Instrumentation
4-20 mA Loop Resistance Budget
Maximum loop resistance a two-wire transmitter can drive from a given supply voltage, and whether the loop you have described has enough headroom.
Calculated reference results
- Maximum loop resistance
- 650ohm
- Actual loop resistance
- 256.4ohm
- Headroom
- 393.6ohm
- Wire resistance in the loop
- 6.40ohm
- Burden at 20 mA
- 5.13V
- Voltage left at the transmitter
- 18.87V
- HART capable
- Yes, at least 230 ohm present
Comfortable margin.
500 ft, both conductors
HART communication generally needs a minimum loop resistance of about 230 ohm.
The arithmetic
- Available voltage = 24.0 V - 11.0 V = 13.0 V
- Maximum resistance = 13.0 V / 0.020 A = 650 ohm
- Wire = 500 ft x 2 conductors x 6.40 ohm/1000 ft = 6.40 ohm
- Total = 250 input + 0 other + 6.40 wire = 256.4 ohm
What this does not account for
- Full scale is the worst case. A loop with marginal headroom reads correctly at low values and misbehaves as the signal rises, which makes the fault look like a process problem.
- Intrinsic safety barriers and isolators add significant resistance. Include every series device.
- Immunity to wire resistance is not immunity to noise. Cable routing, shielding, and single-point shield grounding still matter.
How this is calculated
Maximum loop resistance equals the supply voltage minus the transmitter minimum operating voltage, divided by 0.020 amperes. Full scale is the worst case because the voltage burden is highest there. Running out of headroom produces a loop that tracks correctly at low readings and clips or goes nonlinear near full scale.
Formulas
R_max = (V_supply - V_transmitter_min) / 0.020
- 0.020 A — full scale, where the voltage burden is highest
Assumptions and limitations built into this calculator
- Two-wire loop-powered transmitter.
- Default wire resistance figures are for copper at room temperature. Resistance rises with temperature.
- Both conductors of the run are counted.
Frequently asked questions
- What is the maximum distance for a 4-20 mA signal?
- Distance is limited by total loop resistance, not by a fixed length. Work out the available voltage, divide by 0.020 amperes, subtract the receiver and every series device, and the remainder is the wire budget. Several thousand feet of ordinary instrument cable is routinely fine.
Direct contact
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