CompoBus/D Communications Line Design | Chapter 2 | |
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Right Side Calculation
Node 4: (40 0.015 + 1 0.005) 0.15 = 0.09075 (V)
Node 5: (80 0.015 + 2 0.005) 0.25 = 0.3025 (V)
Node 6: (120 0.015 + 3 0.005) 0.15 = 0.27225 (V)
If 0.09075 + 0.3025 + 0.27225 = 0.6655 V x 4.65 V, the conditions are satisfied.
2-3-4 Step 3: Splitting the System into Multiple Power Supplies
Go to Step 3 if the best location for the nodes cannot be calculated from the formulae. In the third step, there are multiple power supplies and the power supply system is split.
HSplitting the Power Supply System
•Be sure to use a Power Supply Tap for each power supply when the Network is supplied by two or more power supplies.
•Remove the fuses in the Power Supply Tap to split the power supply system.
Once the power supply system is split, return to Step 1 or 2, and determine the best location of the nodes in each system.
HPower Supply Tap Configuration
Connector C
Fuse A | Fuse B |
Power supply cable
Cable A
Model |
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Specification Power supply tap (with a grounding terminal and reverse current prevention function )
Manufacturer
Cable B
Connector A Connector B
HInternal Circuitry of the Power Supply Tap
Power Supply Tap
Fuse A | Fuse B |
| V+ | |
CAN H | ||
Shield | ||
on side A | ||
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| CAN L | |
| V– |
Schottky diode
Ground V– V+ terminal
Power supply device on side C
V+
CAN H
Shield
CAN L
V–
Fuses used: Littel fuse 312008 Rated amperage: 8 A Rated voltage: 250 V 6.35 Φ x 31.75 mm