
0–10 V vs 4–20 mA: Which Analog Proximity Sensor Output Should You Choose?
Choose 4–20 mA for demanding cable routes and useful loss-of-signal monitoring, provided the loop can drive its total load. Choose 0–10 V for a controlled local connection to a suitable voltage input. Neither guarantees better accuracy: the sensor, PLC input, wiring and scaling must work together.
Which analog output should you choose first?
Start with the controller input you can actually use. If both options are available, favor 4–20 mA for a demanding cable route or useful loss-of-signal monitoring. Favor 0–10 V for a local, controlled connection to a suitable voltage input. Either can work well when the entire measurement chain is matched.
| Your situation | Practical starting point | Decisive check |
|---|---|---|
| A local fixture with a voltage input | 0–10 V can avoid an additional input module or converter. | Input resistance, signal reference and required accuracy. |
| A longer route near drives or motors | Evaluate 4–20 mA first. | Available output voltage, total load, routing and isolation. |
| Loss of signal must be distinguishable from minimum position | 4–20 mA offers a live zero. | The sensor’s fault behavior and the input’s enabled diagnostics. |
| Sensor and controller have different ground potentials | Resolve grounding and isolation before selecting either. | Permitted common-mode voltage and the actual current-return paths. |
A short cable beside a variable-frequency drive can be more troublesome than a longer, well-routed cable. Neither “voltage below 10 m” nor “current above 100 m” is a dependable rule for all products.
Compare the installed cost, too: input hardware, power, isolation, cabling and commissioning can change the decision more than the sensor’s purchase price.
What changes when the sensor sends voltage instead of current?
The electrical signal carries the measured position to the controller. It does not make an unsuitable target measurable or remove errors already present in the sensor. Target material, mounting, measuring range, linearity and temperature behavior still matter.
0–10 V depends on the receiving reference
The input measures a voltage difference between signal and reference terminals. A high input resistance draws less current, reducing loading of the sensor output. But noise on the signal or movement of the reference potential can change the value the controller sees.
For illustration, a 0.10 V reference-related error equals 1% of a 10 V span. Using thicker cable may reduce conductor drop, but it does not by itself fix an incorrectly shared reference or a ground loop.
4–20 mA depends on enough voltage to regulate the current
In a single series path, the same current passes through each component. Added cable resistance consumes voltage; it does not directly reduce the commanded current while the output remains within its operating limits. This makes current transmission useful for demanding field wiring.
The available voltage for driving that path is often called compliance voltage. Once it is exhausted, the output can no longer reach the requested current. Current transmission also does not provide electrical isolation by itself.
Neither signal is automatically more accurate or faster. Compare the sensor’s specified errors, analog output performance, input-module accuracy, conversion time and filter settings. More ADC bits do not correct target-related error; stronger filtering may smooth a reading while adding delay.
Will the output match your PLC input and wiring?
Match both the signal range and the electrical circuit. An input labeled “analog” may accept voltage, current, or selectable ranges with different terminals. PNP/NPN descriptions of a separate switching output do not establish analog compatibility.
Check input range and load in the correct direction
- For voltage: the input resistance must be high enough for the sensor’s drive capability. A 0–5 V-only input cannot directly measure the complete 0–10 V span. A suitable ±10 V input can accept a 0–10 V signal, but its hardware range, resolution, common-mode limits and scaling still need checking.
- For current: add the receiver’s burden—the load it presents—to the cable and other series loads. The total must remain within the output’s allowed load at the actual supply voltage.
A real datasheet shows why one resistance rule is not enough
Documented model example: the Pepperl+Fuchs IA6-12GM50-IU-V1 is a four-wire analog inductive sensor. Its datasheet lists a 15–30 V supply, a current-output load of 0.4 kΩ at 15 V and 1 kΩ at 30 V, and a voltage-output load current of no more than 15 mA.
The lesson is the direction of the limit: current-output resistance must not be too high; voltage-output current demand must not be too high. Do not substitute a fixed “400 Ω” or “2 kΩ” rule for both interfaces. These values describe this model, not an xsz sensor product. See the manufacturer’s output specifications, page 2.
Identify who supplies the current loop
A two-wire 4–20 mA transmitter normally takes its power from the measurement loop. A three- or four-wire sensor may have a separate supply and an active current output. 4–20 mA does not mean two-wire.
Original conceptual diagrams, not connector pinouts or installation drawings. Input isolation, protection and exact supply/reference terminals are not shown.
A passive input receives current without supplying loop power. An active input also powers a compatible transmitter. Read the terminal diagrams rather than relying on those labels alone. Two passive devices need an appropriate supply; an active output must not be connected to an active input unless the manufacturer explicitly permits that arrangement.
If the ranges do not match, a suitable signal converter can help, but it adds its own error, delay, power and load requirements. A resistor is not a full 4–20 mA-to-0–10 V converter: an ideal 250 Ω shunt produces 1–5 V, not 0–10 V. Any approved shunt arrangement also needs suitable input loading, scaling and loop headroom.
Why can a 4–20 mA signal stop short of 20 mA?
One possible cause is insufficient voltage across the transmitter after the other loop components take their share. A loop may work at 4 mA yet fail near full scale because resistive voltage drops increase with current.
For a compatible two-wire, loop-powered circuit, the supply must cover the transmitter’s minimum operating voltage, the current-dependent drops and any other series-device drops. For separately powered outputs, use the manufacturer’s load curve or compliance specification instead of assuming the same equation.
A loop that works at minimum position can still fail at full scale
Illustrative calculation, not a measured installation. Assume a minimum loop supply of 20.4 V, a transmitter needing at least 12 V, a 250 Ω input and 100 Ω total cable resistance, including both conductors. No other series devices are fitted.
| Component | Calculation | Voltage |
|---|---|---|
| Transmitter | Assumed minimum | 12 V |
| Input burden | 0.020 A × 250 Ω | 5 V |
| Cable pair | 0.020 A × 100 Ω | 2 V |
| Total required | 12 + 5 + 2 | 19 V |
The assumed supply leaves 1.4 V above that requirement. Now add an approved series indicator with a 100 Ω burden: it needs another 2 V at 20 mA. Total demand becomes 21 V, exceeding the available 20.4 V.
At 4 mA, those same 450 Ω of combined resistance drop only 1.8 V. That explains why a minimum-position check can look satisfactory while full-scale operation fails. Recalculate before adding a display or changing the cable. The 1.4 V remainder is not a universal design margin; actual tolerances, hot-cable resistance and any supported overrange current must also fit.
How do you turn the electrical signal into the correct distance?
Use the sensor’s configured endpoints, not an assumed zero-distance origin. For a linear, rising output, first calculate the fraction of span and then apply the physical range.
0–10 V: fraction = voltage ÷ 10 V
4–20 mA: fraction = (current − 4 mA) ÷ 16 mA
Distance = start distance + fraction × (end distance − start distance)
Illustrative example: a sensor is configured so that 0 V or 4 mA represents 2 mm, and 10 V or 20 mA represents 8 mm. A 5 V or 12 mA reading is halfway through the span: 2 + 0.5 × (8 − 2) = 5 mm. It is not 3 mm, and 12 mA is not 60% of a 4–20 mA span.
Confirm the curve before assuming linear scaling
Some sensors have teachable endpoints or a reversed output direction; others have a fixed characteristic that depends on the target. A linear equation is appropriate only for the documented or calibrated relationship. Teaching endpoints does not remove nonlinearity between them.
For a documented configuration example, ifm’s OMH551, OMH553 and OMH555 allow selection of current or voltage output and teaching of analog start/end points. That capability must not be assumed for every analog proximity sensor.
If the PLC reports raw counts, use the input module’s count values for the selected electrical range. Keep fault/status information separate from the distance calculation, so an invalid signal is not silently turned into a believable position.
What does a zero, low or unstable reading actually tell you?
It narrows the investigation, but rarely identifies the failed component by itself. Compare the target position, raw input value and scaled display before changing sensitivity or PLC scaling.
Live zero helps distinguish a valid minimum from no current
In a normal 4–20 mA measurement span, 4 mA represents the configured starting value. Approximately 0 mA is outside that normal span, but may result from lost power, an open connection, wrong wiring or a device fault. A value slightly below 4 mA is not automatically a broken wire: supported underrange, alarm thresholds and diagnostic settings differ.
For 0–10 V, zero volts may be a valid minimum. A disconnected input may read zero, drift or respond to interference depending on its input circuit. Do not treat either signal as a complete fault-detection system without checking the sensor and receiver behavior.
| Observed symptom | Check first | What changes the next step |
|---|---|---|
| Current plateaus below 20 mA | Output span, supply and total burden. | Insufficient headroom points to the circuit; adequate headroom shifts attention to the sensor range, target and input. |
| Voltage reads too low | Input mode, load and signal reference. | A correct sensor-terminal value but a low receiver value points toward the connection, reference or receiver. |
| Reading jumps with drive operation | Raw signal, supply, routing and bonding. | A correlated raw-signal disturbance needs electrical investigation; a stable raw value with a jumping display points toward processing. |
| Stable signal, wrong distance | Endpoints, slope, target and calibration. | Correct electrical readings with incorrect converted units indicate a scaling problem, not automatically a defective sensor. |
Follow the specified signal-cable routing and shield termination for either output. Changing to current will not correct poor protective bonding, an overloaded output or a wrong common connection. The separate sensor cable shielding guide covers that installation decision.
Safe measurement matters. Qualified personnel should make electrical measurements using the equipment manuals and site isolation procedure. De-energize before rewiring. Voltage is measured across the specified points; an ordinary meter in current mode belongs in the current path, not across the supply. Do not lift protective earth or introduce live shorts to diagnose an analog fault.
What should you confirm before ordering and commissioning?
Before ordering, give the supplier enough information to match the sensor to the input and target. Before commissioning is complete, show that the same combination works throughout the required measurement span.
- Measurement: target material and size, start/end distances, mounting, required error limit and response.
- Electrical match: exact sensor and input-module numbers, output range, terminal diagram, supply limits, load or input resistance, and isolation requirements.
- Installed route: cable type and length, conductor resistance, added indicators/isolators and nearby switching equipment.
- Commissioning record: raw signal and scaled distance at both endpoints and intermediate positions, actual filter settings and behavior during normal machine operation.
- Fault response: documented behavior for lost supply, invalid signal and loss of connection, checked through the approved commissioning method.
The better output is the one that preserves the required measurement at the controller. Use 4–20 mA where its transmission and live-zero advantages solve a real need. Use 0–10 V where the receiving input and route make it a sound, simpler connection. Finish the choice with the load, reference and scaling checks—not the signal label alone.
Sources and method references
- Balluff — Analog signals: 0…10 V vs. 4…20 mA. Transmission trade-offs and the meaning of live zero.
- NI — Current-loop fundamentals, system design and setup. Series current, voltage budgets, receiver shunts and ground-loop isolation.
- NI — C Series specifications explained. Analog input range, impedance, resolution, common-mode limits and input delay.
- Pepperl+Fuchs — IA6-12GM50-IU-V1 datasheet. Supply-dependent load specifications, four-wire connection and target-dependent output curves; issue 2026-02-26.
- Phoenix Contact — Digital and analog signal conditioning. Active/passive inputs, conversion and isolation functions.
- ifm — OMH device configuration. Model-specific output selection and analog endpoint teaching.
- Fluke — 233 Getting Started Manual. Meter safety and the distinction between voltage and series-current measurements.
Numerical scenarios are illustrative calculations, not laboratory or customer test results. Manufacturer examples are not xsz sensor specifications. The hero is an AI-generated conceptual illustration; the wiring sketches are original simplified explanations.