4–20 mA Loop Load Calculator

Can a 2-wire 4–20 mA loop supply drive the 250 Ω input, cable and indicators? Voltage left at the transmitter, maximum load resistance and cable length.

Loop supply (V)Transmitter minimum voltage (V)Maximum current to check (mA)Input resistor (Ω)

Other resistance (Ω) — indicators, isolatorsOther voltage drop (V) — barriers, loop displays

One-way cable length (m)Wire sizeWire temperature (°C)

Results
Cable loop resistance
Total loop resistance
Transmitter terminal voltage at max current
Voltage margin
Maximum load resistance
Minimum supply voltage needed
Maximum one-way length with this wire
Voltage across the input resistor (4 mA · 20 mA)

The defaults (24 V, 12 V minimum, 20 mA, 250 Ω, 0.75 mm² over 100 m) are an example. Take the transmitter's minimum voltage and the current of its load diagram from the «supply voltage» or «load» section of its manual.

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The supply is shared by the transmitter, cable and input resistor. At 20 mA, 250 Ω takes 5 V and 1 km of 0.75 mm² cable (about 46 Ω out and back) about 0.9 V, leaving about 18.1 V for the transmitter.

Formulas

Cable loop resistance = 2 × ρ × one-way length ÷ cross-section (copper ρ20 = 1/58 Ω·mm²/m)
Loop resistance = input resistor + cable + other resistance
Transmitter voltage = supply − max current × loop resistance − other voltage drop
Maximum load = (supply − transmitter minimum − other voltage drop) ÷ max current

Example: with a 24 V supply, a 12 V transmitter minimum and 20 mA, the maximum load is (24 − 12) ÷ 0.02 = 600 Ω. After a 250 Ω input resistor, 350 Ω is left for cable and indicators.

Why check at maximum current

The voltage lost in the loop resistance grows with current, so the transmitter has the least voltage at the highest current — hence the check at 20 mA. If the transmitter is set to signal failures at 21 mA or more (NAMUR NE 43), the loop must carry that current too for the failure signal to arrive, so try that value. Also check which current the load diagram in the transmitter manual assumes.

Symptoms of too little voltage

Readings are right at the low end but stop rising near the top (stuck at, say, 18 mA).
The top of the range disappeared after lengthening the cable or adding an indicator or isolator.
Values are clipped only when the supply sags (when other loads switch on).

FAQ

Why does a 250 Ω resistor give 1–5 V?

Ohm's law: 4 mA × 250 Ω = 1 V and 20 mA × 250 Ω = 5 V. It is how a voltage-only input reads a 4–20 mA signal.

How far can I run a loop on 24 V?

It depends on the transmitter minimum voltage and input resistor. With 12 V minimum, 250 Ω and 20 mA the cable may use up to 350 Ω; with 0.75 mm² copper (about 0.046 Ω/m out and back) that is about 7.6 km one way, so indicators and barriers usually run out first.

Does this apply to 4-wire transmitters?

A 4-wire (active output) transmitter has its own supply and drives the current. Compare the total loop resistance with the «maximum load» in its output specification instead of a supply voltage.

To convert the current to engineering units or PLC counts, use the 4–20 mA Scaling Calculator.