Thread Content
The 4-20 mA current loop may seem \"old-fashioned,\" but it continues to be widely used in the digital age, and there are many \"little secrets\" behind it that even experienced engineers may not be fully aware of. Below, I’ll pick out the most practical and easily overlooked tips, as a sort of “unboxing”. “4 mA is essentially the \"fuel\" for field instruments. This value of 4 mA wasn’t chosen arbitrarily as a starting point; it is used first to power the two-wire transmitters: the sensor, the A/D converter, the linearization circuit, the cold-junction compensation circuit, the linear compensation circuit, the isolation circuit, and the V/I conversion circuit… All of these functions rely on the energy supplied by those 4 mA. Only the remaining 0–16 mA is actually used to “communicate”. Therefore, for values below 3.6 mA, most transmitters will ‘stop working’ rather than ‘output 0’ – this is the first clue to determine whether the instrument has ‘run out of power’. Open circuit = 0 mA, which the PLC can identify immediately. The current loop is most vulnerable to open circuits, but 0 mA is an “invalid value” in the 4–20 mA range. PLC/DCS will report an \"open circuit\" as long as it reads < 2 mA, which is much simpler than having to add a separate open-circuit detection circuit for voltage signals. Many systems also consider values below 3.5 mA as a \"pre-alarm,\" while 21–22 mA is regarded as \"above the upper limit\" – 20 mA is not the final limit, but rather the \"safe ceiling.\" Current signals can also be powered via a “two-wire” system, reducing the need for half as many cables. In this two-wire configuration, 24 V and the signal are combined on the same wire: a constant 24 V voltage at the PLC end → a constant current of 4–20 mA on that same wire → and a stable 3–9 V voltage at the transmitter end for its own use. As long as the total resistance of the circuit is ≤ the power supply tolerance (typically 600 Ω at 24 V), it is possible to transmit signals as well as supply power; no additional power cables are needed on-site, and the explosion-proof barrier can be simple – one safety barrier is sufficient. Current is not affected by wire resistance, so its accuracy remains unchanged even over distances of a kilometer. A voltage signal (0–10 V) loses 1 mV for every 1 Ω of cable resistance, whereas the internal resistance of a current source is essentially “infinite”; as long as there are no branch circuits in the loop, a current of 12 mA is maintained regardless of whether the wire resistance ranges from 0 Ω to 1 kΩ – thus, 3 V is still obtained at the sampling resistor of 250 Ω at the far end. In other words, temporarily replacing the 1.5 mm² cable with a 0.5 mm² one does not require any change to the calibration coefficients, as long as the total resistance remains within the supply capacity. 20 mA is the upper limit for a \"safe spark.\" Back then, the IEC set 20 mA as this upper limit because 24 V × 20 mA = 0.48 W, and the spark energy generated in the event of a fault short circuit is below the minimum ignition energy required to ignite flammable gases (Grade II C ≤ 1.3 W). Therefore, 4–20 mA two-wire devices can be directly used in Zone 0 as long as their intrinsically safe parameters are matched – this is a \"pass\" that the 0–20 mA and 10–50 mA standards do not provide. Current loops are inherently resistant to EMI; twisted pair is sufficient. In industrial environments, there are factors such as 50 Hz frequencies, frequency converters, lightning, and walkie-talkies… all of which generate common-mode voltages. The current loop receiver is only concerned with the \"current flowing through itself\" and is inherently immune to common-mode voltage ； Adding twisted pair turns the noise into differential mode, causing the instrumentation amplifier to fail as soon as it is applied. Actual measurements show that when wiring is placed next to a 100 A busbar, the fluctuation for 4–20 mA is < 0.01 mA, whereas for 0–10 V, the drift is already 200 mV. A 250 Ω \"universal port\" for 1–5 V – do I need to use it to connect a 4–20 mA signal to the voltage input module of a PLC on-site? A 250 Ω±0.1 % precision resistor is sufficient: 4 mA→1 V, 20 mA→5 V, with linear variation in between. As long as the resistance temperature drift is ≤ 25 ppm, the accuracy remains at 0.1% across -40 ℃ to 85 ℃ — which is ten times cheaper than buying an isolated converter. Make sure the resistor is placed on the PLC side, not at the field location; otherwise, a short circuit at the field site will cause 24 V to be fed directly into the PLC’s analog inputs. Current can also be used to transmit signals – HART does this by using a signal between 4 mA and 20 mA, to which is added a ±0.5 mA FSK signal at 1.2 kHz/2.2 kHz; this combination constitutes HART. There is no need to replace the wiring; by using a handheld controller or PLC module to \"eavesdrop\" on the current loop, it is possible to read secondary variables, set ranges, and perform diagnostics. Secret: A coupling capacitor of just 0.1 µF is sufficient to extract high-frequency signals, while leaving the DC component unchanged. Wrong connection of passive/active components can cause immediate damage to the other side; active meters (four-wire type) have their own 24 V supply, and the PLC also provides 24 V. When these two supplies are connected in opposition to each other, a current of 40 mA+ is generated instantly, which can either cause the fuse to blow or damage the ADC chip. Remember the mnemonic: “Two wires, passive; connected in series in the circuit” ； Four-wire active; the PLC only samples data without providing power. ” Before making the connection, use a multimeter to measure the open-circuit voltage; if it is > 25 V, it can generally be determined that the other side is active. 4–20 mA can also be “wireless” – first, the current is converted into a wireless transmitter powered by 3G/4G batteries; the sensor signal is then scaled to 0–16 mA, after which the loop is disconnected. An internal 3.6 V lithium battery is used to supply power to the loop, and power is cut off immediately after sampling. The average current is less than 50 µA, allowing a 19 Ah battery to last for 5 years. The secret lies in the \"duty cycle\" – it sends a signal for only 0.5 seconds per minute, with the circuit in sleep mode for the rest of the time, thus ensuring both accuracy and longevity. Keep these “little secrets” in mind; the next time you encounter 4–20 mA debugging, selection, or troubleshooting, you’ll be able to see through it all like an experienced expert. Wish you no circuit explosions, stable signals, and a successful pass on the inspection!