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From: hull%hp-lsd@janus.Berkeley.EDU (Christopher Hull)
Newsgroups: rec.audio.high-end
Subject: Re: Quasi-second order crossover
Message-ID: <9614@uwm.edu>
Date: 18 Feb 91 13:53:31 GMT
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Approved: tjk@csd4.csd.uwm.edu
Originator: tjk@csd4.csd.uwm.edu

In article <9528@uwm.edu> you write:
>
>I am planning to build my first set of speakers, and
>in my research I have come across a line (VMPS kits)
>that use what they call a quasi-second order crossover.
>I have not seen this term anywhere else.  It claims
>to keep the same phase relationships as a first-order
>network, but with steeper rolloffs, or something
>along those lines.  It also said something like
>using 6dB slopes for the low frequencies and 12dB
>for the highs.     
>
>Does anyone know exactly what this is?  What exactly
>does such a circuit look like, maybe I can figure out
>for myself what it does.
>

The quasi second order is really a first order network with        
steeper slopes in the x-over region.  The x-over configuration
consists of hooking the drivers up in series, and then
putting an inductor in parallel with the tweater, and
a capacitor in parrallel with the woffer.  For a first
order newtwork:

L = R/(2 Pi f)        C=1/(2 Pi R f)

For a Quasi-second order I recall that

L = R/(4 Pi f)        C = 1/(Pi R f)


Actually there is a coninum of available x-over choises

L = R/(2 x Pi f)         C = x /( 2 Pi R f)         1<= x <= 2

The Quasi-second order has 120 degrees phase shift between
the drivers at the x-over frequency.  This can be a problem
since a mere 60 degree phase error in the drivers (or due
to driver aligment) can cause a cancelation at the x-over
frequency.  You might try differen values of x to see 
what sound best.  By the way, R is the impedance of the
two drivers, and it is best that they both be close to
the same value (within say 30%).     

Chris Hull
hull@janus.berkeley.edu