I have a PLC that is powered with 3.3V and a sensor powered with 12V that outputs a 4-20mA signal. In order to allow the signal to be readable by my PLC while isolating the current in the 12V loop from the 3.3V loop I need to use an optoisolator which I have never used before. I am hoping for some help figuring out the appropriate circuit.

The isolator I am looking at is the hcnr201 datasheet here

In the datasheet it gives a bunch of rather confusing wiring diagram examples that are not clear about where the optocoupler is in the diagram. I am wondering how I could hook it up in reality so that the circuit connected to my PLC is producing the same current as is going through the LED on the 12V side. My thinking is that the LED would be driven by my 4-20mA 12V loop (pins 1&2) and pins 5 & 6 I would connect to a 3.3V source that runs through the optocoupler and is then connected to a resistor with an ADC measuring the voltage drop across as seen in my diagram below:

Does my thinking make sense? It seems overly simple compared to the example circuits I was looking at in the datasheet and I am thinking I may need to use an opamp with the 4-20mA loop to even turn on the LED and the down-regulate it back to 4-20mA on the other side before passing it through the resistor... Thanks!

I have a PLC that is powered with 3.3 V and a sensor powered with 12 V that outputs a 4-20 mA signal. In order to allow the signal to be readable by my PLC while isolating the current in the 12 V loop from the 3.3 V loop I need to use an opto-isolator which I have never used before. I am hoping for some help figuring out the appropriate circuit.

The isolator I am looking at is the HCNR201.

In the datasheet there is a bunch of rather confusing wiring diagram examples that are not clear about where the optocoupler is in the diagram. 

I am wondering how I could hook it up in reality so that the circuit connected to my PLC is producing the same current as is going through the LED on the 12 V side. My thinking is that the LED would be driven by my 4-20 mA, 12 V loop (pins 1 and 2) and pins 5 and 6 I would connect to a 3.3 V source that runs through the optocoupler and is then connected to a resistor with an ADC measuring the voltage drop across as seen in my diagram below:

Does my thinking make sense? It seems overly simple compared to the example circuits I was looking at in the datasheet and I am thinking I may need to use an op-amp with the 4-20 mA loop to even turn on the LED and then down-regulate it back to 4-20 mA on the other side before passing it through the resistor.

I have a PLC that is powered with 3.3V and a sensor powered with 12V that outputs a 4-20mA signal. In order to allow the signal to be readable by my PLC while isolating the current in the 12V loop from the 3.3V loop I need to use an optoisolator which I have never used before. I am hoping for some help figuring out the appropriate circuit.

The isolator I am looking at is the hcnr201 datasheet here

In the datasheet it gives a bunch of rather confusing wiring diagram examples that are not clear about where the optocoupler is in the diagram. I am wondering how I could hook it up in reality so that the circuit connected to my PLC is producing the same current as is going through the LED on the 12V side. My thinking is that the LED would be driven by my 4-20mA 12V loop (pins 1&2) and pins 5 & 6 I would connect to a 3.3V source that runs through the optocoupler and is then connected to a resistor with an ADC measuring the voltage drop across as seen in my diagram below:

Does my thinking make sense? It seems overly simple compared to the example circuits I was looking at in the datasheet. Thanks!

I have a PLC that is powered with 3.3V and a sensor powered with 12V that outputs a 4-20mA signal. In order to allow the signal to be readable by my PLC while isolating the current in the 12V loop from the 3.3V loop I need to use an optoisolator which I have never used before. I am hoping for some help figuring out the appropriate circuit.

The isolator I am looking at is the hcnr201 datasheet here

In the datasheet it gives a bunch of rather confusing wiring diagram examples that are not clear about where the optocoupler is in the diagram. I am wondering how I could hook it up in reality so that the circuit connected to my PLC is producing the same current as is going through the LED on the 12V side. My thinking is that the LED would be driven by my 4-20mA 12V loop (pins 1&2) and pins 5 & 6 I would connect to a 3.3V source that runs through the optocoupler and is then connected to a resistor with an ADC measuring the voltage drop across as seen in my diagram below:

Does my thinking make sense? It seems overly simple compared to the example circuits I was looking at in the datasheet and I am thinking I may need to use an opamp with the 4-20mA loop to even turn on the LED and the down-regulate it back to 4-20mA on the other side before passing it through the resistor... Thanks!