How the loop budget works
In a two-wire loop, everything is in series: the power supply, the receiver (input card or HART resistor), the cable, and the transmitter. Every resistance drops voltage in proportion to the loop current (V = I × R). Whatever is left over has to be at least the transmitter's minimum operating voltage from its datasheet.
Rloop max = (Vsupply − Vtransmitter min) ÷ Imax
Worked example
24 V supply, 250 Ω HART resistor, transmitter minimum 12 V, checked at 20 mA:
- Maximum loop resistance = (24 − 12) ÷ 0.020 = 600 Ω
- Voltage at the transmitter with only the 250 Ω resistor = 24 − 0.020 × 250 = 19 V
- That leaves 350 Ω for cable, barriers and indicators.
Why check at the alarm current, not 20 mA
If the transmitter is configured to fail upscale, it has to drive about 21–23 mA to signal a fault (NAMUR NE 43 uses ≥ 21 mA). A loop that only just works at 20 mA can starve the transmitter at the alarm current. The high-alarm signal then never reaches the control system, which is exactly when you need it.
Things that eat voltage
- The input card's sense resistor, or a 250 Ω resistor added for HART communication
- Intrinsic safety barriers (a zener barrier can add a few hundred ohms) and isolators
- Loop-powered indicators and any other series devices
- Long cable runs, especially in smaller wire gauges (counted out and back)
Converting the loop current to a process value? Try the 4–20 mA scaling calculator.