When to use: Use this when designing or troubleshooting a 4–20 mA current loop to verify the power supply has enough compliance voltage to drive the full-scale 20 mA signal through the wire, barriers, indicators, and the transmitter itself. The maximum total loop resistance is set by the difference between the supply voltage and the transmitter's minimum operating voltage. Use the span/zero converter to translate a measured mA value into percent of span and engineering units.
The 4–20 mA current loop is the dominant analog signal standard in industrial process control, transmitting sensor measurements as a current that is independent of wire resistance over distances up to several thousand feet. Engineers use this calculator to verify the power supply has sufficient compliance voltage to drive 20 mA through all series resistances in the loop, and to convert a measured current into a percentage of span and engineering-unit value.
A 4–20 mA loop consists of a transmitter, a power supply, and series load elements (wire resistance, isolators, barriers, and receiver input resistors). The power supply must maintain at least the transmitter's minimum operating voltage at the transmitter terminals even at full-scale 20 mA. The maximum loop resistance is calculated as R_max = (V_supply − V_tx_min) / 0.020 A. If the total series resistance exceeds R_max, the transmitter cannot source 20 mA and the signal will be clamped below full scale.
The live-zero convention — 4 mA representing 0% of span rather than 0 mA — is critical for diagnostics: a reading below 4 mA indicates an open circuit or failed transmitter, not simply a zero-process reading. Span conversion is: % of span = (mA − 4) / 16 × 100. Engineering unit value = PV_min + (% span / 100) × (PV_max − PV_min).
ISA-50.1 (now withdrawn but historically referenced) and IEC 60381-1 define the 4–20 mA current signal standard. HART (Highway Addressable Remote Transducer) protocol — defined by the FieldComm Group — superimposes a digital signal on the 4–20 mA loop at 1200 baud using FSK modulation; HART requires a minimum 230 Ω loop resistance for the handheld communicator or multiplexer to decode the signal. Intrinsic safety (IS) barriers per IEC 60079-11 add resistance to the loop (typically 250–500 Ω) and must be included in the loop resistance budget calculation.
Always verify the loop resistance budget at the design stage by summing all series elements: wire resistance (Ω per 1000 ft × cable length / 1000 × 2 for both conductors), barrier resistance, isolator input impedance, and AI card input resistance. For 24 V DC supplies with a transmitter minimum of 10.5 V, the maximum loop resistance is (24 − 10.5) / 0.020 = 675 Ω, which provides headroom for a 250 Ω HART resistor plus wire resistance.
Cable capacitance is a secondary concern for HART: excessive capacitance (>0.3 µF) can attenuate the HART signal and cause communication errors. For long runs exceeding 3000 ft, consider a higher supply voltage (30 V DC) or a loop-powered isolator to maintain compliance voltage.
Enter the supply voltage (typically 24 V DC), the transmitter minimum operating voltage (check the data sheet — typically 10–12 V), and all series resistances (wire resistance plus any barrier or isolator input impedance). The calculator shows the maximum allowable loop resistance, the remaining headroom, and whether the current supply voltage is adequate. To convert a measured mA reading to process value, enter the span limits (value at 4 mA and 20 mA) and the engineering units, then enter the measured current.
The live-zero convention uses 4 mA as the 0% signal so that an open circuit (0 mA) is clearly distinguishable from a valid zero-process reading. This enables continuous diagnostic monitoring: any reading below 3.6 mA indicates a wiring fault, open circuit, or failed transmitter, which would be indistinguishable from a true zero process value if 0 mA were the zero signal.
HART requires a minimum 230 Ω in the loop for the FSK signal to be decodable by a handheld communicator or multiplexer. If the total loop resistance is below 230 Ω, a 250 Ω HART resistor is added in series. The HART resistor adds to the loop resistance budget, so it must be accounted for in the compliance voltage calculation.
Wire resistance = (resistance per 1000 ft) × (cable length in ft / 1000) × 2. The factor of 2 accounts for both the forward and return conductors. For 18 AWG copper, resistance is approximately 6.4 Ω/1000 ft, so a 500 ft run has 6.4 Ω. For 22 AWG, it is approximately 16.1 Ω/1000 ft.
When total loop resistance exceeds R_max, the transmitter cannot maintain 20 mA at full scale because the power supply compliance voltage is insufficient. The actual loop current will be clamped below 20 mA, causing the measured signal to read low. The error is proportional to the excess resistance. The fix is to increase supply voltage, reduce wire resistance (larger conductor), or remove series elements.
Yes, HART is designed to coexist with the 4–20 mA signal. HART superimposes a 1200 baud FSK digital signal (±0.5 mA) on top of the DC 4–20 mA current. The average current is not affected because the FSK signal is symmetric. The only requirement is that the loop resistance is at least 230 Ω for the HART modem to extract the signal, and loop capacitance is kept below 0.3 µF.
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