4–20 mA Loop Calculator
4–20 mA Loop Calculator
Convert between loop current and engineering units either way, with the sense-resistor voltage, the ADC counts your resolution gives, and a loop budget that says plainly whether your supply can actually drive the resistance you have put in the loop.
4–20 mA current loop
0–100 units on 4–20 mA, process value 50, 250 Ω sense, 24 V supply, 12 V transmitter, 20 Ω of wire
The span, the sense resistor and the loop budget
Vsense = I × Rsense counts = round(Vsense ÷ Vref × (2n − 1))
loop budget: headroom = Vsupply − Vtx,min − 0.020 × (Rsense + Rwire + Rother) Rmax = (Vsupply − Vtx,min) ÷ 0.020
- I low
- 4 mA for a live-zero loop, 0 mA for a 0–20 mA one
- 0.020
- twenty milliamps — the budget is always worked at the top of the span, because that is where the loop drops the most voltage
- V tx,min
- the transmitter’s compliance voltage: the least it can work on, from its data sheet
- R wire
- both cores of the pair, out and back
- headroom
- what is left over. Negative means the loop cannot reach 20 mA
Worked example
0–100 units on 4–20 mA, process value 50, 250 Ω sense, 24 V supply, 12 V transmitter, 20 Ω of wire
50 of 0–100 is 50% of span, so the current is 4 + 0.5 × 16 = 12.000 mA
Across a 250 Ω sense resistor that is 3 V, which on a 12-bit ADC with a 5 V reference is 2,457 counts of 4,095
One count is 0.004884 mA, or 0.03053 of your units — the resolution floor of the whole chain
Loop budget at 20 mA: the loop holds 270 Ω, which drops 0.020 × 270 = 5.40 V
24 V − 12 V − 5.40 V = 6.60 V of headroom; the largest total loop resistance this supply allows is 600 Ω
A 0–100 range on 4–20 mA, end to end
| % of span | Loop current | Process value | Across 250 Ω | As 1–5 V | As 0–10 V | 12-bit counts (5 V ref) |
|---|---|---|---|---|---|---|
| 0% | 4.000 mA | 0.0 | 1.000 V | 1.000 V | 0.00 V | 819 |
| 25% | 8.000 mA | 25.0 | 2.000 V | 2.000 V | 2.50 V | 1,638 |
| 50% | 12.000 mA | 50.0 | 3.000 V | 3.000 V | 5.00 V | 2,457 |
| 75% | 16.000 mA | 75.0 | 4.000 V | 4.000 V | 7.50 V | 3,276 |
| 100% | 20.000 mA | 100.0 | 5.000 V | 5.000 V | 10.00 V | 4,095 |
NAMUR NE 43 signal levels
| Loop current | What it means | Notes |
|---|---|---|
| Below 3.6 mA | Failure, low | NAMUR NE 43 fault. A broken wire or a dead transmitter reads 0 mA and lands squarely here — which is the whole point of a live zero. |
| 3.6 to 3.8 mA | Under-range | The measurement is below the bottom of the span but the transmitter is alive and saying so. |
| 3.8 to 20.5 mA | Measuring range | Normal operation; 4.0 to 20.0 mA is the span itself, with a little margin either side for tolerance. |
| 20.5 to 21 mA | Over-range | Above the top of the span, transmitter still healthy. |
| Above 21 mA | Failure, high | NAMUR NE 43 fault. Many transmitters drive 21.5 mA deliberately to report a sensor failure. |
Why the zero is at 4 mA, and why the budget matters more than the maths
A 4–20 mA loop carries a measurement as a current, and it does so because current is the one electrical quantity that is the same everywhere in a series circuit. Put a hundred metres of cable, a terminal block and a safety barrier in the way and the current at the receiver is still the current the transmitter set — the voltage lost along the wire simply comes out of the transmitter’s share. A voltage signal, by contrast, is divided down by every ohm of wire resistance between the two ends.
The live zero. The span starts at 4 mA, not 0, so that zero current means something. A cut cable, a pulled terminal, a dead transmitter and a failed power supply all give 0 mA, which is outside the measuring range and cannot be confused with a genuine reading of nothing. NAMUR NE 43 makes that formal: below 3.6 mA or above 21 mA is a fault, 3.8 to 20.5 mA is a measurement, and many transmitters drive 21.5 mA deliberately when their own sensor fails. The 4 mA also gives the transmitter something to run on, which is what makes a two-wire transmitter possible at all — it powers itself from the loop it is modulating, and it can never draw less than the bottom of its own span.
The loop budget is the calculation that bites. Everything in the loop is in series with everything else, so at 20 mA the loop drops 0.020 × the total resistance. What is left after that and after the transmitter’s own minimum operating voltage is your headroom. With a 24 V supply, a transmitter needing 12 V, a 250 Ω sense resistor and 20 Ω of cable, the loop drops 5.40 V at full scale and 6.60 V is left over — comfortable. Raise the sense resistor to 500 Ω, add a barrier, or run the cable twice as far, and it goes negative. When it does, the symptom is cruel: the loop works perfectly at low readings and flattens off near the top of the span, which looks exactly like a sensor that has stopped responding. The largest total loop resistance a supply can carry is simply (Vsupply − Vtx,min) ÷ 0.020 — 600 Ω here. Work the cable resistance out with the AWG wire size calculator or the voltage drop calculator, remembering to count both cores.
The receiver end. A 250 Ω resistor turns 4–20 mA into 1–5 V, which is why that value is everywhere; 100 Ω gives 0.4–2 V and costs less headroom. Whatever you choose, keep 20 mA × Rsense below the ADC’s reference or the top of the span clips. Resolution follows from the reference and the bit count, not from the loop: with a 5 V reference and 12 bits one count is 0.004884 mA — about 0.0305 of a 0–100 range, which is better than any industrial transmitter’s own accuracy, so the ADC is rarely the limit. The 1–5 V and 0–10 V columns are there because plenty of equipment wants a voltage; note that 0–10 V throws the live zero away.
The usual thing at the far end of the loop is a temperature sensor — see the PT100 / PT1000 RTD calculator for a platinum RTD and the NTC thermistor calculator for a thermistor. The sense resistor and the voltage it makes are just Ohm’s law: Ohm’s law calculator.
Frequently asked questions
Why does a 4–20 mA loop start at 4 mA?
So that zero current can mean a fault. A broken wire gives 0 mA, which is below the measuring range and cannot be mistaken for a genuine reading of zero. The 4 mA also powers a two-wire transmitter, which draws its supply from the same pair it signals on.
How do I convert 4–20 mA to engineering units?
Value = low + (I − 4)/16 × (high − low). For a 0–100 °C range, 12 mA is (12 − 4)/16 = 50% of span, so 50 °C. For a 0–20 mA signal the 4 becomes 0 and the 16 becomes 20.
What supply voltage does a 4–20 mA loop need?
Enough to cover the transmitter’s minimum operating voltage plus 20 mA times every ohm in the loop — sense resistor, both cores of the cable, barriers and any other receivers. 24 V is standard because it leaves room for a 250 Ω sense resistor, a 12 V transmitter and a few hundred metres of cable.
Why 250 Ω for the sense resistor?
Because 4–20 mA through 250 Ω is exactly 1–5 V, the other standard analogue signal. It costs 5 V of loop budget at full scale, which is the price. 100 Ω costs only 2 V but gives 0.4–2 V, so the ADC sees a smaller signal.
What is NAMUR NE 43?
The recommendation that standardises what currents outside the span mean. Below 3.6 mA or above 21 mA is a failure signal; 3.8 to 20.5 mA is a valid measurement with margin. It is what lets a control system distinguish “the process is at zero” from “the transmitter has died”.
Does cable length affect a 4–20 mA reading?
Not the reading, only the budget. The current is the same at both ends whatever the wire does, so a long run adds no error — right up to the point where the wire resistance eats the transmitter’s headroom, and then the loop simply cannot reach 20 mA. Count both cores when you work the resistance out.
Related calculators
References
- NAMUR recommendation NE 43, Standardisation of the signal level for the failure information of digital transmitters. Summarised by DIVIZE industrial automation: “When the loop current is below 3,6 mA or above 21 mA this is interpreted as a sensor fault.”
- Building Automation Products Inc. Designing 4 to 20 mA current loops, application note. The worked loop budget — “Adding all the voltage drops together equals: 1.195 + 15 + 5 = 21.195 volts” — is wire drop plus the transmitter’s 15 V minimum plus 5 V across a 250 Ω sense resistor at 20 mA.
- Analog Devices. Two-wire 4-to-20 mA loop-powered transmitters — application material on compliance voltage, the loop budget and why the transmitter’s own supply current must fit under 4 mA.
- Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. §1.3.2 on the decibel and the two factors, and Appendix A on the oscilloscope — what a measured amplitude actually is.