The 4–20 mA current loop is a simple, reliable way to transmit measurement data from a sensor to an indicator, controller or recorder—even across long cable distances.
How a 4–20 mA loop works
The loop uses a variable current signal to represent the measurement transmitted by an instrument. Unlike a voltage signal, which can be affected by cable resistance and voltage drop, a current signal maintains its accuracy over long cable runs with minimal signal degradation.
A typical loop consists of:
- A TD100 transmitter
- A receiver such as a PLC, meter, controller or recorder
- A two-wire cable connecting the devices
One wire supplies power to the loop while the other carries the current signal generated by the transmitter. The transmitter adjusts the current according to the level measured by the probe.
Understanding zero and span
The standard uses a live zero: the signal begins at 4 mA instead of 0 mA.
Full scale = 20 mA
20 mA − 4 mA = 16 mA span
A live zero allows the receiver to distinguish a valid zero reading from a broken wire, provides a method for signalling faults and improves troubleshooting reliability.
For the most accurate span, set the TD100’s 4 mA point at the bottom of the compartment and its 20 mA point at the top of the compartment.
Signal conversion at the receiver
The loop’s return current connects to a specialized receiver input. Inside the receiver, a sensing resistor converts the current into a proportional voltage signal. The PLC, meter, display or recorder measures that voltage to determine the process value.
Designing a reliable system
The key design requirement is sufficient operating voltage for both the receiver and the transmitter. Before installation, determine the receiver’s analog-input requirements by consulting its specifications or measuring input resistance with a multimeter.
Receiver input resistance
| Analog input | Input resistance | Calculation |
|---|---|---|
| 10 V input | 500 Ω | 10 V ÷ 20 mA |
| 5 V input | 250 Ω | 5 V ÷ 20 mA |
TD100 power-supply requirements
When power is supplied to the 4–20 mA module through Pin 4 of the TD100, provide approximately 5 V more than the receiver input requires. This additional voltage margin is called headroom.
PLC input+5 V
TD100 headroom=15 V
minimum supply
If only 12 V is supplied in this example, the transmitter may not have enough voltage to produce the full 20 mA output and the loop may not reach the upper end of its measurement range.
A higher-voltage power supply may therefore be needed at Pin 4. Titan cable harnesses include additional conductors for installations that require extra supply voltage.
Measuring analog-input resistance
Use a multimeter set to ohms to measure resistance between the analog-input terminal and ground. The measurement helps you:
- Verify the receiver specifications.
- Calculate the operating-voltage requirements.
- Determine whether an additional supply is needed.
- Confirm compatibility with the vehicle’s battery voltage.
Advantages of a 5 V PLC input
Configuring the PLC for a 5 V analog input rather than a 10 V input can reduce operating-voltage requirements, simplify installation and improve compatibility with vehicle electrical systems.
