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Old 8th May 2004, 06:19 PM   #1
xcortes is offline xcortes  Mexico
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Default baffle step correction and bypassing it

I built a pair of minimonitors using Jim Griffin's design based on Jordan JX92S drivers. I included a baffle step compensation network as suggested by Jim.

Sometimes I use these speakers in my bedroom and there I don't have the choice of placing them far from the back wall. In that placement the compensation network adds to the the back wall resulting in too much bass response.

What I would like to do is to be able to bypass the network depending on the speakers position. Kind of what the Revel Performa M20 speakers do with a switch to select stand mounting or flush mounting.

Instead of adding a switch (expensive a good one is!) I'm considering adding a third binding post connected directly to the positive connection of the driver. This way I connect the speaker cables choosing one of two binding posts: one goes through the compensation network, the other bypasses it.

Will this work? I'm only concerned that, when bypassing the network it will still be connected to the driver. You could say that the signal would be divided between the network and the driver. Am I explaining myself?

Thanks
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Old 8th May 2004, 06:54 PM   #2
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Assuming the network is a single inductor, your proposal will work.
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Old 8th May 2004, 07:07 PM   #3
xcortes is offline xcortes  Mexico
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it is a single inductor in parallel with a resistor.
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Old 8th May 2004, 09:52 PM   #4
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That should still be OK. Try it out.
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Old 8th May 2004, 10:04 PM   #5
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Im not sure how this is going to be wired up to actually work? Maybe im missing something but if the network is in parallel with the driver then you will need a physical way of disconnecting the circuit from the drive unit.

How would a third binding post achieve this?

See pic? is this how you were thinking of doing it? Red BP one and RED BP two will work in the same way.
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Old 8th May 2004, 10:30 PM   #6
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You're missing something. An inductor and resistor in parallel with the driver wouldn't be a BSC circuit.
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Old 8th May 2004, 11:18 PM   #7
xcortes is offline xcortes  Mexico
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the network is a resistor in parallel with an inductor.

the network is in series with the driver:
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Old 8th May 2004, 11:20 PM   #8
xcortes is offline xcortes  Mexico
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the drawing is wrong:

the network is in series with the +, not with the -.
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Old 8th May 2004, 11:50 PM   #9
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If the drawing is wrong why did you post it?
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Old 8th May 2004, 11:59 PM   #10
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Yes I now realise that you were correct, I didnt see that you couldnt use a parallel BSC circuit as used with filters to compensate for BSC. The idea you suggested to alter between BSC comp and no BSC should work fine.
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