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3.5 Network Power/Grounding
3
Figure 3-14 Example of Solution to an Overload
(a) Total current consumption of section 1 =1.1A+1.25A=2.35A
(a’) Total current consumption of section 2 =0.5A+0.25A+0.25A+0.85A=1.85
(b) Overall power cable length of section 1 = 100m
(b’) Overall power cable length of section 2 = 140m
(c) Maximum current that can be supplied to the cables in section 1 based on Figure
3-10 = 2.93 A
(c’) Maximum current that can be supplied to the cables in section 2 based on Figure
3-10 = About 2.19 A (Obtained by linear approximation of 100 to 150 m)
(d) Since the total current consumption of both sections 1 and 2 is smaller than the
maximum current, power can be supplied to all the nodes by single power unit
central connection.
(e) Install a network power unit with a rated current capacity of 4.2 A or more.
(Select one with an adequate capacity, considering the conditions of use.)
Node 6Node 5Node 4Node 1Node 2Node 3
ネットワーク
電源供給装置
電源
タップ
セクション 1
100 m
V
+−
1.1A 1.25A V
−
0.5A 0.25A 0.25A 0.85A
セクション
2
140 m
Section 1 Section 2
Power
tap
Network power
unit
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