So you think you understand the Maximum Power Transfer Theorem

Discussion in 'General Electronics Chat' started by studiot, Jul 1, 2009.

  1. studiot

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    An audio amp produces 10V RMS and has an output impedance of 0.1 ohms.

    This amp can drive an 8 ohm loudspeaker at just over 12 watts. What is the maximum power this amplifier can deliver to any speaker of any impedance, assuming the output stage is capable of supplying the requisite current?
     
    Last edited: Jul 1, 2009
  2. KL7AJ

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    About 250 watts.
     
  3. steveb

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    You get high marks for bravery. You never know when studiot is setting one of his famous traps.

    Just in case he has a trick up his sleeve, I'm going to say I agree with you so that you are not left alone on the limb.

    We'll have to see what he has planned. :p
     
  4. studiot

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    Presses fire button "This is not an exercise ...repeat...This is not an exercise..."

    I am just concerned that AAC should even consider adding the statement

    "The MPTT does not apply to audio amplifiers"

    rather than simply offering a disproof and a clear statement that it does.
     
  5. steveb

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    I haven't given an opinion about this since audio is not my area. The only thing I will say about this theorem is that it is often misused, even by some very smart people. I recently found an example in the research literature where the idea was applied to a mechanical system. The authors went through a very nice derivation and explanation of how to match the impedances (as complex conjugates) for maximum power transfer. However, the system had coupling between the source and load system such that the source impedance was a function of the load. The end result was that matching did not maximize power for that system. This was a reviewed paper which means that 2 or 3 other experts reviewed the paper and did not catch the mistake.

    The example you showed is very simple and the basic case where the theorem makes sense. You need to have a constant source impedance and variable load. No coupling or variability of the source impedance is allowed, or the equations need to be remaximized and the matching idea is not likely to result.

    I think some word of warning is a good idea whenever this theorem is discussed.
     
  6. studiot

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    SteveI I couldn't agree more strongly.

    this is an extract from my post #26 in the thread that prompted me to start this one

    http://forum.allaboutcircuits.com/showthread.php?t=22936

    You will note I state exactly the same caveat.

     
  7. russ_hensel

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    Jan 11, 2009
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    It is worth noting that when it does apply that half the power is wasted. A .1 ohm speaker would be interesting, most speakers even with low dc resistance would see an effective increase in impedance when it began to produce sound. Since I see no i in impedance ( as used here ) this problem as stated cannot be complex, yes?
     
  8. studiot

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    I'm trying to keep it as simple as I can.
     
  9. The Electrician

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    It would be good to have a counterexample when discussion of this topic arises again. Can you exhibit a simple circuit where the theorem fails because it is not true that one sub network contains all the sources and the other contains all the loads?
     
  10. R!f@@

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    LMAOF :p

    Rifaa
     
  11. studiot

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    Well that's put me on the spot and made me scratch the old noddle.

    The simplest example I can think of, off the cuff, is the charging of a rechargeable battery, as in my sketch.

    As a battery becomes exhausted its internal impedance rises.

    In this case as the load battery charges its internal impedance falls towards that of the supply, but its own terminal voltage rises thus reducing the current through it and thus the power delivered. When the impedances are equal the terminal voltages are equal and zero power is exchanged.
     
  12. The Electrician

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    It seems to me that this restriction:

    "It does not apply to any network where there is interaction between the source and load (eg feedback) other than across the two terminals of connection."

    is not necessarily always true. Imagine that you have a load consisting of a single resistor, and you you have a sensor of some kind measuring the current in the resistor. This signal is then transferred back to the driving side of the circuit with an optoisolator with a 1:1 current transfer ratio. The transferred signal is used to apply feedback. I think the critical thing here is that the feedback must be additive, not multiplicative (or some other non-linear type) to avoid invalidating the MPTT.

    To get an output impedance that is a function of the load, such as steveb mentioned, you would need something like multiplicative feedback. Just the fact that the interaction between the source and load is "other than across the two terminals of connection." isn't enough to do it.

    What do you think?
     
  13. studiot

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    It's good to keep discussing.

    I am posting 5 more examples for discussion, covering the conditions I outlined. As KL7AJ observed the theorem may hold for a given nonlinear circuit.

    You can proove mathematically that the theorem applies to all possible circuits that conform to my conditions. Outside these the theorem may or may not apply, depending upon individual circumstances, but I would be fascinated by any proof that it applies to any general nonlinear function I care to describe.

    Some Notes
    In all cases the output impedance of the source sub network on the RHS of terminals A, B is low and the question arises what happens as we lower a single load resistor R towards this value.

    Fig 1
    Point D is more positive than point C in normal operation. TR2 is off. As R is lowered, C rises until it is sufficiently above D for TR2 to turn on, cutting off TR1.

    Fig 2
    This is a nonlinear pulsed subnetwork operating a relay which supplies R when closed.

    Fig3
    A load network comprising a second source plus load R is added to a simple battery

    Fig 4
    Another version of Fig 3

    Fig 5
    Load R is fed from the output of a standard regulator.
     
  14. studiot

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    Disappointing to see thre are no sons of Rodin here.

    For those who haven't realised the theorem applies to the even numbered examples.

    The theorem fails in the odd numbered examples.
     
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