A transmitter that reads fine on the bench can still fail in the field if the loop simply doesn't have enough voltage left over once every series resistance has taken its share — checked here at 20 mA, the worst case for voltage drop.
Verified against published standard
Supply & Transmitter
Minimum operating (compliance) voltage is on the transmitter's datasheet — typically 10–12 V for a basic 4-20mA device, higher for HART or advanced diagnostics.
Series Loop Resistances
A 250Ω receiver resistance is the common value for a HART-compatible analog input (needed for the HART signal to develop enough voltage to be read); check your actual card's input impedance.
Enter loop data and check the power budget.
About This Calculation
A 4-20mA two-wire transmitter loop is a series electrical circuit: supply voltage must be enough to cover the transmitter's own minimum operating (compliance) voltage PLUS the voltage dropped across every resistance in the loop — wiring, receiver/PLC input resistance, any intrinsic-safety barrier — evaluated at 20mA, the worst case for voltage drop by Ohm's law. A loop that reads correctly on the bench with a short test lead can genuinely fail once installed with the actual, much longer field wiring run, if this budget was never checked.
Frequently Asked Questions
Why is 20mA specifically the worst case for voltage drop, not 4mA?
By Ohm’s law, voltage drop across a fixed resistance is directly proportional to current — the loop’s total series resistance drops the MOST voltage at the HIGHEST current the loop will ever carry, which is 20mA (100% of range) for a standard 4-20mA transmitter, making it the correct condition to check against.
Why does a HART-compatible receiver need a specific minimum resistance, not just enough to be safe?
HART communication is a small AC signal superimposed on the 4-20mA DC current — that signal needs the receiving input to present sufficient resistance (commonly a minimum around 230Ω, though it varies by system) for the HART signal to develop enough voltage swing to be reliably read, which is a genuinely different requirement from simply having "enough" resistance for the DC loop power budget alone.
Does increasing wire gauge always fix a marginal loop power budget?
Usually yes for the wiring’s own contribution, since resistance falls with cross-sectional area for a given length — but it doesn’t address the transmitter’s own minimum voltage requirement or a fixed receiver/barrier resistance, so for a badly marginal loop, raising supply voltage or removing a series device may be the only real fix.