baffle step correction and bypassing it

xcortes · 8th May 2004, 06:19 PM

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

Bill Fitzpatrick · 8th May 2004, 06:54 PM

Assuming the network is a single inductor, your proposal will work.

xcortes · 8th May 2004, 07:07 PM

it is a single inductor in parallel with a resistor.

Bill Fitzpatrick · 8th May 2004, 09:52 PM

That should still be OK. Try it out.

5th element · 8th May 2004, 10:04 PM

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.

Bill Fitzpatrick · 8th May 2004, 10:30 PM

You're missing something. An inductor and resistor in parallel with the driver wouldn't be a BSC circuit.

xcortes · 8th May 2004, 11:18 PM

the network is a resistor in parallel with an inductor.

the network is in series with the driver:

xcortes · 8th May 2004, 11:20 PM

the drawing is wrong:

the network is in series with the +, not with the -.

Bill Fitzpatrick · 8th May 2004, 11:50 PM

If the drawing is wrong why did you post it?

5th element · 8th May 2004, 11:59 PM

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.

8th May 2004, 06:19 PM	#1
xcortes diyAudio Member Join Date: Feb 2004 Location: Mexico City	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

8th May 2004, 11:59 PM	#10
5th element diyAudio Member Join Date: Oct 2002 Location: England	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. __________________ What the hell are you screamin' for? Every five minutes there's a bomb or somethin'! I'm leavin! bzzzz! Droggon Attack!

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8th May 2004, 06:54 PM	#2
Bill Fitzpatrick diyAudio Member Join Date: Jun 2001 Location: Eugene, OR	Assuming the network is a single inductor, your proposal will work.

8th May 2004, 07:07 PM	#3
xcortes diyAudio Member Join Date: Feb 2004 Location: Mexico City	it is a single inductor in parallel with a resistor.

8th May 2004, 09:52 PM	#4
Bill Fitzpatrick diyAudio Member Join Date: Jun 2001 Location: Eugene, OR	That should still be OK. Try it out.

8th May 2004, 10:30 PM	#6
Bill Fitzpatrick diyAudio Member Join Date: Jun 2001 Location: Eugene, OR	You're missing something. An inductor and resistor in parallel with the driver wouldn't be a BSC circuit.

8th May 2004, 11:20 PM	#8
xcortes diyAudio Member Join Date: Feb 2004 Location: Mexico City	the drawing is wrong: the network is in series with the +, not with the -.

8th May 2004, 11:50 PM	#9
Bill Fitzpatrick diyAudio Member Join Date: Jun 2001 Location: Eugene, OR	If the drawing is wrong why did you post it?

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