11 TROUBLESHOOTING

MELSEC-Q

<Calculation example of Example 4>

 

 

 

 

 

 

 

 

 

QX40

 

 

 

 

Leakage current

If a switch with an LED display

 

 

 

 

 

 

 

 

2.33mA

 

 

 

 

Input module is connected to QX40 and current

 

 

 

 

 

 

 

 

 

4.7k

 

 

 

 

 

of 2.33 mA is leaked.

 

24VDC

 

 

 

 

 

 

 

 

Voltage VTB across the terminal and common base is:

VTB = 2.33[mA]5.6[k] = 13[V] (Ignore the voltage drop caused by the LED.) Because the condition for the OFF voltage (11 [V]) is not satisfied, the input does not turn off. To correct this, connect a resistor as shown below.

 

 

 

QX40

Current I

 

R

Input impedance

 

 

 

5.6k

 

 

 

 

 

 

 

 

 

 

4.7k

24VDC

Calculation of current for resistor R

The voltage of QX40 across the terminals must be reduced to 11 [V] or less. The required current

(24-11[V]) ÷ 4.7[k] = 2.77[mA]

Therefore resistor R of flowing current I of 2.77 [mA] or more must be connected.

Calculation of resistance of connected resistor R

11[V] R > 2.77[mA] - 11[V] 5.6[k]

11[V] ÷ R > 2.77-1.96[mA] 11[V] ÷ 0.81[mA] > R 13.6[k] > R

Resistance of the connected resistor R is obtained in the above equations. Suppose that the resistance R is 12 [kW].

The power capacity W of the resistor during activation of the switch is: W = (Applied voltage)2 / R

W = (28.8[V])2/12[k]=0.069[W]

Because the resistance is selected so that the power capacity is three to five times the actual power consumption, a third to a half [W] should be selected.

In this case, a resistor of 12 [k] and a third to a half [W] should be connected across the terminal and COM.

11 - 40

11 - 40