0
\$\begingroup\$

I'm designing a circuit (guitar effect) and have reached a point where I have to switch the signal path between 3 configurations involving a pair of diodes.

The circuit's 'out' path has the option of going:

  1. Straight out (uninterrupted)
  2. Through a pair of parallel diodes (symmetrical clipping diodes)
  3. Via one end of the diode pair, with the other going to ground

I have a dual pole on-off-on switch that I would like to use if possible. The schematic currently looks like this:

enter image description here

Is it possible to switch between these three with a simple switch or will I have to use a rotary switch?

As requested, the three I would like to achieve:

CONFIG 1

enter image description here

CONFIG 2 (note diodes must be NC)

enter image description here

AND CONFIG 3

enter image description here

\$\endgroup\$
1
  • \$\begingroup\$ What does "through a pair of parallel diodes" mean? Normally one end of your D1-D2 pair would be connected to ground so you'd be running "across" the diodes. You could make your question much clearer by adding a schematic (without the switch) showing the three configurations. \$\endgroup\$
    – Transistor
    Commented Dec 19, 2017 at 22:45

1 Answer 1

1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Parallel connection of 0, 1 or 2 diodes across audio signal using a 2-pole ON-OFF-ON switch. (CircuitLab doesn't have a suitable symbol.)


Update after images posted.

schematic

simulate this circuit

Figure 2. Bypass, series and parallel diode switching.

\$\endgroup\$
4
  • \$\begingroup\$ This wouldn't work as it only connects one diode to ground at a time. I have now added images to show what I need. \$\endgroup\$
    – TCassa
    Commented Dec 19, 2017 at 23:02
  • \$\begingroup\$ You're right. D1 should be wired to both contacts of the first pole of the switch. Fixed. \$\endgroup\$
    – Transistor
    Commented Dec 19, 2017 at 23:05
  • 1
    \$\begingroup\$ @TCassa: See the update. \$\endgroup\$
    – Transistor
    Commented Dec 20, 2017 at 7:10
  • \$\begingroup\$ Thanks - I had considered using an additional diode pair but was trying to avoid. This would work though. \$\endgroup\$
    – TCassa
    Commented Dec 20, 2017 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.