Leaderboard

I agree - put the amp on top of the pre-amp... or in it's own "private" space.

If I may make one suggestion, give that amp a bit more breathing room. It'll last longer if it's running cooler.

I will apply delicate attention to the screws this weekend, thank you for the emphasis on minute turns. As things stand now the components hide out of sight behind the left speaker, the True Sub Jr. behind the right speaker.

The effect has been discussed many times, however, the process has not been well defined. As I recall it has something to do with the change in signal path. Rich and BoB P discussed it in a long thread that I can't seem to find. Here are a few random quotes I did find: EDIT: I found this definition in the site "Glossary of Terms": Bridge Tied LoadUsing a stereo amplifier to drive a single channel loudspeaker, by strapping the load across the hot terminals of each channel(one channel is inverted in phase with respect to the other). This results in a higher power output from the amplifier, with some caveats. A standard stereo amplifier is a voltage amplification device, and normally supplies a voltage to two separate 'channels' of amplification. In the illustration below, a stereo amplifier (capable of providing a 45V signal to each channel) will dissipate 250 watts per channel into a nominal 8 ohm load (P=E2/R). Since current is the result of dividing voltage by resistance (I=E/R), the amplifier is able to source 5.5 amps per channel. When a stereo amplifier is bridged, two things happen. First, one of the channels is inverted with respect to the other (the two channels are now 180 degrees out of phase). Second, the 'hot' terminals of each stereo channel are connected to one loudspeaker (the remaining two output terminals are unused). The illustration below presents a bridge tied load configuration of a stereo amplifier. The per-channel output voltage remains the same (45V), and the load remains the same (8 ohms). What has changed, is that each stereo channel's 45V output is both out of phase with the other and tied to the same driver. The result is a 90V differential across the load (double the stereo voltage). Using P=E2/R from the previous example, the power dissipated by the load is now 1012W. Using I=E/R from the previous example, the current required to sustain 1012W is now 11.25A. This seems like a very good situation, and can provide more power to a speaker, provided you follow a simple guideline. The relation I=E/R tells us that as output voltage doubles, so does required current. Also, as driver impedance halves, required current will double. An ideal current source amplifier (often called load invariant) will be able to source the required doubling of current as the load halves (from 8-4-2-1 ohm). Not all amplifiers are load invariant (few are). Using the above formulae for a 4 ohm driver would (ideally) provide 500W into a stereo 4 ohm load (requiring 11.25A per channel), and 2025W when bridging the amp (now requiring 22.5A). The situation becomes harder for the amplifier to tolerate as the load impedance diminishes. At some point, the amplifier can no longer source the current required to provide a doubled voltage level. This is often experienced sonically as a very loud, but 'flabby' sound (especially in the bass). For this reason, most stereo amplifiers that are bridgeable are rated into 8 ohm load drivers only. <pet peeve> It's often repeated that when bridged, each half of the stereo amplifier will 'see' half of the driver load. This is untrue; the load is the load, and hasn't magically sprouted a center tap. What has changed is the current sourcing demand placed on the amplifier due to the doubled output voltage across the load.