EMRFD Message Archive 9691

Message Date From Subject
9691 2014-02-03 06:09:15 miaidea2001 455Khz filter adapting

HI,

I would like to adapt a filter 455khz (Z = 2k/50 ohms) through tapped capacitor. Unfortunately I'm not good in the calculation.

Can anyone help me?

Thanks


73 de IK6GQC  Rocco

9692 2014-02-03 15:25:05 Ed Miller Re: 455Khz filter adapting

I too would be interested in how to use split capacitors for impedance matching. There is a paper at http://www.qrp.pops.net/captap.asp that is supposed to explain how but I am not able to understand it enough to use it, even after reviewing complex algebra.  In the 2008 software of EMRFD there is a program called Zmat08 for LLC T-networks that should work. I need to use a proven circuit and enter the Zin and Zout and compare the resulting values to see if the program is suitable for this use. What little I have been able to determine is the split capacitors act very much like a resistive voltage divider and there will be a voltage change much like a transformer will give, but this has been little help in answering my questions.

           

            I live in Florida where we cannot count votes; how am I supposed to figure out something like this?

                       

HI,

I would like to adapt a filter 455khz (Z = 2k/50 ohms) through tapped capacitor. Unfortunately I'm not good in the calculation.

Can anyone help me?

Thanks

 

73 de IK6GQC  Rocco

9693 2014-02-03 16:31:30 Ephemeral Re: 455Khz filter adapting
If your filter starts with a shunt capacitor you could try replacing it with two capacitors whose series capacitance
has the same value as the original. The ratio of the new shunt capacitance to the original capacitance will be the
square of the impedance ratio. (This is only an approximation). I use a program called AADE Filter Design, which
can be freely downloaded. If you enter the filter into the program you can test it to see if it works and tweak as
necessary. 
 
That is how I do it anyway. Experts may have a more precise formula.
 

9694 2014-02-03 16:35:18 EricJ Re: 455Khz filter adapting
 Look in RF Circuit Design by Chris Bowick. You can get a free pdf of the entire 2nd Edition here:

http://www.pdf-archive.com/2013/06/22/rf-circuit-design-second-edition/

Make sure you use the Download link with the Adobe Reader logo. DO NOT use the green Download button (there or anywhere) unless you collect pointless toolbars as a hobby.

Chapter Two in either the Second or First Edition has a short but very good explanation and Example 204 takes you step by step.

This is a great book. One of the best in my library (EMRFD being first). It sells for $44 on amazon.com, but it's the same as the pdf version.

Eric
KE6US


9696 2014-02-04 03:15:46 miaidea2001 Re: 455Khz filter adapting

hi folks,

thanks for the replies.


The filter is a 455kHz muRata ceramic filter with 3 kHz BW-3db Z (IN / OUT = 2K).

If I'm not mistaken, the Q factor should be about 150 (455000/3000).

More of this I can not do because I confess to being almost a donkey in mathematics.

Someone much better than me, tell me would know the value of C1 and C2?


73 de IK6GQC  Rocco

9697 2014-02-04 04:52:30 Nick Kennedy Re: 455Khz filter adapting
I have a spreadsheet that does many useful ham calculations, one of which is a tapped-C matcher.  The math came from the ARRL Electronics Data Book of 1976, edited by Doug Demaw, page 46.

This matcher has an inductor in parallel with the load and also two series capacitors across the load. The "upper" capacitor from the ungrounded end of the load is C1 and the "lower" one from that capacitor to the grounded end is C2.  The point between C1 and C2  and ground is the input connection and is 50 ohms in this case.

This network is inherently selective and the equations call for a bandwidth as well as input and load resistances and frequency.  So that it doesn't affect the 3000 selectivity of your filter, I picked 30,000 Hz for the network's BW.

The numbers I get for F = 455 kHz, BW = 30 kHz, Rin = 50, Rload = 2000 are:

L = 23.1 uH
C1 = 6300 pF (0.0063 uF)
C2 = 0.0335 uF

73-

Nick, WA5BDU