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Impedance of load

  1. Jul 26, 2005 #1
    Hi,
    I have some problems in understanding how to calculate the input (is it input?) impedance of a small circuit. (I have attached the equiv. circuit with values )

    The circuit is a small loop antenna that resonates at 13.56 MHz and is tuned to 50 ohm.

    From the source point of view: a capacitor in series (Cmatch = 11.74 pF) and then 3 elements in paralell; a capacitor (Cres = 96.76 pF), resistance (Rp = 20050 ohm) and an inductance (L=1.27 microhenry)


    The impedance seen from the source is 50 ohm, however I don´t understand how to reach that answer,as calculating with complex impedances should be straight forward:

    [tex]Z = \frac{1}{\frac{1}{Rp}+\frac{1}{j\omega L}+j\omega Cr} +\frac{1}{j\omega Cm}[/tex]

    Solving this equation with matlab takes me faaaar away from 50 ohm ....


    Am I doing something fundamentally wrong???
    (The circuit and values are from a Texas Instruments Paper, should give impedance = 50 ohm.....???)


    any help is appreciated

    eirik
     

    Attached Files:

    Last edited: Jul 26, 2005
  2. jcsd
  3. Jul 26, 2005 #2

    dlgoff

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    Gold Member

    Your equation is correct. Don't forget omega=2piFreq.
    Regards
     
  4. Jul 29, 2005 #3
    Your equation to calculate the equivalent impedance of the components is correct. But you can clearly see that 20000 ohms will never give you 50 ohms no matter how many inductors and capacitors you hang around it.

    You are trying to design a matching circuit so that your transmitter or receiver see 50 ohms with no reactance. The antenna provides something close to 50 ohms real part at its design frequency. It has some reactance however. The circuit you describe is to cancel out that reactance leaving only the real part of 50 ohms. This will give maximum power transfer to the load.

    The 20000 ohm resistor is to widen the bandwidth of your matching circuit a bit so that the components don't have to be exactly the right values and it so that at other resonant frequencies it will be well behaved. It's called a de-Qing resistor.
     
  5. Aug 2, 2005 #4
    Thanks,

    But where do these 50 ohm come from ("...the antenna provides something close to 50 ohm real part at its design frequency...". The resistance of the coil is much less...

    If the circuit is tuned to 50 ohm, shouldn´t the input impedance be 50 ohm then??

    Any reference to good web sites explaining this topic in detail is highly appreciated

    regards
    eirik
     
    Last edited: Aug 2, 2005
  6. Aug 2, 2005 #5

    The impedance of the antenna itself is a complex calculation that cannot
    be (in general) done analytically. It must be done numerically for arbitrary
    antennas.

    The DC resistance of the coil is low. But once it is an effieicient antenna,
    the energy flys away into space. It looks like a resistor except
    no heat is generated. The coil will have something called "radiation
    resistance" which means it looks like a resistance but it's not- it's radiating
    energy.
     
  7. Aug 2, 2005 #6
    But this is an inductive antenna (for RFID), receiver and transmitter antenna are like a loosely coupled transformer. It does not radiate any energy at all.
    Does the coil still have any "impedance by itself"??

    regards
    eirik

     
  8. Aug 2, 2005 #7
    If there is no resistor connected to the antenna then it must radiate in order to have a real part to its impedance. Transformers only have a real input impedance when the the secondary (output) is connected to a load or a radiating antenna.
     
  9. Aug 2, 2005 #8
    The coil does radiate- in fact it's the only element in the circuit with any appreciable radiation at that frequency.
     
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