Updated: April 21, 2026 | Sino-Inst Engineering Team

The cheapest way to convert a 4-20 mA loop signal into a voltage for a PLC or recorder analog input is a single precision resistor in parallel with the input. A 500 Ω resistor converts 4-20 mA to 2-10 V; a 250 Ω resistor converts 4-20 mA to 1-5 V; a 125 Ω resistor converts 4-20 mA to 0.5-2.5 V. The catch: a plain shunt resistor gives you a 2-10 V offset range, not a true 0-10 V, so when a datasheet says “0-10 V input” and your PLC card lists 0-10 V span, you need either an active signal converter or a scaling change in the PLC. This article walks through the formula, the wiring, when a resistor is enough, and when to buy a dedicated converter.

Contents

• How Do You Convert 4-20 mA to 0-10 V with a Resistor?
• What Is the Formula for 4-20 mA to Voltage Conversion?
• Why Does a 250 Ω Resistor Convert 4-20 mA to 1-5 V?
• When Should You Use a Signal Converter Instead of a Resistor?
• How Do You Wire a 4-20 mA Sensor to a 0-10 V PLC Input?
• What Are the Common Mistakes in 4-20 mA Voltage Conversion?
• Related Products
• FAQ

How Do You Convert 4-20 mA to 0-10 V with a Resistor?

Place a 500 Ω, 0.1% tolerance resistor across the analog input terminals of the PLC. The current loop flows through the resistor, and by Ohm’s law V = I × R, the voltage across the resistor is 2 V at 4 mA and 10 V at 20 mA. That produces a 2-10 V signal — which most modern PLCs accept on a 0-10 V input card and then rescale in software. If the PLC card strictly requires 0-10 V starting at zero, a resistor alone will not give you that; you need an active converter with offset adjustment. A sensor like the SI-300 pressure transducer with 4-20 mA and voltage outputs avoids the conversion step entirely by offering both signal types on the same part.

Pick a resistor with 0.1% tolerance or better and 1/4 W power rating. At 20 mA through 500 Ω, dissipation is 0.2 W — cutting it close for a 1/4 W part. Using a 1/2 W resistor leaves headroom for short-term overcurrent faults and keeps the resistor from drifting with self-heating.

What Is the Formula for 4-20 mA to Voltage Conversion?

The formula is Ohm’s law: R = V_full / I_full, where V_full is the desired voltage at 20 mA and I_full = 0.020 A. Pick the resistor value from this quick table:

Target Voltage Range	Resistor Value	Voltage at 4 mA	Voltage at 20 mA	Power at 20 mA
2-10 V (equiv. 0-10 V)	500 Ω	2.0 V	10.0 V	0.20 W
1-5 V	250 Ω	1.0 V	5.0 V	0.10 W
0.5-2.5 V	125 Ω	0.5 V	2.5 V	0.05 W
0.4-2 V	100 Ω	0.4 V	2.0 V	0.04 W

The 4-20 mA standard was chosen so that 4 mA (the live zero) is measurably non-zero. When you do the resistor conversion, the live zero carries over: 4 mA × 500 Ω = 2 V. This is a feature, not a bug — it lets the receiving PLC distinguish between “sensor reading minimum” (2 V) and “broken wire” (0 V).

Why Does a 250 Ω Resistor Convert 4-20 mA to 1-5 V?

250 Ω is the industry convention because 1-5 V was the original HART-compatible voltage input range, and 250 Ω happens to match both the voltage conversion and the minimum impedance HART modems need to communicate on the loop. Plugging into Ohm’s law: V = 0.020 × 250 = 5 V at full scale; V = 0.004 × 250 = 1 V at live zero. The result is a clean 1-5 V span with live zero preserved.

Two practical notes: first, check the 4-20 mA source’s maximum loop resistance on its datasheet. Most modern HART pressure transmitters handle 250 Ω plus wiring and a PLC barrier without issue, but long cable runs or multiple drops eat into that budget. Second, if you add a 250 Ω resistor to a loop that already has a PLC internal shunt, the resulting parallel resistance is much lower and the voltage drop is wrong. Always remove any existing shunt before inserting a precision resistor.

When Should You Use a Signal Converter Instead of a Resistor?

A smart differential pressure transmitter and a basic analog transmitter behave the same way on the electrical side — both produce a 4-20 mA current and both work with a precision shunt. What changes is when you should invest in an active converter. Use one of these in these four situations:

• You need a true 0-10 V span, not 2-10 V. An active converter scales and offsets the output, so 4 mA = 0 V exactly and 20 mA = 10 V exactly.
• You need galvanic isolation between the sensor loop and the PLC. A resistor provides no isolation; a converter with 1500 V isolation protects the PLC from ground loops and surge events.
• You need a high-impedance output for a long voltage cable run. A resistor-derived voltage has the same source impedance as the resistor (e.g. 500 Ω), which picks up noise on long runs. An active converter outputs a low-impedance voltage.
• The loop has multiple devices on it. Each added shunt drops more voltage and eats into the compliance voltage of the 4-20 mA source. A converter that loops through without consuming loop voltage preserves the budget.

For single-sensor short-run applications with a PLC that accepts 2-10 V (or can be rescaled in software), a resistor is fine and saves the cost of a converter. For anything beyond that — multi-drop, long runs, isolation-required, or true 0-10 V needed — buy the converter.

How Do You Wire a 4-20 mA Sensor to a 0-10 V PLC Input?

Two wiring patterns cover almost all cases. The simple resistor drop method:

1. Confirm the 4-20 mA source type. Two-wire (loop-powered) sensors get their 24 V DC from the same two wires that carry the signal. Three-wire sensors have separate supply and signal.
2. Bring the 24 V supply positive to the transmitter +. Connect the transmitter signal output to the PLC analog input positive (+).
3. Connect the PLC analog input negative (-) back to the 24 V supply negative. This completes the loop.
4. Install the precision resistor across the PLC analog input terminals (+ and -). 500 Ω for 0-10 V card, 250 Ω for 0-5 V card.
5. Check polarity with a multimeter in series before energizing. Current should flow from 24 V+ through the transmitter, into PLC+, through the resistor, out PLC- and back to 24 V-.

For troubleshooting a finished loop, measure voltage across the resistor with a handheld DMM. A stable reading between 2 V and 10 V means the loop is healthy; 0 V means open circuit (broken wire, loose terminal); above 10 V means the 20 mA limit has been exceeded or the resistor is open.

What Are the Common Mistakes in 4-20 mA Voltage Conversion?

Three mistakes account for most failed installations:

• Forgetting to scale 2-10 V back to 0-100% in the PLC. After the resistor, the input reads 20% at minimum, not 0%. Update the PLC analog scaling so 2 V = 0% and 10 V = 100%.
• Using a low-tolerance resistor. A 5% resistor contributes 5% of full scale to the error budget — more than the transmitter itself. Use 0.1% metal-film or wire-wound resistors.
• Exceeding the loop compliance voltage. A 24 V supply with 500 Ω shunt and 200 Ω wiring leaves only ~14 V of compliance for the transmitter. A HART transmitter needs 10-12 V minimum at its terminals; anything less causes the loop to drop out under noise.

Check the loop budget before ordering parts: add the source voltage drop, wiring resistance, precision shunt resistance, and any barrier or protector resistance. Subtract that total voltage drop at 20 mA from the supply voltage. The remainder is what the transmitter sees. Many pressure transducer output signal types specify minimum and maximum loop resistance on the datasheet — respect those limits.

Related Products

FAQ

Can I use a 500 Ω resistor to get a full 0-10 V signal?

A 500 Ω shunt converts 4-20 mA to 2-10 V, not 0-10 V. The 2 V offset is the “live zero” of the 4-20 mA standard. Either rescale in the PLC software or use an active converter with zero offset.

What resistor value converts 4-20 mA to 1-5 V?

250 Ω, 0.1% tolerance. This is the standard HART-compatible voltage span.

Do I need galvanic isolation when converting 4-20 mA to voltage?

Required for long sensor cables, multi-rack installations, or hazardous-area boundaries. Not required for a single sensor on a short cable feeding one PLC within the same cabinet.

Will adding a 500 Ω resistor damage my 4-20 mA transmitter?

No, as long as the total loop resistance stays within the transmitter’s compliance voltage. Most loop-powered transmitters with a 24 V supply handle up to 650 Ω total loop resistance. Above that, the loop starts to saturate and the 20 mA output drops below specification.

How do I check a 4-20 mA to voltage conversion is working?

Measure voltage across the shunt resistor with a handheld DMM. At known 4 mA calibration input, expect the minimum voltage (2 V for 500 Ω shunt). At 20 mA, expect full scale (10 V). Values outside that range point to a bad resistor, open wiring, or a misfiring source.

Need a transmitter or recorder that fits directly into your existing signal scheme? Tell us the PLC model, input card range, and loop topology — we’ll match the right instrument on the first try.

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