Impedance matching - Why?

Discussion in 'General Electronics Chat' started by Abhinavrajan, Sep 6, 2016.

  1. Abhinavrajan

    Thread Starter Member

    Aug 7, 2016
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    I have read the Maximum Power transfer theorem which states that Maximum power will be transfered from the source to the load only when the source resistance is equal to the load resistance.

    I understand.
    But why should the resistances be equal for maximum power transfer ?
    Please explain the concept behind it.

    I have also read that, incase of unequal impedance between the source and load (impedance mismatch) - Reflections of signal from load to source occurs.
    What does this imply?
    Will the source take up the power reflected from the load? Isn't the source just a pumping device? Why will it accept the power pushed by the load to it?

    Analogy or example will give me more clarity.
    Thanks! :)
     
  2. crutschow

    Expert

    Mar 14, 2008
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    You have the Maximum Power Transfer theorem backward.
    It states the maximum power will be transferred when the Load impedance it made equal to the Source impedance (assuming the Source impedance is fixed).
    If the Source impedance can be varied then the maximum power is transferred when the Source resistance is set to zero.
    If he Source impedance is fixed, then if you calculate the load power for various load resistances, you will find that the load power is maximum when the two impedances are equal.
    To prove that you can take the derivative of power versus resistance (dP/dR) for the maximum value.

    Reflections don't occur between a mismatched source and load, they occur between a source or load which is mismatched to the transmission line impedance.
    This is only a problem if the line electrical length (time for the signal to travel from end to end) is comparable to a significant fraction of the signal risetime (for digital signals) or signal wavelength (for a sinewave signal).
    Here's more on transmission lines.
     
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  3. DickCappels

    Moderator

    Aug 21, 2008
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    Easy to look at it:

    You have a battery with a 10 ohm internal resistance.

    If you connect a zero ohm load, there will be zero volts across the load so no power.

    If you connect an infinite resistance load, there will be zero amps through the load, so no power.

    Trying various load resistances you find that the greatest amount of power is dissipated with a 10 ohm resistor (same as the resistance of the battery). This works the same with impedance.
     
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  4. ErnieM

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    Apr 24, 2011
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    The maximum power transfer, while mathematically proveable and quite correct, is sort of a false conclusion.

    While you get the greatest transfer of power when load and source R are equal you also get a 50% efficiency. Now sometimes that is acceptable, but if you look at a power grid for example you will waste a ton of coal for every ton you burn to power the consumers. In this case you want a source R much much lower than the load so most of the power gets out to do something useful.
     
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  5. GopherT

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    Nov 23, 2012
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    The OP is also asking about reflections that may occur if there is an impedance mis-match. Any impact to reflections if the source is so much lower than load?
     
  6. dannyf

    Well-Known Member

    Sep 13, 2015
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    Obviously you don't understand.

    Power delivered to the load is the product of voltage over the load and current through the load.

    For a given source impedance, high load impedance leads to high voltage across the load but lower current through the load. And vice versa.

    So optimal point is ....
     
  7. ErnieM

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    Apr 24, 2011
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    I only addressed the first of two questions. I am not well versed in the theory to answer the second question.
     
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  8. joeyd999

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    Jun 6, 2011
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    Regarding reflections, a picture is worth a thousand words:

     
  9. crutschow

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    Mar 14, 2008
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    That's neat.
    I really miss the old Bell Labs and all the pioneering electronic work they did when they had many of the top scientists in the country working for them (the invention of the transistor probably the foremost, the important theories by Shannon, Nyquist, and Hamming, among many others, along with the CCD, UNIX, radio astronomy, etc.) but I certainly don't miss the monopoly on telephone service they had.
    But I suppose one sort of went with the other. :rolleyes:
     
    Last edited: Sep 6, 2016
  10. KL7AJ

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    Nov 4, 2008
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    A corollary from the radio world.....while performing a conjugate match will indeed result in the maximum power transfer possible....this is not always a desirable condition...especially for linear R.F. amplifiers. And even some high powered broadcast transmitters are operated WELL below optimum loading.....since efficiency is generally a priority over maximum power output. Most broadcast transmitters are designed to "loaf" along at around half their maximum (conjugate matched) power.
     
  11. BobTPH

    Active Member

    Jun 5, 2013
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    The mistake people make with this is that they think you gain something by raising the output impedance to match the input impedance. You do not, you lose. For example, let's look at an audio amp and and 8 Ohm speaker.

    The audio amp puts out 10V RMS and has an output impedance of 0.1 Ohms. The voltage you get across the speaker is thus 10 * 8 / (8.1) or 9.9V. So, let's try to transfer more power by rasing the output impedance of the amp to match the impedance of the speaker. Now you get 5V across the speaker and thus approximately 1/4 of the power in the speaker.

    Matching the impedance only maximizes the power if the output impedance cannot be changed. Lowering the output impedance will always increase the power transferred and raising the output impedance will always lower the power transferred at the same voltage.

    Bob
     
  12. joeyd999

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    In this situation, one would normally use a transformer to match the load to the source. Back of the napkin: An 1:9 transformer will match the 0.1 ohm source to an 8 ohm speaker. At 100% efficiency, ~250W delivered to the load vs. ~12W.

    Again, back of the napkin.
     
  13. KeepItSimpleStupid

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    Mar 4, 2014
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    Bell Labs still exists BTW. I thought it was dead too. http://www.bell-labs.com

    I thought it went the way of the rodent: www.belllabs.com

    Growing up, there were so many announcements about he achievements of Bel Labs. reading the bell telephone technical journals was always a trip. Early on, the telco signalling was in-band via DTMF tones until SS7 (Signalling System 7). https://en.wikipedia.org/wiki/Signalling_System_No._7

    ==

    Generally agree that the max power xferred gets a little wierd. Things behave differently depending on what world your in. As an example, RF is typically carried in silver plated tubes, not wires in some transmitters. Why? Skin depth.

    There are cases where a single resistor will not replace three of the same resistors in series. Capacitance goes down in the series combination.
    SMD components are mounted on edge to lower inductance. There's a lot of what you learn that flies out the window under certain conditions.
    On a system I built at work, I could probes on a piece of paper and measure the paper's resistance (actually conductance). Glass would show conductance depending on how it was cleaned. Wires that move in the Earth's magnetic field create current. Pushing on stuff (the triboelectric effect) generates charge. Fingerprints are detrimental in some cases.
     
  14. crutschow

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    Try the other back. ;)
    That would give 50% efficiency.
    That's why they don't typically match the source to the load with a solid-state audio amp since it significantly increases the output stage dissipation.
    Also they don't need a transformer to get high audio power levels.

    They do normally use transformers to match source and load in tube amps because the tube impedance is so high and you want all the power you can get with tubes.
     
  15. joeyd999

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    The 100% efficiency was in regard to core losses in the transformer, not those wrt the 0.1 ohm source impedance.

    You've got 10 VRMS to work with in series with 0.1 ohm. A dead short is going cause the source to burn 1000W with nothing to the load. An open will burn nothing, still with nothing to the load. A 0.1 ohm load will share a total of 500W between both, or 250W each. So, yes, 50% overall efficiency.

    Remember, we are talking max power transfer, not efficiency. Please demonstrate a topology that provides more than 250W to the load with the given 10VRMS @ 0.1 ohm source. Switching is not allowed -- the source signal is already in its final form.
     
    Last edited: Sep 7, 2016
  16. joeyd999

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    On second thought, I remove the conditions. Given any source, AC or DC, at 10 VRMS with a 0.1 ohm source impedance, please show how > 250W can be delivered to any load.
     
  17. ErnieM

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    QED: Place source in room. Short output terminals.

    Enjoy your 1,000 watt room heater.
     
  18. joeyd999

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    Zero power to the load. Fail.
     
  19. ErnieM

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    No fail. Success.

    I made a room heater. The room itself is the "load" here.

    100% of the power dissipated goes into the room.
     
  20. joeyd999

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    No. This is not the Kobayashi Maru scenario. You may not change the conditions of the test.

    In your scenario, the source impedance is the electrical load. The question is regarding the Maximum Power Transfer Theorem, i.e. transferring maximum power from a real source with an inherent impedance to a load. You are connecting an ideal source (no impedance) to a load. Different problem.

    Besides, any electro-mechanical system is 100% efficient in converting potential energy to heat -- eventually.
     
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