3
\$\begingroup\$

Came across some old Arcam speakers (cut open) and using the parts to build a bluetooth speaker box. Only problem is I can't figure out the wiring on the crossover/filter. The original speakers are pretty old so Google didn't help much.

enter image description here

Image of component side enter image description here

Have a 2-channel audio amp board and trying to split the signal to one tweeter and one mid on each side.This is what I'm trying to achieve: enter image description here

So question is: does anybody know what the cables are, and how can I cut/solder to achieve my goal?

\$\endgroup\$
2
  • \$\begingroup\$ Any idea where the various wires previously connected? \$\endgroup\$
    – mike65535
    Nov 19 '18 at 18:52
  • \$\begingroup\$ @mike65535 Sorry, have no idea where the cables went. All I know is that the speaker case had 4 inputs, so 4 cables went from the input to the crossover \$\endgroup\$
    – ChrVik
    Nov 19 '18 at 19:05
3
\$\begingroup\$

I'd say that connections 1 and 5 ("I") are inputs. There's a notation on the board that the input is "bi-wire" (presumably to support a bi-amp configuration).

All of the connections labeled "E" are common connections.

The connection labeled "T" goes to the tweeter. This half of the board is configured as an LC high-pass filter — capacitor in series, inductor in parallel with the load.

The connection labeled "B" goes to the bass driver. This half of the board is configured as an LC low-pass filter — inductor in series, capacitor(s) in parallel with the load.

So,

  • 1 and 3 are the input from the amplifier for the tweeter
  • 2 and 4 go to the tweeter itself.
  • 5 and 6 are the input for the woofer (bass)
  • 7 and 8 go to the woofer itself.

If you are single-amping this, then you can connect 1 and 5 together, as well as 3 and 6.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.