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I have a 4 wheeler that i converted from a mechanical locking differential to an electric one. the mechanical one had a 2 wire spring loaded switch that as the mechanism rotated it pushed the pin in and closed the circuit and the 4WD light comes on.

Now that i have converted it to electric differential, there's no where for that sensor/switch to go. instead it has an electric plug on it and i isolated 2 wires that indicate if is in 2WD or 4WD. in 2WD it is closed on those 2 wires, in 4WD it is open.

It would be easy if it were closed when in 4WD .. i could just hook the light sensor/switch wires to it. but it's not.

How can i sense if a circuit is open on 2 wires and if so, close a circuit on another 2 wires?

assume i know how to use a multimeter and how to solder... that's about it. :)

Thanks for any help.

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Pass a current through the wire that energizes a relay. A SPDT relay can be used to direct current through two different indicators to show the current mode.

schematic

simulate this circuit – Schematic created using CircuitLab

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You have a few options but we need to bear in mind that losing the 4WD indication can be dangerous as steering response will be affected.

Simple indicators

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Two LED circuits.

  • (a) The LED defaults to ON giving 4WD indication. When in 2WD mode the LED is shorted out. This will result in the continuous 'waste' of 20 mA through R1 (but this is tiny in comparison to the output of your engine.) R1 should be 1/4 W or greater.
  • (b) A slightly more devious circuit: The combination of the red LED and D1 will require about 2.5 V to get a decent light out of it (1.8 V for the LED and 0.7 for the diode). The green LED will require about 2.0 V. If the green is turned on it will drop the voltage at the junction of the two LEDs and the red should go out. If there's still a bit of a glow from the red then add a second diode in series with D1. This circuit has the advantage that one LED should always be on and, should the switch get disconnected, it will fail to red prompting more caution (than is required in the circumstances).

Use high-brightness LEDs.

Relay circuit

schematic

simulate this circuit

Figure 2. Using a relay to invert the switch logic. Note that this circuit does not fail "safe". If the switch or relay coil get disconnected the lamp will never turn on.

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  • \$\begingroup\$ Like the first one 👍🏻 \$\endgroup\$ – JustMe May 27 '16 at 23:30
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from IPad so not easy, but if there is a flow when it is closed, you can do this. Maybe not understood correctly?

enter image description here

enter image description here

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