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I'm not sure if this is the correct place to ask this, please notify me if there is a better place.

I have just created my first guitar pedal and thought i would go simple with a basic passive circuit but it doesn't work correctly.

If I am switched one way both potentiometers control the signal volume but if i am switched the other way there is no signal but there is a noticeable change in noise as i turn the potentiometer. The footswitch is supposed to switch the signal between the two potentiometers to have 2 selectable volume levels

This is how i have it wired currently: enter image description here

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    \$\begingroup\$ Schematic or it didn't happen. \$\endgroup\$
    – PlasmaHH
    Commented Apr 4, 2018 at 14:39
  • \$\begingroup\$ Welcome to the site. Please can you edit your question to show a schematic (not a block diagram). The schematic editor here is a breeze to use. The better the quality of question, the better the quality of the answers you will attract. Again, a warm welcome to the site. \$\endgroup\$
    – TonyM
    Commented Apr 4, 2018 at 14:45
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    \$\begingroup\$ The switch appears as a uniform 3x3 square, and you don't have your inputs and output jacks labeled. If you want the switch to select between the two pots, you should use two poles of the switch to connect/disconnect both of the leftmost two wires of both pots. Otherwise the unused pot will load down the signal. This might be acceptable when using line-level audio signals, but would probably be undesirable when using a guitar. \$\endgroup\$
    – supercat
    Commented Apr 4, 2018 at 14:53
  • \$\begingroup\$ You forgot to explain how it's supposed to work. If it is to allow you to switch between two volume settings by footswitch then please add this into your question. \$\endgroup\$
    – Transistor
    Commented Apr 4, 2018 at 21:54

1 Answer 1

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Usually, the input would go to the one end of the pot, and the output would be taken from the wiper (moving contact).

You can fix your circuit by simply exchanging the "IN" and "OUT" labels.

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  • \$\begingroup\$ I think i have it wired this way but i drew the schematic quickly and im out so i cant modify it \$\endgroup\$
    – Lachlan Jansen
    Commented Apr 4, 2018 at 22:13
  • \$\begingroup\$ My point is that you can make it work correctly by using the jack labelled as "IN" as the output. \$\endgroup\$
    – Peter Bennett
    Commented Apr 4, 2018 at 22:43

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