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A 4–20 mA loop is a simple circuit: a DC supply, a two-wire transmitter that regulates the current, the cable, and a receiver that measures the current — usually as a voltage across a resistor. Nearly every analog signal problem traces back to one of these four elements.
Why current, and why a loop
In a series circuit the current is the same at every point. A transmitter that regulates its current to represent the measurement therefore delivers the same value to the receiver whether the cable is 10 m or 1 km long: cable resistance changes the voltage dropped, not the current — provided the supply has enough voltage to spare. A current signal is also less susceptible to induced noise than a voltage signal over long runs, and a two-wire transmitter draws its power from the same pair that carries it.
At the receiver, the current is usually converted to a voltage by a precision resistor. With 250 Ω, 4–20 mA becomes 1–5 V — one reason that value appears in so many loops.
Live zero: what 4 mA buys you
The signal is scaled so that 4 mA represents 0 % of the calibrated range and 20 mA represents 100 %, a span of 16 mA. The current for any value is 4 mA + 16 mA × (value − LRV) ÷ (URV − LRV). Starting at 4 mA rather than zero does two things:
- A zero reading is distinguishable from a dead loop. An open circuit, blown fuse or lost supply gives 0 mA, which cannot be mistaken for a valid 0 % measurement.
- The transmitter has power at every reading. A two-wire device runs its electronics on the loop current, and the live zero leaves a few milliamps available even at the bottom of the range.
| Percent of span | Loop current | Voltage across 250 Ω | 0–10 bar g transmitter |
|---|---|---|---|
| 0 % | 4 mA | 1 V | 0 bar g |
| 25 % | 8 mA | 2 V | 2.5 bar g |
| 50 % | 12 mA | 3 V | 5.0 bar g |
| 75 % | 16 mA | 4 V | 7.5 bar g |
| 100 % | 20 mA | 5 V | 10.0 bar g |
Reading a value back works the same way: 13.6 mA from a 0–10 bar g transmitter is (13.6 − 4) ÷ 16 = 60 % of span, or 6.0 bar g. The 4–20 mA range itself is standardised in IEC 60381-1.
NAMUR NE 43: telling a fault from a measurement
A live zero detects a broken loop, but not a transmitter that has diagnosed a failure of its own sensor. NAMUR recommendation NE 43 standardises how a transmitter signals that condition by reserving current bands that a valid measurement never uses:
| Loop current | Meaning | Equivalent % of span |
|---|---|---|
| ≤ 3.6 mA | Failure signal (downscale) | ≤ −2.5 % |
| 3.8 mA | Lower limit of the measurement range | −1.25 % |
| 4–20 mA | 0–100 % of the calibrated range | 0–100 % |
| 20.5 mA | Upper limit of the measurement range | 103.125 % |
| ≥ 21 mA | Failure signal (upscale) | ≥ 106.25 % |
Between 3.8 mA and 20.5 mA the transmitter can report slight under- and over-range; the gaps to the failure thresholds are margins, so a receiver never has to guess. Whether the failure signal goes upscale or downscale is a process decision — drive it toward the state that makes the control system respond safely — and is configured in the transmitter.
Caution
Configure both ends
NE 43 only works if the control system uses the same thresholds. Left at default scaling, a 3.6 mA failure signal is displayed as −2.5 % — a plausible, slightly low reading instead of an alarm.
Loop power: the voltage budget
A two-wire transmitter needs a minimum voltage at its terminals to operate, listed on its datasheet. Everything else in the loop drops voltage in proportion to the current: the sense resistor, the cable, barriers or isolators, and any loop-powered indicator. The design condition is:
Engineering note
Loop compliance
Supply voltage − transmitter minimum voltage ≥ loop current × total loop resistance, checked at the highest current the loop must deliver: 20 mA for full scale, and the upscale failure current (at least 21 mA under NE 43) if upscale failure signalling is used.
Step 1: Available voltage
A 24 V supply and a transmitter that needs at least 12 V leave 12 V for the rest of the loop. At 20 mA that allows up to 12 V ÷ 0.020 A = 600 Ω in total.
Step 2: Add up the loop
A 250 Ω HART sense resistor plus 800 m of 1.0 mm² twisted pair — 1.6 km of copper conductor at about 18.1 Ω/km, roughly 29 Ω — gives 279 Ω. At 20 mA that drops 5.6 V, leaving about 18.4 V at the transmitter; at a 21 mA failure current the drop is 5.9 V. Both are comfortably inside the budget.
Step 3: Add a barrier
Now route the loop through a shunt-diode barrier with an end-to-end resistance of, say, 340 Ω. Total resistance becomes 619 Ω, which needs 12.4 V at 20 mA — more than the 12 V available.
Step 4: The symptom
The loop can now deliver at most 12 V ÷ 619 Ω ≈ 19.4 mA, about 96 % of span. The reading is correct at low values but flattens near the top of the range, and an upscale failure signal can never be delivered. The fix is a higher supply, an isolator with a lower voltage drop or less resistance in the loop — not recalibration.
HART: a digital layer on the analog signal
HART (Highway Addressable Remote Transducer) superimposes a digital signal on the same two wires using Bell 202 frequency-shift keying: 1200 Hz represents a binary 1 and 2200 Hz a binary 0, at 1200 bit/s. The signal is a small sine wave of about ±0.5 mA. Because it averages to zero, it does not disturb the analog measurement, which the control system continues to read as before.
Over that digital channel a HART device reports its tag, range, units and damping, the primary variable plus up to three further dynamic variables, device status and diagnostics. It can be configured, re-ranged and trimmed without disconnecting the loop.
- Primary master
- Usually the control system, communicating through HART-enabled analog input cards or a HART multiplexer, and able to collect diagnostics continuously for asset management.
- Secondary master
- Usually a handheld communicator or a laptop with a HART modem, connected across the loop for configuration and troubleshooting. HART allows one primary and one secondary master on a loop.
- Point-to-point mode
- The normal arrangement: one device per loop, with the analog signal carrying the measurement and HART carrying everything else.
- Multidrop mode
- Several devices share one pair, each with its own polling address and its current typically fixed at 4 mA. There is no analog measurement; values are read digitally and update slowly, so multidrop suits monitoring rather than fast control.
HART needs a loop resistance of at least 230 Ω to develop a readable signal — one reason 250 Ω is the conventional sense resistor. Communication can also be blocked by low-impedance devices or filters in the loop, or degraded by very long cable runs with high capacitance.
Troubleshooting a 4–20 mA loop
| Symptom | Likely causes | First check |
|---|---|---|
| 0 mA | Open circuit, blown fuse, no supply, reversed polarity | Supply voltage at the marshalling terminals, then continuity to the field |
| ≤ 3.6 mA or ≥ 21 mA | Transmitter signalling its own failure under NE 43 | Read the device diagnostics over HART |
| Correct at the transmitter, wrong in the system | Range mismatch, square root applied twice, wrong engineering units | Compare the transmitter’s configured range with the control-system scaling |
| Reading flattens near full scale | Insufficient loop voltage | Terminal voltage at the transmitter at high current |
| Noisy or offset reading | Ground loop, shield grounded at both ends, cable routed alongside power | Isolate the input; ground the shield at one end, usually the control-system end |
| Analog signal correct, HART fails | Loop resistance below 230 Ω, a low-impedance device in the loop, two primary masters, long high-capacitance cable | Measure loop resistance and check what else is connected |
Measure the current in series with a calibrated meter, or with a clamp meter made for milliamp loops, and compare three values: the meter reading, what the transmitter reports over HART, and what the control system displays. Where they disagree shows which part of the loop to examine.
Engineering questions
- Why does a 4–20 mA signal start at 4 mA instead of 0 mA?
- The live zero lets the receiver tell a genuine 0 % reading (4 mA) from a broken or unpowered loop (0 mA), and it leaves enough current at the bottom of the range to power a two-wire transmitter’s electronics.
- What is the minimum loop resistance for HART communication?
- HART needs at least 230 Ω of loop resistance to develop a readable signal, which is why a 250 Ω sense resistor is conventional. The maximum is set by the loop voltage budget: supply voltage minus the transmitter’s minimum voltage, divided by the highest loop current.
- Does HART communication affect the 4–20 mA reading?
- No, in a correctly designed loop. The HART signal is a small FSK sine wave that averages to zero, so the analog input continues to read the DC current while HART exchanges configuration, diagnostics and additional variables.
- What happens to the analog signal in HART multidrop mode?
- Each device’s current is typically fixed at 4 mA and the measurements are read digitally by polling each device’s address. The loop no longer carries an analog value, and updates are slower, so multidrop suits monitoring rather than fast control.