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Tweakers' Asylum Tweaks for systems, rooms and Do It Yourself (DIY) help. FAQ. |
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Tweakers' Asylum Tweaks for systems, rooms and Do It Yourself (DIY) help. FAQ. |
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In Reply to: RE: Speaker Impedance question posted by hahax@verizon.net on February 28, 2015 at 20:38:16:
Yep. For a 1st order crossover f=1/(2*pi*C*Z) so the frequency and impedance are related. For a 2nd order crossover f= 1/(2*pi*sqrt(L*C)) and so the frequency is independent of the impedance. However the Q of a 2nd order filter is dependent on the impedance (Q = Z*sqrt(C/L) ).
Therefore if you have 1st order HP you will need to halve the capacitor and if it is a 2nd order the sqrt(C/L) needs to be halved (C/L is 1/4). Halve the cap value and double the L value (the product L*C will remain the same). Of course this assumes that that real impedance of your 8 ohm speaker is actually twice the real impedance of the 4 ohm speaker - and that is unlikely.
"Our head is round in order to allow our thoughts to change direction." Francis Picabia
Edits: 03/02/15 03/02/15
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Follow Ups
- RE: Speaker Impedance question - neolith 09:43:55 03/02/15 (0)
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