Why Use 4-20mA Current for Signal Transmission?
The 4-20mA current signal is a widely used method for transmitting analog signals in industrial field environments. It offers advantages such as strong anti-interference capability, long transmission distance, and two-wire power supply. It is extensively applied in various industrial measurement and control systems for parameters like temperature, pressure, and liquid level.
How the 4-20mA Current Signal Works
The 4-20mA current signal transmission system primarily consists of the following components:
l Transmitter: A device that converts physical quantities such as temperature, humidity, and pressure into electrical signals that are easy to measure and process. This type of device is called a transmitter. The most common method in industry is using a 4-20mA current to transmit analog signals.
l Receiver: Converts the 4-20mA current signal into a voltage signal (or other forms) proportional to the measured physical quantity.
l Connection Wires: Cables used to transmit the 4-20mA current signal.
The transmitter is the core component of the 4-20mA system. It outputs a 4-20mA current signal proportional to changes in the measured physical quantity. For example, when the measured temperature increases, the output current from the transmitter will gradually increase from 4mA to 20mA.
The receiver then converts the 4-20mA current signal from the transmitter into a voltage signal (or other forms) proportional to the measured quantity. For instance, when the receiver gets a 4mA current, the output voltage might be 0V; when it receives 20mA, the output voltage might be 10V.
The connection wires transmit the 4-20mA current signal. Because the 4-20mA signal is relatively insensitive to changes in wire resistance, ordinary twisted-pair cables can be used for transmission, with distances reaching hundreds of meters or even several kilometers.
Why Use 4-20mA Current for Signal Transmission?
The primary reason for using a current signal is its strong anti-interference capability. This is because:
l Current signals are less susceptible to interference: Compared to voltage signals, current signals are less affected by changes in wire resistance.
l High internal impedance of the source: A current source has very high internal impedance. Even if the loop cable has resistance, it does not significantly affect measurement accuracy.
l Long transmission distance: Ordinary twisted-pair cables can transmit signals for hundreds of meters.
l 20mA upper limit relates to explosion-proof requirements: The spark energy generated by switching a 20mA current is insufficient to ignite explosive gases, meeting intrinsic safety needs.
l 4mA lower limit (not 0mA) facilitates open-circuit detection: The current will not fall below 4mA during normal operation. If a line fault or break occurs, the loop current drops to 0.
2mA is often used as an open-circuit alarm threshold for two reasons:
l To avoid interference, setting a non-zero lower limit prevents low-current noise from being mistaken for an open circuit.
l It supports two-wire power supply. The 4-20mA current loop uses two wires that simultaneously carry the signal and provide power to the sensor. The 4mA minimum current provides the sensor's quiescent operating current.
The theoretical transmission distance for a 4-20mA sensor depends on several factors:
1.Excitation Voltage (Vo): The voltage powering the transmitter. A higher voltage allows for longer distances, as it can accommodate the voltage drop across the wires while maintaining the 4-20mA range.
2.Minimum Working Voltage of the Transmitter (Umin): The minimum voltage required for the transmitter to operate correctly. System design must ensure the voltage at the transmitter terminals is above this value.
3.Sampling Resistor (R): The resistor in the receiver circuit that converts the current signal to a voltage. A larger value creates a larger voltage drop but also consumes more voltage from the loop, affecting distance.
4.Wire Resistance (Rwire): The resistance of the signal cable. Longer cables mean higher resistance and greater voltage drop.
Theoretical Calculation
We can calculate the theoretical maximum transmission distance based on these four factors. Assuming we know the following information:
Excitation Voltage: Vo
Minimum Working Voltage of the Transmitter: Umin
Sampling Resistor (Voltage Taking Resistor): R
Wire Resistance per meter: Rwire
According to Ohm's Law, the voltage drop in the circuit equals the current multiplied by the sum of the resistances. The total circuit resistance includes the sampling resistor, wire resistance, and the transmitter's own internal resistance (Rs).
Total voltage drop = Vo - Umin = (R + Rwire * Distance) * I (where I is the current, either 4mA or 20mA).
From this equation, we can solve for Distance:
Distance = (Vo - Umin) / (I * (R + Rs + Rwire * Distance))
Note:
Since the wire resistance increases with distance, this is a nonlinear equation. Accurate results require iterative calculations.
The transmitter's internal resistance (Rs) is typically a known parameter and can be obtained from the sensor's specification sheet.
The above formula is intended for theoretical calculations. In practical applications, factors such as safety margins and signal attenuation must also be considered.
Practical Application
The actual transmission distance of 4-20mA sensors is typically 200 to 500 meters due to various influencing factors. To extend the transmission distance, cables with a larger cross-sectional area can be used to reduce resistance, or signal amplifiers can be employed to compensate for signal attenuation.
Figure 1: Two-Wire Transmitter Current Signal Transmission Circuit
Figure 1 illustrates the basic structure of a 4-20mA current signal transmission circuit. Here, Uo is the power supply voltage for the transmitter, and under full load current (I = 20mA), it must be ensured that Uo ≥ Umin.
Calculating Maximum Cable Resistance
We can calculate the maximum allowable cable resistance when the transmitter operates at its minimum working voltage using the following formula:
l Uo - Umin = (RL + 2r) × I (Where:
n Uo: Excitation Voltage
n Umin: Minimum Working Voltage of the Transmitter
n RL: Sampling Resistor
n r: Cable Resistance
n I: Current, i.e., 4mA or 20mA)
Example Calculation
Assume the following known information:
l Ue = 24V (Supply Voltage)
l I = 20mA (Current)
l RL = 250Ω (Sampling Resistor)
l Umin = 12V (Minimum Working Voltage of the Transmitter)
Substitute into the formula:
r (maximum cable resistance per wire) = [(24V - 12V) / (20mA) - 250Ω] / 2 = 175Ω
Relationship Between Cable Resistance and Length
According to the cable resistance formula:
r = ρ × L / S
Where:
ρ: Resistivity (Copper = 0.017 Ω·mm²/m, Aluminum = 0.029 Ω·mm²/m)
L: Cable Length (in meters)
S: Cross-Sectional Area (in square millimeters)
Note:
Resistance is directly proportional to cable length and inversely proportional to cross-sectional area. Longer cables have higher resistance, while thicker cables have lower resistance.
Example
Taking copper cable as an example with ρ = 0.017 Ω·mm²/m, a 1-meter copper cable with a cross-sectional area of 1 mm² has a resistance of 0.017Ω.For a resistance of 175Ω, the length of a 1 mm² copper cable is:175Ω / 0.017 Ω/m ≈ 10,294 meters.
Theoretical Transmission Distance
Therefore, theoretically, 4-20mA current signal transmission can achieve distances of tens of thousands of meters (depending on factors such as the operating voltage and the minimum working voltage of the transmitter).
Disadvantages of DC Voltage Signals
The national standard "GB/T 3369.2-2008 Industrial Process Control System Analog Signals - Part 2: DC Voltage Signals" stipulates: "Compared with the analog DC current signals specified in GB/T 3369.1-2008, the analog DC voltage signals specified in this part are not intended for long-distance transmission."
The main reasons are:
l DC voltage signals attenuate after long-distance transmission.
l DC voltage signals are more susceptible to interference.
In contrast, the 4-20mA current signal, being unaffected by variations in wire resistance, offers stronger anti-interference capabilities and is more suitable for long-distance transmission.
Summary
The 4-20mA current signal is a simple, reliable, and economical method for analog signal transmission. Its advantages—strong anti-interference capability, long transmission distance, and two-wire power supply—have made it widely adopted in fields such as industrial control, building automation, and medical equipment.
