Mitsubishi Electric NZ2GFSS2 32D S1 User Manual page 132

3.
When the resistor (R) is 2.0k, for example, the power capacity (W) of the resistor (R) becomes 0.41W.
2
(Input voltage)
W = =
R
4.
Because the resistor requires a power capacity that is 3 to 5 times larger than the actual current consumption, the
resistor connected to the terminal should be 2.0k; and 1.5 to 2W.
■Calculation example (NZ2GFSS2-8D-S1, NZ2GFSS2-16DTE-S1)
When the 24VDC external power supply of the safety remote I/O module is on and an LED indicator switch that has a leakage
current of 3.0mA at a maximum is connected
Leakage current 3.0mA
(1) Safety remote I/O module
1.
The leakage current through the safety remote I/O module does not satisfy the condition of the OFF current of 1.3mA or
lower. Therefore, connect a resistor as follows:
3.0mA
I = 1.7mA
R
2.
To satisfy the condition of the OFF current of the input module (1.3mA or lower), the current through the connected
resistor should be 1.7mA or higher. From the formula below, the connected resistor (R) is lower than or equal to 4.1k.
R × I ≥ Z (Input resistance) × I
R
Z (Input resistance) × I
R ≤
I
R
3.
When the resistor (R) is 3.6k, for example, the power capacity (W) of the resistor (R) becomes 0.23W.
2
(Input voltage)
W = =
R
4.
Because the resistor requires a power capacity that is 3 to 5 times larger than the actual current consumption, the
resistor connected to the terminal should be 3.6k; and 0.75 to 1.2W.
A signal incorrectly inputs data
■Cause
Noise is taken as input data.
■Action
Set the input response time longer. ( Page 72 Parameter Settings)
Ex.
1ms  5ms
If changing the setting of the input response time has no effect, take the following two measures:
• To prevent excessive noise, avoid installing power cables together with I/O cables.
• Take noise reduction measures such as connecting surge absorbers to noise-generating devices such as relays and
conductors using the same power supply.
10 TROUBLESHOOTING
130
10.5 Fault Examples with the Safety Remote I/O Module
2
28.8
= 0.41 [W]
2000 [Ω]
(1)
Tn
Input resistance 2.6kΩ
I = 1.3mA
z
Xn
R
COM-
+ V
R ZD
+ V
2.6 × 1.3 + 3.6
Z ZD
= = 4.1 [kΩ]
1.7
2
28.8
= 0.23 [W]
3600 [Ω]
V = 3.6V
ZD