How to interface MCU with open-collector output?

Question

I have a 3.3VDC microcontroller (based on Nordic nRF52840) with digital inputs which I need to interface to a device that outputs slow pulses (1 second) via an "open collector output".

The device's manual describes its output as follows: "The pulse output is an electronic switch that operates much like an ‘open collector’ transistor." The output can be configured as Form A or Form C, and the pins are labeled K (Common), Y (normally open) and Z (normally closed).

The documentation provides the following ratings for these outputs:

1. Outputs are isolated from ground and each other.
2. Outputs are rated for DC only, from 3.0 volts to 30.0 volts. Observe polarity.
3. The circuits will sink up to 15 milliamperes (DC) for Form-C outputs and up to 5 milliamperes (DC) for Form-A outputs.
4. All electronic pulse outputs incorporate opto-isolators to electrically isolate the device circuitry from the devices receiving the pulses.

This is all the info I'm provided by the manual.

I need to know the proper way to connect this output to my MCU.

Here's the way I think it could be done, using a pull-up resistor, but I'm not sure about it. As I understand it, the open-collector output means the output is basically floating when off, allowing the full 3.3V to appear at the MCU connect point. Then when the output turns on, it shorts Y to K, thus dropping the 3.3V across R1, leaving 0V at the MCU connection point.

Is this correct? Am I missing something?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

If your outputs are optically isolated then you have to be aware of which pin on the output is which. The interface would look like this:

^{simulate this circuit – Schematic created using CircuitLab}