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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? enter image description here

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    $\begingroup$ Welcome to 3dPrinting.SE! I'm really not sure what you're asking here ... could you please edit your question and clarify exactly what you are trying to do? $\endgroup$ Aug 11 '19 at 16:29
  • $\begingroup$ I have a tevo tornado gold 24 V . So I want to put this LJ12 A3-4-Z/BX Inductive NPN NO 4 mm with 6-36 V operation current. I do not want to fry my machine by putting in 24 V into the sensor input: What do I have is a : 12V 1 Channel Optocoupler Isolation Module Isolated Board No Din Rail Holder PLC Processors 80KHz PC817 EL817 i want to know if this optocoupler is ok to set up or do i need another in order to prevent me frying my sensor. $\endgroup$
    – 3d Printer
    Aug 11 '19 at 16:33
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    $\begingroup$ Please EDIT this information into your question. You can click the edit link which is just below the tag, which is below the body of the question. $\endgroup$ Aug 11 '19 at 16:46
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    $\begingroup$ If you gave us a link to the actual Optocoupler Isolation Module you refer to, we could advise. It might be that you only need to add a resistor to limit the current to the optocoupler's LED or it might be that it would need extensive modification. $\endgroup$ Aug 12 '19 at 20:06
  • $\begingroup$ You will never fry the sensor, as this can take 6 V to 36 V. You can fry the module as it is rated for 12 V and not 24 V. $\endgroup$
    – 0scar
    Aug 14 '19 at 21:06
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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.

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Not using a 12 V rated module "on its own".

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.

A properly rated one

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.

Trickery with voltage dividers

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.

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  • $\begingroup$ so in short use 2 resistors $\endgroup$
    – 3d Printer
    Aug 16 '19 at 14:03
  • $\begingroup$ As a voltage divider before the input, yes. $\endgroup$
    – Trish
    Aug 16 '19 at 17:46
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    $\begingroup$ or just double the resistance of the OC's LED to keep the current (about) the same at the double voltage; the input side of an opto is literally just an LED, which cares about current, not voltage. a voltage divider approach would have to be somewhat carefully chosen and balanced to not get too hot, waste power, or starve the OC's LED of intensity that would keep the output saturated. $\endgroup$
    – dandavis
    Aug 16 '19 at 21:19
  • $\begingroup$ @dandavis only if you know what type of LED is used and what kind of resistor is already mounted before... one of which you can't know: the LED. $\endgroup$
    – Trish
    Aug 16 '19 at 21:43
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    $\begingroup$ The advice to use a voltage divider is bad. The module requires a certain amount of input current, which will cause additional drop in the voltage divider, making a "50%" voltage divider give a voltage lower than 12V. To get the "right" voltage divider you need to measure the existing LED and resistor. The advice to use a voltage divider is still bad. It's much better to use a single current limiting resistor. The second resistor in the voltage divider does nothing except waste power in this case. $\endgroup$ Aug 17 '19 at 15:02

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