Posted by Wayne Parham [ 65.214.102.163 ] on March 22, 2007 at 12:06:31:

In Reply to: This looks like the opposite of what I'm asking.... posted by Elvis on March 22, 2007 at 10:16:05:

I see. Then connect it like this:

         (+)
          |
          |
     -----------
     |          |
     s          s
     p          p
     k          k
     |          |
     |          |
     s          s
     p          p
     k          k
     |          |
     |          |
     s          s
     p          p
     k          k
     |          |
     |          |
     s          s
     p          p
     k          k
     |          |
     -----------
          |
          |
         (-)

Each speaker is represented by "spk" with the "s" side being positive and the "k" side being negative. The total impedance of the network is double the impedance of a single driver.

Replies:

• thanks - Elvis 19:15:22 03/22/07 (0)