BSC with 2 drivers? - how to

[AndrewUK1990]

Im looking to attenuate some mid/bass drivers at the low frequency range. The drivers are 4ohm impedance and 3.4ohm DC resistance, they are wired in series with one another into the Xover point. Now my question is, when making a simple BSC circuit do i take the DC resistance as the combined resistance of both drivers, ie 6.8ohm, or just one at 3.4ohm? Baffle width is 15". Im assuming i would just have to wire the BSC in series coming from positive from Xover then to driver #1 which is wired in series to driver #2 then back to the negative port on the Xover.

Cheers in advance as always!

[PeteMcK]

You use the impedance of the drivers AT THE BAFFLE STEP FREQUENCY.
If you're using two drivers in series, double it.
This means you need an impedance plot - but if you don't have one, DC resistance will probably get you in the ball park.

An alternative way to do this is to apply BSC to only one driver (making a 2.5 way system), if the woofers are in series you'd do this by adding a cap in parallel with the lower driver...cheaper than an inductor.

[AndrewUK1990]

right ok, thanks for pointing that out! They are scanspeak drivers so i can easily find the impedance plot from the charts on website. They are 3 way speakers so i think ill just stick to the conventional way of doing it with an inductor and resistor, dont want to confuse myself :P

[Inductor]

Originally Posted by AndrewUK1990

Im looking to attenuate some mid/bass drivers at the low frequency range.

Do you mean to attenuate in the upper (low) frequency range?! (BSC)

[DSP_Geek]

You really want to include a Zobel network on the driver, otherwise the impedance rise from its equivalent series inductance will mess up your BSC filtering.

[DSP_Geek]

Originally Posted by PeteMcK

You use the impedance of the drivers AT THE BAFFLE STEP FREQUENCY.
If you're using two drivers in series, double it.
This means you need an impedance plot - but if you don't have one, DC resistance will probably get you in the ball park.

An alternative way to do this is to apply BSC to only one driver (making a 2.5 way system), if the woofers are in series you'd do this by adding a cap in parallel with the lower driver...cheaper than an inductor.

That won't work. The lower driver is indeed bypassed but the higher driver now sees twice the voltage, so with coupling the far-field response doesn't change at all.

There are entirely valid reasons for doing this - vertical dispersion control, for one - but BSC ain't it.

[planet10]

Originally Posted by DSP_Geek

That won't work. The lower driver is indeed bypassed but the higher driver now sees twice the voltage, so with coupling the far-field response doesn't change at all.

There are entirely valid reasons for doing this - vertical dispersion control, for one - but BSC ain't it.

Very true... unless you are using a current amp.

dave

[sreten]

Hi,

As said the parallel capacitor approach will not work with most amplifiers.

The approach that does work is oversizing the the first inductor in what
is usually a 2nd order electrical low pass filter, forget calculators, it
should be modelled to get near the correct values, see the links.

rgds, sreten.

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