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I currently have a flashing LED that works on 12VDC power.

This LED will be used as a pseudo (imitation alarm) deterrent in our family vehicle.

However, I only wish to have the LED receiving 12V when the ignition is in the LOCK (OFF) position, and 0V when the ignition key is moved to other positions. Hope this makes sense?

I understand this could be achieved using a relay, but I'm not sure how this circuit would need to be wired in order to achieve the desired result?

Any help (with diagrams) would be sincerely appreciated.

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    \$\begingroup\$ Use relay's normally closed contact. Coil fed then ignition ON. \$\endgroup\$
    – user263983
    Apr 25, 2021 at 11:07

2 Answers 2

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Use a Single-Pole, Double-Throw (SPDT) 12V relay and connect your flashing LED to the normal-closed contact

schematic

simulate this circuit – Schematic created using CircuitLab

When the ignition is on, the relay pulls in, disconnecting the flashing circuit. When the ignition is off, the relay releases, connecting the flasher. This also does not pull any additional current when the ignition is off, reducing battery drain.

Please be sure to either connect to already fused circuits or add automotive fuses in-line with both feeds. Accidentally shorting a non-fused (or excessively high-current fused) connection to an auto battery is a way to really ruin your day!

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Simple circuit.

  • When IGNITION is on R1 - D1 is shorted out and D1 turns off.
  • When IGNITION is off the normal ignition loads will provide a return path to chassis for the LED current.
  • R1 may be omitted if the current limiting resistor is built into the LED.
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