[Dave_M]

I got a 1/4" jack and can't work out how the switch operates. Here is the schematic from the data sheet:

Here is a picture of the Jack:

(you might recognize it from the graham slee solo

)

 

[Pars]

Well, if I were guessing I'd say that pins 2 and 3 are the L and R channel, respectively, but I'll bet you had already figured that part out

Pin 1 is obviously the ground. I would guess that pins 7-9 are all switched R channel, and pins 4-6 are switched L channel. Strange switch diagram, but I'd guess normally closed, with the switched pins borken upon jack insertion.

I would verify with a plug and a DMM though.

 

[Girlsound]

As stated above, pin 1 is ground. Pins 2 and 3 are left and right. With phones inserted, pins 5 and 6 short as well as 8 and 9. Without phones inserted, pins 4 and 5 short as well as 7and 8.

 

[MisterX]

Quote:

Strange switch diagram, but I'd guess normally closed, with the switched pins borken upon jack insertion.

Those strange symbols in the switch typically indicate "the switch activation mechanism" is electrically isolated from the plug.

Quote:

Without phones inserted, pins 4 and 5 short as well as 7and 8.

It really doesn't look like 7 8 & 9 ever short but that may be a flaw in the schematic (4 5 & 6 do appear to short when the plug is inserted)

 

[PinkFloyd]

Turn the Jalco upside down you'll see that each leg is numbered. Connect left channel to pin 2, right channel to pin 3 and ground to Pin 1. That's you good to go.

Mike.

 

[Dave_M]

With this socket then it should be possible to set it up so it doesn't short when headphones are inserted unlike other sockets. If output of amp is connected to pin 9, and pin 8 to pin 3. Then that should stop it shorting as long as the switch operates only when the plug is fully inserted.

I have one of these somewhere, I will test it with the multi meter.