Live zero
Live zero — A live zero is a lower range value carried by a non-zero signal, so 4 mA means zero measured and 0 mA means the loop has failed; the wire cannot be broken without the fault becoming visible.
Loop headroom
Loop headroom — Loop headroom is the voltage left across a two-wire transmitter after the supply has paid for the drop across the receiver resistor and the cable, and it must exceed the transmitter minimum supply voltage at 20 mA.
| Loop supply | 24 V DC |
|---|---|
| Receiver input resistor | 250 ohm |
| Cable loop resistance, illustrative run | 35 ohm |
| Drop at 20 mA across receiver plus cable | 5.7 V |
| Voltage left at the transmitter terminals | 18.3 V |
| Typical minimum supply for a two-wire transmitter | 11 - 12 V |
| Observed symptom | Most likely class | Discriminating measurement |
|---|---|---|
| 0.0 mA, HMI shows bad value | Open loop or no supply | Voltage at the transmitter terminals with the loop closed: full supply means the break is downstream |
| Stuck near 3.6 mA | Transmitter in fault signalling | Inject a known current with a calibrator; if the HMI follows, the fault is in the sensor, not the wiring |
| Clipped at the top of scale only | Insufficient headroom at 20 mA | Measure transmitter terminal voltage while the loop is driven to 20 mA, not at rest |
| Reading shifts when a second device is earthed | Ground loop | Measure AC voltage between the two earth points; a real potential difference proves it |
| Stable but wrong by a scale factor | Wrong receiver impedance or wrong range | Compare the mA the transmitter actually produces with the engineering value the PLC computes |
| Fails only when the enclosure is warm or wet | Intermittent connection | Measure loop resistance with the enclosure at operating temperature, then wiggle terminals one by one |
- Illustrative loop-current error from a 10 ohm connection resistance at 20 mA
- 0mA
- Source: Derived from Ohm's law using the example resistances on this page. Illustrative sample dataset, not a measured value.
Why does a 4-20 mA loop tolerate cable resistance when a 0-10 V signal does not?
Because the transmitter regulates current rather than voltage: cable resistance consumes part of the available supply voltage instead of altering the signal, so the current arriving at the receiver is the same current that left the transmitter, and distance only shortens the margin left for the transmitter itself.
Should the cable screen be earthed at both ends?
Earthing a screen at both ends turns it into a second path for circulating current whenever the two earth points sit at different potentials, which is what a ground loop is. Bond one end, at the receiver side in most plants, and rely on that bond to drain induced noise; add galvanic isolation where the two grounds genuinely differ.
My loop reads 3.8 mA with the sensor connected. Is the transmitter broken?
Not necessarily. Below the 4 mA live zero a transmitter is usually signalling a detected fault, with NAMUR NE43 practice placing the fault band under 4 mA and above 20 mA; the next step is to read the sensor element resistance directly and compare it with the wiring table, because an open thermocouple or a disconnected RTD produces exactly this signature.
Every current, resistance and voltage on this page is an illustrative example value used to show the arithmetic. Replace them with your own measured values before publishing.
Why the order of testing matters more than the instrument
A 4-20 mA loop has one property that makes it diagnosable with a multimeter: the same current flows through every element in series, so a single measurement taken at any break point tells you what the whole loop is carrying. That property is also why testing in the wrong order wastes an afternoon. If you start by swapping the transmitter, you change two unknowns at once - the device and whatever the wiring did to it - and the symptom may follow the device for reasons that have nothing to do with it.
Work from the signal inward, splitting the loop into three sections at each step: sensor and transmitter, wiring and terminations, receiver and its scaling. Each test below is chosen so that a pass or a fail eliminates a whole section rather than a single component.
Test one: is there current, and is there voltage
Read the loop current first, with a mA clamp if you do not want to break the circuit, or with a meter in series if you do. Then, and this is the step people skip, measure the DC voltage across the transmitter’s own terminals while the loop is closed. The pair of readings separates four different failures:
- Current 0 mA and full supply at the terminals means the break is downstream - a blown fuse, an opened receiver, a terminal that never made contact, or a cut wire.
- Current 0 mA and no voltage at the terminals means the supply never arrived: wrong pair on the terminal block, reversed polarity on a protected input, or a power supply that only feeds part of the marshalling cabinet.
- Current in the 3.6 mA region means the loop is intact and the transmitter is deliberately signalling a fault. Do not recalibrate it; go and read the sensor element.
- Current pinned above 20 mA is the same message from the other end of the scale, or an overrange the process is genuinely producing.
Test two: headroom, measured under load and not at rest
The transmitter needs voltage to survive. Its supply comes from the loop power supply minus the drops across the cable and the receiver resistor, and those drops are largest at 20 mA, which is precisely where a control system most needs a valid reading. A loop whose headroom has run out does not fail at zero; it clips on the upper part of the range and looks like a range problem. This is why a technician who measures the terminals at rest declares the loop healthy and walks away.
Work the arithmetic before you leave the site. With a 24 V supply, a 250 ohm receiver and a cable loop resistance in the tens of ohms, the drop at 20 mA leaves the transmitter comfortably above the minimum supply typical of two-wire devices. Add a second receiver in series, an isolating amplifier, or a long thin run of cable, and the same loop can end up marginal. Note that a HART connection needs a minimum loop resistance of roughly 230 to 250 ohms to communicate at all, so the resistor that sets your voltage budget is also the reason a handheld will not talk to the transmitter across the marshalling cabinet.
Test three: the ground loop, and what bonding one end actually does
Induced noise and ground loops are different faults that get blamed on each other. Induced noise comes from changing magnetic flux linking the loop area, so it falls when the cable is shorter, twisted, or run perpendicular to the power cable, and it appears as an alternating component on the signal. A ground loop is a direct or low-impedance current circulating because two points of the signal path are bonded to earth at places that sit at different potentials; it shows up as a shift of the reading, and the shift changes when other equipment on either earth point switches on or off.
Measure AC voltage between the two earth points with a high-impedance meter. Anything meaningful there proves the potential difference exists, and the fix is to remove one of the bonds or to insert galvanic isolation at the receiver, not to wind the cable differently. If a screen is bonded at both ends, it is part of that circulating path; bonding it at the receiver end only drains induced charge while keeping it out of the current loop.
Test four: right current, wrong number
When the mA in the wire is correct but the plant screen is wrong, the fault is in the conversion, not the loop. Two causes cover most of them: the receiver input is configured for a different range than the transmitter, or the input resistor is not what the configuration assumes - a 250 ohm resistor read by a configuration that expects 500 ohm halves the apparent signal. Compare the mA measured at the terminals against the engineering value the controller computes from the same current; a consistent ratio such as a factor of two, or a 4-to-0 shift, is arithmetic, and it is fixed in configuration.
Test five: intermittent faults must be provoked
Intermittent faults are not found by waiting for them. Reproduce the condition mechanically: with the loop current on display, wiggle each terminal one at a time, warm the enclosure with hot air, and cool the connector until moisture condenses on it. A connection that is high resistance rather than open produces a reading that sags under load, so measuring loop resistance with the circuit de-energised can show a healthy short circuit while the same joint fails the moment 20 mA has to pass through it. Torque and re-seat, then re-measure at the top of the range, where the margin is smallest.