Happy new year to all!

I'm in the midst of a project that involves converting a once-powered subwoofer to a passive version (i.e., just the original enclosure and 10" driver). The original subwoofer had an LED indicator on the front panel (which illuminated when its amplifier was active), and so I'm trying to duplicate this for cosmetic purposes, but then recently had the idea of installing an LED switch. The new switch would duplicate the illumination, but also allow me to quickly enable/disable the subwoofer.

I'm comfortable supplying the power (say, 5v + any required resisters) to drive the lighted switch, but my confusion concerns how these 4-pin switches operate. In this application, would I also be introducing 5v to the speaker signal? The incoming power should function ONLY to light the LED when the switch assembly is depressed, whereas the "switched circuit" should be the (- or +) audio signal coming from the amplifier.

Are these LED push buttons intended to drive the attached load with the same power that drives the LED?

 

You will need a relay- the speaker signal is AC and it's relation to ground is unknown, and cannot be assumed.
A 5V relay will isolate the audio amp output so nothing bad can happen.

This actually makes the project simple, just connect the relay coil as the load for the switch, (with diode backwards across the coil to prevent inductive spike) connect the relay NO contacts in series with the speaker.

 

The Terminals marked N-O are the Normally-Open Contacts,
and are not connected to the LED Terminals.

The Question is, why would You want to create a potential "Point-Of-Failure",
or Switch-Off the Sub-Woofer in the first place ?????

I could maybe see using an existing LED as an "Cone-Over-Excursion" Warning-Light,
that might be considered somewhat useful, although, a little more complex,
and neither are even remotely necessary, just slightly "cool".
.
.
.

 

From the diagram it seems the switch is a single pole thus, and you will not separate the load switching from the the LED switching without extra poles such as a relay.

 

The Question is, why would You want to create a potential "Point-Of-Failure",
or Switch-Off the Sub-Woofer in the first place ?????

For light music listening in the afternoon, when discerning ears are not home and the subwoofer isn't needed. The ability to depress a switch on the cabinet front will be notably easier than the alternatives:
1. Detaching a speaker wire from the back panel;
2. Navigating into the equipment menus to delete the subwoofer from the configuration.

It is, in fact, even a bit easier than the original state, when one needed to reach behind the subwoofer and turn off the amplifier plate.

 

It's not like you're going wear it out somehow.

Sorry but I just don't get the need for ever turning it off in the first place ????????
.
.
.

 

.....I just don't get the need for ever turning it off in the first place

Keeping the neighbours happy, for one.

 

You will need a relay- the speaker signal is AC and it's relation to ground is unknown, and cannot be assumed.
A 5V relay will isolate the audio amp output so nothing bad can happen.

This actually makes the project simple, just connect the relay coil as the load for the switch, (with diode backwards across the coil to prevent inductive spike) connect the relay NO contacts in series with the speaker.

This is very useful. Thanks to you and boostbuck for confirming the need for a relay. I've not used a relay in a project. Could you point me to a real device (or part number) that would suit this particular application? Many thanks in advance

 

Something like this, perhaps:
https://au.element14.com/panasonic/tx2-5v/relay-signal-dpdt-30vdc-2a/dp/1200950

 

If you used a DPDT LED switch you wouldn't need a relay.
One set of contacts would control the LED while the the other (isolated) set of contacts would switch the woofer signal.

 

What @crutschow said. I was just about to suggest the same.