CC2540*ANTENA MATCHING

Distributor,

I followed your discussion on matching.  This is OK for a quick answer to the solutiion.  The difference what we do is also take in account each piece of trace between the components. This shifts the imaginary part enough so the total adds up to the correct values.   Most of these circuits are check with RF software which includes the layout traces. Your numbers came out pretty close which is what I would expect if you ignore the traces.  

Also the component values are often skewed, especially the inductors since their self resonance is about 6GHz with the shorts return paths.  But they always move slightly lower.  Therefore you must adjust the component values (usually the capacitors) to compensate for the inductors going non-linear.    You must remember the components are not ideal. We use 402 packages because their self resonances is higher than 603. 

I've attached a short explanation for designing this circuit.  If your RF expertise is good you can actually plot this with a Smith chart and be reasonablly close.  We simulate with Agilents ADS product and see correlation to 1%.  

Rgds,6761.RF Matching & Layout.docx

RRS,

Can you please tell me why do you need to match the  chip to a 50 ohm antenna? This would be valid only if the customers use an external antenna but most designs use a PCB etched antenna.Even your reference design uses one of there.  In this case you could forgo all the extraneous L/C components and match the antenna by its physical design. This would save a lot of money when large production quantities are involved!

Mark

RRS,

You forgot to mention that when it comes to 2.4G design, the quality and the composition of the PCB becomes important.

The Core material of the PCB varies wildly with the manufacturer so a design made for a FR4 based on an specific CORE composition might fail miserably

when the fabricator switches the PCB source.

Just my 2 c

Mark

Mark,

You are perfectlly correct in what you say,  but you need to take into account all the other parts, specifically the radio.   Remember the radio is working on +3.3 volt supply and is guaranteed to work down to +2.7V.      To swing peak to peak voltage your need to account for the head room of the chip or the drop at the plus and mius loads internally to the chip.  The chip is not a rail to rail device.

In order to avoid distortion this means a certain load requirement.    Some chips have enough current to go lower 33 ohms but for wost case you should use a larger value.   You are right antennas at 1/4 wavelength and less have real impedances of =< than 36.   But if you matched to these values the chip's output would be in saturation and distortion numbers would exceed compliance of FCC.  

If you take the second direviative of the power transfer you see the minimum point of inflectin which is 33 ohms.  The long time debate is what impedance should one select.  Another part of this discussion is the component values.at high frequencies.  To high of load or source impedance limits effects the component values required for matching.  As you go lower in value the discrete component values become smaller.   With the tolerance of the components you end up with a worst satistical variation due to the tolerance than with hgiher impedances.  

A lot of work goes into determine the best solution,  for what seems a simple problem.  But if you are inolved in a 1 milliion piece build of boards most engineers want zero defects.   So the taking all this into account it is not only simple matching its about preformance, production yield and reliability.

Hope this makes more sense to your question.

Rdgs,

Hi Eli,

I follow your discussion, but where does the 73.8 -21j come from?

regards

eli said:

When looking at the single ended matching network and assuming that antenna impedance is 50Ohm I've got single ended impedance looking into L252 in  a region of 73.8-j21.

The radio load requirement is determined from push pull measurements.    At one frequency there is a load that maximizes the power delivered.   A perfect power transfer says that you must cancel the imaginare part.    So when a data sheet provides a load of 73.8+j21 you need to use the conjugate of this number which is 73.8-j21.  Just chang the sign.  When you add the two vectors the imaginary terms go to zero making it a perfect resistor network.  No energy remains in the capacitor or inductor values.  Hence a perfect power transfer.

But often you would like to change the impedance to 50 ohms then a network consisting of inductor and capactiors changes the impedance to its final resistivity value.  This is basic RF circuit theory.   Google topics on the internet and you will find plenty of examples on power matching networks.   Remember we are concerned with power transfer not voltage or resistance separately,  which is what low frequency circuits operate. 

Rgds