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L Impedance matching network problem

  1. Jul 6, 2015 #1

    goodphy

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    Hello.

    The attached image is L impedance matching network, use for impedance matching between source and load impedance.

    The calculation shows that in this circuit, load impedance Rp sees whole impedance (equivalent impedance) of the source impedance and reactive components as matched to itself and the source impedance also see matched equivalent impedance for other impedance components.

    Thus I have a conclusion:

    1. There is no reflection between source impedance and rest of the components because of matching.
    2. There is also no reflection between load impedance and rest of the components because of matching.
    3. Power is fully transfered to the load from source.

    Could you tell me my interpretation is right? Actually I've wonder if there is local standing wave between start point of the reactive components and the load...
     

    Attached Files:

  2. jcsd
  3. Jul 6, 2015 #2

    tech99

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    You do not give numbers, so you need to check that Rs and Rp are the correct way round. The resistor across the shunt reactance needs to be expressed as a parallel resistance, and that in series with the series reactive element needs to be expressed as a series resistance. The parallel one must be the greater. (Any reactances in source or load need to be "added" to the network).
    Your conclusions look basically correct. In Item 3, there is always some loss in the L-network, mainly arising from the losses in the inductor. Regarding standing waves, this is basically correct, although strictly speaking the network is a lumped element network rather than a transmission line network. The network stores reactive energy in the same way as do standing waves on a transmission line. The network could perhaps be viewed as a lumped element line.
     
  4. Jul 6, 2015 #3

    Baluncore

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    @ goodphy. You are generally correct, but making an assumption that your signal is a continuous pure sinewave without any signal bandwidth.
    For wider band signals you should use several stages of L network with intermediate impedances being the geometric mean of the end impedances. By staggering the resonant frequencies of elements you can then correct gain and phase shift over the band.
    There is a circulating current between the L and C that could be seen as the equivalent of a standing wave on an equivalent transmission line. But one L and one C does not a transmission line make. It is better to consider the L network as a flywheel.
     
  5. Jul 6, 2015 #4

    goodphy

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    Oh I thanks. I really appreciate you!

    The last question is somewhat involved to what happens on the extent over impedance matching is done on the circuit. For example, it is known that quart-wave transformer is used for impedance matching between transmission line and the load. This is nothing but additional λ/4 length of small transmission line of certain characteristic impedance determined by impedances of both original transmission line and load. Anyway, the textbook shows graphic that there is standing wave only over this transformer. How is it possible? The presence of standing wave there means there is reflected wave from the load and that wave is confined only over the transformer?

    Do you have any idea of this?
     
  6. Jul 6, 2015 #5

    goodphy

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    Oh..thanks! And...do you mean that there is actually power loss (not energy) mechanism involving energy storing and exchange between L and C although no reflection occurs at both ends of these matching network?
     
  7. Jul 6, 2015 #6

    Baluncore

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    For a forward travelling wave there is a partial reflection from both ends of the L network, but they cancel on the input line because the step size from the input line to the network and from the network to the output line impedance is carefully controlled. The same is true for the reverse wave.
    When a 1/4 wave TL transformer is used, it will have an impedance that is the geometric mean of the terminal impedances. ZGM = √(Z1 * Z2)
     
  8. Jul 6, 2015 #7

    goodphy

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    Thanks!
    But unfortunately I need little bit more assistance. Could you tell me more about "they cancel on the input line because the step size from the input line to the network and from the network to the output line impedance is carefully controlled."? What does input and output line and most importantly, step size mean here? It is very nice if you specify them in a detailed way.
     
  9. Jul 7, 2015 #8

    Baluncore

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    Think of Rs as being a transmission line with impedance Rs. Think of Rp as being a transmission line with impedance Rp. A 1/4 wave line to match those two lines would have an impedance of Rq = √(Rs * Rp).
    The step impedance mismatch from Rs to Rq would be the same ratio as that from Rq to Rp. Rq/Rs = Rp/Rq.
    A forward wave entering Rq from Rs will have a partial reflection at the mismatch. 1/4 wave period later the forward wave will be again partially reflected passing from Rq to Rp. That second reflection will travel back to the Rq Rp junction where it will be 1/2 wavelength late and so the two reflections will have opposite signs and so cancel.
     
  10. Jul 7, 2015 #9

    goodphy

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    Oh...Thanks! Thus eventually local standing wave is possible as 1st reflected wave is canceled out with the 2nd reflected wave which is 1/2 wavelength delayed! I'm still wandered how not only phase (out-of-phase) but amplitude of both reflected waves are matched such that they are exactly canceled out.
     
  11. Jul 7, 2015 #10

    Baluncore

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