Talk:Impedance matching

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MAXIMUM powerEdit

MAXIMUM power transfer is using voltage bridging. But only if you are able to choose a LOW output impedance. You will have to use some kind of fuse for NOT transferring maximum power and overload the source. If you have to use a significant output impedance, then impedance bridging is the solution. You will loose half the power or even more (not precise impedance matching).

--AK45500 (talk) 14:41, 12 July 2018 (UTC)

Not sure what your point is. For low power signals, such a Ethernet, you do not need a fuse for protection. For the AC Mains, you never load it for maximum power transfer; it would melt the wires and probably the load. Constant314 (talk) 14:55, 12 July 2018 (UTC)
Ha ha, it would melt the national grid as well if you actually succeeded. SpinningSpark 16:18, 12 July 2018 (UTC)
My point is: Real MAXIMUM power transfer occurs with voltage bridging (see: AC mains). But you need some kind of protection (fuse). Ethernet has a significant source impedance (and save against overload). You will use impedance matching to minimize reflections. Maximum Power transfer is not important. Impedance Matching can only transfer 1/4 of the power. (1/2 due to attenuation, 1/2 because you will draw double current using Impedance bridging compared to impedance matching at same load impedance).
My point is: Some people think MAXIMUM power transfer is used (or has to be used) for AC Mains. They like not just to melt their ice cream. This article should state the misunderstanding explicitly. And people think whenever relexions are optimized, and perfect impulse transmission is needed it uses Maximum Power Transfer. i.e. AES3 (digital AUDIO) was optimized for perfect impulse retention and group delay. There is no need for energy transfer (no level problems).
--AK45500 (talk) 09:57, 11 September 2018 (UTC)
Still not understanding your point. Maximum power transfer is not very relevant to AC mains. The AC mains article you refer us to does not even mention "characteristic impedance", "voltage bridging", or "impedance bridging". I am not aware of your claimed common misunderstanding. Rather than continuing to try to explain your perceived problem with the article, perhaps you could link to a source that discusses the issue? Without a reliable source, nothing can go in the article anyway. SpinningSpark 14:35, 11 September 2018 (UTC)
In the old days (perhaps the days of Edison and DC), the concept of maximum power transfer misled some engineers, before they came to understand that mains should be voltage sources. This is in books, and may be what he's referring to. I'll try to find it. Dicklyon (talk) 14:39, 11 September 2018 (UTC)
The problem with Edison's DC system was that it was of necessity low-voltage. That made it highly lossy and unsuitable for distribution over long distances. Attempting to achieve max power transfer would only have made its problems worse. It would be interesting to hear if Edison or his contemporaries actually thought that was the way to go. SpinningSpark 17:27, 11 September 2018 (UTC)
Maximum power in this article is not about getting the maximum power out of a generator or distributing maximum power over the grid. It is about what you can do with the load impedance to get the maximum power in the load given that everything else is fixed. It is not about the maximum sustained power you can get without burning anything up or down. It is about the theoretical maximum regardless of the consequences.Constant314 (talk) 17:55, 11 September 2018 (UTC)
@Constant314 ; Thank you for your statement. This (explicitly) is what I am thinking of. Some People are misinterpreting the MAXIMUM. It is just the maximum power in the load given that everything else is fixed. IF I am able to change source impedance the minimum I am able to choose is the better maximum. This is back to my very first sentence. Just a statement to handle these different maximums. The better maximum is 4 times the power maximum of the maximum power transfer (same load).
--AK45500 (talk) 11:04, 12 September 2018 (UTC)
@AK45500 - careful of the assumptions you are also making. Your comment about lowering the source impedance to increase power transfer into the load is true at low frequencies (essentially at baseband/DC). At higher frequencies (eg RF and microwave) what you are suggesting will result in an impedance mismatch between the source and the load that will create reflections over the transmission line that connects the source and the load, which will actually lower the power transferred to the load. At worst case, it will cause a communication line to become unusable. — Preceding unsigned comment added by 2A00:23C8:293:5001:59DF:B7A7:1AAE:AC9F (talk) 15:31, 12 August 2020 (UTC)

L-section diagram issuesEdit

The description for the first diagram in this section discusses the use of complex impedances - inductors/capacitors, but the American resistor symbol is used in these instances. I propose that these should be changed for boxes, as in the diagram below it. Benjamin J. Crawford (talk) 21:56, 6 November 2019 (UTC)

Complex conjugate match for maximum power transferEdit

If someone wants to flesh out the conjugate case, it is easy to understand in the narrow band situation. When operating at a narrow band frequency, any impedance can be modeled as a resister in series with a reactance. Conjugate matching simply means that you set the reactance that you control to be equal magnitude and opposite sign of the fixed reactance that you are matching. Effectively, you are making the fixed reactance go away by resonating it out of the circuit. What you have left is a resistor that you optimize the same way you optimize when the impedances are purely resistive. For example, suppose the characteristic impedance of a coaxial cable is 50 + 2j ohms at 100 MHz. That is equivalent to a 50 ohm resister in series with a 0.8 nF capacitor. The conjugate load would be 50 ohms in series with a 3.2 nH inductor. Constant314 (talk) 14:02, 18 April 2020 (UTC)

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