DCN4-BR4D Omron, DCN4-BR4D Datasheet - Page 120

• 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

Example Configuration 2
Step 2: Evaluating the Configuration with Calculations
Terminating
Resistor
Branch lines:
Thin cable
Group 1
45 mA + 30 mA × 7 = 255 mA
90 mA × 7 = 630 mA
Trunk line:
Thick cable
16-point
Output Unit
16-point
Output Unit
Node
Node
Node
Master
System 1
.
.
.
2. Calculate the voltage drop when the communications power supply pro-
In this example, the power supply is near the middle 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-
2. Calculate the voltage drop when the communications power supply pro-
Group 3:(30 × 0.015 + 3 × 0.005) × 0.30 = 0.1395 V
Group 4:(40 × 0.015 + 4 × 0.005) × 0.15 = 0.093 V
Total voltage drop = 0.0039 + 0.0651 + 0.1395 + 0.093 = 0.3015 V ≤ 4.65 V
In this case, formula 1 is satisfied, so the power supply can supply just the
communications power.
vides both communications power and internal circuit power (formula 2.)
Group 1:(1 × 0.015 + 1 × 0.005) × 0.545 = 0.0109 V
Group 2:(20 × 0.015 + 2 × 0.005) × 0.84 = 0.2604 V
Group 3:(30 × 0.015 + 3 × 0.005) × 1.1 = 0.5115 V
Group 4:(40 × 0.015 + 4 × 0.005) × 0.85 = 0.527 V
Total voltage drop = 0.0109 + 0.2604 + 0.5115 + 0.527 = 1.3098 V ≥ 0.65 V
In this case, formula 2 is not satisfied, so the power supply cannot supply
the communications power and internal circuit power.
7
Units
vides communications power only (formula 1.)
a) System 1 (Left Side)
b) System 2 (Right Side)
vides both communications power and internal circuit power (formula 2.)
a) System 1 (Left Side)
20 m
Group 1:(20 × 0.015 + 2 × 0.005) × 0.255 = 0.0791 V
Group 2:(10 × 0.015 + 1 × 0.005) × 0.3 = 0.0465 V
Total voltage drop = 0.0791 + 0.0465 = 0.1256 V ≤ 4.65 V
In this case, formula 1 is satisfied on the left side.
Group 3:(10 × 0.015 + 1 × 0.005) × 0.15 = 0.0233 V
Group 4:(30 × 0.015 + 2 × 0.005) × 0.15 = 0.069 V
Total voltage drop = 0.0233 + 0.069 = 0.0923 V ≤ 4.65 V
In this case, formula 1 is satisfied on the right side.
4-point Analog
Input Unit
4-point Analog
Input Unit
4-point Analog
Input Unit
Group 2
30 mA × 10 = 300 mA
80 mA × 10 = 800 mA
Node
Node
Node
.
.
.
10 m
Power
Supply
10
Units
10 m
30 m
Group 3
30 mA × 5 = 150 mA
70 mA × 5 = 350 mA
16-point
Input Unit
16-point
Input Unit
16-point
Input Unit
Node
Node
Node
System 2
.
.
.
5
Units
2-point Analog
Output Unit
2-point Analog
Output Unit
Group 4
30 mA × 5 = 150 mA
140 mA × 5 = 700 mA
2-point Analog
Output Unit
Node
Node
Node
.
.
.
Terminating
Resistor
5
Units
Section 3-5
99