I am working on creating an amplifier with multiple input sources. I have implemented a 3P4T rotary switch to be able to select the audio source among 4 different ones : RCA, Phono, and 2 digital sources coming out of the same DAC. There is a multiplexer above the DAC to select which digital source to send to the DAC. Here is a functionnal drawing of my setup :

Functionnal drawing

As you can see, the 3 poles of the switch are used for : 1. Multiplexer source selection, 2. Left channel, 3. Right channel.

Now I need to be able to add 4 LEDs, to display visually which source is activated.

Obviously, the easy solution would be to use a 4P4T switch instead, and use the 4th pole to handle the LEDs, but here comes the constraint : I need the switch to be able to be soldered directly on the board. However, all 4P4T switches I have seen have at least 2 decks such as this one, which seems to mean it's impossible to just solder it on a board, compared to a 3P4T which can, thanks to their 15 THT pins like those

Hence, my question is the following : how would it be possible to install LEDs on the output of a 3P4T when all the pins are already used ? Could I use one of the poles to perform two actions ? Or do you have another solution in mind ? I've been searching for one all day and can't seem to have a lead. Maybe there's an easy solution I have missed by being to stubborn on one idea, I don't know.

I'd like to avoid having to use a µcontroller, and I need components which can be soldered directly to the board... Else you understand that I would have jumped on a 4P4T.

Hopefully I was clear enough. If not, don't hesitate to ask for more information. Big thanks in advance,


1 Answer 1


This should do it (common on switch to +5V):


simulate this circuit – Schematic created using CircuitLab

If your LEDs are different colors you may wish to use individual resistors rather than R1 to match brightness between the LEDs.

  • 1
    \$\begingroup\$ Thank you so much, it's exactly what I needed ! I knew it wasn't that hard and the solution was right there. Thanks again for taking the time to have a look into it ! \$\endgroup\$
    – hleurent
    Commented Aug 16, 2021 at 9:01

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