Справочник Пользователя для Mitsubishi Electronics Q25HCPU

 

 

 

11 - 40                                                                                                                                                       11 - 40 

MELSEC-Q

 

11   TROUBLESHOOTING 

 

<Calculation example of Example 4> 

 

Leakage current 
2.33mA

24VDC

Input module

QX40

If a switch with an LED display 
is connected to QX40 and current 
of 2.33 mA is leaked.

4.7k

 

Voltage V

TB

 across the terminal and common base is: 

V

TB

 = 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

24VDC

Input impedance

R

4.7k

5.6k

 

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] - 

5.6[k   ]

11[V]

 

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.