SensorCatalog

4–20 mA converter

Scale a loop current to a process value and back, and check it against NAMUR NE 43.

Transmitter

Convert

Result

Signal state
In rangeNAMUR NE 43
Voltage across 250 Ω
3.000 V1–5 V input
Calibrated range
0 to 10 barlinear output
Change per 1 mA
0.625 bar

Scaling points

Current (mA)OutputValue (bar)
4.000 %0
8.0025 %2.5
12.0050 %5
16.0075 %7.5
20.00100 %10

Range finder

Value at 4 mA
0
Value at 20 mA
10
Span
10
Change per 1 mA
0.625

Formulas and standards
PV = LRV + (I − 4) / 16 × (URV − LRV)

Square-root output (flow from differential pressure): PV = LRV + √((I − 4) / 16) × (URV − LRV).

NAMUR NE 43 uses 3.8–20.5 mA for measurement, including slight under- and over-range, and ≤ 3.6 mA or ≥ 21 mA as failure signals.

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How the 4–20 mA converter works

A 4–20 mA transmitter maps its calibrated range linearly onto a loop current: the lower range value (LRV) at 4 mA and the upper range value (URV) at 20 mA. The live zero at 4 mA lets the receiver tell a genuine zero reading from a broken wire, and it powers two-wire transmitters from the loop itself.

For a linear output:

PV = LRV + (I − 4) / 16 × (URV − LRV)

The converter links three fields: loop current, percent of span and process value. Type in any of them and the other two follow. There is no mode switch; the field you edited last is the input.

Worked example

A pressure transmitter is calibrated 0 to 10 bar. At 12 mA:

The scaling table lists the five standard check points (4, 8, 12, 16 and 20 mA) for your range. Click any row to load it into the converter.

Square-root output for flow

Differential-pressure flowmeters produce a pressure difference proportional to the square of flow. Many transmitters can extract the square root internally, so that the current is proportional to flow:

PV = LRV + √((I − 4) / 16) × (URV − LRV)

With a 0–100 range and square-root output, 12 mA is not 50 but 70.71, and 8 mA (25 % of current span) already represents 50 % of flow. Select “Square root (flow)” in the Output field to apply this. Below 4 mA a square-root output has no defined flow value, and the calculator says so instead of returning a number.

NAMUR NE 43 signal states

NAMUR recommendation NE 43 standardises how the current signals faults, so a control system can tell a measurement from a failure:

Current Meaning
≤ 3.6 mA Failure signal (downscale)
3.6 to 3.8 mA Below the measuring limit
3.8 to 4 mA Under range, still a valid measurement
4 to 20 mA Normal range
20 to 20.5 mA Over range, still valid
20.5 to 21 mA Above the measuring limit
≥ 21 mA Failure signal (upscale)

The scale under the converter shows these zones. You can drag the marker, or focus it and use the arrow keys, to see how any current is classified. Configure your control system alarms to the same limits as the transmitter.

Finding the range from two readings

If the calibration is unknown, the range finder works it out from two known points. Enter two readings, each a current and the value it represents, and it extrapolates the values at 4 and 20 mA. Example: 7.2 mA at 2 bar and 16.8 mA at 8 bar give a range of 0 to 10 bar. “Apply this range to the transmitter” copies the result to the range fields above. This assumes a linear output; it cannot recover a square-root characteristic.

Common mistakes

Assuming 0 mA is zero. A dead loop reads 0 mA, which is outside the valid range, not the bottom of the scale. Receivers should treat it as a fault.

Scaling a square-root transmitter linearly. The error is largest at low flow: at 8 mA, a linear interpretation gives 25 % where the true flow is 50 %.

Forgetting the HART resistor. HART communication needs a loop resistance of typically at least 250 Ω. Check the voltage budget with the loop power calculator.

Mismatched ranges. If the transmitter is re-ranged in the field but the control system isn’t, every reading is wrong in proportion. Copy the result from this tool and keep it with the loop documentation.

Frequently asked questions

Why 4 mA instead of 0 mA?

The live zero distinguishes a real zero reading from an open circuit, and the 4 mA minimum supplies enough power for a two-wire transmitter to operate.

What is the resolution of a 4–20 mA signal?

The analog signal itself is continuous. Resolution is set by the receiver’s A/D converter and noise; a 16-bit input card resolves the 16 mA span to a few hundred nanoamps.

Can the converter handle reverse-acting ranges?

Yes. Enter a value at 4 mA that is higher than the value at 20 mA, for example 100 to 0, and the conversion works the same way.

Does it work for 0–20 mA signals?

No. The equations assume a 4 mA live zero. For 0–20 mA, scale linearly from 0 to 20 mA.