I have a car ignition switch that has 4 settings and is +12 V DC:

  • 0 = off (no voltage)
  • 1 = accessory (only some items have power) (no voltage will go to pump)
  • 2 = on (all items have power)
  • 3 = start (momentary switch for start power)

I have a pump I want to wire up in the following way:

  • When the switch is on 3 (momentary), the pump gets power and turns on.
  • The same switch then moves to setting 2, where the pump remains on.
  • If the switch is turned to positions 0 or 1, the pump turns off and will remain off until the switch is returned to position 3
  • The pump will NOT turn on in position 2 alone, but will remain on in position 2 AFTER position 3 has activated the pump.

I believe this can be achieved with two SPST relays, but I can't seem to figure out how to do so.

Is this a good way of doing it?

enter image description here

  • \$\begingroup\$ If you achieve this, how will you get it to turn off? \$\endgroup\$
    – Solar Mike
    Oct 21, 2018 at 0:23
  • \$\begingroup\$ 1 SPST relay + 2 Diodes. \$\endgroup\$
    – brhans
    Oct 21, 2018 at 0:26
  • \$\begingroup\$ this is similar to what I am looking to do electronics.stackexchange.com/questions/231713/… \$\endgroup\$
    – Coots
    Oct 21, 2018 at 0:58
  • \$\begingroup\$ Isn't an engine runs indication what you really want? \$\endgroup\$
    – Janka
    Oct 21, 2018 at 1:28
  • \$\begingroup\$ A latching relay circuit, where power first comes from the starter circuit, but doesn't cut out when the switch goes back to on? \$\endgroup\$
    – Passerby
    Oct 21, 2018 at 1:50

2 Answers 2



simulate this circuit – Schematic created using CircuitLab

This will almost work. The idea is that the both the motor and the relay is powered as long as the relay is closed. So we've made a latching relay. In position 3, everything is powered, and hopefully it will stay that way as the switch changes to position 2.

There's nothing magic about the ignition switch. Position 0 has nothing connected to it, and position 3 just springs back to 2 when you let go. Position 1 may be wired such that it is on in position 2, but since we don't use it, it doesn't matter.

The problem is that the relay will drop if it breaks the connection while switching from Start to Run. If Run stays energized between position 3 and 2, you're golden. Alternatively, the motor may supply power (as a generator) briefly during the switch return. A third possibility is to put a capacitor across the relay coil, though a fairly large one would be required.

  • 2
    \$\begingroup\$ Diodes are require here otherwise you would be back feeding things. \$\endgroup\$
    – Passerby
    Oct 21, 2018 at 2:09
  • \$\begingroup\$ Only if his application has more "things" ? \$\endgroup\$
    – gbarry
    Oct 21, 2018 at 2:14
  • 2
    \$\begingroup\$ Like the rest of the car... \$\endgroup\$
    – Passerby
    Oct 21, 2018 at 2:18
  • \$\begingroup\$ @Passerby Correct. I'm sure diodes will be needed. I was going to use an "add a circuit" fuse from any of the fuses coming from wire 6 on the ignition switch. Those are all hot in "start" and "on". See this wiring diagram for more details: drive.google.com/open?id=1wseps9FH888NHrOFlpe8sWA7ZCFqRWuN \$\endgroup\$
    – Coots
    Oct 21, 2018 at 14:18
  • 1
    \$\begingroup\$ @Passerby Is this a better way of doing it? drive.google.com/open?id=1IJt4i0KBaClB-rA8Znm-z7qQvTasvs6x \$\endgroup\$
    – Coots
    Oct 21, 2018 at 15:04

Here is a possible circuit. The self-latching relay gets its "pull-in" power from terminal 3 and gets its "hold" power from terminal 2. The diode prevents voltage at 2 from feeding back to terminal 3.

Latching Relay for Pump

Terminal 2 also feeds the pump, but only after the relay has been turned on by terminal 3.

  • \$\begingroup\$ Due to D1 being reverse bias the pump and relay would never get turned on. \$\endgroup\$
    – Passerby
    Mar 12, 2021 at 5:08
  • \$\begingroup\$ @Passerby you are correct. The diode should be drawn facing the other direction. Then the circuit will work as described. I am used to electron flow theory, so I drew the diode wrong, forgetting that the power lead is positive. \$\endgroup\$
    – John Canon
    Mar 13, 2021 at 12:49

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