Bridging Rotel RB956ax: I don't understand how it works

[schro20]

I just got myself the Rotel RB956ax. 3x2 channels which can be bridged pair wise. Now I am looking at the circuit diagram and cannot for the world figure out just how the bridging works. Following the manual one is supposed to hookup the SE input to one channel of a given pair (the bridging switch then grounds the input of the second channel). The loudspeaker is then supposed to go between the + terminals of the two channels (and the grounds of either channel output are left open). Looking at the circuit diagram I expected to find an inverter... Usually the way I think of bridging is that the first channel gets the signal, the second channel gets the inverted signal, and the outputs have their grounds connected while the loudspeaker goes between the two + terminals of the output. I am missing something here. This circuit diagram seems to indicate that things work differently here. But how? Can someone help me out in understanding how this particular amp goes about its bridging business?

Why do I want to know? Well, for starters, now that I looked at the diagram I have this fundamental question: how is the bridging accomplished here. But my original motivation was to figure out how I might take best advantage of the fact that my pre-out is balanced. For example, can I just leave the bridging switch alone and hook the + of the balanced input to one channel, the - to the other and the ground to the common ground of both channels and then put the loudspeaker between the two + outputs of the two channels?

I am baffled. (No pun intended!)

Peter

[godfrey]

In bridge mode, switch S1 has pins 2 and 3 connected, so the output of channel A (arriving at pin 2 via R639) is routed to the inverting input of channel B (the base of Q686).
It should work as advertised, but your plan to use a balanced signal from the pre-out is good too.

Maybe try both ways and see which sounds better?

[schro20]

Originally Posted by godfrey

In bridge mode, switch S1 has pins 2 and 3 connected, so the output of channel A (arriving at pin 2 via R639) is routed to the inverting input of channel B (the base of Q686).

Doh. Yep. Now I see it. Ok. My world is back in order.

Originally Posted by godfrey

It should work as advertised, but your plan to use a balanced signal from the pre-out is good too.

Maybe try both ways and see which sounds better?

I shall.

Peter