Calculate Input Impedance of a Small Circuit at 13.56 MHz

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SUMMARY

The discussion centers on calculating the input impedance of a small loop antenna resonating at 13.56 MHz, tuned to 50 ohms. The circuit includes a series capacitor (Cmatch = 11.74 pF) and parallel components: a capacitor (Cres = 96.76 pF), resistance (Rp = 20050 ohms), and inductance (L = 1.27 µH). Users express confusion over the discrepancy between theoretical calculations and the expected 50-ohm impedance, emphasizing the complexity of antenna impedance calculations, which often require numerical methods rather than analytical solutions.

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temujin
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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:

Z = \frac{1}{\frac{1}{Rp}+\frac{1}{j\omega L}+j\omega Cr} +\frac{1}{j\omega Cm}

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
 

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Your equation is correct. Don't forget omega=2piFreq.
Regards
 
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:
temujin said:
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


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.
 
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

Antiphon said:
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.
 
temujin said:
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

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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