Folded Dipole Antenna CC2500 impedance matching

Andrei,

The folded dipole is calculated as outlined in your first part in air.  Then everything is scaled by the velocity factor for the board material to a new effective permittivity.  This of course scales with width thickness and height (thickness of the dielectric). See Pozar book on Microwave Design.   If you look carefully you will notice that part of the antenna (feedline is over ground so it becomes a differential  microstrip lines which are used to match the antenna to the radio following normal microstrip theory and scaled for velocity factor.  You should know to change the impedance you need 1/4 wave lengths or none of the theory will work.  You end up with series inductors (stub lengths).  

It is well documented to calculate the folded dipole resonant frequency which is of course 4- quarter wave lengths around the loop.   We oftern treat it as a dipole and calculate the single folded end 1/2 wave radiator and then put it toghter.  Then you can use a two port VNA and connect only 1 port for S11 measurements.  Calibration is very tricky because of the phase increments are samll for unit length so becareful when doing lab measurements since they can introduce more errors.   We use Agilent ADS program and the simulation engine FEM with excellent results.  Our measurement error from lab and chamber is around 1-2 %.  

I would suggest calculate this in two steps one from the radio to the input of the antenna's input for microstrip lines,  then your antenna as placed over dielectric.  Your results should look similar to the plots I've provided. Good luck.   I've attached a few slides to get you started.  I didn't feel I should give you the entire answer since this is a school learning project,  but it should give you enough information to figure out what you need to do.

4572.FD_Ant Design.ppt

Hello,

I still can't figure out why in CC2500 design note 004 the distance between antenna and receiver chip is 11.5mm(this is not 1/4 wave at 2.45GHz). How is the quarter wave transformer scaled because it doesen't result from calculus(1/4 wavelength at 2.45GHz scaled by the velocity factor is aprox 18mm).

Thanks,

Andrei

The differential antenna can be modeled as two 1/2 wavelength  dipoles.    The real part of the impedance should be 72 ohms or differentially 144 ohms. 

The antenna length when plottend in air is of couse the wavelength equation. But when you place it on PCB material the velocity correction must be made which shortens the radiator wavelenth  and effectively on the smith chart shrtens the traces and effects the impedance.   Bending the  two radiators together allows the engineer to do the phase shift which re-adjust the real part to the desired impedance.  

In terms of the differential antenna input.  the phase of the signal should equal zero so the antenna waves start at zero and travel lengths of 1/4 wavelengths which you now have 4 of them for folded dipole. The antenna sums the signal power or effectively doubles the two sources output power P1/2 plus P2/2 from the radio so no balun is required unless used for impedance power matching.     

The front in of this folded dipole now has a phase between the two inputs.   If you add to differential traces to the antenna the phase or XL will change to some new value with a differential value for R in this format   R + jXL.    But most likely it isn't 1/4 wavelenth.   but you can add the matching compoents or stubs to align the phase to zero at the input of the antenna.    Exampe if it is a single dipole antenna with an antenna input impedance of 50 + j0    Then the signal path output should be 50 -j0 or a perfect power match.   

To summarize once you have your antenna impedance real part adjust correctly,  the value differential should be 144 +j90.   You need a network from between the radio chip to the input of the antenna which has the conjugate  value of 144 -j90.       From the chip your impedance it likes to see is 80+j74.        So now the matching network must be translate the 80+j74 to the 144-j90.   You have 3 options in doing this stubs and traces, disctete balun and traces,  or discrete components and traces.   You must include the traces impedances or you want have the cancelation of imaginary part therefore the wave won't be maximum at your 1/4 wavelenths of the antenna. 

I loved to work this out in detail,  but unfortunately I have too many projects that I'm behind with.   As you can see from the above answers their is several ways to complete the design.  You might consider doing more than one and compare the results.     The big problem with doing this with Smith Chart is you need at least 4 decimal places to get reasonalbe accuracy at 2.45GHz.   Mayby more with the stubs and trace approach. 

Respectfully,