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W2DU's Reflections III is now available from CQCommunications,...
On May 25, 12:32*am, (Richard Harrison)
wrote: Cecil Moore, W5DXP wrote: "Seems to me, the highest efficiency would be achieved by a source with zero source impedance." Me too, but zero source impedance does not match the load as required for maximum power transfer. It seems to me that much too much is made of 'maximum power transfer' in the RF world. In the world of 50 and 60 Hz, where significantly more energy is moved, 'maximum power transfer' is never mentioned. Efficiency is much more of interest. For the most part, 'maximum power transfer' is just an interesting ideosyncracy of linear circuit theory. ....Keith |
#2
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W2DU's Reflections III is now available from CQCommunication...
Keith Dysart wrote:
"For the most part, "maximum power transfer is just an interesting ideosyncracy of linear circuit theory." In the world of 50 and 60 Hz, we don`t want all the power plant can supply when we flip on a light switch. The RF world is usually different. Maximum power transfer only occurs when source and load match conjugately, and the match proves the load and source impedances are equals. It is well known and easily shown that a match results in maximum power transfer. If one has a 100-watt transmitter he probably wants 100 watts out of it sometimes and may only be able to do so when his antenna is matched to his transmitter, Maxumum power treansfer is more than an "interesting ideosyncracy". Best regards, Richard Harrison, KB5WZI |
#3
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W2DU's Reflections III is now available from CQ Communication...
Richard Harrison wrote:
Keith Dysart wrote: "For the most part, "maximum power transfer is just an interesting ideosyncracy of linear circuit theory." In the world of 50 and 60 Hz, we don`t want all the power plant can supply when we flip on a light switch. The RF world is usually different. Maximum power transfer only occurs when source and load match conjugately, and the match proves the load and source impedances are equals. It is well known and easily shown that a match results in maximum power transfer. . . . It's also easily shown that it doesn't. Consider a 10 volt voltage source having a 50 ohm source resistance, feeding a 50 ohm resistive load. Power at the load is 0.5 watt, is it not? Reduce the source impedance to 10 ohms. Now what is the power dissipated in the load? Is it greater or less than it was when the source and load impedances were matched? Roy Lewallen, W7EL |
#4
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W2DU's Reflections III is now available from CQ Communication...
On May 26, 6:20*pm, Roy Lewallen wrote:
Richard Harrison wrote: Keith Dysart wrote: "For the most part, "maximum power transfer is just an interesting ideosyncracy of linear circuit theory." In the world of 50 and 60 Hz, we don`t want all the power plant can supply when we flip on a light switch. The RF world is usually different. Maximum power transfer only occurs when source and load match conjugately, and the match proves the load and source impedances are equals. It is well known and easily shown that a match results in maximum power transfer. . . . It's also easily shown that it doesn't. Consider a 10 volt voltage source having a 50 ohm source resistance, feeding a 50 ohm resistive load. Power at the load is 0.5 watt, is it not? Reduce the source impedance to 10 ohms. Now what is the power dissipated in the load? Is it greater or less than it was when the source and load impedances were matched? Roy Lewallen, W7EL But Roy, consider that the source resistance remains constant at 10 ohms. Then what load resistance will absorb the most power? The answer is 10 ohms. Any value of load resistance greater or less than 10 ohms will result in less power delivered. I don't believe it's fair to change the source resistance when dealing with the Maximum Power Transfer Theorem. In your example with a source resistance of 10 ohms and a load resistance of 50 ohms the power delivered will be 1.39 watts. But when the load resistance is 10 ohms with the same source resistance the power delivered is 2.5 watts. As I said above, if the load resistance is either greater or less than 10 ohms the power delivered will be less than 2.5 watts. Thus when the source resistance is constant the maximum power will be delivered when the load is matched to the source. Nes pa? Walt |
#5
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W2DU's Reflections III is now available from CQ Communication...
walt wrote:
But Roy, consider that the source resistance remains constant at 10 ohms. Then what load resistance will absorb the most power? The answer is 10 ohms. Any value of load resistance greater or less than 10 ohms will result in less power delivered. I don't believe it's fair to change the source resistance when dealing with the Maximum Power Transfer Theorem. In your example with a source resistance of 10 ohms and a load resistance of 50 ohms the power delivered will be 1.39 watts. But when the load resistance is 10 ohms with the same source resistance the power delivered is 2.5 watts. As I said above, if the load resistance is either greater or less than 10 ohms the power delivered will be less than 2.5 watts. Thus when the source resistance is constant the maximum power will be delivered when the load is matched to the source. Nes pa? Walt Of course, I know that, and I would hope anyone with even very basic electrical circuit analysis knowledge does. And anyone with that knowledge should state as you have, "FOR A GIVEN SOURCE IMPEDANCE, maximum power transfer occurs when the source and load impedances are matched (i.e., the load impedance is the complex conjugate of the source impedance)," which is true. But the statement which was made was that "Maximum power transfer occurs when the source and load impedances are matched." This is NOT true, as the example demonstrates. It's an important distinction. Instead of declaring what's "fair" and what isn't with regard to changing source and load impedances, the maximum power transfer theorem should be stated correctly, in a way which makes it true. Roy Lewallen, W7EL |
#6
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W2DU's Reflections III is now available from CQCommunication...
Roy Lewallen, W7EL wrote:
"Nut the statement which was made was that "Maximum power transfer occurs when the source and load impedances are matched." Inelegant and bereft of qualifications as it may be, I am the author of that statement, and I stand by it. Roy says: "This is not true, as the example demonstrates." Robert L. Schrader says on page 43 of "Electronic Communication": "To produce maximum power in any load, it is necessary that the load resistance equal the internal resistance of the source." I believe our statements are equivalent. Load a battery of a given internal resistance with the same resistance as a load, and for a short time the battery will deliver more power to that load than to any other resistance larger or smaller. Best regards, Richard Harrison. KB5WZI |
#7
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W2DU's Reflections III is now available from CQCommunication...
Roy Lewallen wrote:
"Reduce the source imperance to 10 ohms." FOUL! In the case of a 10-ohm internal source, the load which extracts maximum power is 10 ohms. Best regards, Richard Harrison, KB5WZI |
#8
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W2DU's Reflections III is now available from CQ Communication...
I stand by my position that when the source is an RF power amplifier,
and when all the available power is transferred from the source to the load, the source impedance is the conjugate of the load impedance. In a similar instance, if the load is a pure resistance, the source resistance equals the load resistance. And referring to a statement Dysart made concerning plate resistance, Rp, it must be understood that in Class AB, B and C amplifiers, Rp is NOT the source resistance. In those amplifiers the effect of Rp is a negative feedback that reduces the effect of plate-current change resulting from a change in grid voltage, thus reducing the power output compared to what the output would be if Rp were absent. Compensation for the power lost to Rp is accomplished by simply increasing the grid drive. Consequently, Rp plays no part in achieving a conjugate match at the junction of the network-output and the load. Although lossless elements are required to achieve a perfect conjugate match in both directions, a perfect conjugate match is obtained in the forward direction with real elements when all available power is being delivered to the load. This condition is verified in data presented in Chapter 24 of Reflections 3. Walt, W2DU |
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