Bicolor common anode led true bypass wiring

[aviherman5]

Hi, I want to make a Bicolor common anode led true bypass pedal. Is this diagram correct? Thank you!

[idy]

Hard to read, but yes.

The pins of the switch are numbered top to bottom left row 1,2,3, middle row 4,5,6, right row 7,8,9.

The first pole of the switch(1,2,3) sends the input jack tip on 2 either to 1, the board input, or 3, jumpered to 9 on the last pole (and then the output jack.)
Second pole connects a ground wire on 5 to either red, 4, or green, 6.
Third pole connects out jack tip on 8 to 7, the board out, or 9, the jumpered input jack.

A frequent variation allows the input of the board to be grounded in bypass, maybe quieter.
input jack tip to 1, board input to 2, ground wire to 3. pin  1 is jumpered (diagonally) to pin 9.
Pin 7 is the board out, pin 8 is the out jack. The middle pole is just as you have it.

[MikeA]

Take a look at your battery jack.  Battery jack ground lug should go to PCB ground point, not signal input jack ring lug/9V battery negative.

[aviherman5]

Hard to read, but yes.

The pins of the switch are numbered top to bottom left row 1,2,3, middle row 4,5,6, right row 7,8,9.

The first pole of the switch(1,2,3) sends the input jack tip on 2 either to 1, the board input, or 3, jumpered to 9 on the last pole (and then the output jack.)
Second pole connects a ground wire on 5 to either red, 4, or green, 6.
Third pole connects out jack tip on 8 to 7, the board out, or 9, the jumpered input jack.

A frequent variation allows the input of the board to be grounded in bypass, maybe quieter.
input jack tip to 1, board input to 2, ground wire to 3. pin  1 is jumpered (diagonally) to pin 9.
Pin 7 is the board out, pin 8 is the out jack. The middle pole is just as you have it.

Like this?

[aviherman5]

Take a look at your battery jack.  Battery jack ground lug should go to PCB ground point, not signal input jack ring lug/9V battery negative.

Otherwise good though? Any chance you could draw out what you mean? Thank you!

[MikeA]

Any chance you could draw out what you mean? Thank you!

Sure!  This will (1) use the presence of an external power plug to disconnect internal battery positive, and (2) use the presence of an input signal plug to connect internal battery negative.  (1) prevents paralleling internal and external power.  (2) is an off-switch to save the battery when the pedal is not connected to the signal chain.  Here' I'm using the input jack shield lug as a common ground point, but you can also use your PCB ground hole or any other common point. 

[aviherman5]

Awesome, thanks so much man! Really appreciate the feedback of both of you!

[amptramp]

There may be one simplification you can make:

The green LED may have a forward voltage of about 2 volts (the last one I measured was 2.05 V) and the red LED may have a forward voltage of 1.9 volts (the last one I measured was 1.88 volts).  You can connect it so the green LED is always powered on without any switch connection then when you want the red LED on, you have a switch contact connect it in parallel with the green LED.  Since the voltage is then held at the red LED voltage, the green shuts off because the voltage is limited to 1.88 and that is not enough to get any current through the green LED.

This uses only two lugs on the switch, not three.

[aviherman5]

Interesting suggestion will definitely keep in mind!