3-3
Section
Sharing the Communications Power Supply
35
If the best locations cannot be determined using either the first or second formu-
lae, 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 formu-
lae even after carrying out the above procedure, a single power supply will not be
sufficient. In that case, proceed to Step 3.
Configuration Example
Nodes Located at Both Sides of the Power Supply (Trunk Line: Thick cable, Drop
Lines: Thin cables)
Communications
power supply
16 Output points
2 Analog inputs
7
units
Group 1
Master
Node
Node
Node
Node
Node
Node
Node
Node
Node
Node
Node
Node
10 units
5
units
5
units
2 Analog inputs
2 Analog inputs
4 Analog inputs
4 Analog inputs
4 Analog inputs
16 Output points
16 Input points
Thin
drop line
Group 3
Group 2
Group 4
System 1
20 m
10 m
30 m
System 2
Thick truck line
10 m
16 Input points
16 Input points
Terminating
resistor
Terminating
resistor
Group 1
Group 2
Group 3
Group 4
Communications power supply:
45 mA + 30 mA x 7 = 255 mA
30 mA x 10 = 300 mA
30 mA x 5 = 150 mA
30 mA x 5 = 150 mA
Internal circuit power supply:
90 mA x 7 = 630 mA
80 mA x 10 = 800 mA
70 mA x 5 = 350 mA
140 mA x 5 = 700 mA
•
Calculate the voltage drop of each group in each system when the network is
supplied by the communications power supply only.
System 1
Group 1:
(20 x 0.015 + 2 x 0.005) x 0.255 = 0.0791 V
Group 2:
(10 x 0.015 + 1 x 0.005) x 0.3 = 0.0465 V
Total voltage drop for System 1 = 0.0791 + 0.0465 = 0.1256 V
4.65 V
Thus, the best location for the nodes can be determined by using the first
formula.
System 2
Group 3:
(10 x 0.015 + 1 x 0.005) x 0.15 = 0.0233 V
Group 4:
(30 x 0.015 + 2 x 0.005) x 0.15 = 0.069 V
Total voltage drop for System 2 = 0.0233 + 0.069 = 0.0923 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 in each system when the commu-
nications and the internal circuit power supplies are the same.
System 1
Group 1:
(20 x 0.015 + 2 x 0.005) x 0.885 = 0.2744 V
Countermeasures
