Process Calculator
(4 to 20) mA
In industrial applications, the 4-20 mA signal has become a reliable standard for transmitting analog information between sensors and control systems. This current range allows for the proportional representation of physical variables such as temperature, pressure, or flow rate, facilitating their interpretation and processing by equipment such as PLCs or SCADA systems.
How to enter the values:
-Maximum and Minimum Range: This is the range in which the equipment operates, for example (0 to 100) °C, enter the lowest operating range and then the highest.
- Unit: the unit in engineering units; for example "°C"
-Current: enter the measured value in mA with a minimum range of 3 mA and a maximum of 22 mA.
The range (4-20) mA can also be calculated if two points are known, and for each point the value in units and the value in mA are known, example:
Knowing that 19.634 °C is equivalent to 5.789 mA and that 128.265 °C is equivalent to 15.687 mA, the range can be calculated as 4 mA = 0.20 mA = 175.6
Why is the 4 to 20 mA range used?
1. Diagnostics: Using 4 mA as the starting point, rather than 0 mA, allows for the differentiation of a device's idle state from a fault due to a power failure, a short in the wiring, a turned-off device, or a damaged device.
2. Robustness: Current signals are much less susceptible to electromagnetic interference than voltage signals, ensuring more accurate transmission in industrial environments and allowing the signal to be transmitted over long distances. The current maintains its value with almost no loss along the circuit, regardless of the wiring resistance, allowing for reliable transmissions even over considerable distances.
3. Reliability and installed equipment: Most industrial sensors are designed to operate with 4-20 mA signals. Although this technology began to be used in the 1950s, it remains in use today due to its reliability. It has also evolved with integrations such as the HART digital communication protocol, which offers advanced diagnostics and the ability to integrate the equipment into a distributed control system, allowing it to visualize and control different variables at the same time.
How do you convert a 4-20 mA signal to a process value?
The 4 to 20 mA signal is linear: 4 mA represents the start of the transmitter calibrated range (LRV) and 20 mA the end (URV). Any current in between is converted with a simple proportion over a 16 mA span:
Process value = LRV + (I − 4) / 16 × (URV − LRV)
Current (mA) = 4 + 16 × (Value − LRV) / (URV − LRV)
% of span = (I − 4) / 16 × 100
Worked example
A pressure transmitter calibrated from 0 to 10 bar outputs 12 mA. Value = 0 + (12 − 4) / 16 × (10 − 0) = 5 bar, which is 50 % of span. If the same transmitter reads 7.2 bar, the expected current is 4 + 16 × 7.2 / 10 = 15.52 mA.
4-20 mA conversion table
0 % of span = 4.00 mA = 1.00 V (with a 250 Ω resistor)
25 % of span = 8.00 mA = 2.00 V
50 % of span = 12.00 mA = 3.00 V
75 % of span = 16.00 mA = 4.00 V
100 % of span = 20.00 mA = 5.00 V
NAMUR NE 43 failure limits
A current below 3.6 mA or above 21 mA indicates an instrument or loop failure, while 3.8 to 20.5 mA is considered valid measurement saturation. Setting these limits in your control system lets you tell an out-of-range value from a failed transmitter.
Maximum loop resistance
R max (Ω) = (Supply V − Transmitter minimum V) / 0.020 A
With a 24 V supply and a transmitter that needs 12 V, the maximum load is (24 − 12) / 0.02 = 600 Ω, enough for a 250 Ω resistor (1 to 5 V signal) plus wiring.
Frequently asked questions
How do I find my instrument range from two points?
If you know two (current, value) pairs, the slope is (V2 − V1) / (I2 − I1); from it you get the value at 4 mA and at 20 mA. The calculator on this page does it automatically.
Does it work for negative ranges, like −50 to 150 °C?
Yes, the formula is the same: at 12 mA the value is −50 + 0.5 × 200 = 50 °C.
How much error does the 250 Ω resistor add?
Its tolerance goes straight into the measurement: a 1 % resistor adds up to 1 % error. For calibration loops use 0.1 % resistors or better.
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