Disconnect the woofer

I agree with Rodman. It depends on the crossover configuration. If it is more than a first order crossover (i.e., an inductor in series with the woofer), you would want to open the INPUT to the low frequency part of the crossover network, not its output which is connected to the woofer.

Regards,
-- Al

Contrary to audiophile wisdom, crossover networks "soak up" almost no power, even when all the drivers are connected. If they did they would get hot, and if mounted in a sealed enclosure stuffed with fiberglass or wool, very hot.

Disconnecting the load on the low pass leg may affect characteristics of the high pass leg.

Contrary to audiophile wisdom, crossover networks "soak up" almost no power, even when all the drivers are connected. If they did they would get hot, and if mounted in a sealed enclosure stuffed with fiberglass or wool, very hot.

Yes, inductors and capacitors do not dissipate significant power. However I would think that there would be some sonic benefit with many amplifiers to disconnecting the woofer's crossover elements, because the reactive (inductive and capacitive) load seen by the amplifier would be minimized.

Disconnecting the load on the low pass leg may affect characteristics of the high pass leg.

I believe that is only true for the relatively small number of speakers which are designed with the crossover legs in series. In those cases, a low pass filter is connected across the high frequency driver, a high pass filter is connected across the low frequency driver, and the two filters (and consequently the two drivers) are connected in series. The unwanted frequencies for each driver are shunted (bypassed) around the driver. But that kind of design is uncommon, I believe. More typically, the multiple crossover legs are in parallel and do not interact.

Koestner -- If you were thinking of leaving the low pass crossover elements unused, I'm not sure how you are planning to make the rolloff of the signal applied to the woofer complementary to the rolloff of the high pass part of the crossover. I think you would have to use an electronic crossover ahead of the low frequency amplifier, set to provide a low pass function for the woofer with rolloff carefully matched to what those crossover elements originally provided.

Regards,
-- Al

Contrary to audiophile wisdom, crossover networks "soak up" almost no power, even when all the drivers are connected. If they did they would get hot, and if mounted in a sealed enclosure stuffed with fiberglass or wool, very hot.

Agreed. They don't soak up much power. However they do act as a bottleneck or choke and any voltage drop across the crossover means less amplifier power reaches the speaker drivers.

Perhaps you've never seen an EQing X-over, such as that in the Rogers, Harbeth, Chartwell, Spendor or KEF LS3/5A, LS5/9 and other BBC approved monitors, the Harbeth HL-P3(with it's 21 element crossover) and so on. There are a number of trim resistors, shunt coils(some that act as autotransformers), and caps that are in the notching/shelving circuits and go to ground from the positive input(the main reason they are so stinking inefficient). If these are not removed from the LF signal path, they WILL dissipate energy as designed, but even more as the woofer will not be in the circuit as a load. That's why I asked what his crossover looked like. There are too many variables to make a blanket(uneducated) statement. ie: What would occur in this system, if the woofer were simply removed: (http://www.syer.net/images/DAHLQUIST_DQ10Schems.jpg) YES- The best/most accurate way to bi-amp is via an active x-over, before the power amps. The second best, is to separate the HF/LF sections of the existing crossover, install two new terminals and use identical amps on both sections(either horizontally or vertically).