Does biwiring change impedence?

Maybe he was talking about strapping two stereo amps to mono and using one amp for each speaker. When you strap a stereo amp to mono the load it sees from the speaker is cut in half.

The system was two stereo amps in bridged mono and the owner of the system had them biwired to his 4 ohm Paradigm speakers. I think it would be better to run the amps in stereo and biamp, but that is the way it was. The amps were H/K 16's upgraded and bridged by a craftsman who makes a living repairing and selling audio gear. The seller (the technician) is the one who told both of us that biwiring cuts in half the impedence. He said the amps were looking at 2 ohms since they were bridged and biwired.

Yes, they were looking at 2 ohms, but not because they were biwired. As Racamuti correctly speculated, since the amps were bridged they would see the speaker impedance (in this case 4 ohms) divided by two. They would see the same 2 ohms even if the connections were not biwired.

Regards,
-- Al

OK. When an amp is bridged, is impedence cut in half absolutely, or is the impedence inversely proportional to the power gained? My Citation 7.1 is rated at 130W x 4 and 400W X 2 bridged stereo. I think it puts out 1KW into 1 ohm.

It is cut in half absolutely, and the power gained is in turn limited by how much power (voltage and current) can be delivered into that reduced impedance while operating in bridged mode.

Basically, what bridged mode does is to put equal but opposite polarity signals into the two output sections of the amplifier (that are normally assigned to each of the two channels), with the speaker being connected between the positive output of one channel and the positive output of the other channel. That results in the voltage across the speaker being twice what it would have been in normal stereo configuration, under the same signal conditions.

By Ohm's Law (voltage = current x resistance) the doubled voltage will cause twice the current to flow through the speaker, and hence through each of the two output amplifier stages (which are essentially in series with each other and the speaker). From the perspective of each amplifier stage, twice the current is flowing that would flow at the same voltage in normal stereo configuration. So from the perspective of each amplifier stage the load impedance has been cut in half, again by Ohm's Law (resistance = voltage/current).

Since power = voltage x current (assuming the voltage and current are in phase, which is to say that the load is assumed to be essentially resistive), which in turn equals (by substitution of the Ohm's Law formula) voltage squared divided by resistance, bridging can potentially provide 4 times the power of normal stereo operation. But that is typically limited to a considerably lower number by peak current capability.

Hope the clarifies more than it confuses!

Regards,
-- Al