A two-wire pressure transmitter is supplied with 24 V DC and sends its measurement signal to a 4–20 mA analog input of the PLC. The measurement works reliably. A local digital display is then to be added directly to the current loop. Since the display does not require a separate power supply, it appears that it can simply be connected in series. Afterwards, however, the measured value only reaches approximately 18 mA even though the process should actually correspond to the upper end of the measuring range.
The cause may be the available loop voltage. Although a loop-powered display does not require an additional supply line, it obtains its operating power from the 4–20 mA loop. This requires a certain voltage. In addition, there is the minimum operating voltage of the two-wire transmitter, the voltage drop across the PLC input, cable losses and, if applicable, other devices in the loop. If the supply voltage is no longer sufficient for the sum of these voltage requirements, the transmitter may no longer be able to drive the required loop current completely.
The decisive factor is therefore not only the nominal supply voltage of, for example, 24 V DC, but the voltage that actually remains available for the transmitter after subtracting all voltage drops within the loop. This becomes particularly critical with long cables, high input resistances, additional displays or isolating modules, and supply voltages operating near the lower end of their tolerance range.
A 4–20 mA current loop should therefore be considered as a voltage budget: supply voltage minus the minimum voltage required by the transmitter minus the voltage drops across the display, analog input and wiring. Only if sufficient reserve remains can the maximum loop current be reached reliably.
Why does a 4–20 mA loop require voltage at all?
With a two-wire transmitter, power supply and measurement signal are transmitted over the same two conductors. The power supply provides the loop voltage, while the transmitter regulates the current according to the process value, typically between 4 and 20 mA. In order for this current regulation to work, the transmitter electronics require a certain minimum voltage at their own terminals.
The remaining supply voltage is available to all other components connected in series. These may include a PLC analog input, a local digital display, a safety barrier or isolating module, as well as the resistance of the connecting cables. Since all components are connected in series, the same current flows through all of them. The individual voltage drops are added together.
The required supply voltage can therefore be simplified as follows:
Usupply ≥ Utransmitter + Udisplay + Uinput + Ucable + Ureserve
The key question is therefore not “Does the power supply provide 24 V?” but “How much of these 24 V remains at the transmitter after all other voltage drops have been subtracted?”
What does load mean in a 4–20 mA signal loop?
In a current signal loop, the load is the electrical resistance that the current output must drive. A PLC analog input may, for example, have an input resistance of 250 Ω. The voltage drop across this load is calculated using Ohm’s law:
U = I × R
At 20 mA and 250 Ω, the result is:
0.020 A × 250 Ω = 5.0 V
| Load | Voltage drop at 4 mA | Voltage drop at 20 mA |
|---|---|---|
100 Ω |
0.4 V |
2.0 V |
250 Ω |
1.0 V |
5.0 V |
500 Ω |
2.0 V |
10.0 V |
600 Ω |
2.4 V |
12.0 V |
This also explains why a current loop may work correctly at low current values and only show problems near the upper end of the measuring range. The voltage required by an ohmic resistance increases proportionally with the loop current. A 250 Ω load requires only 1 V at 4 mA, but already 5 V at 20 mA.
Why does a loop-powered display require voltage?
A loop-powered digital display has no separate 24 V auxiliary supply. It is connected directly in series in the current loop and powers its electronics from the current flowing through the loop. To do this, the device requires a certain voltage drop between its input and output terminals.
This voltage drop must be included in the voltage budget in the same way as the voltage drop across a resistor. If a display requires, for example, 5 V, these 5 V are no longer available for the transmitter, analog input and wiring. The electrical advantage is that no separate supply line is required for the display. However, the loop voltage must be dimensioned accordingly.
When planning the loop, a distinction should be made between an input specified as a resistance and a device with a specified constant or typical voltage drop. An analog input with 250 Ω is calculated using U = I × R. For a loop-powered display, on the other hand, the voltage drop specified in the data sheet is entered directly into the calculation.
Systematically calculating the loop voltage
For a typical current loop with a 24 V supply, two-wire transmitter, local display and PLC input, each voltage contribution can first be determined individually. All values are then added together and compared with the supply voltage actually available.
| Component | Example value | Voltage requirement at 20 mA |
|---|---|---|
| Two-wire transmitter | Minimum operating voltage | 10.5 V |
| Loop-powered display | Voltage drop | 5.0 V |
| PLC analog input | 250 Ω |
5.0 V |
| Wiring | Total loop resistance | 0.7 V |
| Total | 21.2 V |
With an exact supply voltage of 24.0 V, this example leaves 2.8 V of reserve. That may initially appear sufficient. However, if power supply tolerance, voltage drop at the end of a long supply chain or a possible fault current above 20 mA are taken into account, this reserve can quickly disappear.
Why should the calculation be based on maximum current?
For load calculations, the worst-case intended loop current is decisive. Anyone calculating only with 20 mA should check whether the transmitter may deliver, for example, 21.5 or 22 mA in the event of a fault. An ohmic input produces a greater voltage drop at 22 mA than at 20 mA.
For an input with 250 Ω, the voltage drop at 22 mA is:
0.022 A × 250 Ω = 5.5 V
Cable losses also increase proportionally with current. Especially in tightly dimensioned loops, a measured value up to 20 mA may still work, while a defined fault current can no longer be reached correctly. For safety-related or diagnostic applications, the design should therefore be based on the actual maximum intended loop current.
Correctly accounting for cable losses
Every cable has electrical resistance. Since a current loop consists of an outgoing and a return conductor, the total electrical cable length must be considered when calculating resistance. With a distance of 500 m between the control cabinet and transmitter, for example, the current flows through approximately 1,000 m of copper conductor in total.
The cable resistance can be approximated using:
Rcable = ρ × l / A
For copper, a resistivity of approximately ρ ≈ 0.0178 Ω mm²/m can be used. With a total cable length of 1,000 m and a conductor cross-section of 0.5 mm², the result is approximately:
Rcable = 0.0178 × 1000 / 0.5 ≈ 35.6 Ω
At 20 mA, the voltage loss is approximately:
0.020 A × 35.6 Ω ≈ 0.71 V
With short cables, this value is often small. However, in large plants, with small conductor cross-sections or several terminal and connection points, it can become decisive for a current loop that already has little voltage reserve.
Taking supply voltage and tolerance into account
A nominal 24 V supply does not automatically mean that exactly 24.0 V is available to the current loop under all operating conditions. Power supplies have a specified output range, and additional voltage drops may occur across fuses, supply cables, distribution terminals or decoupling elements.
For a robust design, the calculation should therefore use the lowest voltage that can occur at the beginning of the actual current loop. If, for example, only 21.6 V is available under worst-case operating conditions, the previously comfortable-looking voltage budget becomes much tighter.
A good design therefore deliberately includes reserve. This reserve is not intended to compensate for an incorrect setup, but to absorb component tolerances, supply fluctuations, cable losses and small additional contact resistances.
Practical example: 24 V loop with display and PLC
According to the design data, a pressure transmitter requires at least 10.5 V at its terminals. A local loop-powered display requires another 5 V. The PLC input has an input resistance of 250 Ω, and the long connecting cable produces a voltage drop of approximately 0.7 V at 20 mA.
At 20 mA, the total voltage requirement is:
10.5 V + 5.0 V + 5.0 V + 0.7 V = 21.2 V
With an actual supply voltage of 24 V, this leaves 2.8 V of reserve. However, if the actual available supply drops to 21.6 V, only 0.4 V remains. If the transmitter is expected to output 22 mA in the event of a fault, the voltage drop across the 250 Ω input additionally increases to 5.5 V, and the cable loss also rises slightly.
The loop can then reach its operating limit. The transmitter no longer has sufficient terminal voltage to regulate the required current. Instead of, for example, 22 mA, only 19 or 20 mA may be reached. The actual problem is then neither the scaling nor the PLC, but insufficient voltage reserve in the current loop.
This is exactly why an additional loop-powered display should not simply be added retrospectively to an existing 4–20 mA loop without recalculating the voltage budget.
How does excessive total load become noticeable?
Excessive total load often does not result in a complete failure. At low output currents, an ohmic load requires little voltage and the measuring point appears to work normally. Only as the process value increases does the voltage requirement rise until the current output reaches its regulation limit.
| Observation | Possible cause | Useful check |
|---|---|---|
| 4 mA is correct, but 20 mA cannot be reached | Insufficient loop voltage or excessive load | Measure the transmitter terminal voltage at high current |
| Fault occurs only after installing a display | Additional voltage drop caused by the display | Compare voltage budget before and after installation |
| Fault occurs only with long cables | Cable resistance reduces voltage reserve | Calculate loop resistance and voltage drop |
| Measurement works at 24 V but not at a lower supply voltage | Insufficient voltage reserve | Calculate using the minimum permissible supply voltage |
| Signal stops increasing near the top of the range | Transmitter reaches its compliance limit | Measure the voltage directly at the transmitter |
What must be considered for HART communication?
A HART current loop has additional requirements. Suitable loop impedance must be available for communication. At the same time, connected displays, isolators or other devices must not excessively attenuate the communication signal superimposed on the 4–20 mA signal.
A local display used in a HART application should therefore be explicitly suitable for this operating mode or be HART-transparent. Any additional communication resistor that may be required also increases the loop load and must be included in the voltage budget.
The same basic principle therefore remains decisive: every additional component connected in series may require voltage. Communication capability and voltage budget should therefore be planned together.
Systematically designing a current loop
- Determine the actual available supply voltage: Do not use only the nominal value of the power supply.
- Determine the minimum operating voltage of the two-wire transmitter: The specific device version is decisive.
- Add the voltage drops of all loop-powered devices: Include displays, isolators, barriers and other components.
- Calculate ohmic loads using the maximum intended loop current: Include analog inputs and additional resistors.
- Take outgoing and return conductors into account: Calculate cable resistance from length and cross-section.
- Allow for supply tolerances and a reasonable reserve.
- After installation, check the system at high loop current: Measure the voltage directly at the transmitter terminals.
Common mistakes
- Treating a loop-powered display as if it had no voltage requirement: It does not require a separate auxiliary supply, but it does cause a voltage drop in the loop.
- Testing only at 4 mA: Many load-related problems occur only at 20 mA or at the maximum fault current.
- Using nominal 24 V without verification: The actual voltage at the loop connection may be lower.
- Considering only the PLC input as the load: Wiring, displays, isolators and barriers must also be included in the voltage budget.
- Calculating cable length only one way: The loop current flows through both outgoing and return conductors.
- Adding devices in series afterwards without recalculation: Every additional component reduces the voltage remaining for the transmitter.
- Allowing no reserve: A loop that works only exactly on paper is unnecessarily susceptible to real-world tolerances.
Loop-powered displays and power supplies
Different designs are available for local indication directly within a 4–20 mA current loop. Loop-powered field displays are particularly suitable where no separate auxiliary power is available at the installation location. For control cabinet applications, compact panel meters are available that can likewise be connected directly in series in the 4–20 mA loop.
When selecting a display, attention should not be paid only to display size and scaling. For the electrical design of the current loop, the voltage drop required by the display is particularly important. This value must be calculated together with the minimum voltage of the transmitter, the input load of the evaluation system and the cable losses against the available supply voltage.
Suitable power supplies and additional components can be found under Power supply equipment at ICS Schneider. Loop-powered digital displays and displays suitable for 4–20 mA signals are also available in the electrical display and control cabinet instrumentation range.
Conclusion
A loop-powered 4–20 mA display does not require a separate supply line, but it still requires electrical power. This power is taken from the current loop and results in an additional voltage drop. This value must be included in the loop design.
The available supply voltage must at least be sufficient to cover the minimum operating voltage of the transmitter, the voltage drop across the display, the load of the analog input and the cable loss at the same time. For ohmic loads, the required voltage increases with loop current, which means that the most critical condition is typically at the upper end of the current range.
It is also particularly important not to calculate using only the nominal 24 V supply. Supply tolerances, long cables and additional devices connected in series reduce the actual reserve. A tightly designed loop can therefore work perfectly at 4 or 12 mA and only reach its voltage limit at 20 mA or at a fault current.
For a reliable 4–20 mA current loop, the following therefore applies: determine every voltage drop individually, calculate ohmic loads at the maximum intended current, take cable losses and minimum supply voltage into account, and provide sufficient voltage reserve for the transmitter.
FAQ: Load and voltage in the 4–20 mA current loop
What does load mean in a 4–20 mA signal loop?
The load is the electrical resistance against which a current output must drive the loop current. With a 250 Ω input, for example, a voltage drop of 5 V occurs at 20 mA.
Why does a loop-powered display not require its own power supply?
It obtains the required energy directly from the 4–20 mA current loop. However, this causes a voltage drop across the display that must be taken into account when designing the loop.
Can an additional display prevent the loop from reaching 20 mA?
Yes. If the display leaves insufficient voltage available for the transmitter, the transmitter reaches its regulation limit and can no longer provide the full required loop current.
How do you calculate the required loop voltage?
Add together the minimum voltage required by the transmitter, the voltage drops across all devices connected in series, the voltage across ohmic loads and the cable losses. An appropriate voltage reserve should also be included.
How much voltage drops across 250 Ω at 20 mA?
According to U = I × R, the result is 0.020 A × 250 Ω = 5 V.
Why should the load be calculated at the maximum loop current?
The voltage drop across ohmic resistances increases proportionally with current. The highest demand on the voltage budget therefore normally occurs at the maximum intended loop current.
Does cable resistance matter in a 4–20 mA loop?
Yes. Because the current is relatively low, the losses are often moderate, but with long cables and small conductor cross-sections they can consume a relevant portion of the available loop voltage.
Why does a current loop sometimes work at 4 mA but not at 20 mA?
For ohmic loads, the voltage drop at 20 mA is five times greater than at 4 mA. A tightly dimensioned loop can therefore function correctly at low measured values and only reach its voltage limit at the upper end of the measuring range.
How can insufficient loop voltage be identified?
Typical indications include a current signal that does not reach its upper target value or a fault that occurs only at high process values. Measuring the actual voltage directly at the transmitter terminals while the loop current is high is particularly informative.
What else must be considered in a HART loop?
In addition to the voltage budget, a suitable loop impedance must be available for communication. Devices connected in series must also be compatible with the HART signal or sufficiently transparent to it.