Ahmedonics

Automation

4–20 mA signal scaling calculator

Convert a 4–20 mA, 0–20 mA or 0–10 V signal into percent of span, engineering units and a PLC raw count, and back again, with the NAMUR NE 43 fault bands and the voltage the loop current produces across the receiver's sense resistor.

Signal, range and PLC
e.g. 0–10 bar, −50–150 °C, 0–500 m³/h; whatever the transmitter is ranged to.
The loop current from a meter in series, or the mA the PLC displays.
Leave blank to skip the reverse conversion.
Siemens S7 analogue inputs use 0–27 648 for the nominal range; a 12-bit converter 0–4 095; a 16-bit 0–65 535; some cards use 0–20 000 or 4 000–20 000. Take it from the module manual.
250 Ω turns 4–20 mA into 1–5 V; some inputs use 100 Ω or 500 Ω. Ignored for voltage signals.

Scaled value

Enter your values and press Calculate.

What the calculator does

A transmitter turns a pressure, level, flow or temperature into a current between 4 and 20 mA; the PLC turns that current into a number. Between the two sit three straight-line maps: signal to percent of span, percent to engineering units, and percent to the raw integer the analogue input card delivers. The calculator does all three for a measured signal, and runs them backwards for an engineering value, so you can check a loop with a meter, verify a scaling block, or work out what the PLC must display when the transmitter sits at a known point.

For 4–20 mA it also says whether the reading is inside the measuring band, in the under- or over-range band, or at one of the NAMUR NE 43 failure levels, and it gives the voltage the loop current produces across the receiver's sense resistor, which is the easiest thing to measure without breaking the loop.

Formula

Percent of span: % = (I − Ilo) ÷ (Ihi − Ilo) × 100
Engineering value: EU = EUlo + (% ÷ 100) × (EUhi − EUlo)
PLC raw count: raw = rawlo + (% ÷ 100) × (rawhi − rawlo)
Resolution: one count = (EUhi − EUlo) ÷ (rawhi − rawlo)

Reverse: I = Ilo + (EU − EUlo) ÷ (EUhi − EUlo) × (Ihi − Ilo)
Sense resistor: V = I × R

where Ilo and Ihi are the ends of the signal span (4 and 20 mA, or 0 and 10 V), EUlo and EUhi the engineering range the transmitter is ranged to (its lower and upper range values), and rawlo and rawhi the counts the input card delivers at those two ends. All three maps are the same straight line through two points. The PLC's scaling instruction, whatever its name in your vendor's library, is that line with the raw counts as its input points, and the percent of span is the intermediate that makes the arithmetic easy to check by hand.

Worked example

A pressure transmitter ranged 0–10 bar with a 4–20 mA output, wired to a Siemens S7 analogue input (0–27 648 for the nominal range) through a 250 Ω sense resistor. A meter in series with the loop reads 12.8 mA.

  1. Percent of span = (12.8 − 4) ÷ (20 − 4) × 100 = 8.8 ÷ 16 × 100 = 55.0 %.
  2. Engineering value = 0 + 0.55 × (10 − 0) = 5.50 bar.
  3. Raw count = 0 + 0.55 × 27 648 = 15 206.4, which the card delivers as 15 206. One count is 10 ÷ 27 648 = 0.000362 bar, or 0.00362 % of span: far finer than the transmitter's accuracy, so the converter is not the limit.
  4. Voltage across the 250 Ω resistor = 12.8 mA × 250 Ω = 3.20 V, between the 1 V and 5 V that mark 4 and 20 mA.
  5. Reverse: a display of 7.5 bar needs (7.5 − 0) ÷ 10 = 75 % of span, so 4 + 0.75 × 16 = 16.00 mA and a raw count of 0.75 × 27 648 = 20 736.

If the PLC shows 5.5 bar while the meter shows 12.8 mA, the scaling is right. If it shows 6.4 bar, the scaling block has been told the signal is 0–20 mA (12.8 ÷ 20 = 64 %). If it shows 8.8 bar, the transmitter has been ranged 0–16 bar while the PLC still says 0–10. Both are common, and both look plausible on the screen until someone checks with a meter.

Common analogue signal standards

SignalZeroFull scaleDetects a broken wire?Notes
4–20 mA4 mA20 mAYes: 0 mA cannot be a valid valueIEC 60381-1; powers two-wire transmitters; NE 43 failure levels ≤ 3.6 mA and ≥ 21 mA
0–20 mA0 mA20 mANoIEC 60381-1; needs a separately powered transmitter; rare in process plants
0–10 V0 V10 VNoIEC 60381-2; building automation and drive references; sensitive to wire resistance, ground offsets and noise
2–10 V2 V10 VYes: a break reads 0 VBuilding-automation actuators and sensors; equals 4–20 mA across 500 Ω
0–5 V0 V5 VNoIEC 60381-2; equals 0–20 mA across 250 Ω
1–5 V1 V5 VYes: a break reads 0 VEquals 4–20 mA across 250 Ω; the form most PLC current inputs actually digitise

A 250 Ω resistor turns 4–20 mA into 1–5 V, and that is what the input card converts. 100 Ω gives 0.4–2 V and 500 Ω gives 2–10 V; every ohm in the loop adds to the voltage the supply must provide.

Assumptions and limitations

  • Linear scaling only. A differential-pressure flow transmitter produces a signal proportional to the square of the flow unless it extracts the square root itself: flow = span × √(percent of DP ÷ 100). Do that as a separate step in the PLC, or configure the transmitter for square-root output, and never both.
  • The PLC's scaling block may clamp. Many scaling instructions limit the output at the engineering range or set a status bit above the nominal raw span, so a reading of 20.9 mA can show as 10.0 bar in the PLC while the calculator gives 10.56 bar. Siemens cards keep counting above 27 648 into an over-range band; other cards clamp at full scale.
  • Raw counts differ by card and by configured range. 0–27 648 is the Siemens nominal range; a 12-bit converter gives 0–4 095, a 16-bit one 0–65 535 or a signed ±32 767, and some cards deliver 4 000–20 000 to mirror the microamps. Whether the card is configured for 4–20 mA or 0–20 mA also changes which count 4 mA produces. Take both counts from the module manual and set the card's range to match the transmitter.
  • No calibration offset or drift. The calculator maps ideal points. A real loop carries the transmitter's zero and span errors, the tolerance of the sense resistor (which is why 0.1 % resistors are used) and the input card's own accuracy; a loop calibrator at 4, 12 and 20 mA is how those are found.
  • Voltage signals lose to the wire; current signals do not. A current is the same at every point in the loop, so wire resistance does not change what the receiver sees. A voltage signal is divided between the source and wire resistance and the receiver's input impedance, and any ground-potential difference between the two ends adds directly to it. The calculator ignores both effects because they belong to the installation, not the arithmetic.
  • The reading should be the loop current, not the transmitter's display. The transmitter's screen shows what it thinks it is sending; the meter shows what the PLC actually receives.

Frequently asked questions

Why does the range start at 4 mA and not 0?

Two reasons. A two-wire transmitter has no other power supply: it runs its electronics on the loop current, so the minimum must be enough to keep it alive, and 4 mA at the 12 V or so left across its terminals is under 50 mW, which is plenty for a modern sensor. And an offset zero, the "live zero", lets the receiver tell a genuine reading of zero (4 mA) from a broken wire, a dead transmitter or a missing supply (0 mA). 0–20 mA and 0–10 V give up both advantages.

How many devices can share one loop?

As many as the supply voltage can push 20 mA through, in series. A typical two-wire transmitter needs about 12 V at its terminals, so a 24 V supply leaves 12 V for everything else. Each 250 Ω receiver drops 5 V at 20 mA, and a long cable adds a little more, so two 250 Ω receivers (10 V) fit with a small margin and a third (15 V) does not. Indicators, isolators and safety barriers each add their own drop, and HART needs at least about 230 Ω in the loop to develop its signal. Add up the ohms, multiply by 0.02 A, and keep the sum below the supply minus the transmitter's minimum.

What is HART?

A digital signal, frequency-shift keyed at ±0.5 mA, superimposed on the same 4–20 mA loop; it carries the transmitter's range, diagnostics and secondary variables without disturbing the analogue value, because the average of the modulated current over a cycle is zero.

Two-wire or four-wire transmitter?

A two-wire transmitter is powered by the loop: the same pair carries the supply in and the signal out, and the transmitter regulates the current. A four-wire transmitter has its own mains or 24 V supply and sources the 4–20 mA from an active output, which suits devices that need more power than 4 mA can deliver, such as magnetic flowmeters and analysers. The receiver input must match: an active output must not be wired into a PLC input that also supplies 24 V to the loop, or the two sources fight.

How do I measure the loop current without breaking the loop?

Measure the voltage across the receiver's sense resistor and divide: 3.20 V across 250 Ω is 12.8 mA. Most transmitters also have test terminals with a diode across them; a meter on the mA range connected there shorts the diode and takes the loop current through itself without opening the circuit. A DC milliamp clamp meter does the same on the wire; an ordinary AC clamp meter will not read it.

References

  • IEC 60381-1:1982, Analogue signals for process control systems — Part 1: Direct current signals — 4–20 mA and 0–20 mA
  • IEC 60381-2:1978, Analogue signals for process control systems — Part 2: Direct voltage signals — 1–5 V, 0–10 V and the other voltage spans
  • NAMUR NE 43 (2003), Standardization of the signal level for the failure information of digital transmitters — measuring band 3.8–20.5 mA, failure levels ≤ 3.6 mA and ≥ 21 mA
  • ANSI/ISA-50.00.01-1975 (R2012), Compatibility of Analog Signals for Electronic Industrial Process Instruments — formerly ISA-50.1; loop compliance and load classes
  • Siemens S7-1200 System Manual, analogue value representation — nominal range 0–27 648, over-range and overflow bands
  • FieldComm Group (formerly the HART Communication Foundation), HART protocol specifications

Last reviewed 2026-09-20.