In Reply to: ????? posted by Kal Rubinson on March 8, 2007 at 18:32:44:
Unproven and unlikely. Since they come from the same output stage, all biwiring is doing is moving the shunt across the speaker terminals back to the amp terminals.And by doing that, it is no longer a shunt. In a bi-wire configuation, there are two cable lengths of impedance between the high and low crossover sections. It is not the same circuit.
Dave
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Follow Ups
- Re: ????? - Dave Kingsland 05:08:25 03/09/07 (9)
- Re: ????? - Kal Rubinson 07:37:06 03/09/07 (8)
- Re: ????? - Dave Kingsland 08:50:36 03/09/07 (7)
- Re: ????? - Kal Rubinson 09:26:24 03/09/07 (6)
- Re: ????? - Dave Kingsland 10:43:14 03/09/07 (5)
- Re: ????? - Kal Rubinson 11:30:37 03/09/07 (4)
- Re: ????? - Dave Kingsland 13:25:02 03/09/07 (3)
- Re: ????? - Kal Rubinson 14:08:35 03/09/07 (2)
- Here are some graphs of CALSOD examples - dlr 05:40:47 03/11/07 (0)
- Thank you both for that excellent exchange - Craiger56 16:42:47 03/09/07 (0)