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Matching reactive load

Discussion in 'Electronic Design' started by Steve Bower, Jun 19, 2007.

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  1. Steve Bower

    Steve Bower Guest

    Can someone tell me in practical terms, ie. suggested components and
    values, how I would match the output of a power op amp to a reactive
    load? For this example, we can assume a 100KHz squarewave and a load
    impedance of 450 ohms.

    Steve Bower
     
  2. 450 ohms (with no phase angle or other indication of
    reactance) indicates pure resistance. Why do you call this
    a reactive load?
     
  3. Tom Bruhns

    Tom Bruhns Guest

    In addition to what John P. already posted, I have to ask about the
    impedance versus frequency. If you want to "match" the load to the
    amplifier, you need to consider the impedance not only at 100kHz but
    at 300, 500, ... for however many harmonics you want to consider.

    Also, exactly what does "match" mean? An op amp with negative
    feedback applied generally has rather low output impedance, and you
    probably do NOT want to present a load to it equal to the complex
    conjugate of its output impedance. What are you trying to accomplish
    with a "match"? I think you need a clear picture of that before you
    can start to worry about how to achieve it.

    Cheers,
    Tom
     
  4. Phil Allison

    Phil Allison Guest

    "John Popelish"

    ** Gotta hunch the OP is really after a "complex conjugate matching network
    " to place between his " power op-amp " and the mysterious 450 ohm reactive
    load - plus the network needs to be effective from 100kHz to circa
    daylight.

    Can anyone put him onto suitable CAD software and a procedure to do that ?

    Bugger the imaginary need.

    Pun intended.



    ........ Phil
     
  5. Eeyore

    Eeyore Guest

    Well, you certainly can't change the op-amp's characteristics.

    What do you mean by 'match' ? You don't 'impedance match' an op-amp to anything.

    Graham
     
  6. Steve Bower

    Steve Bower Guest

    In addition to what John P. already posted, I have to ask about the
    I am already worried ;-)

    Maybe an op amp is not the best solution. BTW, I only need to match
    one frequency at a time. Not a range, as I understand this is not
    practical. The aim is to maximize power transfer.

    Disregarding the 450 ohm figure, what is the recommended technique and
    signal source for matching a signal to a reactive load? More
    specifically, one that has both resistance and capacitance in
    parallel.

    Steve Bower
     
  7. Eeyore

    Eeyore Guest

    At those frequencies you don't need to match a amplifier to a load !

    Just drive the damn load from the output of the op-amp !

    What exactly are you worried about ?

    Graham
     
  8. kell

    kell Guest

    Too vague.
    What's the app?
     
  9. Even if that overloads the power source? By the way, a
    square wave includes energy at a lot more than one
    frequency, even though it has single number of cycles per
    period of time. This is why frequency is measured in hertz,
    not cycles per second. I suspect that you want to optimize
    power transfer at the fundamental frequency of the square wave.
    Adding either parallel or series reactance of the opposite
    type to resonate with the reactance at a given frequency, is
    very common. Either way, you can hide the energy transfers
    through the reactive part. The difference is what happens
    at other frequencies, like those harmonics in your square
    wave. You might also need a transformer or a more
    complicated form of resonator if you need to shift the
    resistive component to a different effective value (change
    the ratio of voltage to current) to get the desired energy
    transfer from source to this load.
    I would probably first try a series inductance that has an
    equal magnitude impedance as the capacitive part of the load
    at the square wave fundamental frequency (series tuned).
    This has the advantage of looking inductive (rising
    impedance as frequency rises) at the higher frequency
    harmonics of the square wave, so that the opamp doesn't work
    very hard to drive them. Don't be surprised to find that
    the inductor has more voltage across it than the opamp puts out.
     
  10. Rich Grise

    Rich Grise Guest

    His grade on the exam? ;-)

    Cheers!
    Rich
     
  11. The point with power amplifiers is not to "maximize power transfer."
    The point is to maximize efficiency. They are not the same thing.

    "Maximizing power transfer" is a small signal concept, and has to do
    with internal impedances of the devices.

    Maximizing efficiency has to do with extracting as much power from the
    power supply as possible, and dissipating as little as possible in the
    device. It is a large signal concept.

    efficiency := P_out/P_dc

    or for power added efficiency

    efficiency_pae := (P_out - P_in)/P_dc

    Note that in neither is there any discussion whatsoever of "maximizing
    power transfer."

    You transform to the desired "characteristic impedance" (really a
    resistance) and conjugate the reactance.

    There is an infinite set of solutions for a given problem. For
    matching at a single frequency, a minimum of two elements are
    typically needed. Distributed line transformation can do it too.

    Can you use a Smith Chart? A Smith Chart makes it bonehead simple,
    you know.
     
  12. Tom Bruhns

    Tom Bruhns Guest

    It would help, as others have suggested, if you could state your
    problem in terms like "I want to deliver xxx watts of power to a load
    that looks like yyy ohms in parallel with zzz pF. The waveform from
    the source is a square wave at 100kHz, and I (do/do not) care that the
    voltage across the load looks pretty much like a square wave. As a
    source, I have a power op amp that's rated to deliver up to qqq volts
    peak, at up to rrr amps." If you can put the problem in terms like
    those, I suspect we'll be able to give a lot more specific answers.

    Another couple examples of common situations where you do NOT want to
    maximize "power transfer":
    --When you plug a load into a standard wall outlet, the source
    resistance of the power supplied through that outlet is generally
    under an ohm. You'd theoretically get maximum "power transfer" if you
    put a load on it equal to the source resistance. But you'd drop the
    voltage to half the open-circuit voltage for an instant before the
    circuit breaker tripped.
    --An audio power amplifier is typically designed to deliver its power
    into loads around 4 to 8 ohms, even though its output impedance is
    almost always a small fraction of an ohm. If the amplifier didn't
    have built-in current overload protection, you could get a lot more
    power out of it if you put a much lower load resistance on the output--
    but only for a short time till the output stage went into melt-down.

    You may save yourself a lot of trouble if you put your circuit into a
    program like the free LTSpice. You'll be able to check current and
    voltage waveforms in an instant, and try all sorts of different
    matching networks till you get something you're happy with. You can
    even check power dissipations--in the load, in the op amp.

    Cheers,
    Tom
     
  13. No reply, total silence. Another thread stopped dead. :(
     
  14. Steve is thinking hard.
     
  15. [/QUOTE]
    Let's have a WAG then.........

    The quoted 458 ohms load impedance might be something
    like a 1k in parallel with 3nF (at 100KHz sine).

    A power opamp was mentioned so the voltage could be
    in excess of 20V pk-pk across that 458 ohms.

    Sine or square? A 100KHz square wave was mentioned.

    So the problem might be the best way to drive a
    1k//3nF load with at least 20V pk-pk and with (say)
    better than 1uS rise and fall times.
     
  16. Tom Bruhns

    Tom Bruhns Guest

    Let's have a WAG then.........

    The quoted 458 ohms load impedance might be something
    like a 1k in parallel with 3nF (at 100KHz sine).

    A power opamp was mentioned so the voltage could be
    in excess of 20V pk-pk across that 458 ohms.

    Sine or square? A 100KHz square wave was mentioned.

    So the problem might be the best way to drive a
    1k//3nF load with at least 20V pk-pk and with (say)
    better than 1uS rise and fall times.
    [/QUOTE]

    Put about 10.3uH between a 100kHz square wave generator and that
    load...my simple brain is fascinated by the resulting waveform.

    Cheers,
    Tom
     
  17. joseph2k

    joseph2k Guest

    Oh no, John, say it isn't so. I "grew up" with cycles per second, surely
    you are not quite so young to not remember changing to Hertz in honor of
    Heinrich Hertz. Oh that's right it is a Metric unit. See:
    http://www.diracdelta.co.uk/science/source/h/e/hertz/source.html
    Resonance is full of surprises for the uninitiated.
     
  18. joseph2k

    joseph2k Guest

    What in the pluperfect hell do you think you are saying? Are you advocating
    serious mismatches in the kilowatt, megawatt and above ranges?
    What part of P_out must not be transferred efficiently to the load?
    Smith charts are not all that useful or relevant at 100 kHz. Nor for
    wideband signals for that matter.
     
  19. No part. That's why efficiency is the concern and not "maximum power
    transfer." Again, MPT says nothing of converting power supply energy
    into signal energy delivered to a load. It is a small signal model.
    That's why MPT is not a good concept for PAs.
    What is it about a smith chart that makes it not work at 100kHz (or
    1Hz for that matter)?
    You missed the "single frequency" comment. There is no DC to daylight
    solution anyway.
     
  20. Tom Bruhns

    Tom Bruhns Guest

    Why not? They certainly can be for me, for both 100kHz and for
    wideband. Hint: they are at this point primarily a visualization
    tool, not a calculation tool. So long as I'm dealing with networks of
    series and shunt reactive components and transmission lines (ladder
    networks), a Smith chart is a valuable tool. That's as true at 1Hz as
    at 10GHz, though it's less likely that I'll be dealing with such a
    ladder network at audio frequencies than at RF.

    Cheers,
    Tom
     
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