How far can a 4-20MA instrument signal be transmitted without attenuation using a DJYVP2R 1X2X1.0 cable?
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How far can a 4-20MA instrument signal be transmitted without attenuation using a DJYVP2R 1X2X1.0 cable?
The power supply in the control room is assumed to be 24V. The minimum voltage for the field instruments is taken as 24(1-10%). By dividing the allowable circuit voltage drop by 20mA, the allowable line resistance is determined. By dividing this allowable resistance value by the resistance per meter of wire, the length of the wire can be calculated; dividing by 2 then gives the maximum distance the cable can span.
It mainly depends on the load capacity of the 4-20mA current signal. Defining the maximum output voltage that ensures the 4-20mA signal remains unchanged as U, the maximum resistance of the wires is R = U/0.02. The length of the wires, taking into account the resistivity and cross-sectional area of the DJYVP2R 1X2X1.0 cable wires, is R/2 (since there are two wires). That is the desired answer. In practical applications, when considering the sampling resistance of the terminal DCS system, this sampling resistance value should be excluded before calculating the wire length.
The answer on the second floor is correct. Additional information: for the DJYVP2R 1X2X1.0 cable, the resistivity of its conductors is 0.0175 ohms per meter; when calculating the cable length, this value must be multiplied by 2, as current flows in a loop. . . . I remind the building owner to consider two factors: 1. The maximum capacity under load, which is 4~20 MA, that is, the load resistance; 2. The minimum operating voltage of the load
Thank you both, but I still don’t quite understand. Could you provide a specific example? Thank you. How are the calculations done for these two types: the 4-20MA output of the most typical two-wire pressure transmitters, and the 4-20MA output used by DCS systems to control control valves? This post was last edited by hua_fan on 2009-3-4 21:43]