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The 4–20 mA Current Loop: How It Works and How to Troubleshoot It

Signals Β· 9 min read

The 4–20 mA loop is still the most common way to send a measurement from a field transmitter to a controller. It is simple and tolerant of long cable runs. Once you know what to measure, it is also easy to troubleshoot. This note covers how the loop works, how to check the power budget, how to scale the signal and how to find faults.

Why current instead of voltage?

In a series circuit the same current flows through every component. Cable resistance, terminal contact resistance and small voltage drops do not change the current the transmitter sets. A voltage signal sent over hundreds of metres would lose accuracy to those drops. A current signal does not, as long as the loop has enough supply voltage.

The "live zero"

The range starts at 4 mA instead of 0 mA for two reasons:

Basic two-wire loop

A typical loop has a 24 VDC supply, the transmitter, and a receiver input, all in series. The receiver is usually an analog input card with a 250 Ξ© resistor. The transmitter regulates the loop current between 4 and 20 mA in proportion to the process value.

FIELD CONTROL ROOM / MARSHALLING 24 VDC Loop supply + βˆ’ Analog input (PLC / DCS) 250 Ξ© 1–5 V to A/D PT 2-wire transmitter sets 4–20 mA I = 4–20 mA same current everywhere in the loop
Figure 1 β€” A two-wire loop. One current flows through the supply, the transmitter and the 250 Ξ© input resistor. Across the resistor it reads as 1–5 V.

Checking the loop power budget

Every transmitter needs a minimum voltage at its terminals, called the lift-off voltage, to work correctly. It is often around 10–12 V; check the datasheet. The supply has to cover that voltage plus the drop across every resistance in the loop at full current. Allow a little above 20 mA so the loop can still drive alarm currents.

Rmax = (Vsupply βˆ’ Vmin, transmitter) Γ· 0.022 A

Example: 24 V supply, transmitter minimum 10.5 V.

Rmax = (24 βˆ’ 10.5) Γ· 0.022 β‰ˆ 614 Ξ©
Where the 24 V goes at 22 mA (example loop) Transmitter min. 10.5 V 250 Ξ© Β· 5.5 V cable ~1.1 V spare margin 0 V 24 V supply Cable example: 2 km of 1.5 mmΒ² copper β‰ˆ 2 Γ— 12 Ξ©/km Γ— 2 km β‰ˆ 48 Ξ© β†’ about 1.1 V at 22 mA.
Figure 2 β€” The loop voltage budget. If the drops add up to more than the supply, the reading flattens out before 20 mA.

The 250 Ξ© input resistor, the cable resistance (both conductors) and any barriers or indicators in the loop must together stay under about 614 Ξ©. If they do not, the reading will flatten out before 20 mA. The transmitter simply runs out of voltage.

Scaling to engineering units

The transmitter is ranged between a lower range value (LRV) and an upper range value (URV). The current is linear between them:

PV = LRV + (I βˆ’ 4) Γ· 16 Γ— (URV βˆ’ LRV)

Example: a pressure transmitter ranged 0–10 bar reading 13.6 mA:

PV = 0 + (13.6 βˆ’ 4) Γ· 16 Γ— 10 = 6.0 bar

Going the other way, the expected current for a given value is:

I = 4 + 16 Γ— (PV βˆ’ LRV) Γ· (URV βˆ’ LRV)

Fault currents (NAMUR NE43)

Many modern transmitters follow the NAMUR NE43 recommendation. It keeps the normal measuring range separate from failure signals, so the control system can tell "process at the limit" apart from "instrument failed":

β‰₯ 21 mA failure (upscale) ≀ 3.6 mA failure (downscale) 4812 16200 0 %25 %50 % 75 %100 % Process value (% of calibrated span) Loop current (mA) 13.6 mA = 60 % = 6.0 bar (0–10 bar)
Figure 3 β€” Current is linear with the process value. The red bands are the NAMUR NE43 failure signals, kept outside the 3.8–20.5 mA measuring range.
CurrentMeaning
3.8 – 20.5 mAValid measurement (including slight under/over-range)
≀ 3.6 mAFailure signal (downscale)
β‰₯ 21.0 mAFailure signal (upscale)

Whether a transmitter fails upscale or downscale is usually set by a jumper or in its configuration. Make sure it matches how the alarm or trip logic is designed.

Field troubleshooting steps

1. Measure the loop current

Put a meter in series with the loop, or use the transmitter's test terminals if it has them. Many loop calibrators can read current through a test diode without opening the loop. Compare the reading with the value you expect from the process.

2. Measure the voltage at the transmitter terminals

If it is below the transmitter's minimum operating voltage, look for too much loop resistance, a weak supply or a bad connection.

3. Interpret what you see

Measure loop mA 0 mA ≀3.6 or β‰₯21 mA Stuck below 20 mA mA OK, HMI wrong Open circuitBlown fuseNo supply Transmitter isreporting a fault β€”read its diagnostics Too much loopresistance or lowsupply voltage Range mismatchbetween transmitterand control system Noisy or jumping reading? Check shield grounding (one end only), ground loops and terminations.
Figure 4 β€” A quick decision path from the measured loop current to the likely cause.
SymptomLikely cause
0 mAOpen circuit, blown fuse, no supply, reversed polarity on some devices
≀ 3.6 mA or β‰₯ 21 mATransmitter is reporting its own failure β€” check its diagnostics
Reading stuck below 20 mA at high processLoop resistance too high / not enough supply voltage
Correct mA at field, wrong value on screenRange mismatch between transmitter and control system
Noisy or jumping readingShield grounded at both ends, ground loop, poor terminations
Safety: follow your site's permit and isolation procedures before opening a loop, and tell operations first. Opening a loop drives the reading to 0 mA, which can trip interlocks or move control valves.

Next: Pressure Transmitter Calibration: A Five-Point Field Procedure