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I'm following the wiring diagram for a digitech fs3x guitar pedal footswitch.
For the third SPST switch, it's wired from the other two switches connected by 2 diodes, both in the same direction. Can anyone explain why a single diode wouldn't work here (with both wires connected to it)
The two diodes are preventing a high potential on EITHER the Ring or Tip from generating a current flow into the Tip or Ring respectively, when none of the momentary switches are depressed. A single diode, say for example the diode between the Tip and the Up button, would not prevent a potential on Tip from generating a non-negligible current flow into the Ring when Ring was at a sufficiently lower potential than Tip. This potential would be approximated by the forward voltage drop of the N14002.
The two diodes are to that the UP switch actuates both lines, while still allowing those lines to act independently and prevent the other switches on those lines to have the same affect as the UP switch.
Looks to be so you can allow three switch configurations that don't affect one another. Mode pulls down tip, down pulls down ring and up pulls them both down. Without the diodes you couldn't use the same spst switch for up. Notice how they tie tip and ring together through the diodes, so they never short to each other, but pressing up pulls them both down at the same time.
