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How does an equivalent impedance consume a real power?

  1. Jan 29, 2017 #1

    goodphy

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    Hello.

    Please looks at the attached image first.

    Equivalent impedance in LCC network.jpg

    The left image is LCC impedance matching network (one of T-type impedance matching networks) and the right is the corresponding equivalent circuit when CT and CL are adjusted such that an equivalent impedance of the sub-circuit surrounded by dashed-line becomes RS.

    You don't need to take care of what ZL really is, that was not important for my question I would like to ask you here. In the left circuit, there is only one resistance, RL, located in ZL so that's the only point that a real power (active power) is consumed. But in the right, Zeq itself is a purely resistive. In order to make sense in terms of power consumption, a real power at Zeq, I2Zeq = I2RS is same to I12RL, I think. But..RL can be a lower than RS and I1 is lower than I. As a result, two consumed powers are not equal.

    How can I solve this apparent contradiction? I think reactive powers for reactive components have some role to match these two power. For example, reactive components receive power over a quarter cycle and release it to the RL for another quarter cycle, not to the source. But I don't know how such an energy flow occurs.

    Could you please help to clarify this?
     
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  3. Jan 29, 2017 #2

    sophiecentaur

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    That diagram may be more complicated than necessary to resolve this dilemma.
    You can think in terms of a simple, ideal transformer., which avoids the need to do complex calculations and the operative word here is 'transformation'. The load RL is presented with a voltage V2 (the secondary volts). The secondary current will be V2/ RL. and the power delivered will be (V2)2/RL. The power that flows into the transformer primary will be V1I1, which must be the same as the power dissipated in the load RL. So the resistance presented to the source is V1/I1, which will be RL T2 where T is the turns ratio.
    The voltage across the load is not the same as the voltage across the primary winding and, if you want to impedance match, the source resistance Rs will not be the same as RL
    The different matching network you use in your diagram will still behave in the same way, with RL being presented as a different value to the source. It's what you would expect. If you are bothered about the "Energy Flow" then you can consider the energy that is not dissipated in the load as being stored within the reactive components. There is stored energy but no Power dissipation. (Perhaps that last sentence may be enough explanation)
     
  4. Jan 29, 2017 #3

    BvU

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    You can solve this by not jumping to conclusions as you appear to do. Check the proper expressions including the phases. For complex numbers the > and < aren't directly applicable or useful if you forget about the phases.
     
  5. Jan 29, 2017 #4

    goodphy

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    Hello. Thanks for giving a quick answer:)

    I'm sorry but I'm afraid I couldn't get your point yet. I know how the transformer works and it seems that you're equating any complex network with a load ZL = RL + iXL to some ideal transformer with a load RL in its secondary side. I don't know how this correspondence is made.

    You use an expression of power as voltage squared divided by an impedance while I use it as current squared multiplied by an impedance, they are both equal. With sticking on my power expression, I have to say these: the power consumed by the sub-circuit surrounded by the dashed-line in left diagram must be P1 = I2RS , according to the right equivalent diagram and it is an active power as the load RS is purely resistive (the active power is the power consumed by a resistance, not reactance). Obviously, the active power of P1 must be consumed by the sub-circuit somehow. I think P1 should be consumed by RL in the left diagram as it is the only resistance in the circuit, so I12RL = I2RS where I1 is the current going through RL. If RL is lower than RS , then I1 is higher than I. However..how this is possible? I is split into I1 and I2 so I1 is lower than I.

    I'm sorry not to understand your comment, but I need to get a little more help to follow your sense. What did I miss?
     
  6. Jan 29, 2017 #5

    goodphy

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    Hello.

    I'm sorry but phase? Do you want me to write phase of ..impedance? I'm talking about general case so a specific phase is not relevant. And what do you mean > and < ?
     
  7. Jan 29, 2017 #6

    sophiecentaur

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    I think that statement should read "I am familiar with what a basic ideal transformer does". A transformer that you would recognise as such is only one of many ways to achieve its function of Impedance Transformation.
    A transformer can be regarded as a set of impedances, connected in such a way as to change volts. current and resistances. Yes, there is Mutual Impedance involved in a common or garden transformer but, only when the transformer is 'ideal' (where the various capacitances and self inductances play a minor part in its operation) can we assume it's the only element involved. Then we can make do with the simple 'transformer equations' that we start off with. My example of a familiar thing like a transformer was one which you (I mean you and not "one") felt you 'understand'. But many other arrangements of impedances can produce the same effect as a transformer with an iron core and two windings. In all of them, there is EM energy stored, temporarily, in reactances and that energy can influence the power being dissipated in the resistive elements. That's the philosophy behind reactive circuits and how they can change the apparent resistance values.
    It's a pain in the butt to do the full analysis of such circuits but you could choose appropriate reactances such that the V and I referred to the point where the circuit emerges through that box such that V/I = RS. You could actually produce any resistance you wanted.
    Is this the first complex circuit that you have considered? If so, it would be better to start a bit further back in the learning process and do some easier examples.
    If you plot X and R on an Argand diagram, the angle between them and the resultant Z can be referred to as Phase.
    A "specific phase" is relevant when it's referred to the input signal.
    I think you need to go back to basics for this.
     
  8. Jan 29, 2017 #7

    jim hardy

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    maybe i misunderstand the question... but here's how i would answer the question as i understand it..

    Of course RL can be lower than RS and the two consumed(dissipated?) powers therefore unequal. But then you have mismatched impedances, and the only downside to that is you have not optimal transfer of power.

    Observe that with matched impedance your efficiency is 50% because half the power gets dissipated inside the source.
    I assure you that's unacceptable for a multimegawatt central station generator. We want to sell ALL the power not just half of it.
    But it's fine for a low power RF signal line, and even desirable because it suppresses reflections.

    old jim
     
  9. Jan 29, 2017 #8

    jim hardy

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    I well remember struggling with these concepts.

    Yes reactive components have some role. You can deduce it intuitively.

    Reactance opposes flow of current, so minimizing reactance will maximize current flow. When reactances of load and source have opposite sign and equal magnitude they'll add to zero and that's pretty well minimized, eh ?
    Therefore for optimum power transfer you want Zload to be complex conjugate of Zsource.

    That logic is easy enough to remember ?
     
  10. Jan 30, 2017 #9

    goodphy

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    Hello. Thanks for giving comments.

    I think I need to think this problem more carefully, before asking further and further.

    However, in order to start right reasoning, I need to confirm one thing: In impedance matching by adjusting CT and CL so that Zeq = RS, a power dissipated in Zeq is P1 = I2RS, which is a pure active power (non-zero net power transfer). As Zeq is an equivalent impedance of the sub-circuit (presented by the dashed-line in my figure) and the power on Zeq is the active power, P1 must be consumed by the sub-circuit without going back to the power source (unlikely to the active power, a reactive power goes back to the source after being received by a reactive circuit or component so net power transfer is zero). In the sub-circuits, RL is the only one resistive component so whole amount of power P1 is dissipated on RL, eventually.

    Could you tell me whether these statements I made are correct or not ? If these are correct, I'm on a right track. If not, there is something big I missed so I need to check up what I was totally wrong with.
     
  11. Jan 30, 2017 #10

    jim hardy

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    I don't know. Still trying to parse them. :smile:
    Also your power flows right to left,, unconventional but i finally figured it out.

    equivalent-impedance-in-lcc-network-jpg.112245.jpg

    hmm latex doesn't quote well ?
    Regardless of that, yes that'll be the power transferred when impedances are matched.
    I'm not sure what your term "pure active" means, though. Are you suggesting the currents I and I1 are in phase ? I don't see that as a condition , yet.

    I think i agree with that one, because in order for Zeq to match a resistive source impedance Zeq must appear resistive also. Complex conjugate of R +j0 is R-j0, and that's resistive.

    did you mean "(unlikely to the REactive power,.." ?
    Yes that's what reactance does, shuttles energy back and forth between components that store it.

    Sounds right to me. P1 = I12RL and RL is the only circuit element that dissipates power.
    Current through RL ≠ current through RS though; as Sophie observed your matching circuit behaves like a transformer.

    We're speaking of course about average not instantaneous power. That's what you meant when you said "net power" , Correct ?


    I think you're on the right track. I hope above was some help. Sophie and the radio guys here are better with the maths than i am.

    A word on style:

    When writing technical stuff i try to break it up into single stepwise thoughts and place one thought per sentence, starting each sentence on a fresh line..
    It requires a lot of editing because i'm prone to run-on sentences..
    But the payoff is i can read it a week later and probably figure out what i was talking about.
    I think also it makes it easier for others to follow my thoughts.
    Try it, you might like it. After all, we do our maths that way don't we ?

    old jim
     
  12. Jan 30, 2017 #11

    goodphy

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    Please never mind. "Pure active" just means that the power on Zeq has no reactive power, it only has an active power.

    The original sentence I made was right. I was trying to emphasize a difference between an active power and a reactive power.

    Thanks for giving me some advice. I'll consider your suggestion of writing style when I have a new post.

    All right. It seems my original reasoning is right, but I still have to figure out how I1 can be larger than I. Instanteously, I1 can be smaller than I but on average, it must be higher than I in order to match up powers, P1,net = <I2RS> and P2,net = <I12RL>. Maybe I1 can be larger in other instantaneous time when currents from reactive components reinforce it.

    I appreciate your kind help. You made me felt more clear view.
     
  13. Jan 30, 2017 #12

    jim hardy

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    Hmmm i was thinking the other way, your matching network acts as a step up transformer not step down .

    or maybe you'd need a different circuit topology.
    Wikipedia suggests that step-down uses reactance in parallel not series with the source


    The amateur radio buffs here are versed in practical side of impedance matching.


    http://www.hamuniverse.com/kb1lqc41balun.html
     
  14. Jan 30, 2017 #13

    sophiecentaur

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    Yes. That point was made to me the first time the maximum power theorem was introduced.
    But the problem of reflections is solved with a good match into the load, at the other end of the feeder line. There are then, no reflected signals hitting the source (=transmitter). This applies to high powered RF transmitters. When appropriate, they are made with high efficiency / very non-linear output stages and they, again, cannot afford to be only 50% efficient. So impedance matching is only vital at the antenna end of the transmission line. For a high power valve transmitter, the output stage needs to be 'tuned' to present the right impedance for the efficient operation of the valve rather than to match the Z0 of the line
    Audio amplifiers seldom have an 8Ω output impedance - often, they are 'Voltage sources" with an output impedance of less than 1Ω. You can expect a (basically) emitter follower style of output stage.
    There are many occasions when matching at the source end is important - particularly with measuring equipment, where a standing wave can be a huge embarrassment.
     
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