How Does Open Circuit Detection on Functional Safety Digital Outputs Work?

Updated May 24, 2019

Reported In

Hardware

  • NI-9350
  • NI-9351
  • C Series Functional Safety Module

Issue Details

  • I want to use the Open Circuit Detection on an NI-9350 Functional Safety Module. However, when I activate it, it always detects an open circuit, although it is not. What am I doing wrong?
  • Should I have a pulldown resistors on our outputs from the NI-9350 to avoid open load detection? I added them on all the inputs per the manual but didn’t see anything for the outputs.
  • The Datasheets of NI-9350 and NI-9351 state three different current values for Open Circuit Detection, namely 11 mA, 2 mA minimum, and 20 mA maximum. What do they mean?

Solution

The current values given in the datasheet describe under what conditions the Open Circuit Detection inside the NI Functional Safety Modules will detect and report an open circuit. That is when the load pulls only around 11 mA. More specifically, typically any value between 2 mA and 20mA triggers the open circuit fault. Therefore, if you want to use Open Circuit Detection, your externally connected equipment should draw more than 20 mA.

If it does not, you need to take measurements to draw additional current. This could e.g. be additional pull-down resistors. These additional pull-downs on the output to avoid open load detection are not called out explicitly in the NI-935x's manuals. So if the load on your Digital Output pulls less than 20 mA and your application needs open circuit detection, then a pull-down resistor should be added to ensure the current sourced through the Digital Output when the output is high is always >= 20mA.

Alternatively, if your application does not require detection of open circuits, or you achieve this through a different diagnostic, then you can just ignore the open circuit faults.

Additional Information

The Digital Outputs of the NI-9350 and NI-9351 modules source a voltage of about 24 V. Using Ohm's law (V = R * I) you can determine the resistance your additional resistor needs. If you need to pull additional 20 mA, this would be:
R = V / I = 24V / 20mA = 1.2kΩ

Please note that this resistor dissipates almost 500 mW (P = V * I = 24V * 20mA = 480mW). Please make sure your resistor is specified for this load.

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