Checks whether a 2-wire, loop-powered 4-20mA transmitter will have enough terminal voltage to operate reliably at the far end of a long cable run, by budgeting the DC supply voltage against the worst-case (20mA) voltage drop across the wiring, the receiver/input resistance, and any other series devices (barriers, isolators, indicators) in the loop.
Standards followed
Standard 2-wire loop-powered transmitter voltage-budget arithmetic (Ohm's law)
What it can do
- Round-trip wire resistance from AWG wire gauge and one-way cable length (ft or m).
- Total loop resistance, voltage drop in the wiring, and voltage available at the transmitter at 20mA (worst case).
- Pass/fail check against the transmitter's minimum rated terminal voltage.
- Reverse calculation: maximum loop resistance budget and maximum cable length for a given supply/transmitter-minimum/receiver combination.
Example problem
24V supply, transmitter needs 10V minimum, 250Ω receiver, 500ft of #22 AWG wire.
- Wire resistance per 1000ft (#22 AWG) = 16.14Ω → round-trip = 2 × 500 × 16.14/1000 = 16.14Ω.
- Total loop resistance = 16.14 + 250 = 266.14Ω.
- Voltage at transmitter at 20mA = 24 − 0.02 × 266.14 = 24 − 5.32 = 18.68V.
- Headroom = 18.68 − 10 = 8.68V ≥ 0 → pass.
18.68V available at the transmitter, 8.68V of headroom — comfortably passes.
Want the full step-by-step version?
Wire resistance table is standard annealed-copper AWG data at 20°C (published wire tables).