14.3 Two-Wire, Four-Wire, and Loop Calculations

Once the pair is classified and the tray is quiet, the remaining PE question is whether the loop can still regulate 20 mA at the far end. That is a voltage, current, impedance, and power inventory — not a brand preference between “2-wire” and “4-wire” nameplates.

Two-wire, four-wire, and buses

A two-wire (loop-powered) transmitter has two field conductors. They carry operating power and the 4–20 mA signal. The transmitter inserts itself in series with the analog input and a DC supply and steals a slice of loop voltage for its electronics. At 12 V and 20 mA that slice is only 0.24 W. That is enough for a modern pressure, DP, or temperature transmitter. It is not enough for a heated sample line, a zirconia cell furnace, a nuclear density source controller, or a radar gauge with a hungry processor. Two-wire is the default for ordinary process transmitters because the home-run count is lowest and the analog input’s 24 V is already there.

A four-wire (externally powered) device has a power circuit (120 VAC or 24 VDC) plus a separate signal circuit (active 4–20 mA, 0–10 V, pulse, or digital). The analog output is a current source that must have enough compliance voltage to push 20 mA through the analog-input resistance, any barrier, and the cable. Four-wire is what you specify when the sensor needs real watts, or when the output must stay alive while the analog loop is open.

Three-wire devices share a common return between power and signal. They are not fully isolated; the shared common is a ground-loop invitation if the power supply common and the analog common are not the same point.

Buses change the inventory again. HART is still a 4–20 mA loop with a phase-shift keyed signal superimposed; the DC budget in this section still applies, plus a loop-resistance window so the FSK has a voltage to ride on (commonly about 230–600 Ω of total loop resistance; 250 Ω is the analog-input value that lands in that window). FOUNDATION Fieldbus H1 and PROFIBUS PA are bus-powered: a fieldbus power supply and conditioner drive the segment, devices draw current from the bus, and there is no 250 Ω analog input. PROFIBUS DP, Modbus RTU, and Ethernet are separately powered communications; do not treat them as loop-powered 4–20 mA.

Voltage, current, impedance, and power

A 4–20 mA current loop is a high-impedance current source driving a relatively small burden. The analog input converts current to voltage with a precision resistor, classically 250 Ω, so that 4 mA → 1.0 V and 20 mA → 5.0 V for a 1–5 V ADC. Some cards use 50 Ω or 100 Ω internally and scale in software; the exam problem will state the resistance if it matters. Voltage signals (0–10 V, 1–5 V) need a high-impedance input and a low-impedance source; they pick up noise more readily on long runs, which is why CSE analog home-runs stayed on current.

Impedance mistakes: placing two 250 Ω inputs in series without checking compliance; paralleling a recorder across the analog input and changing the burden; or inserting a HART resistor on a loop that already has 250 Ω and a 300 Ω zener barrier, then wondering why the transmitter will not start.

Power on a two-wire device is whatever is left after every series drop: P_tx ≈ V_tx × I_loop, and V_tx is whatever the supply did not spend on the input, barrier, and cable. On a four-wire analog output, power in the signal circuit is small; the watts are on the separate power feed, which gets the voltage-drop treatment from Section 14.2.

Compliance voltage, barrier drop, and the 250 Ω input

Compliance is used two ways. For a two-wire loop, it is the supply’s ability to keep enough voltage at the transmitter after every drop. For a four-wire analog output, it is the maximum voltage that output can develop while still regulating current: R_load_max = V_compliance / I_max. A 12 V-compliant four-wire output can drive 20 mA into 600 Ω. Two 250 Ω inputs plus a 300 Ω barrier is 800 Ω and will not regulate.

Series drops at loop current I:

• Analog input: V_ai = I × R_ai. For 250 Ω at 20 mA, V_ai = 5.00 V. At 22 mA overrange (the honest design current for many cards and NAMUR high-scale), V_ai = 5.50 V.
• Barrier: either a manufacturer’s end-to-end resistance (300 Ω × 20 mA = 6.0 V) or a stated 6 V lumped drop. Zener barriers are resistive; the drop scales with current. A “6 V barrier” quoted at 20 mA is 6.6 V at 22 mA if it is really 300 Ω.
• Cable: V_cable = I × R_round_trip (Section 14.2).
• Extra series devices: I/I splitters, lightning SPDs with series impedance, test diodes, HART resistors.

The inequality to satisfy at the highest current you care about is:

V_supply ≥ V_tx_min + V_barrier + I_max × (R_ai + R_cable + R_other)

Worked remaining cable budget. Supply 24.0 VDC. Analog input 250 Ω. Barrier drop 6.0 V at the calculation current. Transmitter minimum 12.0 V (a typical 2-wire smart transmitter floor; some HART devices want more).

At 20 mA:

• Input drop = 0.020 A × 250 Ω = 5.00 V
• Voltage already spoken for = 12.0 + 6.0 + 5.0 = 23.0 V
• Remaining cable budget = 24.0 − 23.0 = 1.0 V
• R_cable_max = 1.0 V / 0.020 A = 50 Ω round-trip
• On 18 AWG at 6.5 Ω/kft, one-way length max = 50 / (2 × 6.5) = 3.85 kft (about 3850 ft)

That 3850 ft looks generous until you design at 22 mA or you believed the barrier drop was constant:

At 22 mA, still treating the barrier as a fixed 6.0 V (as the problem stated):

• Input drop = 0.022 × 250 = 5.50 V
• Spoken for = 12.0 + 6.0 + 5.50 = 23.50 V
• Cable budget = 0.50 V
• R_cable_max = 0.50 / 0.022 = 22.7 Ω round-trip
• 18 AWG one-way max ≈ 1750 ft

If the 6.0 V barrier was actually 300 Ω, barrier drop at 22 mA is 6.6 V and the cable budget is 24.0 − 12.0 − 6.6 − 5.5 = −0.1 V. The transmitter will not start at overrange, and it may not start at 20 mA on a long run either. That is the PE trick: always close the budget at the highest current and the highest series resistance, not at 12 mA “because that is where we operate.”

If the transmitter minimum is 16 V (some HART / high-power 2-wire devices) on the original 24 / 6 / 250 Ω loop: 16 + 6 + 5 = 27 V, which a 24 V supply cannot provide. Fixes that actually work: raise the bulk supply (26–30 V if the I/O card allows it), change to a galvanic isolator with a smaller drop than 6 V, or keep HART with a carefully placed resistor while reducing other series ohms — not silently deleting the 250 Ω and hoping the ADC still reads 1–5 V.

I/I splitters that are not isolated add another series burden. Isolated splitters (and isolated analog outputs) copy the current without stacking another 250 Ω in the same metallic loop.

Isolated outputs and when to isolate

Channel-to-channel isolation on analog outputs or inputs means each loop’s common can sit at a different potential. Use it when:

• Two transmitters bond to different vessels or different building earths.
• A four-wire analyzer output is powered from a grounded 120 VAC supply whose neutral/earth is not the DCS analog common.
• A thermocouple or pH loop already has a process ground through the wet electrode or the thermowell.
• You would otherwise series-connect two analog inputs (DCS plus a local recorder) and create a second earth.

Two-wire convenience is a metallic bond from the analog common, through the barrier, to the field device. If that bond is also a second earth path, isolation is cheaper than chasing a 60 Hz offset for the life of the plant.

/practice/pe-control-systemsPractice questions with detailed explanations