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Impendance matching and power transfer

  1. Mar 30, 2009 #1
    What is the physical explanation for the power transfer in a circuit being maximum when the
    resistance of the system is equal to the internal resistance?

    maybe heat loss?

    any help would be appreciated as i understand the mathematical proof but do not know much about the physical causes.
     
  2. jcsd
  3. Mar 30, 2009 #2

    MATLABdude

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    If you assume a fixed source impedance, you now need to figure out what size load gets the highest power transferred to it. With the mathematical (Calculus) proof, you know that there's a maximum in the power when the load resistance is equal to the source resistance.

    As to why... Anything below this number, and the majority of power is dissipated in the source resistance (it's a simple voltage divider, and the bigger impedance always drops the most power), despite there being more power pulled from the (ideal) voltage source. Anything higher than this number, and, while there's a greater voltage drop across the load resistor, the increased total resistance means less current is available and hence less power dissipated in the load.

    Note that the max power condition isn't used when the power company sends power to you; they want as low of a source resistance as possible. Otherwise, they'd lose half the power in the transmission lines! It is used in electronics to maximize delivery of a low-power signal (e.g. audio and RF) and the complex variant of it (load having the conjugate impedance) is used in order to reduce your apparent power and power bill (though this usually only applies if you're running a business with lots of inductive loads, and not to residential customers).
     
  4. Mar 30, 2009 #3

    berkeman

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    Nice explanation by MATLABdude. One additional point that is often overlooked... When the source impedance is complex, you get the maximum power transfer to the load when the load is the complex conjugate of the source impedance, not equal to it. So basically you are matching the resistive part, but using the opposite Imaginary part.
     
  5. Apr 1, 2009 #4
    thanks, one other thing i dont understand is when there is a constant power supply cost does not matter eg the sun for solar panels why is efficiency taken into consideration if it doesnt matter that power is lost in the circuit.
     
  6. Apr 1, 2009 #5
    If you are transferring a high frequency signal from a coax cable such as RG-8 (50 ohm impedance) into a termination, the maximum power transfer occurs when the termination impedance is 50 ohms. If it is higher or lower than 50 ohms, some of the signal is reflected, and the power transfer is less.
     
  7. Apr 1, 2009 #6

    berkeman

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    Not sure I understand the follow-up question. Power lost is bad in energy applications because you lose money and/or lose capability at the final consumption points. Power lost in communication links means lower data rates for the same error rates.
     
  8. Apr 1, 2009 #7

    MATLABdude

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    Like Berkeman, I also don't know if I understand your question. However, efficiencies (conversion of, say, electricity to mechanical work, or incident solar power into output electrical power in the case of your solar cells) are done to give engineers a way to compare the performance of various things.

    Is a 1% (incident power to electrical power) solar cell better than a 40% solar cell? Would you use it in your product? What if the 40% efficient cell costs 400 times as much to make? (These numbers are used purely as examples, and are not grounded in real life.

    EDIT: As another example, the other way around... Your typical light bulb has an efficiency (converting electrical power to light) of around 5%. Your typical CFL (compact fluorescent light) is around 20% efficient at converting electrical power to light. The variable cost (cost of electricity) is lower, and the fixed cost (cost for the actual bulbs themselves, given that you need to be replacing incandescents more often) is comparable.
     
  9. Apr 2, 2009 #8
    thanks for the replies.
    to clarify the question, when using solar panels unlimited source of energy why is the external load made to bigger to increase the efficiency of the circuit when there is no cost from the sunlight?
     
  10. Apr 2, 2009 #9
    Sunlight is cheap, but the space to put solar cells in is not. If you need 10 kW peak output, how much more would you pay for 22% efficient solar cells than for 15% solar cells? For transferring solar cell power to a grid, you need to use a PWM (pulse width modulation) converter, because the solar cell voltage output depends slightly on the insolation (solar light intensity).
     
  11. Apr 2, 2009 #10

    berkeman

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    ... And read about the "maximum-power point" of operation at this wikipedia page:

    http://en.wikipedia.org/wiki/Solar_cell#Maximum-power_point

    .
     
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