The transmission distance of the signal?
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Hello, everyone! How far can a typical 4--20MA signal be transmitted? What factors influence this distance, and how can the quality of such signals be improved? Let’s discuss it together
It’s around 600 meters, and it depends on your circuit wiring and the supply voltage. Choose better cables; I’m not sure if there are any other good solutions. Looking forward to suggestions from those with more experience.
I remember it was 800–1500 meters; better wire materials were used. To achieve a greater communication distance, fiber optics had to be employed
There is definitely no problem at distances below 500 meters. For distances between 500 and 1000 meters, it is also fine to use cables with a gauge of 2X2.5mm2. For even higher levels, it is recommended to use optical cables.
Under 500 meters, I guess; if long-distance transmission is needed, it’s best to use a signal amplifier
1. The 4–20mA current signal in a two-wire system, provided that the load meets the requirements, experiences no loss due to line resistance during transmission; it does not attenuate, and it has strong interference resistance – this goes without saying. 2. The load-carrying capacity of a 4–20mA output-type transmitter in a two-wire system is calculated using the formula R = 50×(V – u), with the unit being ohms. Here, V represents the actual supply voltage, while u is the lowest supply voltage permitted by the transmitter. Factors such as voltage fluctuations must be taken into account, especially in cases where the voltage is less than 24VDC. 3. The theoretical transmission distance for a 4–20mA current signal in a two-wire system is calculated using the formula L = 0.5×(R – A)÷B, with the unit being kilometers. Here, A refers to the sampling resistance of devices at the signal receiving end, such as acquisition cards or secondary meters; common values for this resistance are 250Ω and 100Ω； B is the resistance value per kilometer for each core of the cable ； In practical applications, it is necessary to leave a margin; it is advisable to estimate this based on 50% of the theoretical transmission distance L. 4. For the transmission of DC signals, it is best to use twisted pair cables with a shielding density of over 80% ； I suggest leaving the shielding layer suspended and not grounded! There are too many reasons; I won’t go into detail. 5. For the power supply, batteries are the best option, followed by transformers (linear power supplies), with switch-mode power supplies being the worst choice! The market is currently flooded with low-quality switching power supplies, which are 100% hidden killers!
The 4–20MA current signal used for instruments is a constant-current signal; the transmission distance depends mainly on the impedance of the load (including the cables), and generally, a value of less than around 500 ohms is sufficient. It varies depending on the instrument. So it depends on the value of the load resistance plus the cable resistance.
A repeater can be added, which is essentially a signal amplifier. However, for greater distances, a digital bus is usually more appropriate. For example, the single segment distance of a 485-based industrial bus (twisted pair) can reach 1500 meters. Just add an interface card, which saves a lot on wiring, as there is usually more than one signal path. Together with the other advantages of digital buses, it’s well worth it.