

Crystal Filter  RF Cafe Forums

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*RsRs^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*RsRs^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*Q2Q2^2);
%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=C2C2p; elseif
X2s>0 %inductive C2=(C2*wo*j1/(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
(25j10). Trev
Posted 11/12/2012



