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Control & Instrumentation Engineering DISCUSSION
I am connecting a 0 to 10 bar pressure transmitter to a PLC analog input. The transmitter output is 4-20 mA on two wires, while the cheap sensors I have used before give 0-10 V. The PLC card accepts either.
Why is 4-20 mA the normal choice in plants, why does it start at 4 mA rather than zero, and what is the formula to turn the measured current back into pressure?
A current signal is not affected by the resistance of the cable. The transmitter regulates the loop current, so the same 12 mA flows through every point of the loop whether the cable is 5 m or 500 m, as long as the supply has enough voltage to drive it. A voltage signal loses some of its value across the cable resistance and is easily disturbed by induced noise, because the receiver input is high impedance. The current loop is low impedance and much harder to disturb.
The 4 mA bottom is called a live zero and it does two jobs. First, a two-wire transmitter has to run its own electronics from the loop, and it does that on the current below 4 mA. Second, it separates a real zero reading from a fault: 4.00 mA means 0 bar, while 0 mA means a broken wire, a blown fuse or a dead supply.
Many transmitters follow the NAMUR NE 43 convention, where the valid measurement range is 3.8 to 20.5 mA and a current at or below 3.6 mA or at or above 21 mA signals a device fault. Set the PLC alarm limits accordingly.
Scaling is linear: PV = LRV + (I - 4 mA) / 16 mA × (URV - LRV), where LRV and URV are the lower and upper range values. For your 0 to 10 bar range, 13.6 mA is (13.6 - 4) / 16 × 10 = 6.0 bar, and 12 mA is exactly mid-range, 5 bar.
A frequent mistake is scaling as if the signal were 0-20 mA, which reads 4 mA as 20 percent instead of 0. In the PLC, check what raw count the card gives at 4 mA and at 20 mA and scale between those two counts.
Check the loop voltage budget, because every device in series takes a share of the supply. The transmitter needs a minimum terminal voltage, given in its datasheet, and the rest is available for the load: R_max = (V_supply - V_min) / I_max. With 24 V, a 12 V minimum and 22 mA to cover the fault-signalling current, R_max = 12 / 0.022 = 545 Ω. A 250 Ω input, which turns 4-20 mA into 1-5 V, plus cable fits easily; adding a loop-powered display and an isolator may not. The symptom of too much load is a reading that is correct at low current and stops rising near the top.
Wire the loop as supply positive, transmitter, input, supply negative, using shielded twisted pair with the shield earthed at one end only.
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