**Origin of 4-20mA sensor**

The reason of using current signal is that it is not easy to be disturbed, and the internal resistance of current source is infinite. The wire resistance in series does not affect the accuracy of the circuit.

The reason for using current signal is that it is not easy to be disturbed, because the amplitude of noise voltage in industrial field may reach several V, but the noise power is very weak, so the noise current is usually less than nA level, so the error of 4-20mA transmission is very small; the internal resistance of current source tends to be infinite, and the resistance of conductor is not connected in series in the circuit. Because of the large internal resistance and constant current output of the current source, we only need to place a 250 ohm resistance to the ground at the receiving end to get 0-5V voltage. The advantage of the low input impedance receiver is that the input current noise of nA level only produces very weak voltage. Noise.

The upper limit of 20mA is due to explosion-proof requirements: the spark energy caused by the current switching off of 20mA is insufficient to ignite gas. The reason why the lower limit is not 0 mA is to be able to detect the break line: when the transmission line breaks due to a fault, the loop current drops to 0. 2mA is often taken as a broken line alarm value. The current type transmitter converts physical quantity to 4~20mA current output, and there must be external power supply for it. The most typical is that the transmitter needs two power lines, plus two current output lines, a total of four wires, called four-wire transmitter. Of course, the current output can be shared with the power supply by a common line VCC or GND, which can save a line, called a three-wire transmitter. In fact, you may notice that the 4-20mA current itself can supply the transmitter. The transmitter is equivalent to a special load in the circuit. The special point is that the power consumption current of the transmitter varies from 4 to 20 mA according to the output of the sensor. The display instrument only needs to be connected to the circuit. This transmitter only needs 2 external wires, so it is called a two wire transmitter. The lower limit of the industrial current loop is 4mA, so the transmitter has at least 4mA power supply as long as it is within the range.

Therefore, 4-20mA signal output is generally not easy to be disturbed and safe, so the two-wire 4-20mA power output signal is widely used in industry. However, in order to better process the sensor signal, there are more other forms of output signal: 3.33MV/V; 2MV/V; 0-5V; 0-10V and so on.

**How far can 4-20mA travel?**

How far can the two wire 4 to 20mA current signal travel? Today, Xiaobian is going to give you a popular science.

1. how far can the current signal of the two wire 4 to 20mA transmitter be transmitted?

Interference factors:

(1) related to the excitation voltage.

(2) related to the minimum operating voltage allowed by the transmitter;

(3) it is related to the size of the pressure resistance used by the card equipment to collect the current.

It is related to the size of wire resistance.

Through these four quantities, the theoretical transmission distance of 4 to 20mA current signals can be calculated.

2. to make the 4 to 20mA signal transmit without loss in the two wire loop, we must satisfy Ohm's law.

That is to say, (excitation voltage-minimum operating voltage allowed by transmitter) > output current * total resistance of current loop

When the output current I=20mA is 0.02A, the upper form takes the equal sign.

The total resistance of the current loop is equal to 0.03 (minimum operating voltage allowed by the excitation voltage-transmitter). The calculated value is denoted as R.

That is, r= (the minimum operating voltage allowed by the excitation voltage transmitter) x 50, unit Omega.

This r, the industry known as the current signal load resistance, that is, the maximum carrying capacity of the current signal.

3. why did the industry specifically give the formula for this r?

That is due to the fact that the current signal from 4 to 20mA can be transmitted far away, which is actually the problem of the actual resistance and the R ratio.

When the total resistance of the loop is greater than r, the transmitter can not output 20 mA current even if the transmission distance is 0.

When the actual total resistance of the loop equals r, the transmitter outputs 20 mA current and the transmission distance can only be 0 meters (except superconducting).

When the actual total resistance of the loop is less than r, the transmitter can output 20 mA current to transmit several meters effectively in the loop.

The resistance of the conductor determines the length of the transmission distance because the resistance of the collector is a fixed value.

The smaller the conductor resistance, the farther the signal transmission distance is;

(3) If the conductor is superconducting and the resistance is 0, it is not a problem for the current to be transmitted to the United States, nor to Mars.

In conclusion, the total resistance R of the current loop must satisfy R < r, otherwise the 4-20mA signal can not be transmitted normally.

4. The total resistance R of the current loop is composed of the piezoelectric resistance R1 and the conductor resistance R2, which collects the current signal from the card equipment.

The resistance voltage R1 is 250 ohms, 150 ohms, 100 ohms and 50 ohms. Now it is popular for 100 ohms to 40 ohms and other small resistors.

Conductor resistance R2 = conductivity * total length of conductor * cross-sectional area of conductor * resistance per unit length * total length of conductor * resistance per unit length * transmission distance * 2

5, compiling a small problem and calculating it.

The theoretical transmission distance of 4-20 mA current signal is L (per kilometer). A adopts 2.5 square millimeter twisted pair wire, and the resistance of each 1000 meter is 7.5_. The self-made SC322 zero drift pressure transmitter with this wire is used to transmit signal A. The minimum allowable excitation voltage is 10 VDC. The piezoelectric resistance R1 for collecting current on DCS board is 100_, and the supply on DCS board is 10_. What is the maximum transmission distance of the current signal when the voltage is 24VDC?

The calculations are as follows:

First, calculate the load resistance of the transmitter R, r= (24-10) x 50=700 ohm.

Second, write the formula of total resistance R of the loop.

R=R1+R2=100+7.5 * L * 2=100+15L unit Omega

(3) because R is less than R, that is, (100+15L).

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