LC impedance matching question

[sparks]

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....
http://bwrc.eecs.berkeley.edu/Research/RF/projects/60GHz/matching/ImpMatch.html
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

 

[sparks]

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

 

[Andrew Holme]

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....
http://bwrc.eecs.berkeley.edu/Research/RF/projects/60GHz/matching/ImpMatch.html
to get values for 2 matching networks, to match 50->910 ohms and 910-
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

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.

 

[Andrew Holme]

Transforming to the equivalent series form:

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

Correction / clarification:

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

 

[sparks]

Hi there Andrew,
Thanks very much for the help with understanding the complex impedance
part of the puzzle, and for working the example.
It was many years ago that I last tacked such stuff - and other scary
things like Smith Charts.
Sorry its been a while since I last posted. Been of town for a week.
All the best.
JEFF