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Old 14th July 2006, 02:38 PM   #1
Tenson is offline Tenson  United Kingdom
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Default Passive Shelf Filter

I have built a pair of standmount speakers and got the components for the passive crossovers the other day. The crossover sounds great, but the notch filter I had planned is way too aggressive. However I found coincidently, that the treble control on my Yamaha amp which I was using to test them worked perfectly for taming the treble peak.

How do I build a shelf filter for the tweeter? It is a B&G Neo8 which has a flat impedance of 3.6Ohms.

When I set the amps treble control to -2dB it sounds great. I took a measurement of the amps output and have attached it below. How can I duplicate this in the crossover?

From the Rod Elliot site there is a baffle step compensation circuit but it is for line level. I guess I need a resistor and a cap in parallel with the tweeter but Iíve no idea how to determine the values.

Top measurement is no treble control adjustment and bottom is the treble control at -3dB, but I have decided its better at -2dB.

Click the image to open in full size.

Thanks for your help peeps!
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Old 14th July 2006, 05:19 PM   #2
owdi is offline owdi  United States
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Have you considered a simple LPAD, it will accomplish close to the same thing.

Dan
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Old 14th July 2006, 08:17 PM   #3
Tenson is offline Tenson  United Kingdom
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It would do nearly the same but not quite. I think the slope being at (looks like) 3KHz helps fix a slight suck out around the crossover point as it brings the rest down closer to its level which is nice. I don't have the right value resistors to do an L-Pad right now so I might as well do the shelf if someone knows how to work it out for the impedance I have here?

What I want must be the same as passive baffle step compensation but at a higher frequency. Isn't thins a normal kind of circuit?
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Old 14th July 2006, 08:24 PM   #4
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Sure is - usually consisting of an inductor and parallel resistor. See http://www.quarter-wave.com/General/BSC_Calculator.xls for one way to caluclate the values, there are others.
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Old 14th July 2006, 08:39 PM   #5
Tenson is offline Tenson  United Kingdom
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Hi,

I found a formula for doing a passive baffle step correction but it is for doing -6dB and you set the -3dB point. How can I change this for getting only -2dB as the final attenuation of the shelf?


Quote:
TrueAudio.comFind the network required to compensate the spherical diffraction loss of a 4 Ohm speaker system with a 0.25 meter wide baffle. R = 4 (the nominal impedance of the system) (this resistor should have a power rating something like a quarter of the system power rating) L1 = .25 x 4 / 1.021 = 0.979 mH (1 mH will be close enough)
I see most of these suggest using an inductor. Is it not possible to use just a resistor and a cap in parallel across the signal line like rod Elliot does for his active baffle step circuit? I don't like inductors

Thanks guys!
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Old 14th July 2006, 09:06 PM   #6
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You could, using a standard Lpad layout except instead of the resistor across the tweeter terminals use a resistor and capacitor in series.

I don't know of an easy to use calculator for it, though. Probably the easiest thing to do would be model it in Speaker Workshop or other XO program and play with the values until you get what you want.
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Old 14th July 2006, 09:27 PM   #7
Tenson is offline Tenson  United Kingdom
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Okay scrap that last message...

I just downloaded The Edge. It has a baffle step calculator for circuits using two resistors and a cap. But it does -6dB compensation. Anyone know how I can get it to only give me -2dB?

Thanks
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Old 14th July 2006, 10:00 PM   #8
Svante is offline Svante  Sweden
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Quote:
Originally posted by Tenson
Okay scrap that last message...

I just downloaded The Edge. It has a baffle step calculator for circuits using two resistors and a cap. But it does -6dB compensation. Anyone know how I can get it to only give me -2dB?

Thanks
If you want a passive circuit you should use the upper one with the inductor. if you don't want the full 6 dB step, you cvan manually adjust the f1 frequency to about 80% of f2. Use the nominal impedance of the speaker as R1.
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Old 14th July 2006, 11:00 PM   #9
Tenson is offline Tenson  United Kingdom
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Thanks. Why do you suggest the one with an inductor? I have always felt inductors do far worse things to the sound than a cap in parallel.

Maybe this is a silly question but I'm silly... will the load presented to the rest of the crossover still be 3.6Ohms with this just before the tweeter?

By the way, The Edge is a great program. I have not needed to use it much but I know my friend does and it has been a lot of help here!
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Old 15th July 2006, 08:49 AM   #10
Svante is offline Svante  Sweden
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Quote:
Originally posted by Tenson
Thanks. Why do you suggest the one with an inductor? I have always felt inductors do far worse things to the sound than a cap in parallel.

Maybe this is a silly question but I'm silly... will the load presented to the rest of the crossover still be 3.6Ohms with this just before the tweeter?
Well, the lower circuit is active, ie it needs to be inserted before the power amplifier. A passive circuit with a parallel capacitor would require a series resistor, and this would steal at least half of the power. No such circuit is suggested by The Edge.

The loading issue is not resolved in The Edge. It simply assumes that the load is R1, which is an approximation. Also, if the compensation is inserted after the crossover, the impedance seen by the crossover will be larger for lower frequencies.

Really, if BSC is to be performed thoroughly, the simple calculator in The Edge is too simple. It should be integrated with the crossover and loudspeaker design, and simulated as a whole.
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