Mitsubishi Electric MELSEC iQ-R Series User Manual page 123

■Case 4
Cause Even if the switch with LED indicator is turned off, there is a leakage current rising above the OFF current of the input module.
Action Connect an appropriate resistor so that the current flowing into the input module would fall below the OFF current.
RX40C7
Iz=2.0mA
Z=3.3kΩ
Z: Input impedance
Calculation This column gives a calculation example of the resistance value of a resistor to be connected.
■Assuming that a switch with LED indicator where a leakage current of 2.82mA flows by applying a 24VDC power supply is connected to
example
the RX40C7
Check the following items referring to the specifications of the module.
• OFF current: 2.0mA
• Input resistance: 3.3k
I (Leakage current) = I
I = I - I = 2.82 - 2.0 = 0.82[mA]
R Z
To satisfy the condition that an OFF current of the RX40C7 should be lower than 2.0mA, connect a resistor R through which a current of
more than 0.82mA will flow. The resistance value, R, of a resistor to be connected is given by the following:
I : I = Z: R
R Z
I
Z
R ≤ × Z =
I
R
 The obtained result is: Resistance value R < 8.05k.
[Checking the connected resistor by calculating power capacity]
If the resistance R is 6.8k, the power capacity, W, of the resistance R is given by the following:
• V: Input voltage
28.8
V 2
W = =
R
6.8 [kΩ]
Because the power capacity of a resistor needs to be 3 to 5 times as large as the actual current consumption, a resistor to be connected to
the terminal concerned should be 8.2k and 1/2W.
In addition, when the resistance R is inserted, the OFF voltage is given by the following:
1
1
+
6.8 [kΩ]
This value, therefore, satisfies the condition that an OFF voltage of the RX40C7 should be lower than 8V.
■Case 5
Cause There is a sneak path allowing current to flow because of the use of two power supplies.
Input
module
Action • Use one power supply.
• To prevent the sneak path, connect a diode as shown below:
DC input
Input
module
2.82mA
R
I =0.82mA
R
24VDC
(OFF current of the RX40C7) + I
Z
2.0
× 3.3 = 8.05 [kΩ]
0.82
2
= 0.122 [W]
× 2.82 [mA] = 6.27 [V]
1
3.3 [kΩ]
E2
E1
E1 > E2
E2
E1
(Current flowing to the connected resistor)
R
9.2 Input Circuit Troubles and Corrective Actions
9 TROUBLESHOOTING
121
9