Delta ASD-A2 Series User Manual page 42

Chapter 2 Installation
Servo Drive
(kW)
0.75 ECMA-L△1305
ECMA-L△1313
1.5
3.0 ECMA-L△1830
Medium
ECMA-J△1330
–High 3.0
Inertia
ECMA-L△1845
4.5
ECMA-L△1855
5.5
ECMA-L△1875
7.5
ECMA-L△1308
1.0
High
Inertia
ECMA-M△1309
1.5
2
Eo= J * wr /182 (joule), Wr: r/min
Assume that the load inertia is N times to the motor inertia and the motor decelerates from
3000r/min to 0, its regenerative energy is (N+1) x Eo. The consumed regenerative resistor is (N+1)
× Eo - Ec joule. If the cycle of back and forth operation is T sec, then the power of regenerative
resistor it needs is 2× ((N+1) x Eo - Ec) / T.
Followings are the calculation procedure:
Steps
Set the capacity of regenerative
1
resistor to the maximum
Set T cycle of back and forth
2
3 Set the rotational speed wr
4 Set the load/motor inertia ratio N
Calculate the maximum regenerative
5
Set the absorbable regenerative
6
Calculate the needful capacitance of
7
Revision February, 2017
Rotor Inertia
Motor
J (× 10-
4kg.m2)
54.95
77.75
99.78
142.7
11.18
Item
operation
energy Eo
energy Ec
regenerative resistor
Regenerative power
from empty load
3000r/min to stop
Eo (joule)
13.1 16.20
23.6 29.18
67.93
12.7 15.70
96.12
123.35
176.41
17.1 84.56
55.29
Calculation and Setting Method
Set P1-53 to the maximum value
Enter by the user or read via P0-02
Enter by the user or read via P0-02
Refer to the above table
2 x ((N+1) x Eo – Ec) / T
The maximum
regenerative power of
capacitance
Ec (joule)
42.43
42.43
42.43
42.43
51.17
57.41
62.40
42.43
57.41
Enter by the user
2
Eo= J * wr /182
ASDA-A2
2-13