Loop Resistance and Power Supply Calculator
Every resistance in a two wire 4 to 20 mA loop takes voltage away from the transmitter. Check that the transmitter still gets its minimum voltage at the highest loop current, and see how long the cable can be.
- Cable resistance = 2 × 0.0172 × 300 ÷ 0.75 = 13.8 Ω
- Total load = 250 + 0 + 0 + 13.8 = 263.8 Ω
- Maximum load = (24 − 10.5) ÷ 0.022 A = 614 Ω
How to Calculate the Maximum Loop Load
In a two wire loop the power supply, transmitter and all loads are in series. The transmitter needs a minimum terminal voltage (lift off voltage) to work, so whatever is left over can be dropped across the loads.
Rcable = 2 × ρ × L ÷ A (ρ copper = 0.0172 Ω·mm²/m)
Use the highest current the loop can carry, not 20 mA. Transmitters signal faults at up to 21 or 22 mA (NAMUR NE 43) and HART devices may saturate even higher.
Worked example
A 24 V supply drives a transmitter needing 10.5 V. At 22 mA the maximum load is (24 − 10.5) ÷ 0.022 = 614 Ω. A 250 Ω input card leaves 364 Ω for cable, enough for several kilometres of 0.75 mm² cable.
Frequently Asked Questions
What is transmitter lift off voltage?
It is the minimum voltage the transmitter needs across its terminals to operate. It is typically 10.5 to 12 V for standard transmitters and higher for HART or display versions. Check the datasheet.
Why is a 250 Ω resistor used in 4 to 20 mA loops?
It converts 4 to 20 mA into 1 to 5 V for voltage input cards, and HART communication needs at least about 230 Ω of loop resistance to work.
What happens if the loop resistance is too high?
The transmitter works at low currents but cannot drive the loop to its full value, so readings clip or freeze below 20 mA at high process values and fault alarms may not be reached.
Learn more on the blog
4 to 20 mA current loop explained
A practical, field focused guide on instrumentationblog.in that explains the theory behind this calculator and how engineers apply it on real plants.
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