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Section
Step 2: Calculating the Best Location of the Actual Nodes
6-5
In:
The consumption current required for the communications power
supply and the internal circuit power supply for each node
0.005 Ω = The contact resistance of the taps
6-5-2 Countermeasures
If the best locations cannot be determined using either the first or second formulae, follow the procedure as shown below.
• Locate the communications power supply in the center of the network and the
nodes to both sides of it.
• If the nodes are already located at both sides of the power supply, move the
power supply in the direction that requires the larger current capacity.
• If thin cable is being used, replace it with thick cable.
• Move the highest current consumption node nearer the power supply.
If the best locations still cannot be determined using the first and second formulae even after carrying out the above procedure, a single power supply will not be
sufficient. In that case, proceed to Step 3.
Configuration Example 1
Nodes Concentrated to One Side of the Power Supply (Trunk Line: Thick cable,
Drop Lines: Thin cables)
Communications
power supply
40 m
30 m
20 m
1m
Thick truck line
Terminating
resistor
Terminating
resistor
Thin
drop line
Master
16 Output points
Node
16 Input points
Node
Group 1
Group 1
Node
4 Analog inputs
7
units
5 units 16 Output points
Node
2 Analog inputs
Node
16 Output points
Node
16 Input points
4 Analog inputs
Node
Node
Group 2
Group 2
Node
2 Analog inputs
10
units
4 Analog inputs
Node
5
units
2 Analog inputs
Node
Node
Group 3
Group 4
Group 3
Group 4
Communications power supply:
45 mA + 30 mA x 5 = 195 mA
30 mA x 7 = 210 mA
30 mA x 10 = 300 mA
30 mA x 5 = 150 mA
Internal circuit power supply:
70 mA x 5 = 350 mA
90 mA x 7 = 630 mA
80 mA x 10 = 800 mA
140 mA x 5 = 700 mA
• Calculate the voltage drop of each group when the network is supplied by the
communications power supply only.
Group 1:
(1 x 0.015 + 1 x 0.005) x 0.195 = 0.0039 V
Group 2:
(20 x 0.015 + 2 x 0.005) x 0.21 = 0.0651 V
Group 3:
(30 x 0.015 + 3 x 0.005) x 0.30 = 0.1395 V
Group 4:
(40 x 0.015 + 4 x 0.005) x 0.15 = 0.093 V
Total voltage drop = 0.0039 + 0.0651 + 0.1395 + 0.093 = 0.3015 V 4.65 V
Thus, the best location for the nodes can be determined by using the first
formula.
• Calculate the voltage drop of each group when the communications and the
internal circuit power supplies are the same.
Group 1:
(1 x 0.015 + 1 x 0.005) x 0.545 = 0.0109 V
Group 2:
(20 x 0.015 + 2 x 0.005) x 0.84 = 0.2604 V
65