Calculate Input Impedance of a Small Circuit at 13.56 MHz

In summary: It's a small coil so the radiation resistance is small. The real part of its impedance is mostly due to radiation resistance. It also has a reactive part due to inductance and capacitance, but the radiation resistance is the main contributor to the real part of its impedance. In summary, the circuit described is a small loop antenna that resonates at 13.56 MHz and is tuned to 50 ohm from the source point of view. The input impedance is calculated using complex impedance calculations and a matching circuit is designed to cancel out the reactance and achieve maximum power transfer to the load. The 50 ohm resistance comes from the antenna's radiation resistance, and the coil itself has a low DC resistance but contributes
  • #1
temujin
47
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
 

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  • #2
Your equation is correct. Don't forget omega=2piFreq.
Regards
 
  • #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.
 
  • #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:
  • #5
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.
 
  • #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

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.
 
  • #7
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

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.
 
  • #8
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.
 

1. What is input impedance?

Input impedance is the measure of the opposition that an electrical circuit presents to the input current or voltage. It is a complex quantity that includes both resistance and reactance.

2. How is input impedance calculated?

To calculate input impedance, you will need to use the formula Z = V/I, where Z represents the impedance, V represents the voltage, and I represents the current. This formula can be applied to any frequency, including 13.56 MHz.

3. Why is it important to calculate input impedance at 13.56 MHz?

13.56 MHz is a commonly used frequency in many electronic devices, including RFID systems and wireless charging. Knowing the input impedance at this frequency allows engineers to design circuits that are optimized for efficient operation at this frequency.

4. What factors can affect the input impedance of a small circuit at 13.56 MHz?

The input impedance of a small circuit at 13.56 MHz can be affected by the type of components used, the length and thickness of conductors, and the layout and design of the circuit. Other factors such as temperature and parasitic capacitance can also impact the input impedance.

5. How can the input impedance of a small circuit at 13.56 MHz be adjusted?

The input impedance of a small circuit at 13.56 MHz can be adjusted by using different components, changing the layout and design of the circuit, and adding matching networks such as inductors or capacitors. Simulation tools can also be used to analyze and optimize the input impedance of a circuit.

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