How far can the two-wire 4-20mA current signal pass? Today, Xiaobian will give you a popular science!
1. How far can the current signal of the two-wire 4-20mA transmitter be transmitted?
Interference factors: 1 is related to the excitation voltage; 2 is related to the minimum operating voltage allowed by the transmitter; 3 is related to the size of the voltage-receiving resistor used by the board device to collect current; 4 is related to the size of the wire resistance. Through these four related quantities, the theoretical transmission distance of the 4-20 mA current signal can be calculated.
2. To make the 4~20mA signal transmit without loss in the two-wire loop, Ohm's law must be met.
That is: (excitation voltage - the minimum operating voltage allowed by the transmitter) ≥ output current × current loop total resistance
When the output current I=20mA, ie 0.02A, the upper formula takes the equal sign, then: the total resistance of the current loop = (excitation voltage - the minimum operating voltage allowed by the transmitter) ÷ 0.03, the calculated value is denoted as r, That is r = (excitation voltage - the minimum operating voltage allowed by the transmitter) × 50, the unit Ω r, the load resistance of the industry called the current signal, that is, the maximum load capacity of the current signal.
3. Why should the industry deliberately give this r calculation formula?
That is because: 4 ~ 20mA current signal can travel far, the actual is the actual resistance, and the size ratio of r. When the actual total resistance of the loop is > r, even if the transmission distance is 0, the transmitter can not output 20mA current; when the actual total resistance of the loop = r, the transmitter outputs 20mA current, and the transmission distance can only be 0. Meter (except superconducting); when the actual total resistance of the loop is <r, the transmitter outputs 20mA current to effectively transmit several meters in the loop. 1Because the voltage-receiving resistor used for the current collection of the board is a fixed value, the wire resistance determines the length of the transmission distance; 2 the smaller the wire resistance, the farther the signal transmission distance; if the wire is superconducting, the resistance is ≈0. It is not a problem to pass the current to the United States. It is no problem to pass to Mars.
In summary, the total resistance R of the current loop must satisfy R ≤ r, otherwise the 4 -20 mA signal cannot be transmitted normally.
4. The total resistance R of the current loop is composed of a voltage-receiving resistor R1 for collecting a current signal on the card device, and a wire resistance R2.
The voltage-receiving resistor R1 is more common in 250Ω, 150Ω, 100Ω, and 50Ω, and is now popular with small resistors such as 100Ω to 40Ω. Wire resistance R2=Conductivity×Total wire length÷Wire cross-sectional area=Unit length resistance×Total wire length=Unit length resistance×Transmission distance×2
Make a small calculation
4~20mA current signal theoretical transmission distance L (units), A uses 2.5 square millimeters twisted pair, resistance per 1000 meters = 7.5Ω, use this wire to transmit signal A self-developed SC322 zero drift pressure Transmitter, the minimum allowable excitation voltage = 10VDC, the voltage-receiving resistor R1=100Ω for collecting current on the DCS board, the supply voltage on the DCS board=24VDC, what is the maximum transmission distance of the current signal?
The calculation is as follows: 1 first calculate the load resistance of the transmitter r, r = (24-10) × 50 = 700 Ω 2 and then write the total resistance of the loop R:
R=R1+R2=100+7.5×L×2=100+15L unit Ω
3 Calculated by R≤r, ie (100+15L)≤700: L≤40 km, ie the maximum transmission distance is 40000 m
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