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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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