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Okay so this is extremely specific and I've searched around but I might be using the wrong terminology. I have two automotive foot pedals that I can attach microswitches (or anything I guess) to, either when depressed or not to know its state.

I've drawn it out many ways but can't wrap my mind around making a circuit to where it's always on unless only one condition is met: brake pedal is released and accelerator is pressed. This is the condition breaks the circuit and any other combination completes the circuit.

However I'm willing to concede that if this is not possible, then the bare minimum circuit would require it to be always off but ONLY on if the brake pedal is pressed and accelerator is not. But I don't know if this is possible either.

I wonder if there isn't some additional thing I'd need apart from a microswitch on each pedal to detect its state. I'm hoping I can get this to work.

Thank you for any answers!

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    \$\begingroup\$ Your description of what you want is confusing. Make a logic table \$\endgroup\$
    – Passerby
    Dec 10, 2017 at 6:58
  • \$\begingroup\$ hint: put the two switches in parallel ... don't forget that if you use an SPDT switch, then it can be wired as normally on or normally off \$\endgroup\$
    – jsotola
    Dec 10, 2017 at 7:14

1 Answer 1

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Table 1. Required logic.

Brake   Accel   Output
up      up      on
up      down    off
down    up      on
down    down    on

This can be simplified to

Brake   Accel   Output
up      up      on
up      down    off
down    X       on

where 'X' is "don't care".

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Circuit to achieve required logic.

Note that there will be a momentary break in continuity when the brake pedal changes state due to both N.O. and N.C. contacts being disconnected for a moment.

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  • \$\begingroup\$ So both of these switches would be mounted on the front side of the pedals then for when the pedals are at rest and up, do I have that right? Thanks everyone for introducing to logic tables, don't know how I'd never come across them. That helps big time in understanding and in my attempts I was missing the NC brake back to compete the circuit also. Thank you! \$\endgroup\$
    – splendid
    Dec 10, 2017 at 14:15
  • \$\begingroup\$ The switches are shown in their released / normally closed (N.C.) / "up" position. I've added 'UP' and 'DN' to the schematic for added clarity. \$\endgroup\$
    – Transistor
    Dec 10, 2017 at 15:06
  • \$\begingroup\$ Thank you for the help. In my research trying to find my own answer I learned that schematics are drawn in the default resting state, but the switches I saw had the line going to the lower circle. Thanks again, I'm gonna give this layout a go. \$\endgroup\$
    – splendid
    Dec 10, 2017 at 17:25
  • \$\begingroup\$ They can be drawn either way - normally closed on the top side or lower side. I chose this way since you mentioned foot-switches and the switches would normally be "up" until stood upon. \$\endgroup\$
    – Transistor
    Dec 22, 2017 at 19:55

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