make the receiving and transmitting systems symmetrical with respect to power: If the transmitter is linear (like the antenna is linear), i.e. Class-A, 1/2 or more of the generated power will be lost in the source. In a linear resonant system, about 1/2 of the power sourced reaches the antenna and about 1/2 of the received power makes it to the receiver. It's the old maximum power transfer theorem at work. A receiving antenna must be resonant to enable full acceptance of available energy, and it must be matched to avoid re-radiation of more than 50% of the energy it is able to grab. If off-resonance, the receiving antenna has too-high impedance for significant induced current. Of course, we have such good receivers we can do without good efficiency. A properly tuned antenna tuner ensures that the *antenna system* is resonant for both transmit and receive (assuming the receiver's input impedance is the same as the transmitter's output impedance). Note that an off-resonant antenna *wire* is integrated into a resonant antenna *system* through the use of an antenna tuner. Chapter 7 in _Reflections_II_ explains how even though it might better have been titled, "My Transmatch Really Does Tune My Antenna" *SYSTEM*. -- 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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Cecil, W5DXP wrote:
"If the transmitter is linear (like the antenna is linear), i.e. Class-A, 1/2 or more of the generated power weill be lost in the source." True, that would be an equalizer between reception and transmitting system efficiencies of antennas, but Class A isn`t the only way to get linear amplification, Hi-Fi nuts to the contrary not withstanding. Class B is often used to combine efficiency with high undistorted output capability. Class B amplifiers are biased to cut-off so they draw no current when there is no signal input. A class B amplifier may have 60% efficiency at full power output, for example. Such an amplifier will have only about 30% efficiency at 1/2 of its maximum power output. Turman writes on page 354 of his 1955 edition: "With the largest signal that the (Class-B) amplifier can be expected to handle satisfactorily, Emin/Eb will be small, and the actual efficiency at full power is commonly of the order of 60%." The receiving antenna can never be more than 50% efficient due to re-radiation which I don`t seem to be able to explain. Sorry. Best regards, Richard Harrison, KB5WZI |
Richard Harrison wrote:
The receiving antenna can never be more than 50% efficient due to re-radiation which I don`t seem to be able to explain. Sorry. It's because receiving antennas are linear devices which I don't seem to be able to explain. :-) -- 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! =----- |
Don, K6MHE wrote:
"Where did you come up with that one?" (A response to my statement that a receiving antenna must be resonant to enable full acceptance of available energy) I`ve tweaked antenna trimmers which dramatically boosted the signal when reasonance was reached. I`ve seen grounded 1/4-wave structures near a broadcast station detuned, thus eliminating the distortion they had caused in the station`s radiation pattern. If they`re not resonant, they don`t accept enough energy to make any difference in the station`s pattern. !/2-wave wires in free-space are resonant. Resonance is defined as unity power factor, that is, XL=XC. At resonance, reactance is balanced out and only resistance is left to oppose current in a wire. Usually the wire has a radiation resistance which is large as compared with its loss resistance in practical antennas. At frequencies below first resonance, the ungrounded wire is less than a 1/2-wavelencth. It has a low radiation resistance and a high capacitive reactance. We can add inductance to tune the wire to resonance. At frequencies above first resonance, the ungrounded wire is more than a 1/2-wavelength, and if it is not much longer, the wiire has an inductive reactance. The phase flip-flop at resonance is abrupt and the reactance is an impediment to the current on either side of resonance. The correct series capacitor can be placed in series with the roo-long wire to tune out its excess inductive reactance. A mechanical analog is the vibrating-reed frequency meter used at power frequencies. All the reeds are in the power-frequency field. Only the reed of resonant length has so little opposition to the excitation that it vibrates freely. A versatile antenna tuner can insert either inductive or capacitive reactance in series with an antenna to correct its power factor (tune it to resonance) so it can accept maximum excitation. Best regards, Richard Harrison, KB5WZI |
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"Richard Harrison" wrote in message ... teve Nosko wrote: "BTW--what is your line, Richard?" I apologize for a critical tone in my response to Steve`s posting. An ancient previous discussion of dissipationless resistance in this newsgroup leaves me primed to comment when it appears unappreciated. I only found it mildly critical. Your tone was not interpreted as hostile in any way, just driven to add. Dissipationless resistance is the stuff which allows a Class C amplifier exceed 50% efficiency. See comment on this later... I won`t say I`ve been teaching X years, as I`ve never had that role. Long ago, I found my patience and temperament unsuited to tutoring. I am a long retired electrical engineer and find entertainment in the Ahhh! I say, with an air of new found understanding. newsgroups. Best regards, Richard Harrison, KB5WZI |
Danny,
You can't ever discard the factor Q in any discusion with respect to antenna efficiency or any calculation for that matter. Q is intrinsic in any calculation that determines efficiency especialy when considering what the object of an antenna is. Cecil's aproach to what it is the 'object' is to provide a total system , not a system that is led around by its nose by a predetermined antenna structure is an example. The idea of designing a house around a workable door that is pre-supplied is what we do today with respect to communication, and is why I use a different antenna to the norm. When I am confident that personal attacks come to a halt per Antennex statement I will be happy to explain more in depth. If you are content with what you have then that is understandable as humans always resist change, including myself. You being an antenna guru I understand even more the resistance to accept the possibility of advancement from one who is less educated in the field than oneself. Regards Art "Dan Richardson @mendolink.com" ChangeThisToCallSign wrote in message ... On Sun, 22 Feb 2004 13:07:46 -0600 (CST), (Richard Harrison) wrote: A receiving antenna must be resonant to enable full acceptance of available energy, Where did you come up with that one? I suggest you revisit capture area. Danny, K6MHE |
"Richard Harrison" wrote in message ... Reg, G4FGQ wrote: "What allows a class-C amplifier to exceed 50% efficiency is a small operating angle." While this is too vague, Richard tries to add detail, but mis-steps just a bit... and Steve goes into an extended "You ain't quite correct blurb..." Exactly, and during the majority of the degrees it`s switched completely off. It draws no current and suffers no "IsquaredR loss" during the amplifier off-time. Impedance is approximately E/I, but I is the average I, which is much less than the bursts of I during the conduction angle. We must be careful with the word "average" here. First, my "class C" model is a follows: I liken it to digital or "switched modes". While I have never scoped the plate to observe this... When the tube is cut off for a large part of the cycle, there is a high voltage on the tube (I believe it swings higher than the supply dou to the "ringing" of the plate tuned circuit), but no current. Hence, ExI=0. When the tube is on, it is slammed hard on by the "high" grid signal and there is a high plate current, but the plate voltage is very low (anybody know how low and if I am all wet? ... tubes aren't quite like transistors in the digital mode)--therefore ExI=somthing, but since the E is low, it is lower than in class A during that part of a cycle. There may also be some effect due to the fact that the plate tank is swinging low allowing the plate voltage to be even lower. Did you know that in class A, the plate power dissipated goes DOWN by the amount that is delivered to the load??? Cool! huh? Isn't physics neat! Second, it is the RMS current through the tube which will waste power, so it is what we must be concerned with. Yes, if the tube is off the current is zero at that time, but the RMS must be considered and it does not go down as fast you might think. As an example, for the same current pulses, but spaced out to half the duty cycle, the average drops to half, but the RMS only drops to .707. There is a square root in there. [[Anybody see the "AC Watt meter article in QST]] It is an OOPS! Most power supplies don't draw sine wave current. It is pulses. I have been in contact with both Bob Shrader (the author) and Stu Cohen (Tech editor) and I just finished an analysis and am going to make more measurements to verify, but the numbers Bob published can be as much as 1/3 the true power values (depending upon the DVM he used and the current waveform of the supplies he measured. -- Steve N, K,9;d, c. i My email has no u's. wa-da ya blokes think. The switched-off time makes the I in the denominator of E/I very small indeed and the solution to Ohm`s law is a high impedance without the dissipation of a resistance that remains in place continuously while agitating the atoms of a poor conductor to limit current. Instead, we have a low-resistanc in high conducton for short spurts. On-time is limited, instead of conduction, to produce a certain effective resistance. Another way of saying just wjat I did above, but "effective resistance" is one way of thinking about it and this resistance must be calculated using the RMS values. An automobile Kettering ignition system may use a dwell-meter to indicate how much of the time the points are closed. An ohmmeter indicates the resistance between its test prods. I'd be willing to place a bet (knowing how an analog ohm meter works, that the *diflection* of the two meter pointers is the same (see below). Both meters respond to the averacge current through them and both will show full scale when the points are open (I think thta is the correct polarity). Here's the "below": There is, however, the confusion added by the coil/cap waveform for which the ohm meter is not equiped to limit - whereas, I believe the dwell meter, if well designed, will have something to limit so as to remofe it as a complicatin. -- Steve N, K,9;d, c. i My email has no u's.. The two test circuits are almost the same although limitation of the deflection of the dwell-meter is different from limitation of the deflection of the ohmmeter due to the difference between limited conduction angle ignition points, and the continuous conduction through a current-limiting resistor. There`s an analogy between Class C and Class A amplifiers in there somewhere. Best regards, Richard Harrison, KB5WZI |
On Mon, 23 Feb 2004 16:51:43 GMT, "aunwin"
wrote: You can't ever discard the factor Q in any discusion with respect to antenna efficiency or any calculation for that matter. Q is intrinsic in any calculation that determines efficiency especialy when considering what the object of an antenna is. Hi Art, Q is NOT the arbiter of all that is efficient. In fact high Q can lead to very poor communication links. The most efficient and simplest antenna, the dipole, exhibits a very low Q for the very obvious reason: it is built to lose power by design. The loss to radiation resistance, Rr, is indistinguishable to Ohmic loss when computing Q. This in itself directly states that maximizing Q is inimical to transmitting power if you do not separate out the two losses. Terman treats this inferentially in his discussion of Power Amplifiers and their Plate Tank's Q. To select one that exhibits too high a value is to risk very poor operation. He suggests that a Q of 8 to 15 is a reasonable value. This confounds many who seek to peak their designs and fail to come to terms with unloaded and loaded Q valuations. It is the Q's relation to power loss to heat that makes the difference, not to the power curve in isolation of this loss. Small antennas suffer from high Q for this very reason - too little thought is given to the radiation resistance's correlation to the Ohmic loss of the system. As Richard has pointed out, one Ohm loss within the structure is hardly a loss leader for an antenna with 73 Ohms Rr. To achieve 50% efficiency requires your antenna to exhibit less than this same value of Ohmic loss (however, let's be generous in comparisons to 1/1000th that value). The rage of "High Q" antennas is in various loops of small diameters. Let's look at small loops' Rr for various sizes in tabulated form: Fo 1M diameter Efficiency with 1 mOhm loss 160M 29 µOhms 2.8% 80M 500 µOhms 33% 60M 1.5 mOhms 60% 40M 7.5 mOhms 88% 30M 24 mOhms 96% 20M 120 mOhms 99% Let's examine the validity of that generous assignment of 1 mOhm loss and see if it is reasonably warranted. Skin effect is the single largest contributor to this loss as a source (aside from poor construction techniques). Using the 1M diameter loop as being a practically sized construction, and if were using 2.54cM diameter copper wire/tubing we find: Fo skin effect loss 160M 13.8 mOhms 80M 20 mOhms 60M 23 mOhms 40M 28 mOhms 30M 33 mOhms 20M 39 mOhms Well, 1 mOhm was too generous and if we look at those loops' Rr once again against a robust, thick loop element: Fo 1M diameter Efficiency with skin effect loss 160M 29 µOhms 0.2% 80M 500 µOhms 2.4% 60M 1.5 mOhms 6% 40M 7.5 mOhms 21% 30M 24 mOhms 42% 20M 120 mOhms 75% These "High Q" loops are NOT efficient, they are convenient. The two terms are not the same at all and yet in common discussion they are confused to mean the same thing. I would point out further, that commercial vendors do not use 1 inch tubing (as the numbers above would force to even bigger conductors). Instead they use much larger tubing; but even here, one vendor uses a flat strap which is a very poor substitute as the skin resistance is defined at the edges and the face of the flat strap is far less conductive. The physics of conduction forces current to seek the smallest radius (the edge) to the exclusion of the broad surface (this is why we use tubular conductors and not flat ones). I will leave it to the student to reverse-engineer the required conductor size to obtain the same 1 mOhm results of the first table above. Even then, it will be seen that the common dipole still reigns supreme in efficiency. 73's Richard Clark, KB7QHC |
Dan Richardson wrote:
"The fact is resonance or not is not the determining factor." Resonance of the antenna system is the determining factor in the performance of a standing-wave antenna. This is an amateur group, so you may check the "ARRL Antenna Book". My 19th edition has resonant antennas on page 9-2. Fig 2 is a series RLC circuit representation of the typical standing-wave antenna. Ohm`s law should be noncontroversial (I=E/Z). To maximize I with a given voltage, Z must be minimized. Z in the series circuit is the phasor sum of R and X. R has probably been established firmly in an antenna by its construction and placement but we can tune the antenna system to make it resonant so that we eliminate X to get maximum current into the antenna and to thereby get maximum performance out of the antenna. Best regards, Richard Harrison, KB5WZI |
Art, KB9MZ wrote:
"You can`t ever discard the factor Q in any discussion with respect to antenna efficiency or any calculation for that matter." Kraus writes in his 1950 edition of "Antennas" on page 299: "The Q of an antenna, like the Q of any resonant circuit, is proportional to the ratio of the energy stored to the energy lost (in heat or radiation) per cycle." (in heat or radiation) are Kraus` words, not mine. It means the R of the antenna used in the Q operations is formed of the sum of Rr+Rloss. Efficiency is Rr/Rr+Rloss The Q=X/R, where R is the sum of Rr+Rloss. If R is heavily weighted toward Rr, the antenna is efficient. If R is heavily weighted toward Rloss, the antenna is inefficient. Q as an indicator of efficiency is baloney. Best regards, Richard Harrison, KB5WZI |
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Dan it is wise to remember that Richard "cherry picks" statements from
assorted books and disregards the chapter heading and what is discussed in that chapter. He only looks for a statement that suits his present state of mind and suggests authority and ignores the underlying subject of the thread. In that same chapter it is discussed how the losses mount up as you move away from the resonant point and shows where the losses could amount to 5dB no less. There is a proviso in that the antenna books that I have precede the one that he is refering to so there may be some evidence that somebody has pushed the envelope beyond Maxwell and others. I think the actual chapter refers to broadband antennas rather than those of a narrow bandwidth so it is always advisably to check so called "book quotes" rather than accept what people interprete what is relavent in their own minds. Regards Art "Dan Richardson @mendolink.com" ChangeThisToCallSign wrote in message ... Let me try this one more time. You had posted earlier and I commented on this: "A receiving antenna must be resonant to enable full acceptance of available energy, and it must be matched to avoid re-radiation of more than 50% of the energy it is able to grab." I commented on the first portion of your statement (above). My only point is that it makes no difference if an antenna's is resonate or not in determining how much energy it grabs. That's it, nothing more. 73 Danny, K6MHE On Mon, 23 Feb 2004 13:43:55 -0600 (CST), (Richard Harrison) wrote: Dan Richardson wrote: "The fact is resonance or not is not the determining factor." Resonance of the antenna system is the determining factor in the performance of a standing-wave antenna. This is an amateur group, so you may check the "ARRL Antenna Book". My 19th edition has resonant antennas on page 9-2. Fig 2 is a series RLC circuit representation of the typical standing-wave antenna. Ohm`s law should be noncontroversial (I=E/Z). To maximize I with a given voltage, Z must be minimized. Z in the series circuit is the phasor sum of R and X. R has probably been established firmly in an antenna by its construction and placement but we can tune the antenna system to make it resonant so that we eliminate X to get maximum current into the antenna and to thereby get maximum performance out of the antenna. Best regards, Richard Harrison, KB5WZI |
Dan, K6MHE wrote:
"My only point is that it makes no difference if an antenna`s is resonate or not in determining how much energy it grabs." Not in the "grand scheme of things", perhaps, but a resonant antenna extracts more energy when swept by a traveling wave than an off-resonance conductor does at the same distance from the transmitter. On some occasion you may have adjusted the antenna trimmer on a receiver and found a signal peak. That adjustment balanced out the reactance in the antenna system leaving only the antenna resistance to oppose current flow in the antenna. Proper adjustment maximizes antenna current induced by the wave. The mechanism is simple. The antenna is a series RLC circuit. Current in the antenna is a function of the field strength divided by the impedance of the RLC circuit. Impedance to antenna current is least when all of the reactance has been eliminated. You have probably experienced peaking the signal this way with your ears, a meter, or an oscilloscope. Let`s look at an oft repeated type of true story by John E. Cunningham in his "Complete Broadcast Antenna Handbook": "Not every tall structure will affect the pattern of a broadcast station. The effect on the pattern depends on the height of the structure. In one case, a microwave tower was erected, one section at a time, near a 4-tower directional antenna system. As construction progressed, the current in one tower began to drop as each new tower section was added to the microwave tower. The current continued to drop until it nearly reached zero. But, as more sections were added to the microwave tower, the current began to rise. When the microwave tower reached its final height, all of the tower currents of the broadcast antenna, as well as the pattern, were normal. Resonance, or lack of resonance, makes all the difference! Best regards, Richard Harrison, KB5WZI |
Art, KB9MZ wrote:
"Dan it is wise to remember that Richard "cherry picks" statements from assorted books and disregards the chapter heading and what is discussed in the chapter." Forget the books! Surely some time in your long career you`ve peaked an antenna trimmer and it worked. Why? Best regards, Richard Harrison, KB5WZI |
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Steve Nosko wrote:
"First, my "class C" model is as follows: I liken it to digital or "switched modes"." I do too. "----Second, it is the RMS current through the tube which will waste power, so that is what we must be concerned with." I don`t believe current through a Class C amplifier consists of an ordinary sine wave. I think it consists of short unidirectional pulses. The tuned "tank circuit" is the source of sine waves. RMS is the effective value, not the average value, of an a-c ampere. It is defined as 0.707X the peak value of the waveform. It is derived from the average of the squared current over a half cycle, as the heating value of an ampere is proportional to the current squared. Speaking inversely, the ratio of maximum to effective value for a sine wave is 1.414, which is the square root of 2. Ordinarily, with nonsinusoidal currents, the ratio of maximum to effective value is not the square root of 2. Best regards, Richard Harrison, KB5WZI |
Dan Richardson wrote:
So according to you, as an example, a resonate 1/4 or 1/2 monopole will extract more energy than a non-resonate 5/8-wave monopole. A non-resonant length physical 5/8WL monopole is made into an electrically resonant monopole with the addition of the base loading coil. Without that resonating coil at the base, not much energy would be delivered to a 50 ohm receiver or accepted from a 50 ohm transmitter. -- 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:
I don`t believe current through a Class C amplifier consists of an ordinary sine wave. I think it consists of short unidirectional pulses. The tuned "tank circuit" is the source of sine waves. Yep, the Class C amp is like the energy pulse from a pendulum clock spring. The tank/filter circuitry is like the pendulum. -- 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! =----- |
Dan, K6MHE wrote:
"So according to you, as an example, a resonate 14 or 1/2 monopole will extract more energy than a non-resonate 5/8 monopole." A 5/8-wavelength monopole has a directive gain over a 1/4-wavelength self-resonant monopole. However, the 5/8-wavelength monopole must be resonated externally to remove its reactance to take advantage of its extended length so that the volts per meter it is exposed to can make a large current flow in the antenna. Best regards, Richard Harrison, KB5WZI |
On Tue, 24 Feb 2004 06:17:01 -0800, Dan Richardson
wrote: On Mon, 23 Feb 2004 17:47:15 -0600 (CST), (Richard Harrison) wrote: Not in the "grand scheme of things", perhaps, but a resonant antenna extracts more energy when swept by a traveling wave than an off-resonance conductor does at the same distance from the transmitter. So according to you, as an example, a resonate 1/4 or 1/2 monopole will extract more energy than a non-resonate 5/8-wave monopole. Enjoy your dreams, I'm QRT 73 Danny, K6MHE Hi All, As a test of modeling and actual, real data, I put this to the test: 1.) I established two full quarter wave antennas; 2.) each with a set of 48 radials; 3.) tuned to 1.7 MHz; 4.) separated them by 20KM (100 wavelengths); 5.) set one to have a source of 1000W; 6.) loaded the other at the base with a 50Ohm resistor; all in EZNEC with enough segments to hit the 500 limit. When I called for the load data, EZNEC calculated it to be: Power = 0.0008166 watts (already confirmed accurate to FCC Field trials) THEN, I increased the same antenna's length to 5/8th Wavelength and observed its load's response to the identical field: Power = 0.0001185 watts Through the simple observation of power ratios, the untuned 5/8th Wavelength antenna suffers to the tune of: -8.38dB Now, if this discussion revolves on the myopic consideration of resonant antennas, devoid of other issues like tuners, matching, lines that connect and so forth; then it is painfully obvious that NO ANTENNA absorbs ANY POWER (by virtue of the load having been defined out of the problem). This is further confirmed in the daily experience of a million yagi owners whose passive elements offer just exactly that lossless coupling. When interfaced to a real load, any SWLer will confirm through simple tuning (and through modeling) that the received power peaks through adjustment. The 5/8th antenna is no different. 73's Richard Clark, KB7QHC |
Richard Harrison wrote:
However, the 5/8-wavelength monopole must be resonated externally to remove its reactance to take advantage of its extended length so that the volts per meter it is exposed to can make a large current flow in the antenna. In other words, the linear system transfers the most energy when the source (the antenna) is matched to the load (the receiver). It cannot transfer more than 1/2 the energy, but it can transfer less. It is conceivable that a receiver's input impedance could be optimized to the feedpoint impedance of a 5/8WL monopole but I don't know of any receiver like that. -- 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! =----- |
Hi Richard...
"Richard Harrison" wrote in message ... [...] "----Second, it is the RMS current through the tube which will waste power, so that is what we must be concerned with." I don`t believe current through a Class C amplifier consists of an ordinary sine wave. And I didn't say that it does nor do I believe it does. I'm inclined to take my 100MHz storage scope to to the 6146's of my TS830s and see for myself. Your words imply (at least I infer) you are thinking that only a sine wave has an RMS value. Every wave of any shape has an effective or RMS value - its heating or "power causing" value. [...] I think it consists of short unidirectional pulses. The tuned "tank circuit" is the source of sine waves. This certainly has to be correct. The tank will most likely cause some sine-like VOLTAGE waveform, but the tube current has to be pulses of some shape. This is a very timely discussion in view of the AC power meter QST article and the extensive investigation I just completed on several pulse shapes.. RMS is the effective value, not the average value, of an a-c ampere. I will differ here. The RMS value is more appropriately described as the power producing value of ANY wave form. Pulses can produce heat just as well as sine wave AC. We all know this from a practical view since tubes can only conduct in one direction and the plates DO get hot. ...as the heating value of an ampere is proportional to the current squared. This is actually a simplification. P=ExI Power is the product of voltage and current *only*. Because this is a second order effect, in a resistance it can be related to either voltage squared or current squared... because that captures the second order character. Maybe there's a better way to say it mathematically, but I don't know it. When we get to non sine shapes, then we have to fall back on the actual definition. root [avg of square] ...with the integral and all. http://www.ultracad.com/rms.pdf [...snip...] Ordinarily, with nonsinusoidal currents, the ratio of maximum to effective value is not the square root of 2. Best regards, Richard Harrison, KB5WZI Doing the math for pulses with the shape of sine, triangle (a single slope with sudden end) and trapezoid (a sudden start to one level then a slope to a peak and a sudden end), I decided to look at the RMS to AVERAGE ratio since average is what a common meter will measure in Bob Shrader's article (AC watt meter Jan 04 QST). I was particularly interested in the sine-shaped pulses of various duty cycle because the current of common power supplies occurs in short pulses with a sine-like shape that are near the peak of the voltage waveform. It was interesting that for all these shapes, this ratio was very similar. One relatively simple thing to understand which came out of the analysis was that the average value is directly proportional to the duty cycle as you might reasonably postulate. Where duty cycle is the ratio of "on" time to off time. Where "on" time is the time that ANY current flows. Whereas the RMS is proportional to the Square root of the duty cycle. e.g. drop the duty cycle to half and the RMS drops to .707. I have to do some verification, but it sure looks as though Bob's numbers can be as much as three times what he quoted, depending on the waveshape and some measurements I made. http://www.irf.com/technical-info/an949/append.htm Trapezoid=rectangular. Also for the phase controlled sine, the things that look like tau and a small n are both pi i.e. sin [pi x (1-D)] cos [pi x (1-D)] and denominator of 2 x pi Some average & RMS values here. http://home.san.rr.com/nessengr/techdata/rms/rms.html More (better) average formulas: http://www.st.com/stonline/books/pdf/docs/3715.pdf NOW I know where the average value of a sine wave comes from = (2/pi) The Greek delta = d. A calculator for RMS: http://www.geocities.com/CapeCanaveral/Lab/9643/rms.htm |
I wrote:
"RMS is the effective value, not the average value of an a-c ampere. Steve Nosko responded: "I will differ here. The RMS value is more appropriately described as the power producing value of ANY wave form.." A periodic function is not necessarily a sine wave but when I look up alternating current in my dictionary, I find the diagram of a sine wave and it is tagged: "alternating current". By fourier series, any complex waveform may be resolved into a fundamental plus a finite number of terms involving its harmonics. My electronics dictionary says: "The rms value of a periodic quantity is equal to the square root of the average of the squares of the instantaneous values of the quantity taken throughout one period. If that quantity is a sinewave, its rms amplitude is 0.707 of its peak value." My dictionary also says: "Effective current---That value of alternating current which will give the same heating effect as the corresponding value of direct current. For sine-wave alternating currents, the effective value is 0.707 times the peak value." The average value for one alternation of a sine wave is 2/pi or 0.637 times the maximum value. I make lots of mistakes, but fortunately, I don`t think I made any of consequence in this thread. I probably didn`t go into enough detail, but I was just an engineer, not a professor. Best regards, Richard Harrison, KB5WZI |
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Richard Harrison wrote:
If we have a Class C amplifier feeding power to the same antenna and enjoying a conjugate match, we can have a source that takes less than 50% of the available energy. So, the transmitting antenna system can be more efficient than the receiving antenna system, it seems to me. Comparing a Class C amplifier to a linear receive antenna source is comparing apples and oranges. If one makes the transmitter Class-A linear (like the receive antenna is linear) then the antenna is reciprocal for transmit and receive. (A Class C or Class B or even Class AB *single stage* is not linear). Putting two of them in push-pull doesn't make the individual single stages linear. -- 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 in message
... I wrote: "RMS is the effective value, not the average value of an a-c ampere. Steve Nosko responded: "I will differ here. The RMS value is more appropriately described as the power producing value of ANY wave form.." Richard, I'm not sure what your point of contention is here for most of these comments. Perhaps I was not clear in my lone objection. All I was correcting was your reference to "a-c ampere". My intent was to point out that pulses to one side of zero also have an RMS value. If you do the "square root ...average...squares..instantaneous values..quantity..throughout one period." thing, you will get it. So therefore, tube's current pulses have some RMS value; and that's what'll heat the tube and cause the efficiency stuff. That's where I was going. If we can define or approximate the shape, then we could calculate an RMS. I wonder if a "true RMS" DVM can handel this? A periodic function is not necessarily a sine wave but when I look up alternating current in my dictionary, I find the diagram of a sine wave and it is tagged: "alternating current". Sure. I'm ok with that. A sine is indeed one example of a periodic function. By fourier series, any complex waveform may be resolved into a fundamental plus a finite number of terms involving its harmonics. Also ok, but not sure how it plays into the RMS discussion. See below. My electronics dictionary says: "The rms value of a periodic quantity is equal to the square root of the average of the squares of the instantaneous values of the quantity taken throughout one period. If that quantity is a sinewave, its rms amplitude is 0.707 of its peak value." No issue here. (from me, that is) The "average" turns out to be an integration and I had to drag out the son's calc book to remember how to do it to see some of the answers myself. My dictionary also says: "Effective current---That value of alternating current which will give the same heating effect as the corresponding value of direct current. For sine-wave alternating currents, the effective value is 0.707 times the peak value." Still none. The average value for one alternation of a sine wave is 2/pi or 0.637 times the maximum value. Yea. untill I went through the trouble of doing the calculation, I had never known the source of the 0.637 number. Now I know. Pretty cool to actually do it! The integration for this is pretty simple. Took me a few looks to figure out where the pi came from. It's the old indefinite to definite integral thingy. I make lots of mistakes, but fortunately, I don`t think I made any of consequence in this thread. I probably didn`t go into enough detail, but I was just an engineer, not a professor. I'm not a "professor", just an Enginer and "adjunct faculty" for two community colleges for some years. (that's a fancy name for part time teacher) 73, Steve |
"Richard Clark" wrote in message
... The one I noted (mistake) was in your reference: It is derived from the average of the squared current over a half cycle which necessarily forces both a doubling, and a symmetry that is not demanded of native RMS determinations. It then follows that the commonplace illustration of the mains Sine wave completes the illusion. Few EE students migrated beyond this simplicity because the world is nasty place to measure power. RMS by and of its mathematical nature through the squaring operation negates any requirement for "half cycle" determinations (no issue of negatives). It also preserves the natural order (of two forced by the half). If you think about it, any biased sine wave impinging upon a load imparts the power loss of the bias at the 180° portion of the Sine cycle. RMS copes with this, the notion of half cycles does not. I assumed Richard's intent here is that you only have to do the calculation for one full period of the periodic component to derive the RMS value - and if it is symmetrical, then only one half period will suffice. This all assumes a symmetrical AC shape. .. I see no reason why this would not be true. However a DC biased periodic shape requires another squaring and root operation if you capture all the components. It gets a bit more harry The simple determination of RMS is the graphical integration of the area under the curve. There are as many "correction factors" for RMS as there are shapes, and they all derive from this simple concept. Here I'll take issue with the ONE WORD "graphical". You can integrate if you can describe the function of the wave shape mathematically. When the computational horsepower requirement becomes enormous (there are many here that give up too easily with complexity); it is the provence of the "Old School" to suggest that since RMS is all based on the notion of power, you simply measure the caloric result and ignore shape altogether. This may be done with thermo-electric piles or other measurable property transformers that perform the complexity of integration through physics*. I can anticipate those who dearly embrace the complexity that they shudder to face (such contradictions of their love-hate relationships) when I hear Crest, or pulse/power factor (or duty cycle) uttered. Clearly the problem will have migrated from Power to some other consideration, but is dressed as an RMS debate. 73's Richard Clark, KB7QHC Yikes! Not sure where you went on that last bit, Richard C... Now, I ask. Do the power meters on the outside of our houses take all those factors into consideration and REALLY show TRUE watt hours? I have one in the basement and I think I figured out why I was seeing twice the reading I should have (letting a light bulb sit on for awhile) ... I counted the teeth to get the ratio of the gear train, just to find that it is printed (somewhat cryptically) on the face) (I made a two wire / three wire connection error) 73, Steve K9DCI |
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