Industrial Sensor 4–20 mA Loop Tool
Convert loop current into a process value, size a receiver shunt resistor, and check whether the supply can support your transmitter, cable, and input burden.
Define your loop
Linear 4–20 mA scaling. The optional supply check models one passive, loop-powered 2-wire transmitter and a low-side receiver shunt.
Your loop results
Try a starting point, then enter the values for your actual sensor and receiver.
Current interpretation
Receiver & resistor checks
Five-point calibration reference
| Current | Process value | Shunt voltage |
|---|
Nominal linear values, not calibration measurements. Current-source error, resistor drift, ADC errors, and input loading are not included.
Supply voltage budget
| Requirement at check current | Voltage |
|---|
The summary includes inputs, assumptions, and checks. No sensor reading or circuit is simulated.
One current, several voltage drops
In a passive 2-wire loop, the supply powers the transmitter through the same conductors that carry the signal. Every series burden uses part of the available voltage. A high-impedance voltage input reads the shunt in parallel.
Connection and compliance-voltage background: Texas Instruments: Two-wire 4–20 mA transmitters.
Three checks before choosing components
1. Match the process scale
Use the sensor’s configured endpoints. A pressure sensor set to 0–100 bar gives 50 bar at 12 mA; a reverse scale works in the opposite direction. Alarm currents are not trustworthy process values.
PV = PV4 + (I[mA] − 4) × (PV20 − PV4) / 16
I[mA] = 4 + 16 × (PV − PV4) / (PV20 − PV4)2. Leave room for alarm current
A 250 Ω shunt produces 1–5 V over the normal range, but 22 mA produces 5.5 V before tolerance. An input that measures only up to 5 V cannot capture that high-current condition without clipping.
V shunt = I[A] × R[Ω]
P shunt = I[A]² × R[Ω]
R ceiling = (V input,max − headroom) / (I check × (1 + tolerance))3. Keep the transmitter powered
Use minimum supply voltage and worst-case cable and load values. The remaining voltage at the transmitter must exceed its minimum requirement. A positive margin can still be smaller than your chosen reserve.
V required = V transmitter,min + V fixed + I check × (R shunt,high + R cable + R other)
Margin after reserve = V supply,min − V required − reserveShunt conversion and loop-load background: Analog Devices: 4–20 mA current control loops. Input protection requires separate review: Analog Devices: Protecting ADC Inputs.
Read the current band first
The NE43-style option is a diagnostic guide, not proof of a particular fault. Check the transmitter’s supported standard, configured alarm direction, receiver thresholds, and hysteresis.
| Current used by this guide | Interpretation |
|---|---|
| ≤ 3.6 mA | Low-alarm region |
| > 3.6 to < 3.8 mA | Low transition band |
| 3.8 to < 4 mA | Extended underrange |
| 4 to 20 mA | Calibrated signal span |
| > 20 to 20.5 mA | Extended overrange |
| > 20.5 to < 21 mA | High transition band |
| ≥ 21 mA | High-alarm region |
Reference implementation: Endress+Hauser Fieldgate FXA320 operating instructions, section 8.1.3. Boundary assignments above define this tool’s display; actual devices can implement different thresholds and behavior.
Investigate the right constraint
- Near 0 mA: check supply, polarity, terminals, and loop continuity. Do not interpret it as the lower process endpoint.
- Correct at 4 mA, low at 20 mA: inspect transmitter terminal voltage at the high-current end. More current causes more voltage loss in the same series resistance.
- Voltage input saturates: compare shunt voltage at the configured alarm current with the actual measurement ceiling. Review protection independently of the normal range.
- A consistent scale error: check both process endpoints, resistor value and tolerance, input loading, and receiver scaling settings.
- Unstable readings: review grounding, common-mode limits, isolation, shielding, filtering, and nearby switching loads. This DC model does not predict noise.
- Automatic sizing gives no solution: identify whether input headroom or loop voltage is exhausted. Changing resistor series alone cannot restore missing supply voltage.
Measure loop current with the appropriate instrument and approved procedure. A current meter is normally inserted in series; connecting it directly across a supply can create a short circuit.
What the calculation does—and does not—cover
Resistance and voltage screening
Automatic sizing chooses the largest E24 or E96 nominal resistance from 1 Ω to 100 kΩ that stays below both enabled ceilings. The theoretical option returns the continuous ceiling within that range. The selected tolerance is used to calculate the highest shunt voltage and dissipation.
Cable resistance is entered directly or calculated as twice the one-way length multiplied by conductor resistance. Temperature effects must already be reflected in that resistance.
Not a full sensor or ADC model
The model assumes a linear signal, negligible receiver loading, and steady DC current. It does not include sensor accuracy, resistor temperature drift, ADC reference and gain errors, input common-mode range, transient behavior, or protection clamp current.
If the supply budget fails, the calculated voltage and power values are hypothetical values at the requested current. The tool does not predict the current that the real loop will settle at.
4–20 mA loop questions
Why does the signal start at 4 mA instead of zero?
The live-zero level lets a loop-powered transmitter operate while representing the bottom of its measurement span. It also separates normal zero-scale operation from very low current conditions, though the actual fault must still be diagnosed.
Does a shunt resistor convert 4–20 mA into 0–5 V?
Not on its own. A 250 Ω resistor converts 4–20 mA into 1–5 V. Producing 0–5 V requires offset removal and gain, or digital rescaling after conversion. Input protection and any analog signal-conditioning stage need their own design check.
Should I always use a 250 Ω resistor?
No. Choose the resistance from input range, maximum possible current, accuracy, power, and loop voltage constraints. Some communication systems also impose load requirements. This calculator does not verify HART compatibility.
Can I use this with a PLC that already has a 4–20 mA input?
The scaling calculation still applies. For a loop budget, use the PLC’s specified input burden as a series load and avoid counting it twice. The shunt-sizing calculation is intended for a separate voltage-sensing receiver; it does not tell you to add a resistor to an existing current-input channel.
Does this check an active 3-wire or 4-wire transmitter?
No. Current-to-process conversion and shunt voltage calculations remain useful, but disable the 2-wire supply check. An active output needs the manufacturer’s output-compliance or maximum-load specification and the correct grounding or isolation arrangement.
Why use 22 mA for a 20 mA measurement?
A transmitter may drive above 20 mA during overrange or a configured high alarm. The 22 mA default is an example design value, not a universal requirement. Replace it with the maximum relevant to the chosen transmitter, including tolerance. A higher entered current automatically raises the electrical check current.
Does meeting the supply budget mean the complete circuit is approved?
No. A satisfied DC budget only means that the entered values meet the modeled constraint. Verify actual terminals, input protection, environmental ratings, accuracy, isolation, and fault behavior before commissioning.
Source components for your sensor interface
Send YURUNOX the part numbers or BOM, quantity, input range, supply conditions, package, and required grade. Include the shunt value and accuracy target when requesting receiver components.