CompoBus/D Communications Line Design | Chapter 2 | |
|
|
|
Basically, in the CompoBus/D Network the permissible maximum voltage drop within the system can be specified at 5 V for a power supply line (+V or
Of the permissible
The following formulae are applicable when power is supplied independently for communications and the internal circuit. For details on voltage drop and formulae when the communications power supply and internal circuit power supply are shared, refer to the CompoBus/D (DeviceNet) Opera- tion Manual.
HFormulae
Try to calculate the best location for each node using the formula below. If the best location for each node can be determined using the formula, the specifications for the power supply to each node can also be met. Do not exceed the maximum current capacity of the cable (Thick Cable: 8 A and Thin Cable: 3 A).
{(L1 ⋅ RC + N1 ⋅ 0.005) ⋅ l1} + {(L2 ⋅ RC + N2 ⋅ 0.005) ⋅ l2} + ..... + {(Ln ⋅ RC + Nn ⋅ 0.005) ⋅ ln} x 4.65 V
Li: | The distance (m) of the trunk line between the power supply and node i. |
Rc: | Maximum cable resistance for approx. 1 m |
| (Thick Cable: 0.015 Ω /m, Thin Cable: 0.069 Ω /m) |
Ni: | The number of |
Ii: | The consumption current required for the communications power supply for node i. |
0.005 Ω = The contact resistance of the
Note If there are nodes on both sides of the power supply, the formula is used to calculate the best location in each direction, and if the conditions are satisfied, then the locations are valid. The conditions are satisfied if the following equations are true.
Voltage drop (V) on trunk line at left side x 4.65 V
Voltage drop (V) on trunk line at right side x 4.65 V
DCalculation Example
Terminating Resistor | Trunk line | Trunk line | Terminating Resistor |
|
|
3 m max.
Node | Node | Node | Communications | Node | Node | Node | |
power supply |
| ||||||
|
|
|
|
|
|
| |
0.1 A | 0.25 A | 0.2 A |
|
| 0.15 A | 0.25 A | 0.15 A |
| 40 m | 40 m | 40 m | 40 m | 40 m |
| 40 m |
Left Side Equation
Node 1: | (120 | 0.015 + 3 | 0.005) 0.1 | = 0.1815 | (V) | |
Node 2: | (80 | 0.015 + 2 | 0.005) | 0.25 | = 0.3025 | (V) |
Node 3: | (40 | 0.015 + 1 | 0.005) | 0.2 = 0.121 (V) |
If 0.1815 + 0.3025 + 0.121 = 0.605 V x 4.65 V, the conditions are satisfied.