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■Calculation example
When a module-and-sensor power supply (24VDC) of the input module is on and a switch with LED indicator that has a
leakage current of 3mA at a maximum is connected
Leakage current 3mA
1.
The OFF current through the input module is not 1.7mA or less. Therefore, connect a resistor as shown below.
3mA
I
= 1.3mA
R
24VDC
Z: Input impedance
2.
To satisfy the condition, the current through the connected resistor should be 1.3mA or more. From the formula below,
the value of the connected resistor is lower than 4.31k.
I
:I
= Z: R
R
Z
I
1.7
Z
R ≤
× Z =
× 3.3 = 4.31[kΩ]
I
1.3
R
3.
When the resistor (R) is 3.9k, for example, the power capacity (W) of the resistor (R) becomes 0.179W.
÷
÷
2
2
W = V
R = 26.4
3900 = 0.179[W]
V: Input voltage
4.
Because the resistor requires the power capacity of 3 to 5 times as large as the actual current consumption, the resistor
connected to the terminal should be 3.9k and 1 to 2W.
5.
OFF voltage when the resistor (R) is connected becomes 5.36V. This satisfies that the OFF voltage of the input module
is 8V or lower.
1
× 3[mA] = 5.36[V]
1
1
+
3.9[kΩ]
3.3[kΩ]
Input module
24VDC
Iz = 1.7mA
Resistor
Z = 3.3kΩ
11 TROUBLESHOOTING
11.6 Examples of Troubles with the I/O Module
11
135
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Troubleshooting
