Current Loop Voltage Drop Calculator

Will your transmitter power up? Enter supply voltage, loop devices and cable details — get the voltage available at the transmitter and a clear pass / fail.

from the datasheet ("lift-off")
DCS input + indicator + barrier
21.5 covers NE 43 fault current
1.5 mm² Cu ≈ 12.1 · 1.0 mm² ≈ 18.1 · 0.75 mm² ≈ 24.5
Voltage at transmitter (at max current)
V
loop resistance Ω · drop V · margin V

Why loops fail to power up

A two-wire transmitter is powered by the same 4–20 mA loop that carries its signal. Every ohm in that loop — the DCS input resistor, a loop-powered indicator, an intrinsic-safety barrier, and both conductors of the cable — drops voltage in proportion to the current. At full-scale current the drop is largest, and if what remains at the transmitter terminals falls below its minimum operating ("lift-off") voltage, the loop saturates early or the transmitter drops out entirely. It's one of the most common commissioning faults, and one of the cheapest to prevent with thirty seconds of arithmetic.

The formulas

R_cable = 2 × L(km) × r(Ω/km)

The factor 2 counts both conductors — the current travels out and back.

V_transmitter = V_supply − I_max × (R_devices + R_cable)

and the design check:

R_loop,max = (V_supply − V_min) ÷ I_max

Transmitter datasheets draw exactly this last equation as the familiar "load line" graph of allowable loop resistance versus supply voltage.

Worked example

24 V supply, transmitter needing 10.5 V minimum, a 250 Ω DCS input, 500 m of 1.5 mm² copper cable (12.1 Ω/km per conductor), checked at 21.5 mA fault current:

  1. Cable: 2 × 0.5 km × 12.1 = 12.1 Ω → total loop = 262.1 Ω
  2. Drop: 0.0215 A × 262.1 Ω = 5.64 V
  3. At transmitter: 24 − 5.64 = 18.36 V — comfortably above 10.5 V. ✔
  4. Maximum allowable loop resistance: (24 − 10.5) ÷ 0.0215 = 628 Ω, so there is room for an indicator or barrier later.

Field notes

  • Check at fault current, not 20 mA. NAMUR NE 43 high-alarm current is up to ~21.5 mA; a loop sized only for 20 mA can lose its alarm signal exactly when you need it.
  • HART needs resistance too. Communication typically requires at least 250 Ω in the loop — a loop can have too little resistance as well as too much.
  • IS barriers are hungry. A zener barrier can add 100–340 Ω and its own voltage drop; always use the barrier's end-to-end specification.
  • Supply tolerance. A "24 V" supply at the end of a long marshalling run may deliver 23 V or less — use the worst-case value.

Frequently asked questions

What is the minimum voltage a 4-20 mA transmitter needs?

It varies by model — typically 9 to 12 V DC at the transmitter terminals, stated as the "lift-off" or minimum operating voltage on the datasheet. HART communication often requires the loop to also contain at least 250 ohms of resistance.

Why calculate voltage drop at 20 mA and not 4 mA?

Voltage drop across loop resistance is highest at maximum current. If the transmitter still has enough voltage at 20 mA (or ~21.5 mA fault current), it will work everywhere else in the range.

How much resistance does a typical 4-20 mA loop contain?

A common loop has a 250-ohm input resistor at the DCS/PLC, plus cable resistance (both conductors), plus any indicators or safety barriers in series. Total loop resistance frequently lands between 250 and 600 ohms.

Provided for reference and education. Verify independently before use in safety-critical work. See our disclaimer.

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