YASKAWA MEMOCON GL120 User Manual page 266

Here, a dummy resistor can be connected in parallel with the input load to correct
the problem.
AC Output Module
Common
Assuming that the load will operate properly with a leakage current of 1.5 mA or
less, then the value of the dummy resistor R can be computed as follows:
R
3 mA × < 1.5 mA
R + 6 kΩ
∴ R < 6 kΩ
Thus, the value for R should be 6 kΩ or less.
If the resistance is too small, the amount of heat generation will increase, and a high
wattage will be required. Here, we will compute the wattage for a dummy resistor of
6 kΩ.
(Power supply voltage)
W = = = Approx. 1.7 W
R
Normally, about three times the computed value is used to allow surplus wattage.
A 5-W resistor would thus be used.
7) Connecting Solenoids with Diodes
Some solenoids used as the load of the AC Output Module may contain built-in
diodes.
Solenoids with diodes have an advantage in that they are driven by half-wave recti-
fication and thus require a lower activation current.
Abide by the following precautions when such diodes are used as a load for an AC
Output Module.
a) Overvoltage can be applied to the load when the output is OFF. The rectifying diode
must therefore be able to withstand a reverse voltage of 2
AC Output Module
Fig. 6.19 Solenoids with Diode (1)
Load device
Output
Dummy
resistor
Load power supply
(100/200 VAC)
2
2
(100 V)
6 kΩ
Output
Leakage current = 5 mA
Common
Load power supply
(100/200 VAC)
6-39
6.4 Precautions on Wiring
Load impedance
ZL
E or greater.
2
Solenoid
Rectifier diode
Flywheel
diode
6