Current-Loop Voltage Budget

Method & assumptions

1. Model one DC series loop with a two-wire loop-powered transmitter, resistive receiver and total outgoing-plus-return cable resistance. Use the lowest available supply and worst-case device data for your selected current.
2. Convert mA to A by dividing by 1000. Receiver and cable voltage drops are I × R. Include additional series device drops separately without counting the same resistance twice.
3. Minimum supply = transmitter minimum + receiver drop + wire drop + other drops. Voltage margin = available supply − minimum supply. Negative margin means the requested current cannot be supported under these assumptions; this tool does not predict the actual saturated current.
4. Additional resistance headroom = voltage margin / I. A negative result is a deficit, not an allowable resistance. Receiver and cable dissipation each equal I²R; select physical ratings with the applicable derating.
5. Original example: 24 V supply, 12 V transmitter minimum, 20 mA, 250 Ω receiver, 100 Ω total wire and 1 V other drop require 20 V and leave 4 V margin. Include both cable conductors.
6. Check the greatest operating or alarm current you require, which may exceed 20 mA. Also check transmitter maximum voltage at low current, receiver ratings, supply tolerance/current capacity, temperature effects, isolation and grounding. A positive margin is not a complete device or hazardous-area approval.

Sources

• NI: Current-loop system design and power supply selection ↗