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Crystal Filter - RF Cafe Forums
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SDO
Post subject: Crystal Filter Posted: Fri Oct 31, 2008 7:42 am
Captain
Joined: Tue Aug 26, 2008 5:18 am
Posts:
22
Location: UK
Hi all,
I am new to low frequency design
and presently designing channel selective module. I am using crystal
filter and a gain stage in my design along with other components. Gain
stage is matched to 50 ohm. Crystal filters are high impedance (around
900 ohm). IF frequency is 21MHz. What is normal practice to match crystal
filter to 50ohm matched gain block?. I've tried lumped element matching
net work and it seems to be working.
Thanks.
SDO
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markapexwireless
Post subject: Posted: Fri Oct 31, 2008
5:53 pm
Captain
Joined: Tue Jul 08, 2008 7:15 pm
Posts:
11
Location: Boulder Colorado
Hi,
One common method is
to use "tapped capacitor" transformer which is essentially an inductor
and two series caps in parallel with inductor. Amp output goes to tap
between caps and filter goes to inductor. Cap ratio is square root of
impedance transformation which in your case is 4.2. Composite L/C is
tuned to 21.4 MHz. This network will provide additional BP filtering
particularly away from 21.4 MHz IF.
Solid State Radio Engineering
by Krauss, Bostian and Raab does a great job on these types of networks,
but you can probably find detailed design guide on the internet.
Good Luck!!
- Mark -
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darcyrandall2004
Post subject: Posted: Mon Nov 03, 2008 1:10 am
Colonel
Joined: Tue Feb 27, 2007 6:16 am
Posts: 46
Hello,
I have some Matlab code that calculates the component values required
for a Tapped Cap Match. Perhaps this will be of some help. Always check
the results. Cheers
function [L,C1,C2]=Tapped_cap_match(r1x,r2x,Q,fo)
format long
%r1x and r2x are the source and load impedance respectively.
%no need to enter the conjugate values for max power transfer. Calcs
automatically.
%r1x > r2x
%r1x-|-------------
% | | |C1
% |L| ---
% | | ---
% | |--------r2x
% | C2---
%
| ---
% | |
%change r1x to a parallel combination of reactance
and resistor
%NO VALUES SHOULD BE NEGATIVE!
wo=2*pi*fo;
Rs=real(r1x);
Xs=imag(r1x);
if Xs<0 %capacitive
C=abs(1/(Xs*wo));
Rin=(1+wo^2*C^2*Rs^2)/(wo^2*C^2*Rs);
Xc=(Rin*Rs-Rs^2)^.5;
Cp=C/(1+(Rs^2/Xc^2));
elseif Xs>0 %inductive
Ls=Xs/wo;
Rin=(Rs^2+wo^2*Ls^2)/(Rs);
Lp=(Rs^2+wo^2*Ls^2)/(wo^2*Ls);
else Rin=real(r1x)
end
Rs=real(r2x);
X2s=imag(r2x);
if X2s<0 %capacitive
C=abs(1/(X2s*wo));
R2=(1+wo^2*C^2*Rs^2)/(wo^2*C^2*Rs);
Xc=(R2*Rs-Rs^2)^.5;
C2p=C/(1+(Rs^2/Xc^2));
elseif X2s>0
%inductive
Ls=X2s/wo;
R2=(Rs^2+wo^2*Ls^2)/(Rs);
L2p=(Rs^2+wo^2*Ls^2)/(wo^2*Ls);
else R2=real(r2x)
end
L=Rin/(wo*Q); %calculate the
inductance
%R2s= R2/(Q2^2+1)=Rin/(Q^2+1)
Q2=sqrt( (R2/Rin)*(Q^2+1)-1
);
C2=Q2/(wo*R2);
C1=C2*(Q2^2+1)/(Q*Q2-Q2^2);
%account for reactance of source and load impedances.
if Xs<0
%capacitive
L=1/(wo*j)*(1/(wo*L*j)-wo*Cp*j)^-1;
elseif Xs>0
%inductive
L=1/(wo*j)*(1/(wo*L*j)-1/(wo*Lp*j))^-1;
end
if X2s<0 %capacitive
C2=C2-C2p;
elseif X2s>0 %inductive
C2=(C2*wo*j-1/(wo*L2p*j))/(wo*j);
end
a=1/r1x+1/(wo*L*j)
b=1/a+1/(wo*C1*j)
c=1/b+wo*C2*j
d=1/c
r2x
_________________
Regards, Darcy Randall, Perth, Western Australia
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SDO
Post subject: Thank youPosted: Mon Nov 03, 2008 6:46 am
Captain
Joined: Tue Aug 26, 2008 5:18 am
Posts:
22
Location: UK
Thank you gentle men for your help.
Much
appreciated.
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nubbage
Post subject: Posted:
Mon Nov 10, 2008 5:58 am
General
Joined: Fri Feb
17, 2006 12:07 pm
Posts: 218
Location: London UK
Hi Darcy
To design a transformer, I need to translate the routine you wrote,
because I do not have Matlab. I use Fortran, CalcEd and Mathcad for
complex math.
What are the a, b, c and d variables at the end?
The Fortran equivalent compiled OK, but on running the *.exe file
it throws up an "arithmatic error". I was running a 50MHz calc, with
a Q of 5, with input Z of (75+j15) and an output Z of (25-j10).
Trev
Posted
11/12/2012
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