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I'm creating a logic based circuit and below is my schematic.

The circuit first steps down a 24 VDC input to about 3.7 VDC using a 150K ohm resistance and a 5.1V Zener diode.

This 3.7 VDC is fed to the input of optocoupler PC817 pin 1 and 2.

On the output, I wanted to read the high and low through an Arduino connected to pin 3 and 4 of the optocoupler. I was planning to pass in 5 VDC from the Arduino to one of its digital I/O pin through the optocoupler.

Is there anything wrong here? I'm not able to read the high on the output of the optocoupler. I checked that the voltage around pin 1 and 2 is 3.7v.

Schematic showing the circuit with Optocoupler

Update

I reduced the resistance to about 4K since i didn't have the exact 4.7K as mentioned in the comments below. I also tried 5.6K, however seems like the optocoupler doesn't work in any case.

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    \$\begingroup\$ Welcome to EE.SE! Have you calculated the LED current going to your 817 and compared it with the values listed as recommended operating conditions in the datasheet? \$\endgroup\$ – winny Jan 4 at 11:10
  • \$\begingroup\$ After your schematic is still showing reverse supply connection to the LED as pointed out in the comments below. If you really want useful answers you need to show your actual circuit with correct pin-out and component values. It's not clear how you think a 5.1 V Zener will protect an infrared LED with a typical forward voltage of about 1.4 V. \$\endgroup\$ – Transistor Jan 4 at 14:01
  • \$\begingroup\$ How are you connecting the Optocoupler output to the Arduino? Your description is not very clear. \$\endgroup\$ – StarCat Jan 4 at 21:02
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The input to optocoupler is wrong, pin 1 is the anode (positive) and pin 2 is the cathode (negative). Also 150k resistor does not provide enough current to light up the optocoupler LED, change the resistor to 4.7 kilo-ohms to provide about 5mA at 24V. You can also throw away the zener, it is useless.

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  • \$\begingroup\$ I changed the input polarity but this time the voltage across pin 1 and 2 is 0.6v. \$\endgroup\$ – DigitalNewbie Jan 4 at 11:35
  • \$\begingroup\$ Then you should update the schematic. A wrong question will get wrong answers. \$\endgroup\$ – Transistor Jan 4 at 11:39
  • \$\begingroup\$ I updated my answer, LED current is obviously too low for it to work. You might want to draw how you connected it to Arduino, as you might have that wrong as well. \$\endgroup\$ – Justme Jan 4 at 11:55
  • \$\begingroup\$ 0.6V is correct if you reversed the 24V input. Think about it. You're now forward biasing the zener, \$\endgroup\$ – SteveSh Jan 4 at 11:56
  • \$\begingroup\$ Why do you think you need the zener? All you really need is the right value resistor, though good design practice usually adds a couple of more components for robustness and protection. \$\endgroup\$ – SteveSh Jan 4 at 11:58
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Something like this should work in most situations (with low switching frequencies, up to a few KHz or so):

schematic

simulate this circuit – Schematic created using CircuitLab

This produces a non-inverted output.

If you need the output to be inverted, connect pin 3 to GND, connect the Arduino input to pin 4 and put a pull-up between pin 4 and +5V. I'm assuming your Arduino runs on 5V.

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