DCN4-BR4D Omron, DCN4-BR4D Datasheet - Page 119

• Manufacturer Part Number: DCN4-BR4D
• Description: Thin Cable-Flat Cable Conv Con
• Manufacturer: Omron

Specifications of DCN4-BR4D

• Lead Free Status / Rohs Status: Lead free / RoHS Compliant

Step 2: Evaluating the Configuration with Calculations
3-5-2
Example Configuration 1
98
Modifying the Configuration
Communications power supply
Terminating
Resistor
Internal circuit power supply
Power
Supply
Branch lines:
Thin cable
1 m
Group 1
45 mA + 30 mA × 5 = 195 mA
70 mA × 5 = 350 mA
Formula 2: Calculating the Voltage Drop in the Trunk Line
Ln: The distance between the power supply and node n (not including the
Rc: Maximum cable resistance
Nt: The number of Taps between node n and the power supply
In: The total current (communications current and internal circuit current)
If the result of formula 1 or formula 2 indicates that power cannot be supplied
properly, use the following procedure to modify the communications power
supply configuration.
If formula 1 or formula 2 is still not satisfied after making the changes listed
above, the network’s power cannot be supplied by a single power supply. Pro-
ceed to Step 3.
In this example, the power supply is at one end of the trunk line. The trunk line
is thick cable and the branch lines are thin cable.
1. Calculate the voltage drop when the communications power supply pro-
20 m
• Move the communications power supply towards the center of the net-
• If there are already nodes on both sides of the power supply, move the
• If thin cable is being used, replace it with thick cable.
• Move the nodes with higher current requirements closer to the power sup-
Trunk line:
Thick cable
Σ [(Ln × Rc + Nt × 0.005) × In] ≤ 0.65 V
length of the branch line)
(Thick cable: 0.015 Ω /m, thin cable: 0.069 Ω /m])
(0.005 Ω = The contact resistance of a Tap)
required by node n
work so that there are nodes to both sides of it.
power supply in the direction that requires the higher current.
ply.
vides communications power only (formula 1.)
Group 1:(1 × 0.015 + 1 × 0.005) × 0.195 = 0.0039 V
Group 2:(20 × 0.015 + 2 × 0.005) × 0.21 = 0.0651 V
Node
Node
16-point
Input Unit
16-point
Input Unit
Node
Master
.
.
.
30 m
5
Units
Group 2
30 mA × 7 = 210 mA
90 mA × 7 = 630 mA
40 m
16-point
Output Unit
16-point
Output Unit
16-point
Output Unit
Node
Node
Node
.
.
.
7
Units
Group 3
30 mA × 10 = 300 mA
80 mA × 10 = 800 mA
4-point Analog
Input Unit
4-point Analog
Input Unit
4-point Analog
Input Unit
Node
Node
Node
.
.
.
10
Units
Group 4
30 mA × 5 = 150 mA
140 mA × 5 = 700 mA
2-point Analog
Output Unit
2-point Analog
Output Unit
2-point Analog
Output Unit
Node
Node
Node
.
.
.
Terminating
Resistor
5 Units
Section 3-5