I have a Tevo Tornado Gold 24 V. I want to use this LJ12 A3-4-Z/BX Inductive NPN NO 4 mm with 6-36 V operation current as a Z probe. I do not want to fry my machine by putting in 24 V into the sensor input.
What do I have is a 12 V, single channel optocoupler isolation module.
I want to know if this 12 V optocoupler module can be used with a 24 V power supply, or do I need another module in order to prevent me frying my sensor.
If I do need another what would I need?
You can safely use the module with 24V.
The input side shows a red LED, optocoupler and 1k resistor in series. The LED and optocoupler probably have a voltage drop in the neighbourhood of 3.1-3.5 V put together, so for a 12 V input you will get a current of approximately 9 mA-.
For a 24 V input voltage the increased current will cause a slightly higher voltage drop, but even if the voltage drop remains as low as 3.1 V the current will still only be 21 mA. This is well within the rating of the optocoupler (similar optocouplers are often rated for 60 mA) and slightly pushing the rating of the LED (similar LEDs are usually rated for 20 mA) but it will probably be fine.
For extra peace of mind you could connect an additional resistor in series with the input. The "ideal" value (that is, to keep the current identical to that at 12 V) would be 1.3 kΩ, though any small value resistor (above 100 Ω) would be fine.
Using a 12 V/5 V optocoupler to try to connect 24 V to the 5 V circuit is running the optocoupler outside of its rating, meaning you will destroy it, either immediately or after a short time.
To shield the 5 V against the maximal 24 V from the probe without extra parts, you will need to use a 24 V/5 V optocoupler.
With a 50 % voltage divider made from two properly rated resistors, you could limit the voltage to the optocoupler, which in turn would turn the 24 V signal into a 12 V signal, which would protect our optocoupler and the board beyond.
