Adding a Second Controller to a 4–20 mA Loop
A 24 Vdc supply (adjustable 24–28 V, ±5 % at whatever it is set to) feeds a two-wire pressure transmitter whose datasheet requires at least 12 V at its terminals at 20 mA, a 600 m run of 0.75 mm² copper (≈47 Ω/km for the loop), and a 250 Ω input resistor at the PLC. A panel indicator that must be inserted in series drops 2.0 V. Operations now wants a second controller on the same loop, which means a second 250 Ω sense resistor in series with everything else. The transmitter's maximum supply is 36 V and its over-range current is 22 mA. Before parts are ordered, the engineer must show whether the loop still complies, propose a fix that survives supply tolerance, and settle the shield-grounding question the second panel raises.
1. Compute the worst-case voltage available at the transmitter terminals with and without the second sense resistor. Does the loop comply?
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Loop resistance = 0.6 km × 47 Ω/km = 28 Ω. At 20 mA:
Present loop: 250 Ω × 20 mA = 5.00 V, cable 28 Ω × 20 mA = 0.56 V, indicator 2.00 V → 7.56 V total. Available at the transmitter = supply − 7.56. At the 22.8 V worst case that is 15.24 V ≫ 12 V. Compliant.
With the second 250 Ω: sense resistors 500 Ω × 20 mA = 10.00 V, plus 0.56 V cable, plus 2.00 V indicator = 12.56 V. Available:
- 24.0 V nominal → 11.44 V
- 22.8 V worst case → 10.24 V
- 25.2 V best case → 12.64 V
The loop is non-compliant: the transmitter needs 12 V and gets 10.2 V in the worst case. Worse, the over-range condition is the one that matters for ranging and diagnostics: 528 Ω × 22 mA = 11.62 V + 2.00 = 13.62 V, leaving only 9.2 V at 22.8 V supply. The marginal 'it works on the bench at 24 V' observation is exactly the trap - it does not work at −5 %.
2. Propose two compliant fixes with numbers and choose one.
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Option A - raise the supply to 28 V. Worst case 26.6 V: 26.6 − 12.56 = 14.04 V at 20 mA, and 26.6 − 13.62 = 12.98 V at 22 mA. Both exceed 12 V, and 28 V is well under the transmitter's 36 V maximum. Cheap: a screwdriver. Check the indicator's and PLC input's voltage ratings and the loop's isolation, but at 28 V there is no issue.
Option B - loop-powered isolator/splitter with a burden under 1 V feeding the second controller. Total drop = 5.00 (PLC 250 Ω) + 0.56 + 1.0 = 6.56 V → 17.44 V available at 24 V nominal, 16.24 V at worst case. Large margin regardless of supply setting.
Choice: B if the second controller's input is single-ended, which it almost certainly is. Two grounded 250 Ω inputs in one series loop tie two separate earth references together through the loop return; any potential difference between the panels drives current through the loop and the shield, producing exactly the offset and noise that a current loop is supposed to be immune to. The isolator breaks that path and costs one device. Option A is the correct answer only if both inputs are genuinely floating/differential - then it is compliant, cheaper, and needs no new failure point.
3. The installer proposes bonding the cable shield at both the transmitter junction box and the second panel, citing 'better shielding'. How should the engineer respond?
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Refuse it. Bonding both ends creates a closed loop of shield and earth; the earth potential difference between two panels in an industrial plant is routinely hundreds of millivolts and can be volts during switching. That difference drives a shield current that couples capacitively into the signal pair - the classic 50/60 Hz plus switching-transient interference.
The current loop itself is largely immune in terms of accuracy (a 0.8 V earth difference does not change a 20 mA current), but the consequences are real: the shield current appears as common-mode voltage at the receiver input, and if the input is single-ended that leaks into the reading; the second controller's ground reference also differs from the PLC's, so the two readings disagree by the IR drop in the loop return.
Correct practice: bond the shield at one end only - normally the supply/PLC end - and leave the far end isolated (insulated from the junction box). If the second panel is a separate earth, use the isolated splitter from the previous task, which removes the ground path entirely. If plant practice insists on two-point bonding for safety, then the signal pair must be galvanically isolated at one end; an unisolated shield grounded at both ends is the one arrangement to rule out.