Loop-Powered Transmitter Starves After Indicator Addition
A 24 VDC loop-powered pressure transmitter is wired through a 300 Ω intrinsic-safety barrier, a 250 Ω PLC analog input, and 25 Ω total cable resistance. The transmitter nameplate states a minimum terminal voltage of 12 V at 20 mA, and the HART communicator requires at least 250 Ω in the loop. The loop worked reliably until a 100 Ω local indicator was added in series at the marshalling cabinet. Now the DCS value tracks correctly up to about 18 mA, then the transmitter output drops to 0 mA and the PLC input shows under-range. At 12 mA the loop measures normally. The area is Class I, Div. 1, and the barrier certificate lists a maximum supply voltage of 26 VDC and an allowed barrier resistance of 300 Ω. The technician proposes raising the supply to 28 VDC and grounding the cable shield at both ends to reduce noise.
1. Calculate the total loop resistance and the transmitter terminal voltage at 12 mA, 18 mA, and 20 mA before and after the indicator was added.
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Before the indicator: R_loop = 300 + 250 + 25 = 575 Ω At 12 mA: Vtx = 24 - (0.012 × 575) = 24 - 6.9 = 17.1 V At 18 mA: Vtx = 24 - (0.018 × 575) = 24 - 10.35 = 13.65 V At 20 mA: Vtx = 24 - (0.020 × 575) = 24 - 11.5 = 12.5 V
After adding the 100 Ω indicator: R_loop = 675 Ω At 12 mA: Vtx = 24 - 8.1 = 15.9 V At 18 mA: Vtx = 24 - 12.15 = 11.85 V At 20 mA: Vtx = 24 - 13.5 = 10.5 V
The transmitter needs at least 12 V. With the indicator, it drops below 12 V above about 17.8 mA: I = (24 - 12) / 675 = 0.01778 A = 17.8 mA That matches the symptom: normal at 12 mA, failure as the current approaches 18 mA.
2. Evaluate the technician's two proposals. What is wrong with each, and what are the acceptable alternatives?
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Raising the supply to 28 VDC is not acceptable in this hazardous location. The barrier certificate limits the supply to 26 VDC, and the intrinsic-safety entity parameters must not be exceeded. Doing so invalidates the IS installation and creates an ignition risk.
Grounding the shield at both ends is also incorrect. It creates a shield ground loop that can carry stray current, inject noise, and defeat the single-point grounding scheme. In an IS loop, follow the plant and barrier instructions, typically grounding the shield at one end only.
Acceptable alternatives include removing the series indicator and using a high-input-impedance parallel display, using a lower-resistance approved barrier, or raising the supply only to the approved 26 VDC maximum after checking all ratings. At 26 V and 20 mA with the indicator, the transmitter would see 26 - 13.5 = 12.5 V, which is just above the minimum; verify margin and entity parameters. HART still sees more than 250 Ω.
3. What minimum supply voltage would be needed to keep 12 V at 20 mA with the indicator, and what would you document?
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The minimum supply is: Vmin = 12 + (0.020 × 675) = 12 + 13.5 = 25.5 V
A 26 VDC supply would provide only 0.5 V margin at 20 mA. That is tight, so check the transmitter's actual minimum voltage over temperature, barrier resistance tolerance, and cable resistance. It is better to remove the extra series resistance than to operate at the limit.
Document the loop calculation, barrier certificate limits, transmitter minimum voltage, HART resistance requirement, and the as-left wiring. Update the loop drawing and calibration records, and process the change through management of change. Verify the PLC input scaling and alarm response after the repair.