How to use this on a job
The transmitter can regulate current only while it has enough voltage left to operate. Every series resistance spends part of the supply. A loop that behaves at 4 mA can run out of voltage near the top of the range.
- 01
Use the lowest supply voltage expected at the loop and the transmitter’s minimum operating voltage from its data sheet.
- 02
Enter the design current, total receiver resistance and complete outgoing-plus-return cable resistance. If a device specifies a fixed voltage drop rather than a resistor, this simple resistive model is not sufficient for that device.
- 03
Check the remaining transmitter voltage and headroom. A negative margin cannot support the chosen operating point in this model. Test at the maximum current the transmitter may produce, including a configured alarm current when applicable.
Work through one example
margin = Vₛ − I × (Rload + Rcable) − Vminimum
At 20 mA, a 250 Ω receiver and 50 Ω complete cable loop drop 6 V. A 24 V supply leaves 18 V at the transmitter. If the transmitter needs 12 V, the headroom is 6 V. Reduce the supply to 15 V and the headroom becomes −3 V; the loop cannot sustain that current under these assumptions.
Before you trust the answer
Enter both cable conductors, not just the one-way length. Temperature, supply tolerance, barriers and isolators can change the real budget. This is a DC resistive calculation; it does not establish intrinsic safety, isolation suitability or HART impedance compliance.