what effect does a 4 driver have on an 8 ohm crossover?

[60ndown]

in general terms, when using a 4 ohm driver in a system that has an 8 ohm x over, what does that do to x over points/volume/anything else?

in general terms ?

[tinitus]

in general ?

depends on whether its a tweeter or woofer, or mid
in other words, high pass or low pass

series inductor reacts as if twice as big, thus cuts lower

series caps on a tweeter reacts the opposite, and like it was smaller
and so on

simpler
low impedance results in smaller series inductor(low pass)
high impedance results in smaller series cap(high pass)

[60ndown]

lol, i *almost* understood that

im using one of these in a recent build,

DIAMOND 2 WAY STEREO PASSIVE SUBWOOFER CROSSOVER S @NEW | eBay

when i connected my (4) subwoofers, each pair was about 3.9 ohms.

i think they should have been 8 ohms / pair?

sounds good, just wondering whats actually xed over where ??

Originally Posted by tinitus

in general ?

depends on whether its a tweeter or woofer, or mid
in other words, high pass or low pass

series inductor reacts as if twice as big, thus cuts lower

series caps on a tweeter reacts the opposite, and like it was smaller
and so on

simpler
low impedance results in smaller series inductor(low pass)
high impedance results in smaller series cap(high pass)

[planet10]

In general terms it just doesn't work... have you got an off the shelf generic XO? Usually they just don't work.

dave

[chlorofille]

Say you have a 4ohm driver and use a 2 way 4ohm crossover which rolls off at 12db/oct @ 2000Hz. If you use the same crossover for an 8ohm speaker, both the tweeter and woofer will start rolling off earlier (eg 4000Hz and 1000Hz) respectively so you will be left with a "hole" in the midrange.

[60ndown]

Originally Posted by chlorofille

Say you have a 4ohm driver and use a 2 way 4ohm crossover which rolls off at 12db/oct @ 2000Hz. If you use the same crossover for an 8ohm speaker, both the tweeter and woofer will start rolling off earlier (eg 4000Hz and 1000Hz) respectively so you will be left with a "hole" in the midrange.

so doubling/halving the ohms, kinda doubles/halves the x-over point?

im guessing theres plenty of variables (impedance rise etc) in there, but thats a good simple general explanation,

thank you.

so looking at the crossover im using (posted above) it looks like the subwoofer x-over is 150 @ 12db with an 8 ohm driver, im using (4) 6 (approx) ohm drivers, each pair wired for 3.9,

so the sub is now crossed at 75 hz not 150, and i have a gap from 75-150 approximately + variables.

im going to wire the subs for 12 ohms tomorrow and see what that sounds like,

its easier for me to listen then think

[Jay]

Originally Posted by 60ndown

in general terms, when using a 4 ohm driver in a system that has an 8 ohm x over, what does that do to x over points/volume/anything else?

Xo point will be lower. How much lower cannot be predicted, as these numbers are only approximation. Theoretically one octave but empirically will be smaller than that. This is not a problem. I would probably prefer this as the integration/overlap between subwoofer and main speaker become "minimal".

Volume will be higher (thus better subjective quality). This is because the power is approximately doubled. Formula are V=IR, P=VI. V is the signal from your amp. With the same V, if you half the R, I will be doubled, hence also P.

System minimum impedance will be lower, probably around 1 ohm. Lets say if with 8-ohm subwoofer minimum impedance is 4 ohm, with 4-ohm subwoofer it could reach 3 ohm. Your amplifier will work harder. Probability to fail (at high volume) is higher.

[60ndown]

very nice, thank you.

Originally Posted by Jay

Xo point will be lower. How much lower cannot be predicted, as these numbers are only approximation. Theoretically one octave but empirically will be smaller than that. This is not a problem. I would probably prefer this as the integration/overlap between subwoofer and main speaker become "minimal".

Volume will be higher (thus better subjective quality). This is because the power is approximately doubled. Formula are V=IR, P=VI. V is the signal from your amp. With the same V, if you half the R, I will be doubled, hence also P.

System minimum impedance will be lower, probably around 1 ohm. Lets say if with 8-ohm subwoofer minimum impedance is 4 ohm, with 4-ohm subwoofer it could reach 3 ohm. Your amplifier will work harder. Probability to fail (at high volume) is higher.

[tinitus]

Originally Posted by Jay

Xo point will be lower.

that is correct
not saying it will function 100% correctly tho
it will start rolling off a bit earlier
but maybe also more 'shallow'
due to paralel caps now smaller, in relative terms

but for a sub filter, it might be even better this way

anyway, very low cut passive xo tend to roll off later than 'expected'
woofer/box low ressonance make it almost impossible to achieve very low xo point with passive xo
meaning, trafo inductors would need to be massive
and looking at actual inductors used here, the 150hz claimed is very unlikely to happen
but the chance is definately better now, with low impedance woofers

[60ndown]

sometimes i just get lucky