Using an optocoupler to separate a 4-20 mA loop

Question

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.

Answer 1

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.

For this, see HCNR201 Datasheet’s Fig.16:

As you see, if your transmitter (“sensor” + 4-20mA conversion) can be powered by a 12V Power Supply (but better if a bit higher, as 18V), the 4-20mA current loop (+Iin & -Iin) powers the Receiver-side of the linear optocoupling circuit (top left).
The other part (top right) would be powered by the PLC side. These two Op Amps are not galvanically connected, of course.
The good linearity comes from the optical feedback the PhotoDiode PD1 does on the Left side that corrects intrinsic opto non-linearities, while the PD2 on the right side gets this linearized signal, amplified by its Op Amp.

In your schematic, you also plotted another 4-20mA on the PLC side, using a 3.3V rail.
It does not work like in your schematic, starting from a higher voltage needed by the Right-OpAmp and PD2.
Indeed, Figure 22 of same datasheet shows a more practical circuit (not just the concept in Fig.16) that uses at least 5V on either side of the assembled circuit. As you see, the output at right is a voltage, not a current loop. And for this, the OpAmp as LM158 (as LM358) is used because of its capability to operate from single supply, as low as 3V, then 5V has a good headroom for this application. Observe just One OpAmp is used per PD side, exactly to have galvanic isolation.
See highlighted parts in red:

Answer 2

More of a question than an actual answer because I am not a EE. (Somebody please delete if this is not appropriate.)

Are you certain you need isolation? What if you just did this?

^{simulate this circuit – Schematic created using CircuitLab}