What's wrong here? Xover calculators...

[jjdche]

I'm trying to figure out what I need to highpass my paralleled TB W3-871s'. 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?

[K-amps]

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 re-enter 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.

[jjdche]

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.

[Svante]

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 Linkwitz-Riley filters for even order crossover filters. In the second-order 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 non-ideal amplitude and phase response of the drivers near f0.

HTH

Edit: LP filters give exactly the same values for L and C.