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I have a car ignition switch that has 4 settings and is +12v 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 can't seem to figure out how to do so.

EDIT: Uploaded diagram

[![ignition fuses][2]][2]

EDIT: Is this a good way of doing it: enter image description here

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  • \$\begingroup\$ If you achieve this, how will you get it to turn off? \$\endgroup\$ – Solar Mike Oct 21 '18 at 0:23
  • \$\begingroup\$ 1 SPST relay + 2 Diodes. \$\endgroup\$ – brhans Oct 21 '18 at 0:26
  • \$\begingroup\$ this is similar to what I am looking to do electronics.stackexchange.com/questions/231713/… \$\endgroup\$ – Coots Oct 21 '18 at 0:58
  • \$\begingroup\$ Isn't an engine runs indication what you really want? \$\endgroup\$ – Janka Oct 21 '18 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 '18 at 1:50
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schematic

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.

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    \$\begingroup\$ Diodes are require here otherwise you would be back feeding things. \$\endgroup\$ – Passerby Oct 21 '18 at 2:09
  • \$\begingroup\$ Only if his application has more "things" ? \$\endgroup\$ – gbarry Oct 21 '18 at 2:14
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    \$\begingroup\$ Like the rest of the car... \$\endgroup\$ – Passerby Oct 21 '18 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 '18 at 14:18
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    \$\begingroup\$ @Passerby Is this a better way of doing it? drive.google.com/open?id=1IJt4i0KBaClB-rA8Znm-z7qQvTasvs6x \$\endgroup\$ – Coots Oct 21 '18 at 15:04
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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.

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