Controlling one relay with three LEDs - energize relay if any one of the three LEDs are lit

Question

I have three 12V indicator LEDs which are the simplest way to see if one of the three circuits is 'ON'.

Only ONE of the LEDs can be ON at any given time. They all share GND and a controller sends POS to each, as it's activated.

I need to sink a GPIO pin on an ESP32S3 to GND when any one of the three LEDs are ON.

My thought was to use a 12V relay, with the OUTPUT connected to the GPIO pin and GND, so when the relay is energized it would sink that GPIO pin to GND.

On the relay input side my thought was to use the GND from the three LEDs, which is a common ground for the relay negative, and then connect each of the LEDs' positive through a diode, to the relay positive.

Anytime an LED was lit, it would energize the 12V relay connecting the output pins and effectively sinking the GPIO pin to GND.

Does anyone see any issues with this?

Will the diodes be sufficient to keep the other LEDs from being lit when one is ON?

Is there a better way to do this?

I also have access to a single 120V circuit and a 120V solenoid, that are live when any one of the lights are ON. I tried using a magnetic reed switch to sense the energizing of the solenoid but the reed was too sensitive or something and fluctuates rapidly on/off when the solenoid is activated. I was trying to stay as non-invasive as possible.

UPDATED - SCHEMATIC ADDED

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

A single 120 V AC relay, with gold-plated contacts, should do.

Here's the schematic.

Answer 2

The simplest way to do this probably is as follows:

^{simulate this circuit – Schematic created using CircuitLab}

Assuming the lamps are individually driven by 12V, any one that is lit will turn on Q1, which will give a logic low signal to the ESP32. I show incandescent lamps, but LEDs will work the same.

Answer 3

You need NOR logic gate which can be really made with diodes. LEDs are also diodes but "light emitting diodes"; so we can made them perform both logic and visualizing functions. Therefore, the PStechPaul's circuit can be simplified.

^{simulate this circuit – Schematic created using CircuitLab}

Usually, the GPIO has an internal "pull-up" resistor that should be enabled in this case; so R3 can be omitted. R1 can be also omitted but it is better to have it.

I tried using a magnetic reed switch to sense the energizing of the solenoid but the reed was too sensitive or something and fluctuates rapidly on/off when the solenoid is activated.

A reed switch or a reed relay? I can help you to realize this idea...

Answer 4

I think your circuit looks good, the diodes are a nice touch. You did forget the obligatory diode across the relay coil to keep it from killing everything when it switches off.

Instead of powering a relay coil, though, a quieter, less power hungry solution would use a transistor:

^{simulate this circuit – Schematic created using CircuitLab}

This assumes you are able to join the grounds of the two systems together. If not, or if you are unsure, then isolate the two systems using an opto-isolator, such as the 4N25:

^{simulate this circuit}