Why loop power needs checking
A two-wire 4–20 mA transmitter takes its operating power from the same loop that carries its signal. Everything else in the loop (the input resistor of the PLC or indicator, barriers, isolators and the cable itself) drops voltage in proportion to the current. What remains at the transmitter terminals must stay above the transmitter’s minimum operating voltage at the highest current the loop will ever carry. If it doesn’t, the transmitter cannot drive the full current, and the reading saturates below the real value without any error message.
The equations
The calculator uses Ohm’s law throughout:
- Cable resistance (both conductors): R_cable = 2 × ρ × L / A, with ρ = 0.0172 Ω·mm²/m for copper at 20 °C
- Maximum loop resistance: R_max = (V_supply − V_min − V_other) / I_max
- Terminal voltage at the transmitter: V_t = V_supply − V_other − I × (R_load + R_cable)
- Voltage margin: V_t − V_min
It also works out the minimum supply voltage for your load and cable, the largest load you could add with this cable, and the longest cable you could run with this load.
Worked example
A 24 V supply feeds a transmitter that needs at least 10.5 V. The receiver has a 250 Ω input resistor (common for HART), and the transmitter is 300 m away on 0.75 mm² copper. The transmitter can drive up to 22 mA in alarm.
- Cable resistance: 2 × 0.0172 × 300 / 0.75 = 13.76 Ω
- Maximum loop resistance at 22 mA: (24 − 10.5) / 0.022 = 614 Ω
- Total load + cable: 263.8 Ω, 43 % of the maximum
- Terminal voltage at 22 mA: 24 − 0.022 × 263.76 = 18.20 V
- Margin: 18.20 − 10.5 = +7.70 V
- Minimum supply for this loop: 16.30 V
- Maximum cable length with this load: 7928 m
At 1000 m the margin drops to 6.99 V: still fine. With a 12 V supply on the same 300 m loop, however, the transmitter would be 4.30 V short at 22 mA.
Why size for 22 mA, not 20 mA
Under NAMUR NE 43, a transmitter signals an upscale failure with 21 mA or more, and many transmitters saturate at 20.5 to 23 mA during over-range. If the loop is sized for 20 mA, the alarm current may never be reached, and a fault looks like a high but valid reading. Use the maximum current from the transmitter datasheet; 22 mA is a safe default when unknown.
What goes into “Other drops”
Add every series voltage drop that is not a resistor you entered as load:
- Intrinsic-safety barriers and isolators, which often specify a fixed drop or an equivalent resistance
- Loop-powered displays and indicators, typically 1 to 4 V
- Diodes and reverse-polarity protection
Fixed drops belong in “Other drops”; resistive elements (input resistors, fuses with significant resistance) belong in the load.
Common mistakes
Using the transmitter’s minimum voltage at 4 mA. Some datasheets give a minimum that rises with current, or a load diagram. Use the value at maximum current.
Forgetting both conductors. The current flows out and back, so cable resistance counts twice. The calculator includes both.
Ignoring temperature. Copper resistance rises about 0.39 % per °C. A cable run through a hot area has more resistance than the 20 °C value; leave margin.
HART without a resistor. HART needs typically at least 250 Ω in the loop for the communicator to see the signal. If you remove the resistor to gain margin, HART stops working.
Frequently asked questions
What if the margin is negative?
Lower the load resistance, use a thicker or shorter cable, raise the supply voltage, or choose a transmitter with a lower minimum voltage. The “Minimum supply” and “Maximum load” results show how far each change needs to go.
Does the calculator work in feet and AWG?
Yes. Select feet as the length unit (the value converts automatically) and choose an AWG size from the conductor list.
Does it apply to four-wire transmitters?
No. Four-wire transmitters are powered separately and actively drive the current; check their maximum load specification instead.
How do I check the scaling of the same loop?
Use the 4–20 mA converter to convert between current, percent and process value.