billt@chips.com (Bill Tuohy) (02/13/91)
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. While I'm at it, I have another question about my design. I was thinking of using two 5 1/4" midrange cones per speaker, for increased power handling, but am unsure how to connect them. I think I would like to keep the impedance at a nominal 8 ohms. My first thought was to connect two 4ohm drivers in series, but then I read that series connections can cause problems if the drivers don't match exactly and the amp has a high damping factor (or was that a low damping factor?) Anyway, if I connect two 8ohm drivers in parrallel, the impedance is 4ohms. Can I put 8ohm resistors in series with the drivers, making it a parallel connection of 16ohm drivers, giving 8ohms? If I have an L-pad for the midrange, should I leave the parallel connection at 4ohms and adjust with the L-pad? My Carver amp recommends using only 8ohm speakers when A and B speakers will be used together. I guess the A and B are connected to the amp in parallel. Is there any other benefits or problems with using two midrange drivers like this? I assume that dual-woofer theory applies to base-mid cones as well. Would two drivers give any better imaging or anything, or should I try to find a single driver that can do what I want power wise? Sorry for all the questions, but as anyone who has designed speakers knows, there are a lot of variables. Thanks, Bill
hull%hp-lsd@janus.Berkeley.EDU (Christopher Hull) (02/18/91)
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