4–20 mA loop with a Zener barrier: voltage budget, IS limits and an earth-loop offset
A two-wire gauge pressure transmitter (range 0–16 bar, 4–20 mA, minimum terminal voltage 12 V at 20 mA, maximum permissible loop resistance 600 Ω) is connected to a PLC analogue input card with a 250 Ω shunt. The signal passes through a Zener barrier whose series resistance is 300 Ω. The field cable is 350 m of 1.5 mm² twisted pair, 12.1 Ω/km per conductor. The loop is powered from a 24 V DC supply at the control panel. Readings are acceptable at mid-scale, but the transmitter saturates near 19.2 mA when the vessel is full, and the operator now wants to add a loop-powered local indicator with a 2 V maximum drop. Separately, at zero pressure the PLC reads 0.55 bar instead of 0 bar, the offset drifts with plant load, and the shield of the field cable is bonded to the local earthing bar in the field junction box as well as at the PLC panel. A 2.2 kW VFD for the nearby pump sits 1.5 m from the signal cable tray.
1. Compute the loop resistance and transmitter terminal voltage at 20 mA, explain the 19.2 mA ceiling, and determine what adding the local indicator would do.
Read the worked answer
Loop resistance. Rloop = shunt 250 + barrier 300 + cable 2 × 0.350 × 12.1 = 250 + 300 + 8.5 = 558.5 Ω
Voltage at full scale. At 20 mA the drop is 0.020 × 558.5 = 11.17 V, so the transmitter sees 24 − 11.17 = 12.83 V against a 12 V minimum - only 0.83 V of margin, equivalent to 41 Ω of extra loop resistance.
The 19.2 mA ceiling. At 19.2 mA the drop is 10.72 V and the transmitter would have 13.28 V, which is above its minimum - so on paper it should reach 20 mA. The fact it does not means the budget is worse than the nameplate suggests: measure the 24 V supply at the panel terminals under load (a nominal 24 V unit with a shared load may sit at 23.0 V or lower), and measure the actual barrier resistance rather than trusting the label. At 23.0 V, 20 mA would leave 11.83 V - below the minimum, and the transmitter current-limits at about 19.4 mA, matching the observation. Look also for extra terminations, a fuse in the loop, or a second device in series.
Adding the indicator. With 2 V of indicator drop, the available current becomes (24 − 2 − 12)/558.5 = 17.9 mA, so the top 21% of the range disappears and the PLC will show about 13.5 bar at full vessel. Not acceptable.
Fixes, in order of preference. (1) Raise the supply to 28 V - permitted only if the barrier certificate's maximum supply voltage allows it and the transmitter's maximum terminal voltage is respected; 28 V gives 28 − 2 − 11.17 = 14.83 V, adequate. (2) Replace the 300 Ω Zener barrier with a galvanically isolated barrier of 50–100 Ω series resistance, which also solves the earthing issue below. (3) Choose an indicator with a 1 V drop and re-verify the budget at the worst-case supply. Do not simply reduce the shunt - that changes the scaling and the noise immunity.
2. Verify the intrinsic-safety compatibility of the field cable with the barrier and state the installation rules that must be enforced.
Read the worked answer
Entity check. The barrier certificate gives permissible cable parameters, for example Cc = 0.083 µF and Lc = 4.2 mH for the gas group in use. The 350 m of instrumentation cable contributes roughly 0.1 nF/m → 0.035 µF of capacitance and about 1 µH/m → 0.35 mH of inductance, both well inside those limits. Add the transmitter's own Ci and Li (usually negligible for a 2-wire unit) before confirming. The transmitter's entity parameters must also be satisfied by the barrier output: Ui ≥ 28 V, Ii ≥ 93 mA, Pi ≥ 0.65 W, all of which a standard 28 V / 300 Ω barrier meets.
Installation rules that are actually enforced at site.
- The barrier sits in the safe area and its earth terminal is bonded to the dedicated intrinsic-safety earth bar with a low-resistance, separately identified conductor; the barrier earth must never be removed while the circuit is live.
- The field cable is identified as an IS circuit (blue sheath or permanent marking), is segregated from power and from non-IS signal cables, and shares no multicore with them.
- The field junction box is clearly labelled as containing IS circuits; the screen is not used as a return path.
- The maximum cable length for the loop's resistance and capacitance is documented on the loop drawing, so a future extension of the run cannot silently break the IS certificate.
- A fuse in the barrier's supply is the correct protection; do not add an upstream device with a higher rating than the certificate permits.
3. Diagnose the 0.55 bar zero offset and drifting reading, and specify the correct shield and routing practice.
Read the worked answer
Magnitude. 0.55 bar on a 16 bar span is 3.4% of span, i.e. 0.55 mA. It is not a calibration error, because a genuine zero error would be stable - this one moves with plant load, which is the signature of interference, not miscalibration.
Mechanism. The shield is bonded at both ends, forming a loop that encloses the VFD motor cable. Capacitive coupling from the drive's PWM output drives common-mode current along the shield, and the potential difference between the two earth points appears as a series voltage across the loop. The PLC's single-ended input converts that common-mode voltage into a reading error and, at high enough levels, into noise on the analogue value.
Correct practice.
- Earth the shield at one point only - normally the PLC panel end, on the signal earth bar, not the power earth bar. At the field end, cut the drain back, insulate it, and terminate nothing.
- Use twisted pair for the loop, and route it at least 300 mm from the VFD motor cable, crossing at 90° where it must cross, and never in the same tray for a parallel run.
- Use shielded twisted pair for the drive motor cable with the shield bonded 360° at both ends - the drive cable shield is a different case and should be earthed at both ends, which is why the two rules are often confused.
- Check that the transmitter's output is isolated from its case; if it is mounted on a bonded, earthed pipe and the output is not isolated, an isolated barrier is the robust answer.
Verification. Simulate 4 mA with the VFD off and then at full speed: if the offset vanishes when the drive is stopped, the diagnosis is confirmed. Also check the AI card's common-mode range and that its common is referenced as the manufacturer requires.