How can you detect a fault in an instrumentation sensor that causes an open in a 4-20 mA loop?

• A. The loop current drops toward 0 mA; verify transmitter supply, inspect wiring, and test with a known-good resistor or loop simulator.
• B. The loop current rises above 20 mA.
• C. The loop voltage becomes zero.
• D. The loop power supply must be removed.

Answer: (A)

In a 4-20 mA loop, the signal is carried as current. If the instrumentation sensor or its wiring opens the loop, there is no complete path for current, so the loop current collapses toward zero. That near-zero current is the diagnostic clue that something in the loop is open.

The best way to verify and locate the fault is to start from the symptom and work outward: confirm the transmitter is powered and the loop supply is present, inspect all wiring and connections for breaks or loose terms, and use a known-good resistor or a loop simulator to test the path. A known-good resistor (for example, a precise 250 Ω) lets you convert current to a measurable voltage to confirm current flow and isolate whether the issue is in the transmitter, wiring, or the loop supply. A loop simulator can substitute the transmitter to verify the loop path without relying on the actual sensor, helping pinpoint where the break is.

Other options don’t fit an open-in-the-loop scenario. An over-range current (above 20 mA) suggests a fault like a short or another condition, not an open. Zero loop voltage isn’t a reliable diagnostic for an open in a current loop, since the transmitter and wiring can still present voltage while no current flows. Removing the loop power supply isn’t a diagnostic test and can create unsafe conditions or bring the system offline without identifying the fault.