Use Rmax = (Vsupply − Vtransmitter,min) ÷ Imax. Every resistance in series — receiver input, barrier, indicator, cable — must add up to less than Rmax. The maximum cable run is L = (Rmax − Rreceiver − Rother) × A ÷ (2 × ρ), with ρ = 0.017241 Ω·mm²/m for copper at 20 °C.
Loop resistance and cable length calculator
Copper resistivity 0.017241 Ω·mm²/m at 20 °C (annealed copper standard), temperature coefficient 0.00393 per °C. Cable resistance counts both conductors (2 × run length). Take Vmin and the receiver resistance from the transmitter and input-card datasheets.
How do you calculate maximum loop resistance?
Subtract the transmitter's minimum operating voltage from the supply voltage and divide by the highest current the loop must carry. A two-wire transmitter takes its power from the loop, so whatever voltage the series resistances drop is no longer available at its terminals.
Maximum loop resistance: Rmax = (Vsupply − Vmin) ÷ Imax
Loop check: Rreceiver + Rbarrier + Rcable ≤ Rmax
Cable resistance: Rcable = 2 × L × ρ ÷ A, ρCu, 20 °C = 0.017241 Ω·mm²/m
Minimum supply: Vsupply,min = Vmin + Imax × Rtotal
Use 20 mA for a plain measurement loop. If the transmitter must also drive an upscale alarm current (the NAMUR NE 43 convention puts high alarm at ≥ 21 mA), design for that current instead — at 22 mA the 600 Ω example falls to 545 Ω.
Where the voltage goes in a 2-wire loop
How long can a 4–20 mA cable be?
As long as the total loop resistance stays below Rmax. The current itself does not weaken over distance; the cable only adds resistance. The table uses the same example loop: 24 V supply, 12 V transmitter minimum, 20 mA, 250 Ω receiver (350 Ω cable budget), and the same loop with a 280 Ω intrinsic-safety barrier (70 Ω budget; example value — use your barrier's datasheet).
| Conductor | Loop resistance per 100 m (both wires) | Max run, 350 Ω budget | Max run with 280 Ω barrier |
|---|---|---|---|
| 0.5 mm² | 6.90 Ω | ≈ 5 070 m | ≈ 1 010 m |
| 0.75 mm² | 4.60 Ω | ≈ 7 610 m | ≈ 1 520 m |
| 1.0 mm² | 3.45 Ω | ≈ 10 150 m | ≈ 2 030 m |
| 1.5 mm² | 2.30 Ω | ≈ 15 220 m | ≈ 3 040 m |
| 24 AWG (0.205 mm²) | 16.82 Ω | ≈ 2 080 m | ≈ 420 m |
| 22 AWG (0.326 mm²) | 10.58 Ω | ≈ 3 310 m | ≈ 660 m |
| 20 AWG (0.518 mm²) | 6.66 Ω | ≈ 5 260 m | ≈ 1 050 m |
| 18 AWG (0.823 mm²) | 4.19 Ω | ≈ 8 350 m | ≈ 1 670 m |
Calculated values, copper at 20 °C, not measurements. At 60 °C conductor temperature resistance is about 16 % higher, so lengths shrink by the same proportion. In practice, noise pickup, cable capacitance (for HART) and installation rules limit long runs before resistance does.
Worked examples
| Loop | Calculation | Result |
|---|---|---|
| 24 V, 12 V min, 20 mA | (24 − 12) ÷ 0.020 | Rmax = 600 Ω |
| Same, designed for 22 mA alarm | (24 − 12) ÷ 0.022 | Rmax ≈ 545 Ω |
| 250 Ω input + 800 m of 0.5 mm² | 250 + 2 × 800 × 0.017241 ÷ 0.5 = 250 + 55.2 | 305 Ω ≤ 600 Ω ✓ |
| Minimum supply for that loop | 12 + 0.020 × 305.2 | ≈ 18.1 V |
| 18 V supply, 250 Ω + 280 Ω barrier | (18 − 12) ÷ 0.020 = 300 Ω < 530 Ω | ✗ fails — raise the supply |
What happens when loop resistance is too high?
The transmitter runs out of terminal voltage before it reaches full-scale current. Low readings still look right, but the signal flattens below 20 mA near the top of the range, and an upscale alarm current may never be reached. Check it by measuring the voltage at the transmitter terminals with the process at full scale (or the output forced to 20 mA): it must stay at or above the transmitter's minimum voltage.
- 1Read Vmin from the transmitter datasheet and the receiver input resistance from the PLC or display card.
- 2Add every series resistance: receiver, barrier or isolator, loop indicator, both cable conductors.
- 3Compare with Rmax at the design current; keep some headroom for supply tolerance and hot cable.
- 4If it fails, raise the supply within the transmitter's rating, use a larger conductor, or remove a series device.
HART communication additionally needs at least 230 Ω in the loop (250 Ω is the usual choice), so the receiver resistance cannot simply be removed on a HART loop. SEGMENsensor 4–20 mA pressure transmitters are loop powered from 12–36 V DC with a maximum loop resistance of 600 Ω at 24 V; submersible level transmitters also run on 12–36 V DC. For wiring see the 4–20 mA wiring guide and 2-wire vs 3-wire vs 4-wire transmitters; to scale the signal use the 4–20 mA calculator.
Frequently Asked Questions
R = (24 V − Vmin) ÷ 0.020 A. For a transmitter that needs 12 V that is 600 Ω; for one that needs 10 V it is 700 Ω. Take Vmin from the transmitter datasheet.
Until the loop resistance reaches the limit. With 24 V, a 12 V transmitter and a 250 Ω input, 350 Ω is left for cable, which is about 5 km of 0.5 mm² copper pair at 20 °C (calculated value). A barrier in the loop shortens this a lot.
R = 2 × L × ρ ÷ A, counting both conductors. For copper at 20 °C, ρ = 0.017241 Ω·mm²/m, so a 0.5 mm² pair adds about 6.9 Ω per 100 m of run.
At least Vmin + I × Rtotal. A 12 V transmitter with 305 Ω of total loop resistance needs about 18.1 V at 20 mA; a 24 V supply leaves comfortable headroom.
Not while the loop stays within its resistance limit: the same current flows through every part of a series loop, so cable resistance does not change the reading. It only matters when the transmitter runs short of voltage.
Use the highest current the transmitter can output, if alarm currents above 20 mA must reach the receiver. NAMUR NE 43 places the high alarm at 21 mA or more; designing for 22 mA reduces the 600 Ω example to about 545 Ω.