How to replace a balun in a design ?

[yellowtooth]

How to replace a balun ?

In my design, there is a RF switch, but I donot want to use a balun.

So I must use some lump components or microstrip line to act as a balun.
But I donot know how to design the components or microstrip line 's value .

For example, **broken link removed** , on page 26 and 27.
The input port is differential with Z=2-j4, only four components (C12 ,C13,C16 and L2) match it to 50 ohm sigle ended port.

How to design the value of the components ?

 

[flatulent]

two stage process

You can do it in a two stage process. First convert the impedance level using the trial version of Zmatch downloadable from https://www.zzmatch.com/

then use a bridge type balun with inductors and capacitors which have a reactance of X ohms at your operating frequency where X is the impedance level the balun is operating between. The bridge should be drawn as a diamond shape with the top apex connected to the unbalanced direction and the bottom apex grounded. The left and right corners are the balanced port. The capacitors are in the top left and bottom right legs between the corners and the inductors are in the top right and bottom left sides. You should end up with a loop of L C L C going around on itself.

 

[yellowtooth]

Hi flatulent,
I have DL the soft, it seems can only match single ended port impedance.
Do U mean that in first stage, I should match the RF switch to X ohm ?
Then in the second stage, I transfer single end to differential port ?
But, if so ,how can I know the X ? It is 4 ports network.

PLZ DL the PDF form TI first, then tell me how to design the L and C.

Thank U.

 

[Encrypted]

Read this paper it may help...

**broken link removed**

 

[yellowtooth]

Thank U Encrypted.
The file is very useful to me.
But I still not know how can TI company do the work by so few (only 3)components.
While in the file, there must 4 components at least.

 

[Encrypted]

The TI input circuit is not a balun. They use a phase shifter to drive the differential input of the LNA.

 

[flatulent]

two stages

You first convert from the low complex impedance to the final real only impedance that has to be push pull (balance) with the traditional network types (L, T, pi). The bridge circuit I described is really two phase shifters in parallel. One shifts plus 90 degrees and the other shifts minus 90 degrees. Taking the output between these two makes them 180 degrees apart or push pull or balanced. The X value of impedance is the driving single ended impedance or the push pull side impedance as they are the same.

 

[yellowtooth]

Hi Encrypted,
In the PDF page 10 line 1, it is said that the TI input circuit has 2 task including matching and phase shift.
In spite of this, do U know how to calculate the L an C's value ?

 

[Encrypted]

It's a little hard to explain in a short message (someone else please have a go!) but if you have a specific example i'll try to work out some values.

Note that the LNA input in your example is 500||0.7p. This is about Zin=100-j200 at 900MHz and not as quoted in your first message.

Thought this might help with your matching problem

 

[yellowtooth]

Hi Encrypted,
Why donot U think this is just a example ?

By the way, Zin=100-j200 =(2-j4)*50.