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6 F 3 B 0 3 6 4
53
Node 6Node 5Node 4Node 1Node 2Node 3
Network
power-supply
unit
Power
tap
Section 1
100 m
V
+−
1.1A 1.25A V
−
0.5A 0.25A 0.25A 0.85A
Section 2
140 m
Figure 3.15 Example of Solving the Overload
a) Sum of current consumption in section 1 = 1.1A + 1.25A = 2.35A
a') Sum of current consumption in section 2 = 0.5A + 0.25A + 0.25A + 0.895A = 1.85A
b) Total length extended in section 1 = 100 m
b') Total length extended in section 2 = 140 m
c) Maximum current available on the cable in section 1, based on Figure 3.10 = approx. 2.19A
(Obtain the approximate value between 100 to 150 meter straight cable.)
d) Since both of the sums of current consumption in section 1/2 < maximum current, a single power
unit central connection can supply the power to all nodes.
e) Install a network power unit with a rated current of 4.2A or more.
(Select one with ample current in considering usage conditions.)