Two-wire system 4 ~ 20mA current signal can pass how far?

- Nov 08, 2017 -

Two-wire system 4 ~ 20mA current signal can pass it? Today for everyone to science!

1. Two-wire 4 ~ 20mA transmitter current signal can pass it how far?

Interference factors:

① with the excitation voltage level related;

② with the transmitter to allow the minimum operating voltage;

③ with the card device to capture the current used to take the size of the pressure resistance;

④ and the size of the wire resistance.

Through these four related quantities, we can calculate the theoretical transmission distance of 4 ~ 20mA current signal.

2. To make 4 ~ 20mA signal, lossless transmission in the two-wire loop, must meet Ohm's law.

That is: (excitation voltage - the minimum allowable transmitter voltage) ≥ output current × total current loop resistance

When the output current I = 20mA, that is 0.02A, the above equation to take equal sign, then:

The total current loop resistance = (excitation voltage - the minimum allowable transmitter voltage) ÷ 0.03, the calculation of this as r,

That r = (excitation voltage - the minimum allowable transmitter voltage) × 50, unit Ω

This r, the industry calls the current signal load resistance, which is the maximum carrying capacity of the current signal.

3. Why should the industry specifically give this r formula?

That is because: 4 ~ 20mA current signal can be transmitted far, in essence, the actual resistance, and r than the size of the problem.

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

When the actual total loop resistance = r, the transmitter output 20mA current, the transmission distance of only 0 meters (except for superconducting);

When the total actual ring resistance <r, the transmitter output 20mA current, in order to effectively transfer a number of meters in the loop.

① Since the card pick-up current acquisition resistance for the set value, so the size of the wire resistance, determines the length of the transmission distance;

② the smaller the wire resistance, the farther the signal transmission distance;

③ If the wire is superconducting, resistance ≈ 0, then the current to the United States to go is not a matter, went to Mars no problem.

In summary, the total current loop resistance R, R ≤ r must be met, or 4 ~ 20mA signal, can not be transmitted normally.

4. The total resistance of the current loop R, taken by the card device current signal acquisition resistor R1, and lead resistance R2 composition.

Take the pressure resistance R1, more common 250Ω, 150Ω, 100Ω, 50Ω, and now popular 100Ω ~ 40Ω and other small resistance and pressure.

Conductor resistance R2 = Conductivity × Total conductor length ÷ Conductor cross-sectional area = Unit length resistance × Overall conductor length = Unit length resistance × Transmission distance × 2


Make a small problem count

4 ~ 20mA current signal theoretical transmission distance L (unit km), A using 2.5 mm2 twisted pair, resistance per 1000 meters = 7.5Ω, with this wire signal transmission A self-developed SC322 zero drift pressure Transmitter, the minimum allowable excitation voltage = 10VDC, DCS board current acquisition resistor R1 = 100Ω, DCS board supply voltage = 24VDC, the maximum distance of the current signal transmission distance?


Calculated as follows:

① first calculate the transmitter load resistance r, r = (24-10) × 50 = 700Ω

② write the loop total resistance R formula:


R = R1 + R2 = 100 + 7.5 × L × 2 = 100 + 15L unit Ω


③ As R ≤ r, that is (100 +15 L) ≤ 700

Calculated: L ≤ 40 km, that is, the maximum transmission distance of 40,000 meters

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