4-20mA Loop Power Budget: Calculate Maximum Resistance

A transmitter looks healthy at 4 mA, the receiver is wired correctly, and the supply says 24 V. Then the process climbs and the signal stalls at 18 mA. That is usually not a scaling problem. It is a voltage-budget problem: the loop has spent too much supply voltage in wire and receiver resistance to leave the transmitter operating voltage.

The calculation is short, but the current choice matters. A loop designed around nominal 20 mA can fail exactly when a transmitter needs to report over-range or a high fault. Budget the whole series path—transmitter, cable, burden, and any other receiver load—at the highest current the device must drive.

TL;DR: Calculate maximum total loop resistance as Rmax = (V_supply - V_min) / I_max. Use the transmitter’s over-range or alarm current, commonly 22 mA, rather than nominal 20 mA. With a 24 V supply and 8 V transmitter minimum, the 22 mA allowance is about 727 ohms before splitting it among wire and receiver loads (NotebookLM, Analog Signal Standards notebook, 2026).

Pair this guide with the 4-20mA versus 0-10V comparison to understand why current loops tolerate long runs differently, then use scaling 4-20mA PLC raw counts after the signal reaches the input card.

What Is the Maximum 4-20mA Loop Resistance Formula?

Maximum allowable loop resistance equals the supply voltage left after the transmitter’s minimum terminal voltage, divided by the highest current the loop must drive: Rmax = (V_supply - V_min) / I_max (NotebookLM, Analog Signal Standards notebook, 2026). The result is the combined allowance for every series resistance outside the transmitter, including wire and receiver burdens.

V_supply is the DC loop supply, V_min is the transmitter’s required terminal voltage, and I_max is the highest loop current. A two-wire transmitter draws operating power from that same series current.

The corresponding voltage balance is:

[ V_{supply} \ge V_{min} + V_{wire} + V_{receiver} ]

Start with the transmitter’s published minimum voltage. Add round-trip wire resistance and every series receiver load. If their total exceeds Rmax, the transmitter cannot retain its required terminal voltage at the chosen current.

Use 22 mA when it is the transmitter’s maximum over-range, alarm, or fault current. The research packet identifies 22 mA as a common robust design current and gives 23 mA and 24.4 mA as device-specific alternatives (NotebookLM, Analog Signal Standards notebook, 2026). A nominal 20 mA calculation can leave no compliance voltage for the signal above normal full scale.

Wire drop rises linearly with current. A loop that holds 20 mA may clamp before it reaches a high-fault band; the NAMUR NE 43 fault-current guide explains why that matters.

Citation capsule: Rmax = (V_supply - V_min) / I_max gives the total series-resistance allowance for a 4-20mA loop. Size I_max for the transmitter’s highest required output—commonly 22 mA, with 23 mA and 24.4 mA used by some devices—not nominal 20 mA (NotebookLM, Analog Signal Standards notebook, 2026).

How Does a 24 V Loop Budget Work in Practice?

With a 24 V supply, the transmitter minimum, round-trip cable drop, and receiver burden must share one finite voltage allowance.

A 24 V loop with an 8 V transmitter minimum and 22 mA sizing current has about 727 ohms of total resistance allowance: (24 V - 8 V) / 0.022 A (NotebookLM, Analog Signal Standards notebook, 2026). That figure is not receiver capacity alone; cable resistance consumes part of it before a burden resistor or input card receives anything.

Use the documented cable example. A 1,500-foot physical run of 24 AWG cable becomes 3,000 feet round trip. At 26.2 ohms per 1,000 feet, it has 78.6 ohms of total wire resistance:

[ R_{wire} = 3{,}000\ \text{ft} \times \frac{26.2\ \Omega}{1{,}000\ \text{ft}} = 78.6\ \Omega ]

At 22 mA, that cable drops 1.73 V:

[ V_{wire} = 0.022\ \text{A} \times 78.6\ \Omega = 1.73\ \text{V} ]

The remaining receiver-resistance allowance is therefore:

[ R_{receiver_max} = \frac{24\ \text{V} - 8\ \text{V} - 1.73\ \text{V}}{0.022\ \text{A}} = 648\ \Omega ]

All series receiver loads together must stay at or below 648 ohms in this example (NotebookLM, Analog Signal Standards notebook, 2026). A 250 ohm burden drops 5.5 V at 22 mA. The simplified 24 V, 10 V, 20 mA calculation gives 700 ohms, showing how V_min changes the budget; typical two-wire V_min is 8-12 V DC, but the installed configuration controls (NotebookLM, Analog Signal Standards notebook, 2026).

Citation capsule: In the documented 24 V example, an 8 V transmitter minimum and 22 mA sizing current allow about 727 ohms total. A 3,000-foot round trip of 24 AWG cable uses 78.6 ohms and 1.73 V, leaving 648 ohms for all receiver loads (NotebookLM, Analog Signal Standards notebook, 2026).

How Do Wire Gauge and Cable Length Spend the Budget?

Cable resistance uses round-trip length, so gauge and every installed segment directly reduce the loop's available receiver resistance.

Wire resistance is a series load: it consumes loop voltage by V_wire = I × R_wire and belongs in the same Rmax total as the receiver burden (NotebookLM, Analog Signal Standards notebook, 2026). Calculate cable resistance from round-trip length, not the one-way distance printed on a site drawing.

Current travels out to the transmitter and returns to the receiver. A 1,500-foot installation therefore has 3,000 feet of loop conductor in the resistance calculation. Gauge then sets the resistance per length. At 25 °C, the packet lists 18 AWG at 6.510 ohms per 1,000 feet, 20 AWG at 10.35 ohms, 22 AWG at 16.46 ohms, and 24 AWG at 26.17 ohms per 1,000 feet (NotebookLM, Analog Signal Standards notebook, 2026).

Moving from 18 AWG to 24 AWG increases resistance by a factor of four (NotebookLM, Analog Signal Standards notebook, 2026). Add every loop segment, then multiply the finished resistance by I_max to find its voltage cost.

The useful bookkeeping form is:

[ R_{max} \ge R_{wire} + R_{receiver} ]

Add all series receiver loads before comparing the total to Rmax. If the budget fails, shorten the run, raise the supply voltage, or use thicker wire (NotebookLM, Analog Signal Standards notebook, 2026). Use galvanic isolation and signal isolators for ground-potential differences, not simple resistance.

Citation capsule: Cable resistance is part of the loop’s series resistance and must use round-trip length. At 25 °C, 18 AWG is listed at 6.510 ohms per 1,000 feet and 24 AWG at 26.17 ohms per 1,000 feet; that wire drop rises linearly with the selected maximum current (NotebookLM, Analog Signal Standards notebook, 2026).

What Does Loop Starvation Look Like Above Rmax?

When wire plus receiver resistance exceeds Rmax, the transmitter loses terminal-voltage headroom as loop current rises, even if it looks normal at low current (NotebookLM, Analog Signal Standards notebook, 2026). The usual field signature is top-end roll-off: the output plateaus below its intended value because the transmitter cannot drive the requested current and keep its own electronics powered.

At 4 mA, series voltage drops are small enough that an overstretched loop can appear healthy. As the process rises, I × R rises too. Once transmitter terminal voltage falls below V_min, the output may flatline before 20 mA. The packet gives clipping at 16 mA or 18 mA as examples of this premature saturation (NotebookLM, Analog Signal Standards notebook, 2026).

High-fault currents can range from 21.0 mA to 22.8 mA, with some manufacturer limits reaching 24.4 mA; a 20 mA calculation can clamp before the controller sees that band (NotebookLM, Analog Signal Standards notebook, 2026). Severe under-voltage can also reboot the transmitter microprocessor or disrupt current modulation. For a noisy but stable signal, start with 4-20mA loop-noise troubleshooting; for a top-end failure, calculate compliance voltage first.

Citation capsule: Excess resistance causes transmitter starvation as current rises: output can roll off or clip below 20 mA, and severe under-voltage can cause erratic behavior. The packet gives 16 mA and 18 mA as examples of premature clipping and identifies 21.0-22.8 mA as a typical high-fault band (NotebookLM, Analog Signal Standards notebook, 2026).

How Does a 250 Ohm HART Load Fit the Budget?

A 250 ohm burden converts the loop signal while consuming 5.5 V at 22 mA, so it belongs in the resistance budget.

A 250 ohm burden converts the nominal 4-20mA signal to 1-5 V and drops 5.5 V at 22 mA, so it must be counted as a substantial series load in the power budget (NotebookLM, Analog Signal Standards notebook, 2026). It is often a receiver input burden, not an extra resistor added without checking the existing load.

HART superimposes 1200 Hz and 2200 Hz FSK tones on the analog loop, and the symmetric waveform averages to zero without changing the analog DC measurement (NotebookLM, Analog Signal Standards notebook, 2026). The packet says 250 ohms is typically used for a 1-5 V receiver conversion but does not establish a universal 250 ohm HART-modem minimum. Confirm the communicator’s required impedance in its documentation, include the actual burden in R_wire + R_receiver, and see what HART protocol adds to a 4-20mA loop.

Citation capsule: A 250 ohm burden creates a 1-5 V signal from nominal 4-20mA and drops 5.5 V at 22 mA. The bound packet confirms HART’s 1200 Hz and 2200 Hz FSK tones but does not establish the physical basis for a universal 250 ohm modem-minimum claim (NotebookLM, Analog Signal Standards notebook, 2026).

How Do You Check a Loop Budget Before Commissioning?

Before accepting the field run, check the loop at the transmitter’s documented minimum voltage and maximum required output current, then compare all series resistance against Rmax (NotebookLM, Analog Signal Standards notebook, 2026).

1. Record V_supply and the installed transmitter’s V_min. Typical two-wire V_min is 8-12 V, but the device specification controls (NotebookLM, Analog Signal Standards notebook, 2026).
2. Use the transmitter’s maximum over-range, alarm, or fault current as I_max; 22 mA applies only when it matches the device requirement.
3. Convert every cable section to round-trip resistance, then add the receiver burden and all other series loads.
4. Confirm R_wire + R_receiver does not exceed Rmax and, as a cross-check, calculate remaining transmitter terminal voltage at I_max.

Do not change PLC scaling before checking a top-end clamp. Verify loop current and transmitter terminal voltage at high demand, then return to PLC raw-count scaling only after the electrical budget holds.

Citation capsule: A loop-budget check uses the transmitter’s specified V_min, its maximum required current, round-trip wire resistance, and every series receiver load. The governing comparison is R_wire + R_receiver ≤ (V_supply - V_min) / I_max; with a 24 V supply, 8 V minimum, and 22 mA, the total allowance is about 727 ohms (NotebookLM, Analog Signal Standards notebook, 2026).

Frequently Asked Questions

What is the maximum loop resistance formula for a 4-20mA loop?

Use Rmax = (V_supply - V_min) / I_max. The terms are the loop’s DC supply, the transmitter’s minimum operating voltage, and the highest current it must drive; Rmax covers the combined wire and receiver resistance (NotebookLM, Analog Signal Standards notebook, 2026).

Should I calculate a 4-20mA loop at 20 mA or 22 mA?

Use the transmitter’s maximum over-range, alarm, or fault current rather than nominal 20 mA. The packet gives 22 mA as a common design point and identifies 23 mA and 24.4 mA as examples from other devices (NotebookLM, Analog Signal Standards notebook, 2026).

How much voltage does a 250 ohm burden resistor drop?

A 250 ohm burden drops 1.0 V at 4 mA and 5.0 V at 20 mA, producing a 1-5 V signal. At a 22 mA design current, the same burden uses 5.5 V of loop headroom (NotebookLM, Analog Signal Standards notebook, 2026).

How do I include cable resistance in a loop power budget?

Calculate resistance over the round-trip conductor length and include it as R_wire. In the documented example, 3,000 feet of 24 AWG wire totals 78.6 ohms and drops 1.73 V at 22 mA (NotebookLM, Analog Signal Standards notebook, 2026).

What happens when a 4-20mA loop exceeds maximum resistance?

The transmitter can lose terminal voltage as current rises, leading to brownout, premature top-end roll-off, clipping below 20 mA, missed high-fault current, or erratic behavior. The packet gives 16 mA and 18 mA as example clipping levels (NotebookLM, Analog Signal Standards notebook, 2026).

Citation capsule: The loop-budget FAQ resolves to one check: total wire and receiver resistance must remain within Rmax at the transmitter’s highest required current. At 22 mA, a 250 ohm burden alone drops 5.5 V, so long cable runs and receiver loads must share the remaining voltage deliberately (NotebookLM, Analog Signal Standards notebook, 2026).

Conclusion

A loop power budget is a compliance-voltage calculation: calculate Rmax = (V_supply - V_min) / I_max, then allocate it across round-trip wire and every receiver load (NotebookLM, Analog Signal Standards notebook, 2026).

The 24 V, 8 V, 22 mA example allows about 727 ohms total. Its 24 AWG cable consumes 78.6 ohms, leaving 648 ohms for receiver loads; a 250 ohm burden costs 5.5 V at 22 mA (NotebookLM, Analog Signal Standards notebook, 2026). For related diagnosis, see 4-20mA versus 0-10V, 4-20mA noise troubleshooting, and NAMUR NE 43 fault currents.

Citation capsule: A 24 V loop with an 8 V transmitter minimum and 22 mA maximum current supports about 727 ohms of total series resistance. Design within that allowance—after including round-trip cable and receiver burdens—to prevent top-end roll-off and preserve high-current signaling (NotebookLM, Analog Signal Standards notebook, 2026).