Weidmuller SAI Active Universal Pro Series, Руководство пользователя

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Project Planning with the Universal Pro
Figure 4 Example of system structure
The cable length of the sub-bus line:
between the modules
1 and 2 = 10 m = L1
2 and 3 = 5 m = L2
3 and 4 = 0.3 m = L3
4 and 5 = 0.3 m = L4
5 and 6 = 4m = L5
6 and 7 = 0.3 m = L6
7 and 8 = 0.3 m = L7
8 and 9 = 10 m = L8
9 and 10 = 1 m = L9
The max. residual current on the sub-bus line is the
sum of all currents and in this example is 0.89A.
Thus the max. current-carrying capacity of 3A is not
reached.
Voltage drop L9 = 0.011 V
110 mA x (0.05 Ohm/m x 1 m x 2)
Voltage drop L8 = 0.22 V
(110 mA + 110 mA) x (0.05 Ohm/m x 10 m x 2)
Voltage drop L7 = 0.0093 V
(90 mA + 110 mA + 110 mA) x (0.05 Ohm/m x 0.3 m
x 2)
Voltage drop L6 = 0.0126 V
(110 mA + 90 mA + 110 mA + 110 mA) x (0.05
Ohm/m x 0.3 m x 2)
Voltage drop L5 = 0.204 V
(90 mA + 110 mA + 90 mA + 110 mA + 110 mA) x
(0.05 Ohm/m x 4 m x 2)
Voltage drop L4 = 0.0186 V
(110 mA + 90 mA + 110 mA + 90 mA + 110 mA +
110 mA) x (0.05 Ohm/m x 0.3 m x 2)
Voltage drop L3 = 0.0219 V
(110 mA + 110 mA + 90 mA + 110 mA + 90 mA +
110 mA + 110 mA) x (0.05 Ohm/m x 0.3 m x 2)
Voltage drop L2 = 0.39 V
(50 mA + 110 mA + 110 mA + 90 mA + 110 mA +
90 mA + 110 mA + 110 mA) x (0.05 Ohm/m x 5 m x
2)
Voltage drop L1 = 0.89 V
(110 mA + 50 mA + 110 mA + 110 mA + 90 mA +
110 mA + 90 mA + 110 mA + 110 mA) x (0.05
Ohm/m x 10 m x 2)
Total voltage drop = 1.7774 V
0.89 V + 0.39 V + 0.0219 V + 0.0186 V + 0.204 V +
0.0126 V + 0.0093 V + 0.22 V + 0.011 V
Voltage at module 9 = 21.3226 V
24 V – 0.9 V – 1.7774 V
Voltage at module 9 is greater than 18V DC
→
Okay
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