Current wins distance and noise: a current loop loses nothing to wire resistance and flags a broken wire at 0 mA. Voltage wins simplicity: 0-10 V reads directly at the terminals and can feed several receivers in parallel. Past about 15 m of cable, or beside power wiring, choose current; inside that, with one receiver, voltage is the simpler circuit.
Two transmitters can read the same vacuum and still produce two different commissioning days, and the difference is usually the signal type ordered. The decision comes down to the cable run. Order the 0-10 V version when the transmitter sits within about 15 m of the input, the route stays clear of power wiring, and one or two receivers read the signal. Order the 4-20 mA version when the run is longer, when it shares a tray with motor or power cables, or when a broken wire has to be obvious at the panel. The sections below give the arithmetic behind that split, the load limit to read on the transmitter datasheet, and the fault readings each signal produces.
The short answer
Past about 15 m of cable, or anywhere the route runs beside VFD or power wiring, choose 4-20 mA. Inside 15 m, with a single receiver and a quiet route, choose 0-10 V and keep the meter-at-the-terminals simplicity.
The 15 m line has arithmetic behind it. A voltage signal loses voltage equal to the current the receiver draws times the cable resistance, so at 15 m of 24 AWG pair the round trip is 30 m × 0.084 Ω/m = 2.5 Ω, and at a 1 mA draw that is 2.5 mV, or 0.025% of a full-scale span. Sag still grows slowly past that point; what grows faster is the chance the route shares a tray with power or drive cables, and that risk is what moves the decision. A 1 MΩ input on 18 AWG copper carries a voltage signal well past 100 m with sag too small to measure, so input impedance and routing set the line, not distance alone.
Why distance treats the two signals differently
A voltage transmitter acts as a voltage source. The receiver draws current, and that current times the cable's loop resistance is voltage that never reaches the input. Standard AWG copper tables put 24 AWG at about 0.084 Ω per metre (about 26 Ω per 1000 ft), so a 60 m run, out and back, puts about 10 Ω in series with the signal.
A current-loop transmitter instead regulates the loop current. The input card converts that current back to volts, usually across a 250 Ω sense resistor, which turns the 4 to 20 milliamp range into 1 to 5 volts; 500 Ω on other cards turns the same loop into 2 to 10 volts. This range is the standard process signal, and ANSI/ISA-50.00.01 pairs it with the 1 to 5 V read across a 250 Ω input. Cable resistance still drops current-times-resistance, but it lands as a small extra burden on the loop supply, never as a reading error: every point in a series loop carries the same current.
Side by side
| Factor | 0-10 V voltage signal | 4-20 mA current signal |
|---|---|---|
| Run length | Good to about 15 m; error equals input draw times cable resistance | No reading error from resistance at any distance inside the loop budget |
| Noise near VFD or power wiring | Induced voltage lands directly on the input | Low-impedance loop; induced voltage drives almost no current |
| Broken wire | Reads 0 V, the same value as a true zero | Reads 0 mA, under the 3.6 mA NE43 fault line, so the PLC can alarm |
| Sharing with several receivers | Parallel feeds work if the output drives the combined load | One loop, one receiver; add a splitter or isolator |
| Troubleshooting | Read volts at the terminals | Read volts across the input resistance, then divide by the resistance |
| Transmitter power | Self-powered units need no loop supply | Two-wire units power from the loop; budget transmitter minimum plus input drop plus cable drop |
Read the table top down: the first two rows usually decide the order, the rest decide the details around the wiring.
Where voltage wins
Inside about 15 m, voltage is the simpler circuit. A meter at the input terminals reads the process value directly, with no conversion step, which is why troubleshooting feels easy. Receivers hang in parallel, so one output can feed a card and a local meter at once if the transmitter can drive the combined load. Self-powered transducers that output up to 10 volts need no loop supply at all, which keeps the panel wiring short. Run the pair as shielded twisted pair with the shield grounded at one end, and a short, clean voltage run gives years of stable readings.
Where current wins
Four properties put the current loop ahead on a plant floor:
- Resistance cannot touch the reading. A series loop has one current, so 10 Ω or 100 Ω of cable changes the loop supply's burden, not the value. Distance stops being a signal-quality question.
- Noise has little to grip. A current loop is a low-impedance circuit, so induced voltage from a VFD or a contactor coil drives almost no extra current around it. The same coupling onto a high-impedance voltage input lands directly on the reading as offset, and induced charge can lift a floating input far enough that the PLC input sees a high that is not there.
- A broken wire announces itself. The 4 mA live zero means a healthy loop never reads below 4 mA, so 0 mA means broken wire or dead transmitter. NAMUR NE43 keeps readings from 3.8 to 20.5 mA valid and reserves 3.6 mA and below, and 21 mA and above, for fault codes, so the PLC can alarm on the fault instead of trusting a bad number.
- Parts are everywhere. This is the most widely used process signal in industry, so transmitters, isolators and spare analog cards stock cheaply.
Example: a 60 m vacuum transmitter run
A vacuum transmitter sits 60 m from the PLC panel on 24 AWG shielded pair. Round-trip conductor length is 120 m, and the cable resistance is 120 m × 0.084 Ω/m = about 10 Ω.
Voltage version, 0 to 10 volts into a 10 kΩ input (an illustrative impedance; the module manual prints the real one): at full scale the input draws 10 V ÷ 10,000 Ω = 1 mA, and the cable drops 0.001 A × 10 Ω = 10 mV. That is 0.1% of span. Add one more identical receiver and the two inputs parallel to 5 kΩ, the draw doubles to 2 mA, the drop doubles to 20 mV (0.2% of span), and the output must now drive a 5 kΩ load. If the transmitter datasheet specifies a 10 kΩ minimum load, that second receiver overloads it and the output sags or clips near full scale.
Current version, 4 to 20 milliamps into a 250 Ω input: at full scale the input drops 0.020 A × 250 Ω = 5 volts, and the cable adds 0.020 A × 10 Ω = 0.2 volts, so the loop supply must cover 5.2 volts before the transmitter's own minimum. The current is identical at every point of the loop, so the reading error from cable resistance is 0%. A crushed or cut pair reads 0 mA, under NE43's 3.6 mA fault line, and the panel can alarm on it.
Same cable, same distance: the voltage run gives up 0.1% per receiver and inherits whatever the tray induces; the current run gives up nothing to resistance and hands the PLC a broken-wire alarm. At 60 m, order the current-loop transmitter.
Sharing one measurement with two controllers
A current loop serves one receiver cleanly. Two PLC inputs put their 250 Ω sense resistors in series, 10 volts of burden at full scale, which often breaks the loop budget, so the standard fix is a signal splitter or isolator: one loop in, one matched loop out per controller. Voltage shares more easily. Three 10 kΩ receivers in parallel combine to 10 kΩ ÷ 3 = 3.33 kΩ, which works while the combined load stays above the transmitter's minimum load rating, so read that one datasheet number before landing a second pair of wires. Follow each device's terminal diagram for commons and isolation; improvised series or parallel arrangements across different modules produce the exact offsets the rewiring was meant to remove.
Powering the loop
A two-wire transmitter powers itself from the loop and idles on the 4 mA baseline. Budget the supply at full scale: transmitter minimum voltage, plus input drop, plus cable drop, must stay under the supply. With 24 V, a transmitter that needs 8 V to operate, a 250 Ω input and 10 Ω of cable: 8 V + 5 V + 0.2 V = 13.2 V at the top of the range, comfortable. Under-budget the loop and the transmitter starves near full scale, and the reading folds back toward 4 mA.
A separate supply is not automatic. The PLC's built-in 24 VDC sensor supply is acceptable when its manual lists analog transmitter loads within its capacity. The avoid-it advice exists because the same terminals often feed inductive outputs; contactor and solenoid coils put switching transients onto wiring the analog circuits share. Suppress the coils at the coil and keep the signal pair away from those runs, and a shared supply behaves. When the panel does get its own supply, size its headroom the way you would for an HMI; Sizing a separate 24 VDC supply for HMI inrush current shows the capacity arithmetic. A separate supply does not fix a noisy cable; routing, shielding and suppression do that.
If the same panel also feeds a 12 to 36 V DC notice display, its power wiring is a separate job with its own rules; Wiring 12 to 36 V DC power to a signage display covers it.
Measuring each signal at the panel
Voltage: meter from the input terminal to analog common while the process value is steady, and the meter reads the value including any cable sag. Current: meter across the input terminals; on a 250 Ω input, 1 V means 4 mA and 5 V means full scale, and measured volts ÷ 250 Ω gives the loop current. That extra conversion is the whole troubleshooting penalty for current, and it is a small one. Do not break a live loop to insert an ammeter unless the work procedure allows interrupting the circuit; the drop across the input resistance gives the current without opening anything.
From the analog card to a screen the floor can read
Analog wiring stops at the input card, and everything after it is numbers. Raw counts rarely match the transmitter label on the first try; Why the scale value differs from the PLC analog input explains the span and offset behind that, and Scale a 4-20 mA pressure transducer for an HMI display walks the current-loop version end to end. If the vacuum reading also belongs where the whole shift can see it, the analog signal plays no further part: a network-fed shop-floor display is the layout, and Plan an offline sensor status display for your shop floor covers the pieces.
The decision
Both signals are right somewhere; run length and the tray pick the winner.
Recommendation: order the 4-20 mA transmitter whenever the run exceeds about 15 m, whenever the route shares a tray or conduit with power or VFD wiring, and whenever a broken wire should raise an alarm at the PLC; land it on a 250 Ω (1 to 5 V) input and budget the 24 V loop supply as transmitter minimum plus input drop plus cable drop. Choose the 0-10 V version inside 15 m with one high-impedance receiver and a clean route, and when one measurement must reach two controllers, put a signal splitter or isolator in the middle.
Questions
Can I run a 0-10 V signal 100 m?
Yes, if the input impedance is high and the route stays out of noisy trays. At a 1 MΩ input the receiver draws 0.01 mA at full scale, so 200 m of round-trip 24 AWG wire, about 17 Ω, drops about 0.2 mV, which is nothing. What limits the distance is induced noise and shielding cost, not sag. If the run must share a tray with power or VFD cables, the current-loop version removes the problem.
My current loop input reads 3.4 mA. What does that tell me?
That is below the 3.6 mA fault line NAMUR NE43 reserves, so the loop is broken or the transmitter cannot draw its full range. Sum the burden: input resistance times the full-scale 20 mA, wire resistance times the same current, and the transmitter minimum operating voltage. If that total exceeds the loop supply, the transmitter starves near full scale and the current falls out of the valid 3.8 to 20.5 mA window. Restore supply margin, repair the pair, or shorten the cable.
Does cable length change the calibration of a current loop?
No. Current is identical at every point of a series loop, so 10 Ω or 100 Ω of cable shifts the burden on the loop supply, never the reading. The calibration you set is at the input card: 4 mA to the low engineering value and 20 mA to the high one, which for a 250 Ω input is the same as scaling 1 to 5 volts. Verify with a mid-span check: 12 mA should show 3 volts across the input.
Planning information, not legal or engineering advice. Rules differ by place: check codes and permits with your local authority and your licensed trades.