4–20 mA loop voltage budget: will the transmitter reach full output?

Why a loop can fail only at high current

A two-wire transmitter is powered by the loop itself. As the signal current rises toward 20 mA, or to a configured high-alarm current of 21 or 22 mA, the voltage lost across the receiver, cable and other series parts also rises. The transmitter terminal voltage falls by the same amount. If it falls below the transmitter minimum operating voltage, the device may behave correctly at 4 mA and mid-range but fail to reach full scale or alarm current.

The loop voltage budget calculator checks this at the highest current the loop can carry.

Think of the supply as a fixed amount of voltage that must be shared. The transmitter needs a minimum share to operate. Everything else in the series path uses the remainder. Ohm’s law gives each drop: V = I × R. Because the loop current is set by the transmitter, not by the supply, the drops grow in proportion to current and the transmitter terminal voltage is whatever is left over. Fixed drops, such as a diode, do not depend on current in this simple model.

Formulas the calculator uses

• Total series resistance R_t = receiver + cable + other
• Loop drop = I_max × R_t + fixed drops, with I_max in amps (mA ÷ 1000)
• Margin = supply − transmitter minimum voltage − loop drop
• Maximum allowable loop resistance = (supply − minimum voltage − fixed drops) ÷ I_max
• Voltage at transmitter terminals = supply − loop drop

The cable value is the total out-and-back loop resistance, not one conductor. Barrier, isolator or other series resistance goes in the other-resistance field. Diode or fixed voltage drops go in the fixed-drop field. The calculator accepts maximum currents from 4 to 30 mA.

Worked example 1: a loop with comfortable margin

Supply 24 V, transmitter minimum 10.5 V, I_max 20 mA, receiver 250 Ω, cable loop resistance 60 Ω, no other drops.

1. R_t = 250 + 60 = 310 Ω
2. Loop drop = 0.020 × 310 = 6.2 V
3. Margin = 24 − 10.5 − 6.2 = 7.3 V
4. Terminal voltage = 24 − 6.2 = 17.8 V
5. R_max = (24 − 10.5) ÷ 0.020 = 675 Ω

With 310 Ω connected and 675 Ω allowed, about 365 Ω of further resistance could be added at 20 mA before the margin reaches zero. This ignores supply tolerance, which should be considered separately. Another way to read the budget is as percentages: the loop drop is 6.2 ÷ 24 ≈ 26 % of the supply, the transmitter minimum is 10.5 ÷ 24 ≈ 44 %, and the remaining 30 % or so is margin.

Worked example 2: effect of alarm current

Take the same loop but check at a 22 mA high alarm current.

1. Loop drop = 0.022 × 310 = 6.82 V
2. Margin = 24 − 10.5 − 6.82 = 6.68 V
3. Terminal voltage = 24 − 6.82 = 17.18 V
4. R_max = 13.5 ÷ 0.022 = 613.6364 Ω

The margin is smaller than at 20 mA and the permitted resistance falls from 675 Ω to 613.6364 Ω.

A tighter case: supply 24 V, minimum 12 V, I_max 22 mA, receiver 250 Ω, cable 150 Ω, other 100 Ω for an isolator, fixed drop 0.7 V. R_t = 500 Ω, loop drop = 0.022 × 500 + 0.7 = 11.7 V, margin = 24 − 12 − 11.7 = 0.3 V, terminal voltage = 12.3 V and R_max = (24 − 12 − 0.7) ÷ 0.022 = 513.6364 Ω. The result is positive but nearly zero, so a modest change in supply, cable or device value could remove the margin.

Summary of the three cases

To relate the current to the process value, use the 4–20 mA to engineering units calculator or the engineering units to 4–20 mA calculator. For example, 20 mA is the upper range value, and the budget is checked there.

Common mistakes

• Checking at 4 mA or at a mid-range current instead of the maximum.
• Entering one-way cable resistance instead of out-and-back loop resistance.
• Leaving out isolators, barriers, indicators, surge protectors or test points in the series path.
• Using a typical rather than guaranteed minimum voltage from the datasheet.
• Ignoring the extra requirement for communication, where the transmitter datasheet states one.
• Using the nominal supply value when the supply can sag under load.
• Adding a handheld communicator or test instrument in series and not counting its resistance.
• Treating a small positive margin as a pass without allowing for temperature and cable changes.

Checks before you trust the result

• Take the minimum operating voltage from the transmitter datasheet and note any HART or other communication requirement.
• Use the actual cable resistance for the installed length, conductor size and operating temperature.
• Confirm the maximum current including any configured alarm level.
• Verify the supply voltage at the loop terminals under full load.
• Measure the voltage across the transmitter terminals while the loop is driven to the maximum current with a simulator, if the procedure permits, and compare it with the calculated terminal voltage.
• Recalculate whenever a device is added to the loop, a cable is replaced or the supply is changed.
• Treat the result as an estimate and compare it with the manufacturer data and the site design procedure. The calculation does not address hazardous-area, safety or compliance requirements.

Frequently asked questions

What if the margin is negative?

The calculator flags that the transmitter may not operate correctly. Reduce loop resistance, raise the supply voltage within the limits of the equipment, or choose a device with a lower minimum voltage.

Does the budget change with a lower current?

Yes. The drop is proportional to current, so the margin is larger at lower current, which is why the check uses the maximum.

Where does a barrier or isolator go?

Add its series resistance in the other-resistance field, and any fixed drop in the fixed voltage field, using the datasheet value.

Is a positive margin enough?

It is the minimum condition. Many designs keep extra margin for supply tolerance and ageing.

Recording data

Keep, for each loop, the supply voltage measured at load, the transmitter minimum voltage, the receiver and cable resistance, any extra series devices, the current at which the check was made, the computed margin and the date. Recording measured values next to design values makes later fault finding faster, because a loop that once had margin and now does not points to a changed resistance or a lower supply.