Why is 4-20mA chosen for transmitter signals?

There are two possibilities for 0MA: 1: The signal line is open; 2: The signal value is 0. Note that input/output signals in industrial automation instruments are usually DC signals, and the advantages of this are as follows: 1. During the transmission of signals in these instruments, DC signals are not affected by alternating current interference, which makes it easier to address issues related to interference resistance. 2. DC signals are not affected by the inductance and capacitance of the transmission lines, and there is no issue with phase shift, thus simplifying the wiring. 3. DC signals are easy to convert between analog and digital forms, which enables instruments to be connected easily to data processing devices, computers, etc. 4. It is easy to obtain a reference voltage for DC signals, such as the setpoint of a regulator.

The signal from the field controller to the transmitter is 4-20mA, which is achieved using a two-wire system. This is because the transmitter also requires a operating current even when it is in the zero-point range; 24VDC over two wires meets the operating requirements of the transmitter, and this is an international standard.

Our country began to adopt the 4–20 mA DC standard with DDCIII-type instruments； Early Type I and Type II instruments used the 0–10 mV DC signal standard.

For transmitters that output 4–20mA, circuit tuning and data processing are both quite complicated. However, this type of transmitter can determine whether there is a fault in the circuit or whether the transmitter is damaged by checking whether a current within the normal range is present (the minimum value under normal conditions is 4 mA), even when the transmitter circuit is disconnected, short-circuited, or damaged, which is why it is more widely used. PS: Most early transmitters were of the voltage-output type, that is, they converted the measured signal into a 0-5V voltage output. However, in situations where the signal needed to be transmitted over long distances or where there was significant electrical interference in the surrounding environment, the use of voltage-output sensors was greatly limited; their weaknesses included poor resistance to interference, and line losses that affected accuracy. In contrast, two-wire current-output transmitters boast extremely high resistance to interference.

The 4–20mA signal has a long transmission distance and low loss.

4~20mA also makes it possible to supply power to two-wire instruments. If it is 0mA, where does the instrument get its power?

It seems to be mainly to distinguish between the 0 bit and a power-off fault, right?

Why is the signal from ordinary transmitters 4 to 20 milliamps? 1. In industry, it is necessary to measure various non-electrical physical quantities such as temperature, pressure, speed, angle, etc.; these values must be converted into analog electrical signals in order to be transmitted to control rooms or display devices located hundreds of meters away. A device that converts physical quantities into electrical signals is called a transmitter. In industry, 4–20mA current is most commonly used to transmit analog signals. The reason for using a current signal is that it is less susceptible to interference. Furthermore, the internal resistance of the current source is infinite; the resistance of the wires in the circuit does not affect accuracy, and transmission over several hundred meters is possible using ordinary twisted pairs. The upper limit of 20mA is set due to explosion-proof requirements: the spark energy generated by the on-off of a 20mA current is not sufficient to ignite gas. The lower limit is not set to 0 mA in order to enable detection of broken connections: under normal operation, the value will be above 4 mA, but when the transmission line breaks due to a fault, the loop current drops to 0. 2mA is commonly used as the wire-break alarm value. Current-type transmitters convert physical quantities into 4–20mA current outputs, and therefore require an external power supply to operate. The most typical case is that a transmitter requires two power wires and two current output wires, for a total of 4 wires, which is what is known as a four-wire transmitter. Of course, the current output can share a wire with the power supply (shared VCC or GND), which saves one wire; such transmitters are known as three-wire transmitters. In fact, as everyone may have noticed, the 4-20mA current can itself power the transmitter. In a circuit, the transmitter functions as a special load, with the distinction being that its current consumption ranges from 4 to 20 mA and varies depending on the sensor output. The display meter only needs to be connected in series in the circuit. This type of transmitter requires only 2 external wires, which is why it is called a two-wire transmitter. The lower limit for industrial current loop standards is 4 mA; therefore, as long as it is within the range, the transmitter will have at least 4 mA of power supply. This makes the design of two-wire sensors possible. In industrial applications, the measurement points are usually located on-site, while the display devices or control devices are generally found in the control room or control cabinet. The distance between the two may be several dozen to several hundred meters. Calculated on a 100-meter basis, eliminating 2 wires means a cost reduction of nearly 100 yuan! Therefore, two-wire sensors are inevitably the preferred choice in applications. 2. Structure and principle of two-wire transmitters: The principle of two-wire transmitters relies on the 4–20mA signal to supply power to itself. If the transmitter’s own power consumption is greater than 4mA, it will be impossible to output a lower limit value of 4mA. Therefore, it is generally required that the power consumption of a two-wire transmitter (all circuits including the sensor) be no more than 3.5 mA. This is one of the fundamental design principles of two-wire transmitters. In terms of its overall structure, a two-wire transmitter consists of three main components: the sensor, the conditioning circuit, and the two-wire V/I converter. Sensors convert physical quantities such as temperature and pressure into electrical parameters, while conditioning circuits amplify and adjust the weak or nonlinear electrical signals generated by the sensors, converting them into linear voltage outputs. The two-wire V/I conversion circuit controls the total power consumption current based on the output of the signal conditioning circuit ； At the same time, voltage is obtained from the loop and regulated for use by the conditioning circuit and sensors. In addition to the V/I conversion circuit, each component in the circuit has its own current consumption. The core design principle of a two-wire transmitter is to include all currents within the feedback loop of the V/I conversion. As shown in the figure, the sampling resistor Rs is connected in series at the lower end of the circuit, and all current flows back to the negative terminal of the power supply through Rs. The feedback signal obtained from Rs includes the power consumption of all circuits. In two-wire transmitters, the total power consumption of all circuits must not exceed 3.5 mA; therefore, low power consumption of the circuits represents a major design challenge. Next, the principles and key design aspects of each individual circuit component will be analyzed one by one. 3. Two-wire V/I converter A V/I converter is a circuit that can control the output current using a voltage signal. The difference between a two-wire V/I converter and a conventional V/I conversion circuit is that the voltage signal does not directly control the output current, but rather controls the current consumed by the circuit itself. At the same time, a stable voltage must also be extracted from the current loop to power the conditioning circuit and sensors. The attached figure is the basic schematic diagram of a two-wire V/I conversion circuit: in the diagram, OP1, Q1, R1, R2, and Rs constitute the V/I converter. Analyze the negative feedback process: If point A is above 0V for some reason, the output of op-amp OP1 increases, the voltage across Re rises, and the current flowing through Re becomes larger. This results in an increased overall power consumption; the current flowing through the sampling resistor Rs also increases, causing the voltage at point B to drop (to a more negative value). The result is that the voltage at point A is reduced through R2. Conversely, if point A falls below 0V for some reason, it will also be raised back to 0V by negative feedback. In short, the result of negative feedback is that the op-amp OP1 is in a virtual short circuit, and the voltage at point A = 0V. The following analyzes the principle by which Vo controls the total power consumption: Assuming that the output voltage of the conditioning circuit is Vo, the current flowing through R1 is I1 = Vo/R1. The input terminals of the operational amplifier cannot draw current; therefore, all of I1 flows through R2. The voltage at point B is then VB = -I1*R2 = -Vo*R2/R1. When R1 = R2, VB = -Vo. Between the negative terminal of the power supply and the entire transmitter circuit, there are only two resistors, Rs and R2; thus, all of the current flows through Rs and R2. The upper end of R2 is virtual ground (0V), and the upper end of Rs is GND. Therefore, the voltages at both ends of R2 and Rs are exactly the same, both equal to VB. It is equivalent to Rs and R2 being connected in parallel as the current sampling resistors. Therefore, the total current in the circuit is: Is = Vo/(Rs//R2). If R2 >> Rs, then Is = Vo/Rs. In Figure 3, Rs is set to 100 ohms; when the output of the conditioning circuit ranges from 0.4 to 2 V, the total current consumption is between 4 and 20 mA. It’s not a problem if R2 >> Rs is not satisfied, as the value of Rs//R2 remains constant when Rs and R2 are connected in parallel, and there is still a linear relationship between Is and Vo. The error coefficient can be eliminated during calibration. In addition to having a correct circuit design, for this circuit to function properly two additional conditions are required: first, its own power consumption must be as low as possible, with the saved current being supplied to the conditioning circuit and the transmitter. Secondly, the op-amp is required to be able to operate with a single power supply; that is, it should still be able to accept an input of 0V at its inputs in the absence of a negative power supply, and function properly. The LM358/324 are the most common and cheapest single-supply op-amps, consuming 400 uA per op-amp, which is fairly acceptable. When powered by a single supply, the input can operate properly within the range of -0.3V to Vcc-1.5V. If precision amplifiers such as the OP07 are used, they cannot operate in this circuit because the input voltage is not allowed to drop to 0V. R5 and U1 form a reference source that generates a stable 2.5V reference voltage. The LM385 is a low-cost, low-power reference; it can operate with a current of 20 uA or more. The curve provided in the manual is flattest around 100 uA, so the current is controlled to around 100 uA using R5. OP2 forms a non-inverting amplifier that amplifies the reference voltage to power the conditioning circuit and the sensor. Because regulators with a wide input voltage and low power consumption are rare, their cost is high ； Using a reference amplifier as a voltage regulator is an inexpensive solution. For this part of the circuit, off-the-shelf integrated circuits can also be used. Models such as XTR115/116/105, for example, offer better precision and stability compared to those built manually, and they also consume less power (which means more current can be allocated to the conditioning circuit, making it easier to design that part). However, the cost is more than 10 times higher than that of the aforementioned solution. 4. Design of two-wire pressure transmitters: The output signals from pressure bridges and load cells are weak, belonging to the mV range. Such small signals generally require a differential amplifier for the first stage of amplification. Generally, differential amplifiers with low offset and low drift are selected. Additionally, low power consumption is also essential in two-wire applications. The AD623 is a commonly used low-power precision differential amplifier, often employed as an amplification stage before differential output. The AD623 has an offset of up to 200 uV and a temperature drift of 1 uV/degree, which ensures sufficient accuracy for typical pressure transduction applications. R0 applies a voltage of 0.4V to the REF pin (pin 5) of the AD623; with zero pressure, R0 is adjusted to make the output 4mA, and then RG is adjusted to achieve an output of 20.00mA, thereby completing the calibration. When designing the circuit, it should be noted that the pressure bridge sensor functions as a resistor of around 1 kΩ, and it generally consumes a relatively high amount of power. Appropriately reducing the excitation voltage of the pressure bridge can decrease the power consumption current. However, the output amplitude also decreases, requiring an increase in the gain of the AD623. The sensor shown in Figure 6 is powered with a constant voltage; in practical applications, most semiconductor pressure sensors require a constant current supply in order to exhibit good temperature characteristics, and an operational amplifier can be used to create a constant current source to provide the necessary drive signal. 5. Considerations for stability and safety: Industrial environments are characterized by harsh conditions and high requirements for reliability; therefore, the design of two-wire transmitters must include certain measures to ensure protection and enhance stability. 1. Power protection. Reverse power connection, overvoltage, and surges are common power supply problems in industry. Connecting the power supply in reverse is the most common error that occurs during device installation and wiring; connecting a diode across the input terminal can prevent damage to the circuit in case of reversed power connection. If a full-bridge rectifier is added to the input, it can still function properly even if the power supply is connected in reverse. To prevent the transmitter from being damaged by energy such as lightning strikes, static discharge, and surges, a TVS diode can be installed at the transmitter’s input to absorb the energy of sudden overvoltage. Generally, the TVS voltage value should be set slightly lower than the amplifier’s limit voltage in order to provide protection. If there is a risk of lightning strikes, the TVS may not have sufficient absorption capacity; therefore, varistors are also necessary. However, leakage current from the varistors can cause certain errors. 2. Overcurrent protection. During equipment operation, errors such as sensor disconnections or short circuits may occur. Alternatively, the input value itself is very likely to exceed the range; the transmitter must ensure that the output does not rise indefinitely under any circumstances, otherwise it may damage the transmitter itself, the power supply, or the remote display instrument. In the diagram, Rb and Z1 form an overcurrent protection circuit. Regardless of the reason why the OP1 output exceeds 6.2V (the 1N4735 is a 6.2V Zener diode), it will be clamped by Z1, so the base of Q1 cannot exceed 6.2V. Therefore, the voltage across Re cannot be higher than 6.2 – 0.6 = 5.6 V; hence, the total current will not exceed Ue/Re = 5.6 V/200 = 28 mA. 3. Wide voltage adaptability. Generally, two-wire transmitters can handle a wide range of voltage variations without affecting accuracy. This allows it to be used with various power sources, as well as to handle high load resistances. The part most sensitive to the power supply is the reference source, which is also the main component determining accuracy. In the diagram on the 3rd floor, the reference voltage is regulated by R5; when the supply voltage changes, the current through R5 also changes, which has a significant impact on the stability of the reference voltage. In the figure, a constant current source LM334 is used as the reference power supply; when the voltage varies significantly, the current remains essentially constant, thereby ensuring the stability of the reference. 4. Decoupling capacitor: In typical circuit designs, each integrated circuit has a decoupling capacitor at its power supply terminal. When a two-wire transmitter is powered on, the charging of these capacitors can cause a large current instantaneously, which may damage remote instruments. Therefore, each decoupling capacitor generally should not exceed 10nF, and the total decoupling capacitance should not exceed 50nF. A 10nF capacitor at the input is necessary to prevent oscillations in the circuit under inductive loads with long wires. This post was last edited by Mobei Yihai on 2009-3-30 20:22.]

It has a long transmission distance and strong interference resistance; moreover, 20 milliamps is a safe value, with a lower limit of 4 milliamps, which is used to determine whether the circuit is open