4-20 mA & networks formulas

Instrument networks and the 4-20 mA transmitter. 8 formulas for GATE IN Sensors and Industrial Instrumentation, each with when to use it. Each is explained in the topic notes.

All 8 4-20 mA & networks formulas

MEASURED VALUE TO LOOP CURRENT

I = 4 + 16 \frac{x - LRV}{URV - LRV} mA

Linear transmitter, 4 mA at LRV and 20 mA at URV; I in mA.

LOOP CURRENT TO MEASURED VALUE

x = LRV + \frac{I - 4}{16} (URV - LRV)

Inverse of the relation above; valid for 4 mA to 20 mA.

PERCENT OF SPAN FROM CURRENT

Percent of span = \frac{I - 4}{16} \times 100

0 % at 4 mA, 50 % at 12 mA, 100 % at 20 mA.

RECEIVER VOLTAGE

V_{R} = I R_{R}

A 250 Ω receiver turns 4 mA to 20 mA into 1 V to 5 V.

SUPPLY VOLTAGE REQUIREMENT

V_{s} \geq V_{T, m i n} + I_{m a x} (R_{wire} + R_{receivers})

I_max = 20 mA unless an over-range current is stated; all series resistances are included.

MAXIMUM TOTAL LOOP RESISTANCE

R_{L, m a x} = \frac{V _{s} - V _{T, m i n}}{I _{m a x}}

R_L is wire plus receiver resistance in series; with V in volts and I in amperes, R is in ohms.

FLOW FROM DIFFERENTIAL PRESSURE

Q = Q_{m a x} \frac{Δ p}{Δ p _{m a x}}

Head-type primary element, turbulent flow, constant fluid density; Q_max is the flow at full-scale differential pressure.

LOOP CURRENT FROM FLOW, NO SQUARE-ROOT EXTRACTOR

I = 4 + 16 (\frac{Q}{Q _{m a x}})^{2} mA

Same conditions as the flow law; with a square-root extractor the relation becomes linear, I = 4 + 16 Q/Q_max mA.

Where this was asked: 2024 (1). See the 1 previous-year question on 4-20 mA & networks. This shows what was asked, not what will be asked.

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4-20 mA & networks: the notes and 1 solved previous-year question