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4-20 mA loop: load and signal scaling

Enter the supply, transmitter, cable and everything wired in series to see whether voltage is left up to the alarm current, how many receivers still fit and what each mA of signal is worth.

Pre-sizing and estimating tool: results do not replace design, calculations or reports signed by a licensed professional, nor verification against the applicable codes and standards.

Check
Power supply and transmitter
Cable
Series loads

Typical 28 V zener barrier: 300 to 340 Ω end to end. Powered active isolator: enter 0 Ω and treat its output side as a receiver.

Signal and calibrated range
—Voltage margin at the transmitter
—

Voltage available at the transmitter versus current

Supply voltage at minimum tolerance minus the drops in the cable, receivers and indicators. The loop works while the line stays above lift-off.

Tap the chart to see the values.

Converting the mA signal to engineering units

Calibrated range between 4 and 20 mA, with the NAMUR NE 43 failure zones. Point: current read.

Tap the chart to see the values.

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Method, assumptions and references

Equations used

Minimum supply: Vmin = Vsupply · (1 − tolerance); cable: Rc = 2 · L · ρ20 · (1 + 0.00393 · (T − 20)) ÷ A, ρ20 = 0.017241 Ω·mm²/m
Transmitter voltage: Vt = Vmin − n_ind · V_ind − I · (Rc + n · R_rec + R_barrier + R_other); margin = Vt − V_lift-off
Maximum load: Rmax = (Vmin − n_ind · V_ind − V_lift-off) ÷ I, with I = 20 mA and the chosen alarm current
Linear: value = LRV + (I − 4) ÷ 16 · (URV − LRV); square root: Q = LRV + √((I − 4) ÷ 16) · (URV − LRV)
NAMUR NE 43: measurement 3.8 to 20.5 mA; failure ≤ 3.6 mA or ≥ 21.0 mA

Assumptions and limits

  • Two-wire transmitter powered by the loop itself; the lift-off voltage comes from the data sheet and already includes the internal electronics.
  • Annealed copper conductor (IEC 60228 resistivity) with a temperature coefficient of 0.00393 /K; terminal and splice resistance goes under “other resistances”.
  • Loop indicators treated as a constant voltage drop; zener barriers as series resistance. Intrinsic-safety entity parameters are not checked.

Standards and technical documents

  • IEC 60381-1:1982: Analogue signals for process control systems — Part 1: Direct current signals
  • ANSI/ISA-50.00.01-1975 (R2017): Compatibility of Analog Signals for Electronic Industrial Process Instruments
  • NAMUR NE 43: Standardization of the Signal Level for the Failure Information of Digital Transmitters
  • FieldComm Group HCF_SPEC-54: HART FSK Physical Layer Specification
  • IEC 60228:2004: Conductors of insulated cables
  • IEC 60079-11 e IEC 60079-14: proteção por segurança intrínseca e projeto de instalações em atmosferas explosivas

U.S. regulations

  • NFPA 70 (NEC) Article 504: intrinsically safe systems; Article 725: Class 2 power-limited circuits
  • ANSI/ISA-RP12.06.01: Recommended Practice for Wiring Methods for Hazardous (Classified) Locations Instrumentation — Intrinsic Safety
  • OSHA 29 CFR 1910.307: electrical equipment in hazardous (classified) locations

Frequently asked questions

How many 250 Ω receivers fit on a 24 V loop?

With a 24 V supply, a 10.5 V transmitter and a check at 22 mA, maximum load is (24 − 10.5) ÷ 0.022 ≈ 614 Ω. Allowing 10 to 20 Ω of cable, two 250 Ω receivers fit. With a −5% supply tolerance, load falls to about 559 Ω.

Why check at 22 mA instead of 20 mA?

Under NAMUR NE 43, the transmitter signals failure by going to 21 mA or more, and many go to 22 or 22.5 mA. If the loop cannot carry that current, the high alarm never reaches the control system.

What is 12 mA on a 0 to 400 °F range?

It is 50% of span: 200 °F. For flow measured by ΔP without root extraction in the transmitter, 12 mA is 50% of ΔP but 70.7% of maximum flow.

What is the cable resistance on a 1,000 ft loop?

With 18 AWG copper at 104 °F, each conductor has about 6.9 Ω per 1,000 ft. Over 1,000 ft (2,000 ft of conductor both ways), resistance is about 13.8 Ω, small next to a 250 Ω receiver.