4–20 mA Control Loops: Working Principles & a Field-Proven Troubleshooting Playbook - Just Measure it

4–20 mA Control Loops: Working Principles & a Field-Proven Troubleshooting Playbook

1) Why 4–20 mA still wins

In noisy industrial environments, analog current loops are resilient: the signal is immune to long-run voltage drop, easy to scale, and simple to diagnose with a meter. A single loop can carry both measurement and health information (e.g., out-of-range currents to flag faults), which is exactly why 4–20 mA remains the default in process plants.

2) What a 4–20 mA loop looks like

A basic loop consists of:

  • 24 V DC supply powering the loop,

  • A transmitter that drives 4–20 mA proportional to the process variable (PV),

  • An indicator or controller (PLC/DCS AI) that converts loop current back to PV, often via an internal or external 250 Ω shunt to read 1–5 V by Ohm’s law.

Quick math:

  • 4 mA →

    4mA×250Ω=1.0V4\,\mathrm{mA}\times 250\,\Omega = 1.0\,\mathrm{V}

  • 20 mA →

    20mA×250Ω=5.0V20\,\mathrm{mA}\times 250\,\Omega = 5.0\,\mathrm{V}

Tip: If your AI already has an internal 250 Ω, don’t add another external shunt.

3) Key conversions & checks you’ll use in the field

Linear scaling (PV from loop current):

PV=PVL+I416(PVHPVL)PV = PV_\mathrm{L} + \frac{I-4}{16}\,(PV_\mathrm{H} – PV_\mathrm{L})

Loop load / compliance sanity check:

RtotalVsupplyVheadroomImaxR_\mathrm{total} \le \frac{V_\mathrm{supply} – V_\mathrm{headroom}}{I_\mathrm{max}}

Use the transmitter datasheet for minimum headroom (compliance voltage). If RtotalR_\mathrm{total} (wiring + shunts + AI input) is too high, you’ll never reach 20 mA.

1–5 V readout via 250 Ω:

U=I×R=(4 to 20)mA×250Ω=1 to 5VU = I \times R = (4\text{ to }20)\,\mathrm{mA}\times 250\,\Omega = 1\text{ to }5\,\mathrm{V}

Health ranges suggestion (NAMUR-style):

I3.8 mAUnderrange/Fault,I20.5 mAOverrange/FaultI \le 3.8~\mathrm{mA}\Rightarrow \text{Underrange/Fault},\quad I \ge 20.5~\mathrm{mA}\Rightarrow \text{Overrange/Fault}

4) The troubleshooting playbook

Step A — Verify the loop current first

Put a DMM in series (loop mode) or use a mA clamp to read the actual current.
You’re looking for:

  • 0 mA → open circuit / polarity reversed / blown AI fuse

  • ≈4–20 mA steady → loop likely OK (check scaling next)

  • Erratic reading → noise, shielding, or grounding issues (see Step D)

Step B — Do a substitution test at the transmitter tap

Disconnect the transmitter and connect a loop calibrator / process meter at the same terminals. Source 4 → 12 → 20 mA and watch the controller.

  • If it tracks correctly, wiring + supply + AI are fine → issue lies in transmitter/sensor.

  • If it still fails, check the AI input fuse and 250 Ω presence/location.

Step C — Prove the power supply

Measure the loop supply voltage: typically 24 VDC. If in doubt, temporarily power the loop from the calibrator’s 24 V at the same point. If behavior improves, the plant supply is defective or overloaded.

Step D — Hunt noise, shielding, and grounding issues

Usual suspects: broken shields, ground loops, routing near power, marginal supply.

  • Use DMM (AC function) or a handheld scope to see if AC rides on the signal; a healthy loop should show only mV-level ripple.

  • Fix by single-end shield grounding, twisted pairs, separation from high-energy cables, or adding isolators where needed.

5) Quick diagnostic table (HTML, copy-safe)

SymptomLikely CauseQuick ChecksFix
0 mA, controller shows nothingOpen circuit, reversed polarity, blown AI fuseSeries DMM reading; inspect terminals & AI fuseRestore wiring, correct polarity, replace fuse
Loop current OK but value wrongScaling or 250 Ω mismatchConfirm only one 250 Ω shunt; verify range mappingKeep one shunt; correct AI scaling
4/12/20 mA substitution test OK, but process notFaulty transmitter/sensor or mis-rangedSubstitute at transmitter tap; compare PVRe-range or replace transmitter/sensor
Intermittent/jittery PVNoise, shield/ground issue, cable routingDMM AC / scope; check shield land and routingSingle-point shield, twisted pair, reroute, add isolator
Hits top but won’t reach 20 mAInsufficient compliance voltage (loop load too high)Compute \(R_{\text{total}}\); measure supply under loadReduce resistance, raise supply, or change AI input

6) Good wiring patterns to standardize

  • Two-wire (loop-powered) transmitters: +24 V → transmitter “+”; transmitter “–” → AI “+”; AI “–” back to supply “–”.

  • Three-/four-wire (self-powered): Check whether AI is passive (sinking) or active (sourcing); never drive an active AI from an active transmitter without proper interfacing.

  • Keep one—and only one—250 Ω shunt in the measurement path.

7) Commissioning & safety notes

  • De-energize before moving meter leads; many loop failures happen while “just checking something”.

  • Label shield landing points; enforce single-end grounding policy.

  • For hazardous areas, place barriers/isolators correctly in the loop (per intrinsically safe design files).

8) Handy pocket formulas

  • 1–5 V conversion: U=I×R=(4 to 20)mA×250Ω=1 to 5VU = I \times R = (4\text{ to }20)\,\mathrm{mA}\times 250\,\Omega = 1\text{ to }5\,\mathrm{V}

  • PV from current: PV=PVL+I416(PVHPVL)PV = PV_\mathrm{L} + \dfrac{I-4}{16}\,(PV_\mathrm{H}-PV_\mathrm{L})

  • Total loop resistance limit: RtotalVsupplyVheadroomImaxR_\mathrm{total} \le \dfrac{V_\mathrm{supply}-V_\mathrm{headroom}}{I_\mathrm{max}}

Wrap-up

Start with the current reading, prove the wiring/AI with a substitution test, isolate the power supply, and then clean up noise. This disciplined flow solves the majority of loop problems in minutes—and it’s simple enough to apply under pressure on a live plant.

Share This Story, Choose Your Platform!

Contact Us

    Please prove you are human by selecting the tree.
    Translate »