LC impedance matching question

Hi, I need some advice in matching a 10.7MHz crystal filter into a stage that has 50 ohms input and output impedance. The filter termination impedance required is 910 ohms // 25pF. I used the on-line calculator tool at....

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to get values for 2 matching networks, to match 50->910 ohms and 910-

50 ohms (at 10.7MHz).

I got series L values of 3.084uH and shunt C values of 68pF for the input and output matching stages, so to match the filter to 50 ohms at each end. This didnt however take into consideration the 25pF termination value. The dumb question I have is how I can incorporate the 25pF so the filter sees the correct impedance load. I didnt know how to incorporate the 25pF when I used the calculator. Any comments would be greatly appreciated. thanks and regards, JEFF

Reply to
sparks
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ImpMatch.html

If the filter wants to see 25pF in parallel with 910 ohms, just add a

25pF capacitor. I assume that your 68pF cap is on the 50 ohm side -- adding the 25pF cap will make the matching network a "pi" section instead of an "L", but that's no biggie.

You could (probably) design this so the extra cap isn't needed, but if you don't feel comfortable working with complex impedances you probably don't want to go there.

--
Tim Wescott
Control systems and communications consulting
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Reply to
Tim Wescott

Hi Tim, Thanks. The 68pF caps are on either side of the filter to ground, and the inductors are in series to the input and output, making 2 L networks that mirror each other. I did think to add the 25pF across the output

68pF, but thought it would stuff up the response of that filter. The on-line calculator has facility to add reactance into the equation, but wasnt sure what to add there, so the filters would incorporate the 25pF. JEFF
r

ted text -

Reply to
sparks

That online calculator allows you to specify both resistive and reactive components of source and load impedances; however, it requires them in series form i.e. R + jX.

The reactance of a 25pF capacitor at 10.7 MHz is -j595.

You want the filter to see 910 || -j595.

Transforming to the equivalent series form:

1 / ( 1 / 910 + 1 / j595) = 272 - j417

If you specify the series impedance as 272 + j417, the online calculator will do a conjugate match and give you an LC network which looks like 272 - j417.

BTW that calculator gives javascript errors in IE7 because they have Document.XXX instead of document.XXX in a few places. I had to save a local copy, fix this, and update the image paths to make it work.

Reply to
Andrew Holme

Correction / clarification:

1 / ( 1 / 910 + 1 / -j595) = 272 - j417
Reply to
Andrew Holme

Reply to
sparks

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