User`s manual

I/O circuit troubleshooting Troubleshooting
MELSEC System Q, Hardware 11 – 99
Calculation example of the resistance to be connected in Example 4
Connecting a switch with LED display, in which a maximum 4.0 mA leakage current flows when
24 V DC is supplied to the QX80.
In this case, the circuit does not satisfy the condition that the off current of the QX80 is 1.7 mA
or less. Connect a resistance as follows.
The current flowing through the resistor R must be minimum 2.3 mA.
I
R
= I – I
Z
= 4 mA – 1.7 mA = 2.3 mA
The ratio of the resistors is equivalent to the reverse ratio of the currents:
I
R
/ I
Z
= Z / R
The result for resistor R is:
R = (I
Z
/ I
R
) x Z = (1.7 mA / 2.3 mA) x 5.6 kΩ = 4.14 kΩ
The nearest resistor value of the E12 resistor series is 3.9 kΩ . The electric power W of the resist-
ance R can be calculated by the following formula:
W = (Input voltage)
2
/ R = 28.8
2
V / 3.9 kΩ = 0.2 W
The power dissipation of the resistor should be for safety reasons 3 to 5 times higher, as the real
value.
Therefore select a resistor 3.9 kΩ / 1 W for this example.
QH00045C
Fig. 0-1: Signal input switch to module input
QH00045C
Fig. 0-2: Resistor R parallel to input
Input module
Leakage current 4.0 mA
24 V DC
3.6 k
Ω
QX80
Input
impedance
Z = 5.6 k
Ω
I = 4 mA
24 V DC
R
3.6 k
Ω
QX80
I
Z
= 1.7 mA
I
R
= 2.3 mA