How far can a two-wire 4–20mA current signal be transmitted? Let’s do the calculations. Wishing everyone a happy New Year! . . .

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Preface: How far can a 2-wire 4–20mA current signal travel? It’s common to encounter people making arbitrary claims on a \"theoretical\" basis. As the New Year arrives, it’s easy to spend money recklessly; today is a good day. Science requires understanding the underlying principles and conducting calculations. Today, I’ll go into detail and do the calculations. Wishing everyone a happy New Year! . . . . . . 1. How far can the current signal from a 2-wire 4–20mA transmitter actually travel? The pig said: 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 specific formula for calculating this r? That is because the transmission distance of a 4–20mA current signal depends essentially on the actual resistance, which is related to the value of r. When the total actual resistance in the loop exceeds r, the transmitter cannot generate a 20mA current, even if the transmission distance is zero.:lol When the total actual resistance in the loop equals r, the transmitter can output a 20mA current, but the transmission distance can only be 0 meters (except in the case of superconductors). When the total actual resistance in the loop is less than r, the transmitter can output a 20mA current, allowing it to be transmitted effectively over several meters within the loop. ① Since the resistors used by the circuit board to measure current are fixed values, it is the resistance of the wires that determines the transmission distance. ② The lower the wire resistance, the greater the transmission distance. ③ If the wires are superconductive, their resistance is approximately 0; in that case, it’s no problem to transmit the current to the United States or even to Mars.:lol In summary, the total resistance R of the current loop must satisfy R≤r; otherwise, the 4–20mA signal cannot be transmitted properly. 4. The total resistance R of the current loop consists of the resistor R1 used by the circuit board to measure current, and the wire resistance R2. Common values for R1 are 250Ω, 150Ω, 100Ω, and 50Ω; nowadays, smaller resistances such as 100Ω–40Ω are more commonly used. Wire resistance R2 = 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 do a quick calculation to find the theoretical transmission distance L (in kilometers) for a 4–20mA current signal. The 2.5 square millimeter twisted pair cables sold by Shuaike have a resistance of 7.5Ω per 1000 meters. Using these cables to transmit signals, Shuaike’s self-developed SC322 pressure transmitter has a minimum allowable excitation voltage of 10VDC. The resistor 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 = 700 Ω. ② Then, determine the formula for the total resistance R: R = R1 + R2 = 100 + 7.5 × L × 2 = 100 + 15L, with units of ohms. ③ Since R≤r, we have (100 + 15L) ≤ 700. Solving this gives L ≤ 40 kilometers, meaning the maximum transmission distance is 40,000 meters. 6. Hehehe, 40,000 meters—did you read that correctly? As long as there’s enough money for wiring, it’s no problem to transmit signals from Xi’an city to Chang’an County. Happy New Year, 20mA! Wish you prosperity and a long life in the new year! . . .