crossover calc programs that can do this?

[mondogenerator]

I have been designing my own crossovers since i started making loudspeakers, a few years back. My knowledge of the design of a 2nd order crossover for example, has come from reading Weems' book, and Capels' also.

As such i use the root2 or 1 over root 2 ratio in moving from a 1st to 2nd order filter, as described in Weems' book.

Yesterday i found a ready designed crossover for the drivers im using in my current project, and while checking out component values i noticed that the 2nd order LP for the woofer consists of 2.2mH inductor and 4.7uF capacitor. This doesnt fit the usual ratio that ive seen between L and C, and since the HP has a 4.7uF cap in as well i guess this is just to reduce costs by using the same value and messing with the indutor value to achieve the desired crossover point. Also, as the inductor looks very large for anything other than a LOW crossover point, im thinking that maybe the design allows for some BSC.

Also, is there a program out there that will allow me to input the values of L and C and calc the crossover point? OR can i just calc it as normal, using a calculator and a bit of paper, like with normal Xc or Xl reactance type calcs? i accept this may be a silly question, and im really not all that daft, but i havent come across this situation before

[mondogenerator]

ok well im either asking a stoopid question, OR everyone else doesnt know either...

[mjf]

hello.
.......yes, 2.2mH looks a little bit like baffle step correction (in a 2way speaker design).
you can choose the L C values in your crossover as you like,you must not follow the standard calc's in the textbooks............if it works - why not?
have you tried to simulate it (with an electronic sim,workbench,spice....)?
greetings

[Shaun]

Check out the free MS Excel-based utilities at FRD Consortium. It's all there.

[Michael Chua]

Hi mondogenerator

This is a short cut to save a few dollars. Instead of using a BSC and Zobel, a larger inductor is used. With this larger value, the Butterworth crossover is no longer maximally flat. The woofer will roll-off earlier.

The trick is to adjust the inductor so that the rising response in the mid is leveled, leading to a relatively flat response.

Regards
Mike

[Pano]

Mondo - you should definitely check out the FRD stuff that Shaun linked to. The Passive Crossover designer is great. It can teach you a lot about how filters work.

[tsmith1315]

Originally Posted by mondogenerator

OR can i just calc it as normal, using a calculator and a bit of paper, like with normal Xc or Xl reactance type calcs?

Fo = 1/[2*pi*sqrt(LC)]

The response shape near Fo will be determined by the driver's impedance.

[Shaun]

Originally Posted by tsmith1315

Fo = 1/[2*pi*sqrt(LC)]

The response shape near Fo will be determined by the driver's impedance.

The problem with this textbook approach is that it won't answer the OP's questions about why the new crossover values are so different. Best to use actual in-box frequency response and impedance curves. The FRD tools help to do this.

[mondogenerator]

well thanks all. I did think it was possibly a way of BSC. unfortunately, i have the FRC stuff and no excel, and openoffice opens them but i get a lot of errors so either it doesnt like the file or the file had bugs. either way i cant get it to do anything really....but thanks for the suggestion anyway shaun.

basically i was trying to reverse engineer the crossover, so i could maybe establish the crossover point and calculate my own one, around a similar point. From experiments, the crossover works very well and sounds very flat, in room.

as the design im planning -

My visaton TQW tube speaker eventually started!!!

-has a similar width baffle, im not concerned that the BSC will be too much or located in the wrong place as such so i might just go ahead and use it anyway.

Thanks all for the suggestions and advice

[tsmith1315]

Shaun, it's not an approach, it's a direct answer to his second question. I encourage anyone who's doing any sort of manual calculations these days to continue.


I do agree that crosover complexity can't be reduced to a couple of formulas. My statement following the formula was meant to encourage curiousity about the subject.