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4th June 2004, 06:15 PM  #1 
diyAudio Member
Join Date: Feb 2004
Location: MA

What's wrong here? Xover calculators...
I'm trying to figure out what I need to highpass my paralleled TB W3871s'. I was looking to get some component values for 1st and 2nd order (Butterworth) high pass filters. Initially I was using the calculator in WinISD, for 1st order it gave me a value of 199uF for 200Hz and for 2nd order it gave me 141uF and 2.25mH. I decided to check with another calculator and it gave me the same cap values as WinISD for the 1st and 2nd order filters (199uF and 141uF), but instead it output 4.5mH for the 2nd order inductor. Something is wrong here, I don't have the equations to verify it, does anyone know which one is wrong?

4th June 2004, 06:50 PM  #2 
diyAudio Member
Join Date: Nov 2003
Location: Indiana

I am assuming you used the correct impedances cause the second calculation seems to be for half the impedance of the first (4 ohms?) Just reenter with 4 ohms in both calculators, if they both cannot handle impedances... then thats maybe why you got a different answer, i.e. 4.5mH and 2.25mH.

4th June 2004, 07:41 PM  #3 
diyAudio Member
Join Date: Feb 2004
Location: MA

I did enter 4 ohms in both calculators. If I enter 8 ohms in both I get...
2nd order Butterworth for 200Hz high pass WinISD C1=70.3uF L1=4.5mH Calculator C1=70.3uF L1=9.0mH It seems to me that the WinISD calculator has a problem. I checked a few sites and it looks like you just switch C1 and L1 to get the low pass filter for the same frequency, with the values remaining the same. WinISD shows different inductor values for the high pass and low pass for the same impedance drivers. 
4th June 2004, 08:31 PM  #4 
diyAudio Member
Join Date: Feb 2004
Location: Stockholm

For a 2nd order HP filter (C in series with (paralleled L and Rload))
f0=1/(2*pi*sqrt(LC)) and Q=R*sqrt(C/L) For butterworth design (3 dB @ f0) Q should be 0.707. Your values give: 1: C=70.3 uF, L=4.5mH, R=8 > f0=283Hz, Q=1.000 2: C=70.3 uF, L=9.0mH, R=8 > f0=200Hz, Q=0.707 So, the second is a butterworth filter. It is though, at least theoretically, recommendable to use LinkwitzRiley filters for even order crossover filters. In the secondorder case Q should be 0.5 instead of 0.707. For 8 ohms, the values should be: L=R/(2*pi*f0*Q)=12.7mH C=1/((2*pi*f0)^2*L)=49.7uF Then again, in practical filter design often deviates from theory due to nonideal amplitude and phase response of the drivers near f0. HTH Edit: LP filters give exactly the same values for L and C. 
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