Maximum power transfer theorem with counterpoise.

Discussion in 'General Technical Questions and Answers' started by AA7EJ, May 20, 2019.

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  1. AA7EJ

    AA7EJ Ham Member QRZ Page

    [​IMG]


    The attached picture demonstrating maximum power transfer from source to load is widely accepted.
    No (fool) questions it.

    But insert ONE wire of IDEAL transmission line between Rs and Rl and SECOND wire of SAME ideal transmission line between Rl and "source V" things get "complicated".

    Things get totally out of control when Rl is changed from "72 Ohms impedance" to "5000 Ohms impedance " .

    The circuit is still SAME , same IDEAL (real TL is immaterial to demonstrate the issue ) transmission line is used , only Rl (load) value have changed.

    Assigning Rl as "5000 Ohms impedance" somehow made the "transmission line" DISCONNECTED / open from the load. Thus requiring "counterpoise" to complete the circuit.

    I am from Missouri , the "show me state" ( no offense to Missourians met ) , - show me the "counterpoise " in the circuit above.

    73 Shirley
     
  2. W5DXP

    W5DXP Ham Member QRZ Page

    You know, of course, that the lumped-element circuit model is invalid for distributed networks (e.g. antennas and transmission lines). Here's some information about what is wrong with the lumped-circuit model:

    http://www.teslaradio.com/pages/tesla_coils.htm
     
  3. SM0XHJ

    SM0XHJ Ham Member QRZ Page

    That is something that I have never understood. In my mind it is perfectly possible to model transmission lines (and antennas) as lumped circuits. Yes, the circuit get more complicated, and an infinite number of R, L and C is required to make the model perfect. The transmission line (or antenna) need to broken up in small sections and each section modelled as one set of R, L and C. The more sections you break it up in, the more accurate the model will be.

    That being said, the model above of an end fed antenna without any form of counterpoise/ground is still flawed. A better model would be:
    Screenshot_20190520_173520.png
     
  4. AA7EJ

    AA7EJ Ham Member QRZ Page

    Same as before - show me how the added LOAD components of combined value > 72 Ohms DISCONNECT from transmission line AS A LOAD.

    The original circuit said NOTHING about load being an antenna.

    Don't you ever heard off - "read the question - but do not read anything into it?
    Of course I'll not mention anything about "equivalent circuit" .

    Just for entertainment - the NEW circuit does not have ANY ground references.
    AND I do not have a time to waste to remove the "antenna " text form the picture.
     

    Attached Files:

  5. SM0XHJ

    SM0XHJ Ham Member QRZ Page

    Well, the only case where a "counterpoise" would make sense to "complete a circuit" is when talking about antennas.

    No problem, no ground reference needed to transfer energy from source to load. And no "counterpoise" needed, unless one part of the two conductor transmission line is missing.
     
  6. VU2NAN

    VU2NAN Ham Member QRZ Page

    It should not be forgotten that maximum power transfer occurs in an AC circuit only when the load impedance is a complex conjugate of the source impedance!

    73,

    Nandu.
     
    W1BR and WB2UAQ like this.
  7. WB2UAQ

    WB2UAQ Ham Member QRZ Page

    When you "insert a transmission line" does the shield or second conductor in the case of twinlead, connect to the bottom of the ideal voltage source? If the voltage source and the source resistance define a generator, how can you just insert a transmission line in between the ideal voltage source and the source resistance (defining a Thevenin equiv source)? .I guess you could but this transmission line would have to be part of the source impedance Guess I am trying to understand why you are inserting transmission lines in these locations. What is the goal here?
     
  8. W5DXP

    W5DXP Ham Member QRZ Page

    Maybe should say: maximum power transfer occurs in an AC circuit only when the load impedance seen by the source at the source is a complex conjugate of the source impedance! That to accommodate the impedance transformation by unmatched transmission lines.

    Current travels faster than the speed of light in a lumped circuit wire. That is never true in a distributed network.
     
    VU2NAN likes this.
  9. AA7EJ

    AA7EJ Ham Member QRZ Page

    Kindly allow me to repeat / rephrase the statement in question:

    When the load impedance is > 72 Ohms the "load" gets disconnected from the transmission line.

    The statement implies
    Impedance - AC circuit.

    Need for "complex conjugate " is NOT the question of the statement - the DISCONNECT aka circuit is no loner completed / closed is the QUESTION!

    Implies
    does not matter if DC or AC - just " circuit is open " is the QUESTION!
     
  10. AA7EJ

    AA7EJ Ham Member QRZ Page

    Speed of any kind is NOT the question. See my previous post.
     

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