6
\$\begingroup\$

I'm using a SAW filter which has balanced input and output.

In the datasheet there is shown a matching application for 50ohms balanced output.

But plotting the S22 parameters into a Smith Chart and trying to get the same matching circuit is not working for me. I'm guessing the matching circuit is optimized for the center frequency 1090MHz.

Can someone please verify if the datasheet is wrong, or if it is just me (most likely :-) )

Datasheet and S-Parameters are available here: http://www.triquint.com/prodserv/more_info/proddisp.aspx?prod_id=856096

Matching Network

\$\endgroup\$
2
  • \$\begingroup\$ Can you mark where you are trying to determine your reflection on the above picture, this will remove a large amount of confusion for me. \$\endgroup\$
    – Kortuk
    Commented Nov 8, 2011 at 20:11
  • \$\begingroup\$ I'm trying to figure out how the two 3.3pF capacitors and 36nH inductor, are matching the output to 50 ohms? \$\endgroup\$
    – JakobJ
    Commented Nov 9, 2011 at 19:35

1 Answer 1

Reset to default
2
\$\begingroup\$

@JakobJ From the datasheet it looks like the S22 parameter for 1090 mHz is about -8.2 dB at -177 degrees. Also, there are really an infinite number of combinations of values of components that will theoretically work in such a matching network, the same as there are an infinite number of components for an LC filter that will give you the same cutoff. Just because you come up with a different value graphically matching on the Smith chart doesn't necessarily mean that your solution is "wrong" or that the datasheet circuit is "right."

\$\endgroup\$
2
  • \$\begingroup\$ I don't get those figures. S22 = -8.2dB, 177 degrees at 1090MHz (S22 = 0.39 angle 177). That gives me Z = 22 + j0.8 at 50ohms. I know there is a million ways to get the correct matching, but I can't figure out how the matching showed is matched for 50ohms system. \$\endgroup\$
    – JakobJ
    Commented Nov 14, 2011 at 12:11
  • \$\begingroup\$ Sorry, I forgot that the first parameter was in dB when I did my conversion. I shouldn't do this when I'm tired, I'll have to look again! \$\endgroup\$
    – MattyZ
    Commented Nov 14, 2011 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.