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#11
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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 |
#12
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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 |
#13
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Efficiency and maximum power transfer
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#14
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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 |
#15
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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 |
#16
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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 |
#17
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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 |
#18
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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. |
#19
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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 |
#20
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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 |
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