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Efficiency and maximum power transfer
Someone may regard the following question a bit OT, but as it deals with
impedances I have considered that the antenna newsgroup could be the most appropriate one where to post it. Let us regard a transmitter as an ideal RF generator with a resistance in series. It is well known that, for maximum power transfer, the load resistance must be equal to the generator resistance. Under such conditions efficiency is 50% (half power dissipated in the generator, half delivered to the load). To achieve a higher efficiency, the load resistance should be made higher than the generator resistance, although this would turn into a lower power delivered to the load (the maximum power transfer condition is now no longer met). This can be verified in practice: by decreasing the antenna coupling in a transmitter, one obtains a higher efficiency though with a lower output power. That said, now the question. Usually, when a transmitter is tuned for maximum output power, efficiency results to be higher than 50% (typically 60% for class-B, 70% for class-C). This would seem to contradict the above cited fact that, under maximum power transfer condition, efficiency is 50%. Pertinent comments are welcome. 73 Tony I0JX - Rome, Italy |
Efficiency and maximum power transfer
On Jun 6, 2:12 pm, "Antonio Vernucci" wrote:
Someone may regard the following question a bit OT, but as it deals with impedances I have considered that the antenna newsgroup could be the most appropriate one where to post it. Let us regard a transmitter as an ideal RF generator with a resistance in series. It is well known that, for maximum power transfer, the load resistance must be equal to the generator resistance. Under such conditions efficiency is 50% (half power dissipated in the generator, half delivered to the load). To achieve a higher efficiency, the load resistance should be made higher than the generator resistance, although this would turn into a lower power delivered to the load (the maximum power transfer condition is now no longer met). This can be verified in practice: by decreasing the antenna coupling in a transmitter, one obtains a higher efficiency though with a lower output power. That said, now the question. Usually, when a transmitter is tuned for maximum output power, efficiency results to be higher than 50% (typically 60% for class-B, 70% for class-C). This would seem to contradict the above cited fact that, under maximum power transfer condition, efficiency is 50%. Pertinent comments are welcome. 73 Tony I0JX - Rome, Italy Simple: a transmitter is not an ideal voltage source with a resistor in series. I'm playing with a switching power supply design that delivers about a kilowatt at 100 volts. The output is designed specifically to have a negative resistance, so the output voltage increases as the current drawn increases. The output dynamic impedance is about -1 ohms (adjustable, actually). The linear model is a 100 volt battery in series with -1 ohms. With an 11 ohm load, I get 10 amps load current, with the battery thus delivering 1000 watts, the load dissipating 1100 watts, and the -1 ohm resistance dissipating -100 watts. Which shows the absurdity of thinking of a dynamic output resistance being anything like a real resistance. In my switching supply, I can adjust the dynamic output resistance between a small negative value and a rather larger positive value, with very little change in efficiency. Although transmitters MAY have dynamic output resistances similar to the recommended load resistance, that's not a necessary condition, and has little to do directly with efficiency. Cheers, Tom |
Efficiency and maximum power transfer
On Jun 6, 4:12�pm, "Antonio Vernucci" wrote:
Someone may regard the following question a bit OT, but as it deals with impedances I have considered that the antenna newsgroup could be the most appropriate one where to post it. Let us regard a transmitter as an ideal RF generator with a resistance in series. It is well known that, for maximum power transfer, the load resistance must be equal to the generator resistance. Under such conditions efficiency is 50% (half power dissipated in the generator, half delivered to the load). To achieve a higher efficiency, the load resistance should be made higher than the generator resistance, although this would turn into a lower power delivered to the load (the maximum power transfer condition is now no longer met). This can be verified in practice: by decreasing the antenna coupling in a transmitter, one obtains a higher efficiency though with a lower output power. That said, now the question. Usually, when a transmitter is tuned for maximum output power, efficiency results to be higher than 50% (typically 60% for class-B, 70% for class-C). This would seem to contradict the above cited fact that, under maximum power transfer condition, efficiency is 50%. Pertinent comments are welcome. 73 Tony I0JX - Rome, Italy In addition to Tom's comments, an RF Power Amplifier's efficiency is defined as (Pout/Pin)X100%. Pout is RF and Pin is usually DC. So if you pump 1000 watts DC in to a class B RF amp and get 600 watts rms out you are 60% efficient. Your model of a Voltage Source in series with an internal resistance does not apply here. The maximum power theorem gives conditions where power in the load, is equal to internal power in the generator. Not always a good idea. A 50HZ generator capable of Megawatts of power would dissiapate 1/2 in the generator and 1/2 in our houses if they designed them to conform to the MPT. The 50HZ generators would melt. Utilities design their Generators to have nearly 0.0 ohms internal impedance. Good question. Gary N4AST |
Efficiency and maximum power transfer
Antonio Vernucci wrote:
Usually, when a transmitter is tuned for maximum output power, efficiency results to be higher than 50% (typically 60% for class-B, 70% for class-C). This would seem to contradict the above cited fact that, under maximum power transfer condition, efficiency is 50%. The maximum power transfer theorem only applies to linear sources. What is the linear source impedance of a class-C amp? -- 73, Cecil http://www.w5dxp.com |
Efficiency and maximum power transfer
This has been a good example of a common pitfall in modeling. The error
made in this case was to attempt to apply an unsuitable model (a voltage source in series with a resistance) to a system to be modeled (a transmitter). As the OP showed, the attempt leads to an impossible result. The classic example of this is the "proof" that a bumblebee can't fly, based on a flawed model and immediately shown to be false by simply observing that they do, indeed, fly. Yet we see people falling into this trap daily, not only in modeling electrical circuits, but also in modeling such diverse processes as human behavior, economic systems, and roulette wheel numbers. Unfortunately, the bad results of applying unsuitable models aren't always so obvious as they were here. So it's always wise to check to see if the model fits before putting faith in the results. Roy Lewallen, W7EL |
Efficiency and maximum power transfer
On Fri, 06 Jun 2008 20:58:13 -0500, Cecil Moore wrote:
Antonio Vernucci wrote: Usually, when a transmitter is tuned for maximum output power, efficiency results to be higher than 50% (typically 60% for class-B, 70% for class-C). This would seem to contradict the above cited fact that, under maximum power transfer condition, efficiency is 50%. The maximum power transfer theorem only applies to linear sources. What is the linear source impedance of a class-C amp? The source resistance appearing at the output of either a Class B or C amplifier is R = E/I, where E is the peak voltage at the output terminals and I is the peak current at the output. Or RMS values can also be used. Since E/I is simply a ratio, R is also a ratio. And we know that a ratio cannot dissipate power, or turn electrical energy into heat, thus the output resistance R is non-dissipative. I have made many measurements that prove this. It is also the reason why reflected power does not dissipate in the tubes, because it never reaches the tubes. The reflected power simply causes a mismatch to the source, causing the source to deliver less power than it would if there were no mismatch. The input at the pi-network in the xmtr is non-linear, but the fly-wheel effect of the network tank isolates the input from the output, resulting in a linear condition appearing at the output. Except for a very slight deviation from a sine wave due to a small amount of harmonic content, the voltage E and current I at the output are essentially a sine wave, which one can easily prove with a good oscilloscope, proving the output to be linear. I'm speaking for tube rigs with pi-network tanks, not for solid-state rigs. I nearly forgot. The only dissipation in the amp tube(s) is due to the filament-to-plate current as the electrons bombard the plate. The efficiency is determined by the ratio of the DC input power to the RF output power. The maximum power is delivered when the load resistance equals the output resistance R of the source. But since resistance R is non-dissipative it is not a factor in determining efficiency. The only factors in determining efficiency are the RF output power and the dissipation in the tube caused by the electrons striking the plate. The non-dissipative output resistance is the reason Class B and C amps can have an efficiency greater than 50 percent. If the output resistance were dissipative it would be the determining factor in efficiency, which could never be greater than 50 percent if the load resistance was equal to the the output resistance. A report of my measurements will soon be available in Reflections 3, from measurements taken since those reported in Chapter 19 of Reflections 2. Cecil, if I send you a copy of the new chapter that has the report of my newer measurements do you have any way to make it available to the guys on this thread? Walt, W2DU |
Efficiency and maximum power transfer
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Efficiency and maximum power transfer
"Antonio Vernucci" wrote in
: Someone may regard the following question a bit OT, but as it deals with impedances I have considered that the antenna newsgroup could be the most appropriate one where to post it. Let us regard a transmitter as an ideal RF generator with a resistance in series. It is well known that, for maximum power transfer, the load resistance must be equal to the generator resistance. Under such conditions efficiency is 50% (half power dissipated in the generator, half delivered to the load). To achieve a higher efficiency, the load resistance should be made higher than the generator resistance, although this would turn into a lower power delivered to the load (the maximum power transfer condition is now no longer met). This can be verified in practice: by decreasing the antenna coupling in a transmitter, one obtains a higher efficiency though with a lower output power. That said, now the question. But your statements are not true. The model you propose for a transmitter does not apply in general. Whilst it would be possible to build a transmitter like that, most transmitters that hams use are not built like that. So... it is a loaded question of a type, a question premised on a falsehood. Usually, when a transmitter is tuned for maximum output power, efficiency results to be higher than 50% (typically 60% for class-B, 70% for class-C). This would seem to contradict the above cited fact that, under maximum power transfer condition, efficiency is 50%. If the equivalent source impedance is not important, ie it does not need to be fixed by the design, there here is an analysis. If you take the case of a grounded cathode triode in class C with a steady signal, the conduction angle is usually somewhere around 120°. The anode current waveform is a little like a truncated sine wave, but even for the range of grid voltages where anode current is greater than zero, the transfer characteristic is not exactly linear, and the wave will be further distorted. If the nature of the anode load is that it is some equivalent R at the fundamental and zero impedance at all other frequencies, the power output can be determined by finding the fundamental component of the anode current waveform, squaring it, and multiplying it by R. The input power is the average anode current multiplied by the DC supply voltage. Efficiency is OutputPower/InputPower. By varying the grid bias, drive voltage, load impedance and supply voltage for a given triode, different efficiencies will be found, and the maximum could be well over 80%. Nothing in this approach to design attempts to fix the equivalent source impedance, the design is performed without regard to that characteristic. Nevertheless, some argue that the output network performs magic and achieves source matching naturally without designer intervention, and does this irrespective of parameters like the dynamic anode resistance, and the effects of feedback (such as cathode degeneration in grounded grid amplifiers which in turns depends on the source impedance of the exciter). Owen |
Efficiency and maximum power transfer
Owen Duffy wrote in
: .... voltage. Efficiency is OutputPower/InputPower. By varying the grid bias, drive voltage, load impedance and supply voltage for a given triode, different efficiencies will be found, and the maximum could be well over 80%. If you want to explore this approach to design, I have implemented a spreadsheet which in turn implements the methods described in CPI/Eimac's "Care and Feeding of Power Tubes" it is described and can be downloaded at http://www.vk1od.net/RFPATPC/index.htm . The spreadsheet is populated with the example 4CX20000 in Class C from the above publication. The spreadsheet calculates the anode current wave form from an interpolation based on a number of points from the published characteristics, and calculates the fundamental component of anode current using an FFT. It then calculates anode efficiency and overall efficiency given operating parameters. Owen |
Efficiency and maximum power transfer
Walter Maxwell wrote:
Cecil, if I send you a copy of the new chapter that has the report of my newer measurements do you have any way to make it available to the guys on this thread? Walt, it is sure good to hear from you and I hope you are doing well. I can certainly post your new chapter to my web page thus making it available for downloading. -- 73, Cecil http://www.w5dxp.com |
Efficiency and maximum power transfer
On Jun 7, 12:43�am, Owen Duffy wrote:
wrote : ... The maximum power theorem gives conditions where power in the load, is equal to internal power in the generator. �Not always a good idea. �A 50HZ generator capable of Megawatts of power would dissiapate 1/2 in the generator and 1/2 in our houses if they designed them to conform to the MPT. �The 50HZ generators would melt. �Utilities design their Generators to have nearly 0.0 ohms internal impedance. Actually, the AC power distribution system from alternator down has a manged substantial equivalent source impedance. The source impedance serves to limit fault currents, which reduces the demands on protection devices. Sure, the network is not operated under Jacobi MPT conditions, but neither does it have near zero source impedance. Owen Not really sure I agree. A multi-megawatt 60HZ generator by necessity has near zero source impedance. The ones I am familar with require forced air cooling on their output buses. If you are pumping out Mega- watts, then any non -zero source impedance results in serious heat. I^2R. Gary N4AST |
Efficiency and maximum power transfer
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Efficiency and maximum power transfer
Good to see that everyone agrees that a generator with a resistor in series is
an unsuitable model for an RF transmitter.The easy part of the work is done! Now the more difficult part. As, by the Thevenin theorem, any complex circuit comprising resistors, voltage generators and current generators is equivalent to a generator with a resistor in series, evidently the transmitter model must comprise elements other than just resistors, voltage generators and current generators. Can one suggest how such a model looks like? (even a plain one, that does not take into account second- or superior-order effects). 73 Tomy I0JX - Rome, Italy |
Efficiency and maximum power transfer
Since E/I is simply a ratio, R is also a ratio. And we know that a ratio
cannot dissipate power, or turn electrical energy into heat, thus the output resistance R is non-dissipative. I have made many measurements that prove this. Walt, W2DU Hello Walter, thanks for your explanation. I remember having read your excellent articles on QST magazine many years ago, in which you also explained, among many other things, why reflected power cannot dissipate in the final stage of a transmitter. I am not at all opposing your explanation of "non disspative resistance" (on the other hand how may I contradict a person named Maxwell, hi), but I have some difficulties to appreciate it. In my understanding resistance just means that current and voltage are in phase. There are two possibilities for this to occur: 1) dissipative resistance. Example is a pure resistor, in which power is converted into heat. 2) non-dissipative resistance: Example is a DC motor, that converts electrical power into mechanical power. An ideal motor would convert all absorbed power into mechanical power, producing no heat But I am unable to see how the second case could be fitted into the transmitter model. Thanks and 73 Tony I0JX - Rome, Italy |
Efficiency and maximum power transfer
The maximum power theorem gives conditions where power in the load, is
equal to internal power in the generator. Not always a good idea. A 50HZ generator capable of Megawatts of power would dissiapate 1/2 in the generator and 1/2 in our houses if they designed them to conform to the MPT. The 50HZ generators would melt. Utilities design their Generators to have nearly 0.0 ohms internal impedance. Good question. Gary N4AST Whay you write is perfectly true. To maximize efficiency, the ratio between load resistance and generator resistance must be as high as possible. That is the reason why the internal resistance of the Megawatts generator you took as an example is always made extremely low (very little power is so dissipated within the generator). Those generators however do not operate in the maximum power transfer condition (generator resistance = load resistance). As a matter of fact one cannot decrease the load resistance below a certain threshold because of the generator power delivery limitations. In other words the generator can deliver the maximun power it is able to deliver, but not under maximum power transfer conditions. Never mind, what is important is that efficiency is good! 73 Tony I0JX - Rome, Italy |
Efficiency and maximum power transfer
On Sun, 8 Jun 2008 16:16:56 +0200, "Antonio Vernucci" wrote:
Since E/I is simply a ratio, R is also a ratio. And we know that a ratio cannot dissipate power, or turn electrical energy into heat, thus the output resistance R is non-dissipative. I have made many measurements that prove this. Walt, W2DU Hello Walter, thanks for your explanation. I remember having read your excellent articles on QST magazine many years ago, in which you also explained, among many other things, why reflected power cannot dissipate in the final stage of a transmitter. I am not at all opposing your explanation of "non disspative resistance" (on the other hand how may I contradict a person named Maxwell, hi), but I have some difficulties to appreciate it. In my understanding resistance just means that current and voltage are in phase. There are two possibilities for this to occur: 1) dissipative resistance. Example is a pure resistor, in which power is converted into heat. 2) non-dissipative resistance: Example is a DC motor, that converts electrical power into mechanical power. An ideal motor would convert all absorbed power into mechanical power, producing no heat But I am unable to see how the second case could be fitted into the transmitter model. Thanks and 73 Tony I0JX - Rome, Italy Hi Tony, The reason a Class B or C amplifier can have efficiencies greater than 50 percent is because the output source resistance at the output of the pi-network is non-dissipative, as I said in an earlier post. I realize this phenomenon is somewhat difficult to appreciate. However, I have explained it, and proved it with measurements reported in Chapter 19 in Reflections 2. I have explained it further in an addition to Chapter 19 that will appear in Reflections 3, which has additional proof from measurements made since the original Chapter 19 was written. This addition to Chapter 19 is Chapter 19A, which I asked Cecil to put on his web page for all to see. Just scroll down to Cecil's post to see "Chapter 19A from Reflections III", and double click on the url. You will see the entire Chapter 19A. Apparently you don't have a copy of Reflections 2, so I'm going to email you a copy of the original Chapter 19. I hope these papers will help in understanding why the output resistance of the pi-network is non-dissipative. Walt, W2DU |
Efficiency and maximum power transfer
So, let me call this NG rec.radio.true.amateur.
Compliments, sincerely, to all you OM here. For sure in the posts here it is the maximum efficiency to transfer the power of experience and knowledge to anyone subrscribe this NG - and for sure to a novice as i am, with a bit of losses due to things that i don't know when i'm reading something. Anyway, true Radio Amateurs are here. Apologize for mistakes in my poor english. 73, -.-. --.- , Cristiano, Italy. |
Efficiency and maximum power transfer
On Jun 8, 6:52 am, "Antonio Vernucci" wrote:
Good to see that everyone agrees that a generator with a resistor in series is an unsuitable model for an RF transmitter.The easy part of the work is done! Now the more difficult part. As, by the Thevenin theorem, any complex circuit comprising resistors, voltage generators and current generators is equivalent to a generator with a resistor in series, evidently the transmitter model must comprise elements other than just resistors, voltage generators and current generators. Can one suggest how such a model looks like? (even a plain one, that does not take into account second- or superior-order effects). 73 Tomy I0JX - Rome, Italy I believe you have mis-stated the Thevenin theorem. First, it applies only to linear circuits. Fine -- over some narrow range at least, a transmitter does indeed look like a linear circuit. But more importantly, it describes ONLY what you observe at an external pair of terminals, with no other connections, NOT what goes on inside the "black box" containing those elements you mentioned. A very simple example is a voltage source (a perfect battery) and two resistors in series across the battery; the external terminals for this example will be at opposite ends of one of the resistors. Let's say the battery is 2 volts and each resistor is 2 ohms. That will look like a Thevenin equivalent 1V in series with 1 ohm. Note that it also looks like a one amp source in parallel with a one ohm resistor. But it does not behave INTERNALLY like either of those. Consider also the same internal circuit, except drop the voltage source to 1V and add a 1/2A current source across the output terminals, polarity so that there's no drop across the resistor between the voltage and the current source (with no external load). Now figure the internal dissipation for each of those two cases, with no load, with a 1 ohm load, and with a short-circuit load. Cheers, Tom |
Efficiency and maximum power transfer
There's a common misconception that, for a linear circuit, the maximum
efficiency and/or power available from a voltage source occurs when the source resistance equals the load resistance (or, more generally, when they're complex conjugates). But this isn't universally true, as I'll show with a simple example. Suppose we have a 100 volt perfect voltage source in series with a variable source resistance, and a fixed load resistance of 100 ohms. If we make the source resistance 100 ohms, the source delivers 50 watts, 25 of which are dissipated in the source resistance and 25 watts in the load. The efficiency, if you consider the source resistance dissipation as wasted, is 50%. But what happens if we reduce the source resistance to 50 ohms? Now the source delivers 66.7 watts, of which 22.2 is dissipated in the source resistance and 44.4 in the load resistance. The power to the load has increased, and the efficiency has increased from 50 to 66.7%. The efficiency and load power continue to increase as the source resistance is made smaller and smaller, reaching a maximum when the source resistance is zero. At that point, the source will deliver 100 watts, all of which is dissipated in the load, for an efficiency of 100%. The well known and often misapplied rule about maximizing power transfer by matching the source and load impedances applies only when you're stuck with a fixed source resistance and can only modify the load. Roy Lewallen, W7EL |
Efficiency and maximum power transfer
Roy Lewallen wrote:
But what happens if we reduce the source resistance to 50 ohms? Now the source delivers 66.7 watts, of which 22.2 is dissipated in the source resistance and 44.4 in the load resistance. The well known and often misapplied rule about maximizing power transfer by matching the source and load impedances applies only when you're stuck with a fixed source resistance and can only modify the load. What if we assume that we are stuck with the 50 ohm source impedance above? How do we get greater than 50% power transfer? -- 73, Cecil http://www.w5dxp.com |
Efficiency and maximum power transfer
"Roy Lewallen" wrote in message news:JrCdnR01Yp20tNHVnZ2dnUVZ_sednZ2d@easystreeton line... There's a common misconception that, for a linear circuit, the maximum efficiency and/or power available from a voltage source occurs when the source resistance equals the load resistance (or, more generally, when they're complex conjugates). But this isn't universally true, as I'll show with a simple example. Suppose we have a 100 volt perfect voltage source in series with a variable source resistance, and a fixed load resistance of 100 ohms. If we make the source resistance 100 ohms, the source delivers 50 watts, 25 of which are dissipated in the source resistance and 25 watts in the load. The efficiency, if you consider the source resistance dissipation as wasted, is 50%. But what happens if we reduce the source resistance to 50 ohms? Now the source delivers 66.7 watts, of which 22.2 is dissipated in the source resistance and 44.4 in the load resistance. The power to the load has increased, and the efficiency has increased from 50 to 66.7%. The efficiency and load power continue to increase as the source resistance is made smaller and smaller, reaching a maximum when the source resistance is zero. At that point, the source will deliver 100 watts, all of which is dissipated in the load, for an efficiency of 100%. The well known and often misapplied rule about maximizing power transfer by matching the source and load impedances applies only when you're stuck with a fixed source resistance and can only modify the load. Roy Lewallen, W7EL What seems to be overlooked here is that the source resistance at the output terminals of the pi-nework in Class B and C amplifiers is non-dissipative, which is the reason they can be loaded for delivering all available power for a given grid drive, and still have efficiencies greater than 50 percent. One of the myths circulated for years, and still prevelant, is that the reason for Class B and C amps to have efficiencies greater than 50 percent is that the load resistance must be greater than the source resistance. Tain't so. I've proved the above to be true with extensive measurements using laboratory grade instruments. Reports on those measurements are reported in Chapter 19 in Reflections 2, and additional measurements taken after Reflections 2 was published are reported in Chapter 19A, to be published soon in Reflections 3. This additional chapter is listed here in the rraa for your information. If anyone is interested in reading Chapter 19 in Reflections 2 it appears in my website at www.w2du.com. Walt, W2DU |
Efficiency and maximum power transfer
On Jun 7, 7:14Â*pm, Owen Duffy wrote:
wrote : On Jun 7, 12:43�am, Owen Duffy wrote: wrote innews:73353273-a079-499c-89df-c11975b37c78@z66g200 0hsc.googlegroups.com: ... The maximum power theorem gives conditions where power in the load, is equal to internal power in the generator. �Not always a good ide a. �A 50HZ generator capable of Megawatts of power would dissiapate 1/2 in the generator and 1/2 in our houses if they designed them to conform to the MPT. �The 50HZ generators would melt. �Utilities design their Generators to have nearly 0.0 ohms internal impedance. Actually, the AC power distribution system from alternator down has a manged substantial equivalent source impedance. The source impedance serves to limit fault currents, which reduces the demands on protection devices. Sure, the network is not operated under Jacobi MPT conditions, but neither does it have near zero source impedance. Owen Not really sure I agree. Â*A multi-megawatt 60HZ generator by necessity has near zero source impedance. Â*The ones I am familar with require forced air cooling on their output buses. Â*If you are pumping out Mega- watts, then any non -zero source impedance results in serious heat. I^2R. Gary N4AST Gary, you use the terms impedance and resistant as if they were equivalent. Alternators have a designed value of leakage reactance, and they also have resistance. The combination make the equivalent source impedance, and it is sufficient to limit fault current to something typically in the range of 20 to 50 times the rated output current. Transmission lines and transformers in the transmission and distribution networks are usually designed in the same way. It is not zero, and it is not purely resistive. Most supply authorities would not allow you to connect a capacitive load (a leading PF load), so another concept, conjugate matching (in the Jacobit MPT sense) is also not practiced. Understanding the electricity network does not really give an insight into a typical ham radio transmitter, they do not share the same design objectives. Owen- Hide quoted text - - Show quoted text - We are obviously talking about two different things, that as you say have little to do with a Ham Radio transmitter. Gary N4AST |
Efficiency and maximum power transfer
"Walter Maxwell" wrote
What seems to be overlooked here is that the source resistance at the output terminals of the pi-nework in Class B and C amplifiers is non-dissipative, which is the reason they can be loaded for delivering all available power for a given grid drive, and still have efficiencies greater than 50 percent. One of the myths circulated for years, and still prevelant, is that the reason for Class B and C amps to have efficiencies greater than 50 percent is that the load resistance must be greater than the source resistance. Tain't so. ____________ Walt - what is your thinking on the point that untuned, solid-state amplifiers also can have PA DC-to-RF power conversion efficiencies of 70% or more at the device level? In fact the solid-state, analog FM broadcast transmitters supplied by Harris Corporation and others need no tuning to produce their rated output power into a 1.3:1 SWR or less, anywhere in the FM broadcast band 88-108 MHz. Even the harmonic filter needs no changes, and maintains harmonics at -80 dBc or better. They are frequency agile, and can be reset from one carrier frequency to another, anywhere in the FM band with a transition time of a few seconds The overall AC input to r-f output efficiency of these transmitters exceeds 60% (includes the exciter, control system, IPA, and cabinet fans). RF |
Efficiency and maximum power transfer
Richard Fry wrote:
"Walter Maxwell" wrote What seems to be overlooked here is that the source resistance at the output terminals of the pi-nework in Class B and C amplifiers is non-dissipative, which is the reason they can be loaded for delivering all available power for a given grid drive, and still have efficiencies greater than 50 percent. One of the myths circulated for years, and still prevelant, is that the reason for Class B and C amps to have efficiencies greater than 50 percent is that the load resistance must be greater than the source resistance. Tain't so. ____________ Walt - what is your thinking on the point that untuned, solid-state amplifiers also can have PA DC-to-RF power conversion efficiencies of 70% or more at the device level? In fact the solid-state, analog FM broadcast transmitters supplied by Harris Corporation and others need no tuning to produce their rated output power into a 1.3:1 SWR or less, anywhere in the FM broadcast band 88-108 MHz. Even the harmonic filter needs no changes, and maintains harmonics at -80 dBc or better. They are frequency agile, and can be reset from one carrier frequency to another, anywhere in the FM band with a transition time of a few seconds The overall AC input to r-f output efficiency of these transmitters exceeds 60% (includes the exciter, control system, IPA, and cabinet fans). RF I know this question was directed to Walt, but I'd like to mention that I've designed and built solid state class C amplifiers at the 5 - 10 watt level which have measured efficiencies of greater than 85%. Roy Lewallen, W7EL |
Efficiency and maximum power transfer
"Richard Fry" wrote in message ... "Walter Maxwell" wrote What seems to be overlooked here is that the source resistance at the output terminals of the pi-nework in Class B and C amplifiers is non-dissipative, which is the reason they can be loaded for delivering all available power for a given grid drive, and still have efficiencies greater than 50 percent. One of the myths circulated for years, and still prevelant, is that the reason for Class B and C amps to have efficiencies greater than 50 percent is that the load resistance must be greater than the source resistance. Tain't so. ____________ Walt - what is your thinking on the point that untuned, solid-state amplifiers also can have PA DC-to-RF power conversion efficiencies of 70% or more at the device level? In fact the solid-state, analog FM broadcast transmitters supplied by Harris Corporation and others need no tuning to produce their rated output power into a 1.3:1 SWR or less, anywhere in the FM broadcast band 88-108 MHz. Even the harmonic filter needs no changes, and maintains harmonics at -80 dBc or better. They are frequency agile, and can be reset from one carrier frequency to another, anywhere in the FM band with a transition time of a few seconds The overall AC input to r-f output efficiency of these transmitters exceeds 60% (includes the exciter, control system, IPA, and cabinet fans). RF Hello Richard, Sorry, Richard, I have no knowledge of solid-state untuned amps, so my thinking on them is zero, nada. As you'll note, all of my discussion on the subject concerns only tube amps with a pi-network output, and I've specifically stated these conditions. If you've read Chapter 19 and its addition as Chapter 19A, do you agree with my position that the output resistance at the output terminals of the pi-network is non-dissipative? Walt, W2DU |
Efficiency and maximum power transfer
"Walter Maxwell" wrote
The source resistance appearing at the output of either a Class B or C amplifier is R = E/I, where E is the peak voltage at the output terminals and I is the peak current at the output. Or RMS values can also be used.Since E/I is simply a ratio, R is also a ratio. And we know that a ratio cannot dissipate power, or turn electrical energy into heat, thus the output resistance R is non-dissipative. I have made many measurements that prove this. It is also the reason why reflected power does not dissipate in the tubes, because it never reaches the tubes. The reflected power simply causes a mismatch to the source, causing the source to deliver less power than it would if there were no mismatch. __________ If the source resistance of a tuned r-f PA stage was truly non-dissipative, and the tx simply supplied less power into poor matches, how would that explain the catastrophic failures to the output circuit components often seen when high power transmitters operate without suitable SWR protection into highly mismatched loads? Another reality is that r-f power from two co-sited, tuned transmitters on two frequencies in the same band can be present in each others output stage due to antenna coupling, which causes r-f intermodulation between them. The non-linear (mixing) process occurs at the active PA stage. If reflected (reverse) r-f energy never reaches the PA stage as you assert, then how could this IM generation occur? RF |
Efficiency and maximum power transfer
"Richard Fry" wrote in message ... "Walter Maxwell" wrote The source resistance appearing at the output of either a Class B or C amplifier is R = E/I, where E is the peak voltage at the output terminals and I is the peak current at the output. Or RMS values can also be used.Since E/I is simply a ratio, R is also a ratio. And we know that a ratio cannot dissipate power, or turn electrical energy into heat, thus the output resistance R is non-dissipative. I have made many measurements that prove this. It is also the reason why reflected power does not dissipate in the tubes, because it never reaches the tubes. The reflected power simply causes a mismatch to the source, causing the source to deliver less power than it would if there were no mismatch. __________ If the source resistance of a tuned r-f PA stage was truly non-dissipative, and the tx simply supplied less power into poor matches, how would that explain the catastrophic failures to the output circuit components often seen when high power transmitters operate without suitable SWR protection into highly mismatched loads? You must be talking about solid-state tx when you mention SWR protection. I don't know of any tube tx that have such protection. Any catastrophic failures in tube tx with pi-network output circuits due to poor impedance match at the output without retuning to match the the network to the load simply leaves the tx detuned away from resonance. The result is excessive plate current that would be reduced to normal by resonating the tank circuit. No operator in his right mind would allow the tx to be operated with the tank not tuned for the resonant dip in plate current. Another reality is that r-f power from two co-sited, tuned transmitters on two frequencies in the same band can be present in each others output stage due to antenna coupling, which causes r-f intermodulation between them. The non-linear (mixing) process occurs at the active PA stage. If reflected (reverse) r-f energy never reaches the PA stage as you assert, then how could this IM generation occur? RF :You said two tx operating on two different frequencies, which means the RF signals from the two tx are not phase coherent. In this condition the signal from each tx does enter the other. On the other hand, the wave reflected from a mismatched termination is phase coherent with the source wave, resulting in the addition of the reflected wave to the source wave when either the antenna tuner or the pi-network is adjusted to deliver all available power at the desired grid drive. W2DU |
Efficiency and maximum power transfer
"Richard Fry" wrote in message ... "Walter Maxwell" wrote The source resistance appearing at the output of either a Class B or C amplifier is R = E/I, where E is the peak voltage at the output terminals and I is the peak current at the output. Or RMS values can also be used.Since E/I is simply a ratio, R is also a ratio. And we know that a ratio cannot dissipate power, or turn electrical energy into heat, thus the output resistance R is non-dissipative. I have made many measurements that prove this. It is also the reason why reflected power does not dissipate in the tubes, because it never reaches the tubes. The reflected power simply causes a mismatch to the source, causing the source to deliver less power than it would if there were no mismatch. __________ If the source resistance of a tuned r-f PA stage was truly non-dissipative, and the tx simply supplied less power into poor matches, how would that explain the catastrophic failures to the output circuit components often seen when high power transmitters operate without suitable SWR protection into highly mismatched loads? if the load impedance seen at the transmitter terminals is outside the range that it was designed for you end up with either arcing from excessive voltage or meltdown from high currents. remember, the pa output as long as it is connected to a linear load and you consider only the sinusoidal steady state condition can be completely replaced by a lumped load impedance, hence no reflections necessary to figure it out... same load, same result, hence the reflections have no effect on the internals of the pa. Another reality is that r-f power from two co-sited, tuned transmitters on two frequencies in the same band can be present in each others output stage due to antenna coupling, which causes r-f intermodulation between them. The non-linear (mixing) process occurs at the active PA stage. If reflected (reverse) r-f energy never reaches the PA stage as you assert, then how could this IM generation occur? non-coherent input would of course pass through the matching network going into a pa, there is nothing that precludes that. that current would cause mixing in the non-linear tube or transistor and therefore im generation. but that incoming rf is not reflected, it is the incident wave, with nothing coherent to interfer with why would it not pass into the pa? Now, have fun with this one... two transmitters on exactly the same frequency feeding a common load through equal lengths of coax... what happens if you change the phase relationship between them? make it simpler, remove the coax and connect both pa outputs togther with a single load, now no reflections to worry about... does it matter? do the two pa's feed all their power into each other? where do the reflections go?? |
Efficiency and maximum power transfer
Richard Fry wrote:
"If the source resistance of a tuned r-f PA stage was truly non-dissipative, and the tx simply supplied less power into poor matches, how would that explain the catastropic failures to the output circuit components often seen when high power transmitters operate without suitable SWR protection into highly mismatched loads?" The PA is a switch. Almost no voltage across it when it is closed and no current through it when it is open. Some of its impedance is dissipative and some is non-dissipative. A conjugate match to its total impedance is the way to deliver maximum power from the transmitter to its load. Alexander H. Wing wrote on page 43 of "Thansmission Lines, Antennas, and Wave Guides": "If a dissipationless network is insrted between a constant voltage generator of internal impedance Zg1 and a load ZR such that maximum power is delivered to the load, at every pair of terminals the impedances looking in opposite directions are conjugates of each other." An operating transmitter is normally adjusted for conjugate match with its load. Normal plate dissipation occurs when electrons strike the anode and there is little damage to the tube when the current and cooling are within limits. Let an arc strike across the transmission line and it may effectively become a short circuit which may impose an enormous mismatch in an instant to the transmitter. That`s why a d-c supply is often connected in series with a relay coil across the transmission line. The arc completes the d-c circuit energizing the relay which breaks the interlock circuit. The transmitter instantly is shut down until it is manually restarted. Tubes are often destroyed by internal arcs if overloads don`t act in time. Best regards, Richard Harrison, KB5WZI |
Efficiency and maximum power transfer
Richard Fry wrote:
"Walter Maxwell" wrote The source resistance appearing at the output of either a Class B or C amplifier is R = E/I, where E is the peak voltage at the output terminals and I is the peak current at the output. Or RMS values can also be used.Since E/I is simply a ratio, R is also a ratio. And we know that a ratio cannot dissipate power, or turn electrical energy into heat, thus the output resistance R is non-dissipative. I have made many measurements that prove this. It is also the reason why reflected power does not dissipate in the tubes, because it never reaches the tubes. The reflected power simply causes a mismatch to the source, causing the source to deliver less power than it would if there were no mismatch. __________ If the source resistance of a tuned r-f PA stage was truly non-dissipative, and the tx simply supplied less power into poor matches, how would that explain the catastrophic failures to the output circuit components often seen when high power transmitters operate without suitable SWR protection into highly mismatched loads? the output network might be non-dissipative, but that doesn't mean that the voltages and/or currents won't exceed the limitations of the components. Consider a gedanken.. you have an active device acting as a shunt DC regulator that puts out 1 amp at 1000 volts into a matching network, the other side of which is a 50 ohm load. The power supply is a 5000 volt power supply with a 2000 ohm dropping resistor, and 1 amp is also flowing through the active device. The 2 amps total flowing through the dropping resistor results in the 1000 volt output into the network. The implication is that the matching network, with the 50 ohm output, is presenting a 1000 ohm load. Now, say you change the load impedance on the network so that the impedance on the input side is zero ohms. Now, the voltage across the active device is zero volts, the voltage across that dropping resistor is 5000 volts, instead of 4000 volts, and it will get a bit hotter, and possibly fail. Or, say you change the load impedance so that the input of the network presents an infinite impedance. The same 1 amp is flowing through the active device (say it's a controlled current source), but, since there's no load current, the voltage drop across the resistor is now 2000 volts instead of 4000 volts, and the voltage across the active device is now 3000 volts instead of 1000. If the device can only take 2000 volts, it's cooked. |
Efficiency and maximum power transfer
(Richard Harrison) wrote in news:10691-484F0717-
: .... The PA is a switch. Almost no voltage across it when it is closed and no current through it when it is open. ... The OP's question related to Class B and Class C. Have you hopped on another tram? Your statement is definitely not true for Class B RF PAs. In fact it is difficult achieving theoretical Class B efficiency with tetrodes running on relatively low HT (eg 750V as in many transceivers). If what you state was true, you would expect conversion efficiencies in excess of 95%... and that just doesn't happen with practical electron devices. Practical Class C electron devices cannot achieve those efficiencies at maximum continuous output power. The anode voltage waveform in practical continuous Class C amplifiers with resonant loads is hardly the rectangular pulse waveform that you suggest. In fact, if you observed it with a CRO you would be hard put to argue that it wasn't an almost pure sine wave. Owen |
Efficiency and maximum power transfer
"Jim Lux" wrote in message ... Richard Fry wrote: "Walter Maxwell" wrote The source resistance appearing at the output of either a Class B or C amplifier is R = E/I, where E is the peak voltage at the output terminals and I is the peak current at the output. Or RMS values can also be used.Since E/I is simply a ratio, R is also a ratio. And we know that a ratio cannot dissipate power, or turn electrical energy into heat, thus the output resistance R is non-dissipative. I have made many measurements that prove this. It is also the reason why reflected power does not dissipate in the tubes, because it never reaches the tubes. The reflected power simply causes a mismatch to the source, causing the source to deliver less power than it would if there were no mismatch. __________ If the source resistance of a tuned r-f PA stage was truly non-dissipative, and the tx simply supplied less power into poor matches, how would that explain the catastrophic failures to the output circuit components often seen when high power transmitters operate without suitable SWR protection into highly mismatched loads? the output network might be non-dissipative, but that doesn't mean that the voltages and/or currents won't exceed the limitations of the components. I fail to understand how the limitations of the components have anything to do with the output resistanceof the network. Can you please explain how this is relevant? Consider a gedanken.. you have an active device acting as a shunt DC regulator that puts out 1 amp at 1000 volts into a matching network, the other side of which is a 50 ohm load. The power supply is a 5000 volt power supply with a 2000 ohm dropping resistor, and 1 amp is also flowing through the active device. The 2 amps total flowing through the dropping resistor results in the 1000 volt output into the network. The implication is that the matching network, with the 50 ohm output, is presenting a 1000 ohm load. Now, say you change the load impedance on the network so that the impedance on the input side is zero ohms. Now, the voltage across the active device is zero volts, the voltage across that dropping resistor is 5000 volts, instead of 4000 volts, and it will get a bit hotter, and possibly fail. Or, say you change the load impedance so that the input of the network presents an infinite impedance. The same 1 amp is flowing through the active device (say it's a controlled current source), but, since there's no load current, the voltage drop across the resistor is now 2000 volts instead of 4000 volts, and the voltage across the active device is now 3000 volts instead of 1000. If the device can only take 2000 volts, it's cooked. Sorry Jim, I'm not familiar with a 'gedanken'. Is it distant cousin of a Class B or C amplifier? Your discussion above doesn't appear to have any relation whatsoever to those types of RF amplifiers. Are you using your discussion in an attempt to prove that the output resistance of those types of amps is not non-dissipative? Or what are you trying to prove? I don't get it. Walt, W2DU |
Efficiency and maximum power transfer
Owen Duffy wrote:
"Have you hopped on another tram?" Maybe. I agree. Radio power amplifiers have clean sinusoidal outputs, otherwise they would generate unacceptable harmonics. All except Class A amplifiers which have continuous plate current flows and a maximum efficiency of 50%, have plate currents which flow in pulses. In Class B, the plate current in individual tubes flows in pulses of approximately a half cycle. Actual efficiency is about 60%. In Class C, the plate current pulses last less than a half cycle. Practical efficiencies are in the range of 60 to 80 per cent. Best regards, Richard Harrison, KB5WZI |
Efficiency and maximum power transfer
Walter Maxwell wrote:
"Jim Lux" wrote in message ... Richard Fry wrote: "Walter Maxwell" wrote The source resistance appearing at the output of either a Class B or C amplifier is R = E/I, where E is the peak voltage at the output terminals and I is the peak current at the output. Or RMS values can also be used.Since E/I is simply a ratio, R is also a ratio. And we know that a ratio cannot dissipate power, or turn electrical energy into heat, thus the output resistance R is non-dissipative. I have made many measurements that prove this. It is also the reason why reflected power does not dissipate in the tubes, because it never reaches the tubes. The reflected power simply causes a mismatch to the source, causing the source to deliver less power than it would if there were no mismatch. __________ If the source resistance of a tuned r-f PA stage was truly non-dissipative, and the tx simply supplied less power into poor matches, how would that explain the catastrophic failures to the output circuit components often seen when high power transmitters operate without suitable SWR protection into highly mismatched loads? the output network might be non-dissipative, but that doesn't mean that the voltages and/or currents won't exceed the limitations of the components. I fail to understand how the limitations of the components have anything to do with the output resistanceof the network. Can you please explain how this is relevant? Consider a gedanken.. you have an active device acting as a shunt DC regulator that puts out 1 amp at 1000 volts into a matching network, the other side of which is a 50 ohm load. The power supply is a 5000 volt power supply with a 2000 ohm dropping resistor, and 1 amp is also flowing through the active device. The 2 amps total flowing through the dropping resistor results in the 1000 volt output into the network. The implication is that the matching network, with the 50 ohm output, is presenting a 1000 ohm load. Now, say you change the load impedance on the network so that the impedance on the input side is zero ohms. Now, the voltage across the active device is zero volts, the voltage across that dropping resistor is 5000 volts, instead of 4000 volts, and it will get a bit hotter, and possibly fail. Or, say you change the load impedance so that the input of the network presents an infinite impedance. The same 1 amp is flowing through the active device (say it's a controlled current source), but, since there's no load current, the voltage drop across the resistor is now 2000 volts instead of 4000 volts, and the voltage across the active device is now 3000 volts instead of 1000. If the device can only take 2000 volts, it's cooked. Sorry Jim, I'm not familiar with a 'gedanken'. Is it distant cousin of a Class B or C amplifier? Your discussion above doesn't appear to have any relation whatsoever to those types of RF amplifiers. Are you using your discussion in an attempt to prove that the output resistance of those types of amps is not non-dissipative? Or what are you trying to prove? I don't get it. Walt, W2DU I was attempting to discuss the comment above: "how would that explain the catastrophic failures to the output circuit components often seen when high power transmitters operate without suitable SWR protection into highly mismatched loads?" "gedanken" refers to an intellectual exercise describing an experiment with (usually) contrived or idealized circumstances so that the underlying concepts can be understood. Classics are Schroedinger's cat, Maxwell's demon, etc. http://en.wikipedia.org/wiki/Thought_experiment for more details and history than anyone could want. |
Efficiency and maximum power transfer
Richard Fry wrote:
If the source resistance of a tuned r-f PA stage was truly non-dissipative, and the tx simply supplied less power into poor matches, how would that explain the catastrophic failures to the output circuit components often seen when high power transmitters operate without suitable SWR protection into highly mismatched loads? This question has been asked many times (several times by you, as I recall) and answered (several times at least by me) on this forum. The answer is that when a transmitter is terminated in an impedance too far from the one it's designed to see, voltages and/or currents in the transmitter and tank circuit rise to unacceptably high values. This causes the damage or destruction. I don't think this is difficult to understand. So the reason for the need to keep asking is that it's apparently not being believed, presumably because of the deeply compelling need to see waves of "reflected power" entering the transmitter to wreak havoc. But the simple answer is true. This is very much like trying to convince an astrology believer that he stubbed his toe because he didn't watch where he was going rather than because Mars was lined up with Jupiter. Another reality is that r-f power from two co-sited, tuned transmitters on two frequencies in the same band can be present in each others output stage due to antenna coupling, which causes r-f intermodulation between them. The non-linear (mixing) process occurs at the active PA stage. If reflected (reverse) r-f energy never reaches the PA stage as you assert, then how could this IM generation occur? Power from one antenna is coupled to the other antenna by a process known as "mutual coupling". This power enters the other transmitter via normal transmission along the transmission line, where it reaches the final stage for mixing. No mysterious "reverse r-f energy" is needed to explain this well-known and well-understood phenomenon. Roy Lewallen, W7EL |
Efficiency and maximum power transfer
Roy Lewallen wrote:
This question has been asked many times (several times by you, as I recall) and answered (several times at least by me) on this forum. The answer is that when a transmitter is terminated in an impedance too far from the one it's designed to see, voltages and/or currents in the transmitter and tank circuit rise to unacceptably high values. This causes the damage or destruction. But the offending impedance equals (Vfor+Vref)/(Ifor+Iref), a virtual impedance caused by the superposition of the reflected wave and the forward wave. I don't think this is difficult to understand. What is difficult to understand is the denial of the role that the energy content of the reflected wave plays in causing the offending impedance. If it were not for the reflected waves, that offending impedance would never happen. -- 73, Cecil http://www.w5dxp.com |
Efficiency and maximum power transfer
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Efficiency and maximum power transfer
I wrote:
"In Class C, the plate current pulses last less than a half cycle. Practical efficiencies are in the range of 60 to 80 per cent." Owen Duffy wrote: "Again, not rectangular pulses, not nearly." A pulse does not need to be rectangular. According to my electronics dictionary: "Pulse - 1. The variation of a quantity having a normally constant value. This variation is characterized by a rise and decay of finite duration. 2. An abrupt change in voltage, either positive or negative, which conveys information to a circuit. (See also Impulse.)" A rectified sine wave could properly be called a string of pulses being constantly off between pulses then rising and falling between pulses for a short finite duration. Best regards, Richard Harrison, KB5WZI |
Efficiency and maximum power transfer
This illustrates, I hope, only one, small point:
Owen says: For most practical valves in continous mode, they cannot develop their rated maximum power at very small conduction angles without exceeding rated cathode current, so there is often little benefit in operating at conduction angle much below 120 degrees. Recall that valve/tube diodes in rectifier service were not used with capacitive filters because the rated peak to average current ratio was something like 1.5:1. Modern, solid-state diodes can have ratios of 40:1 and can be happy with capacitive filters in rectifier service. Peak current limits of high-voltage, high-power tubes is a real physical limit. The cathode can only emit so much. 73, Mac N8TT -- J. McLaughlin; Michigan, USA Home: |
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