
EDIT.
SHEET
DRAW. NO.
CUST.
TITLE
87
93
DESCRIPTIONDESIG.DATE
Power Failure Backup Module
Descriptions
A-53866E-464
Selection example: If performing retraction in a gear machine in the event of a power failure.
Conditions
Sub module C used is the A06B-6077-H010 (for 200V)
Maximum cutting output on the synchronous axis
P2=10[kW] (spindle: 7[kW], servo axis: 3[kW])
Retract axis
Rotor inertia of the motor Jm=0.0062[kgm
2
]
Load inertia at motor shaft JL=0.012[kgm
2
]
Motor speed during retraction Vm=1000 [min
-1
]
Friction torque at motor shaft TL=1.6[Nm]
Travel distance to a position where the workpiece and the tool do not interfere with
each other d=20[mm]
Travel distance per motor rotation L=2.5[mm/rev]
Selection
1. Determine the energy W2 [J] for the cutting from the time a power failure occurs until
retraction is started.
From expression <3>
W2 = 32
×10 = 320[J]
2. Determine the energy for movement along the retract axis W3 [J]
From expression <4>
W3 = 5.48
×10
-3
× (0.0062 + 0.012) ×1000
2
+ 6.28×1.6×20 ÷ 2.5 = 180[J]
3. The amount of energy required for retraction in the event of a power failure is
W2 + W3 = 320[J] + 180[J] = 500[J]
3. Determine the amount of energy that can be supplied per sub module W1 [J].
From expression <2>
1) If the input voltage before a power failure starts is 200 VAC
W1 = 2.5
×10
-2
× (200
2
-9.8×10
3
)= 755[J]
2) If the input voltage before a power failure starts is 170 VAC
W1 = 2.5
×10
-2
× (170
2
-9.8×10
3
)= 478[J]
4. Determine the number of sub modules C.
From expression <1>
1) If the input voltage before a power failure starts is 200 VAC
(W2 + W3)
÷ (W1×0.7) = 500 ÷ (755×0.7)= 0.94 → 1
2) If the input voltage before a power failure starts is 170 VAC
(W2 + W3)
÷ (W1×0.7) = 500 ÷ (478×0.7)= 1.49 → 2