# How far can a 4-20mA current signal travel in a two-wire system? 4-20MA current ring can be divided into two wire systems and three wire systems. Today there is not so much other content. Please follow the small make up together to calculate it.

1. Distractions

(1) related to the excitation voltage;

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

(3) It is related to the voltage resistance used by board equipment to collect current; ④ It is related to the resistance of the wire.

The theoretical transmission distance of 4-20mA current signal can be calculated by these four related quantities.

2. Ohm's law must be satisfied in order to transmit 4-20MA signal without loss in a two-wire loop.

That is:

(excitation voltage - the minimum operating voltage allowed by the transmitter) ≥ output current × total resistance of the current loop

When the output current I=20mA, that is, 0.02A, the above equation takes the equal sign,

Is:

The total resistance of the current loop = (excitation voltage - the minimum operating voltage allowed by the transmitter) ÷0.02, the calculated value is denoted as R,

That is, r= (excitation voltage - the minimum operating voltage allowed by the transmitter) ×50, in ω.

This R, known in the industry as the load resistance of the current signal, is the maximum load capacity of the current signal.

3. Why does R have to be calculated specially?

How far the 4-20mA current signal can travel is essentially a question of the actual resistance and R ratio.

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

​when the actual total resistance of the loop is equal to r, the transmitter output 20mA current, the transmission distance can only be 0m (except superconducting);

​when the actual total resistance of the loop is <​ r, the transmitter outputs a current of 20mA to effectively transmit several meters in the loop.

① Because the voltage resistance used by the board to collect the current 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, then the current to the United States is not a matter, to Mars is no problem.

To sum up, the total resistance of the current loop, R, must meet R≤ R, otherwise 4~20mA signal can not be transmitted normally.

4. The total resistance R of the current loop is composed of voltage resistance R1, which is used to collect current signals on the board device, and wire resistance R2.

Voltage resistance R1, see more than 250 ω, 150 ω, 100 ω, 50 ω, and now popular 100 ω -40 ω and other small resistance voltage.

Wire resistance R2 = conductivity × total wire length ÷ wire cross-sectional area = resistance per unit length × total wire length = resistance per unit length × transmission distance ×2

5. ​Making up a little problem

4~20mA current signal theoretical transmission distance L (unit kilometers), A using 2.5 square mm twisted pair, resistance = 7.5ω per 1000 meters, with this wire transmission signal A self-developed SC322 no zero drift pressure transmitter, the minimum allowable excitation voltage =10VDC, DCS board current collection voltage resistance R1= 100ω, DCS board supply voltage = 24VDC, current signal maximum transmission distance is how much?

The calculation is as follows:

First calculate the load resistance of the transmitter R, r=(24-10)×50=700 ω

② Write the formula of the total resistance R of the loop:

R = R1 + R2 = 100 + 7.5 * L * 2 = 100 + 15 Ω L units

③ Because R≤ R, that is, (100+15L) ≤700

Calculated: L≤40 km, that is, the maximum transmission distance is 40000 meters.