4-20 mA loop scaling: the 16 mA rule that earns marks on the Red Seal instrumentation exam

Why the 4-20 mA loop is a guaranteed exam topic

The Instrumentation and Control Technician Red Seal exam draws heavily on the measurement and control block of the occupational standard, and no single signal shows up more often than the 4-20 mA current loop. It is the default way a pressure, level, flow or temperature transmitter talks to a PLC or a DCS. If you can scale a loop in your head, you bank several marks in a couple of minutes.

Live zero: why the loop starts at 4 mA and not 0

A 0-20 mA signal cannot tell the difference between a real reading of 0% and a broken wire. The 4 mA live zero fixes that: the transmitter always draws at least 4 mA, so a current below roughly 3.6 mA is a fault, not a measurement.

• Below 3.6 mA – downscale failure (open loop, dead transmitter, lost power).
• Above 21 mA – upscale failure (short, burnout, gross over-range).
• 4 mA = 0% of span; 20 mA = 100% of span.
• 16 mA = the full span, from 4 to 20. This is the number candidates get wrong.

Those fault thresholds come from the NAMUR NE43 convention, implemented in most modern transmitters.

The one formula you actually need

Every scaling question is a two-step proportion. From current to process value:

Process = LRV + ((mA − 4) ÷ 16) × Span

and the other way around:

mA = 4 + ((Process − LRV) ÷ Span) × 16

LRV is the lower range value and Span is URV − LRV, both in engineering units.

Worked example 1 – current to pressure

A pressure transmitter is ranged 0-500 kPa with a 4-20 mA output. The loop reads 13.6 mA. What pressure is it reporting?

• Span = 500 − 0 = 500 kPa, LRV = 0 kPa.
• Fraction of span = (13.6 − 4) ÷ 16 = 9.6 ÷ 16 = 0.60.
• Process = 0 + 0.60 × 500 = 300 kPa.

Worked example 2 – pressure to current

Same transmitter. The process climbs to 425 kPa. What current should the meter show?

• Fraction of span = (425 − 0) ÷ 500 = 0.85.
• mA = 4 + (0.85 × 16) = 4 + 13.6 = 17.6 mA.

Worked example 3 – the suppressed zero that trips people up

A temperature transmitter is ranged −10 to 40 °C. Span = 50 °C and the LRV is −10 °C, not zero. At 12 mA:

• Fraction = (12 − 4) ÷ 16 = 0.50.
• Process = −10 + (0.50 × 50) = 15 °C.

Forget the LRV and you answer 25 °C. That single slip costs easy marks.

Calibration check points

A five-point check at 0/25/50/75/100% of span must read 4, 8, 12, 16 and 20 mA. For the 0-500 kPa transmitter that is 4 mA = 0 kPa, 8 mA = 125 kPa, 12 mA = 250 kPa, 16 mA = 375 kPa, 20 mA = 500 kPa. If the loop reads 4.05 mA at zero, the error is 0.05 ÷ 16 = 0.3% of span.

Reading the loop in volts

Many PLC analogue cards measure voltage rather than current. A 250 Ω resistor in the loop converts 4-20 mA to 1-5 V (0.004 × 250 = 1 V, 0.020 × 250 = 5 V). Loop resistance therefore matters: at 20 mA a 250 Ω resistor alone drops 5 V, so a 24 V supply has to cover the resistor, the intrinsic-safety barrier and the wire on top of the transmitter's own minimum operating voltage. If the total drop leaves less than that minimum, the reading collapses at the top of the range – a classic bench fault question.

Exam traps to watch

• Dividing by 20 instead of 16.
• Treating 4 mA as 4% of span instead of 0%.
• Ignoring the LRV when the range has a suppressed or elevated zero.
• Confusing percent of span with the process value in engineering units.
• Checking a two-wire transmitter with the loop powered down instead of measuring the current in series.

Practice routine

Take any range you know – 0-100%, −40 to 120 °F, 0-1000 L/min – and write the process values for 6, 10, 14 and 18 mA, then the currents for 25%, 50% and 75% of span. Ten minutes a day and the loop scaling questions become free marks on exam day.