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1. How far can the current signal from a two-wire 4–20mA transmitter travel? This expert says: ① It is related to the level of the excitation voltage; ② It is related to the minimum operating voltage permitted by the transmitter; ③ It is related to the value of the current-sensing resistor used by the circuit board device for current measurement; ④ It is related to the resistance of the wires. By using these four relevant factors, it is possible to calculate the theoretical transmission distance of a 4–20mA current signal. 2. To ensure that the 4–20mA signal can be transmitted without any losses over a two-wire circuit, Ohm’s law must be satisfied, that is: (Excitation voltage – Minimum operating voltage permitted by the transmitter) ≥ Output current × Total resistance of the current loop. When the output current I is 20mA, or 0.02A, the equation becomes an equality, and thus: Total resistance of the current loop = (Excitation voltage – Minimum operating voltage permitted by the transmitter) ÷ 0.02. This calculated value is denoted as r, or r = (Excitation voltage – Minimum operating voltage permitted by the transmitter) × 50, with the unit being ohms. This value of r is referred to in the industry as the load resistance of the current signal, which represents the maximum load capacity of that current signal. 3. Why does the industry provide a formula for calculating this r? This is because the transmission distance of a 4–20mA current signal depends essentially on the actual resistance in the circuit, specifically on whether it is greater than or less than r. When the total actual resistance of the circuit exceeds r, the transmitter cannot generate a 20mA current, even if the transmission distance is zero; when the total actual resistance equals r, the transmitter can output 20mA current, but the transmission distance can only be 0 meters (except in the case of superconductors). When the total actual resistance is less than r, the transmitter can output 20mA current, allowing it to be transmitted effectively over several meters within the circuit. ① Since the resistance used by the circuit board to measure current is a fixed value, it is the resistance of the wires that determines the transmission distance. ② The lower the wire resistance, the farther the signal can be transmitted. ③ If the wires are superconductive, their resistance is approximately 0, meaning the current can be transmitted anywhere—even to the United States or Mars. In summary, the total resistance R of the current circuit must satisfy R≤r; otherwise, the 4–20mA signal cannot be transmitted properly. 4. The total resistance R of the current circuit consists of the resistance R1 used by the circuit board to measure current, and the resistance R2 of the wires. Common values for R1 are 250Ω, 150Ω, 100Ω, and 50Ω; nowadays, smaller resistances such as 100Ω to 40Ω are more commonly used. The wire resistance R2 is calculated as conductivity × total length of the wire ÷ cross-sectional area of the wire = resistance per unit length × total length of the wire = resistance per unit length × transmission distance × 2. 5. Let’s calculate the theoretical transmission distance L (in kilometers) for a 4–20mA current signal. The 2.5 square millimeter twisted pair wires sold by Huachuan have a resistance of 7.5Ω per 1000 meters. For the SC322 pressure transmitter developed by Huachuan, the minimum allowable excitation voltage is 10VDC. The resistance R1 used by the DCS circuit board to measure current is 100Ω, and the supply voltage on the DCS circuit board is 24VDC. The calculations are as follows: ① First, calculate the load resistance r of the transmitter: r = (24 – 10) × 50 = 700Ω. ② Then, determine the formula for the total circuit resistance R: R = R1 + R2 = 100 + 7.5 × L × 2 = 100 + 15L, with units of ohms. ③ Since R must be less than or equal to r, we have (100 + 15L) ≤ 700. Solving this equation gives L ≤ 40 kilometers, meaning the maximum transmission distance is 40,000 meters
Then it also needs to be divided by 2, because the signal has to return as well. ;P The signal being transmitted is only 20MA – could we use a cable with a smaller square number rating? Our factory uses 1.5 square millimeter cables; can we use 0.5 square millimeter ones instead? I’m not sure if there are any rules in this industry, but my idea is to use multi-core wires with small cross-sections from the DCS to connect to the explosion-proof wiring boxes in the workshop, and from there to the instruments located in various corners of the workshop. This way, it’s more convenient to lay the cables, and it’s also easier to make changes to the instruments in the future. Smaller cables save money and take up less space; otherwise the cable trays would be full. It’s also easier to find the wires with multi-core wires. We are in a pilot plant, and projects change frequently.
When calculating the resistance of wires, the transmission distance and the resistance value have already been multiplied by 2; therefore, there is no need to divide the total length of the wire by 2. According to standards, signal cables must be shielded cables with a gauge of 2*1.5. Cables that transmit power (those used to supply external solenoids from a DCS) generally use shielded cables with a gauge of 2*2.5, while cables connecting a DCS to an MCC usually use solid wires with a gauge of 1*1.5. If a 0.5 gauge wire is to be used as the signal wire, that’s possible, but a slightly greater distance can have an impact on more sensitive instruments. Because issues such as voltage drop will worsen. The key acceptance process is also quite troublesome.
It’s not far away, no more than 100 meters; in most cases, it’s around fifty meters.
The specifications do not allow it; the diameter of the signal cable should be between 0.75 and 1.5
This post was last edited by Xiao Fu fu on 2024-6-20 09:10. SH/T 3019-2016 and HG/T 20512-2014: These standards specify that the cross-sectional area of the conductors in instrument signal cables should not be less than 0.5 (under normal conditions) or 0.75 (under special conditions). For multi-core cables, the cross-sectional area of the conductors can be adjusted appropriately as long as the requirements regarding circuit resistance are met.
Those who have their own opinions can share them so we can discuss together!