Chapter 2 Installation ASDA-B2
2-14 Revision May, 2018
Servo Drive
(kW)
Servo Motor
Rotor Inertia
J (× 10
-4
kg.m
2
)
Regenerative
power from empty
load 3000r/min to
stop
Eo (joule)
Max.
regenerative
power of
capacitance
Ec(joule)
Medium
Inertia
3.0
ECMA-E△1830
54.95 217.73 28
ECMA-F△1830
54.95 217.73 28
ECMA-E△1835
54.95 217.73 28
Medium-high
Inertia
1.0
ECMA-F△1308
13.6 67.25 18
2.0
ECMA-F△1313
20.0 98.90 21
2.0
ECMA-F△1318
24.9 123.13 21
3.0
ECMA-F△1830
54.95 217.73 28
High
Inertia
0.4
ECMA-G△1303
8.17 40.40 8
0.75
ECMA-G△1306
8.41 41.59 14
1.0
ECMA-G△1309
11.18 55.29 18
Eo = J x wr
2
/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:
Step Procedure Calculation and Setting Method
1
Set the capacity of regenerative
resistor to the maximum
Set P1-53 to the maximum value
2
Set T cycle of back and forth
operation
Enter by the user
3 Set the rotational speed wr Enter by the user or read via P0-02
4 Set the load/motor inertia ratio N Enter by the user or read via P0-02
5
Calculate the maximum
regenerative energy Eo
Eo= J * wr
2
/182
6
Set the absorbable regenerative
energy Ec
Refer to the above table
7
Calculate the needful capacitance
of regenerative resistor
2 x ((N+1) x Eo – Ec) / T
Take 400W as the example, the cycle of back and forth operation is T = 0.4sec, the
maximum speed is 3000r/min and the load inertia is 7 times to the motor inertia. Then,
the needful power of regenerative resistor is 2 × ((7+1) × 1.68 – 8) / 0.4 = 27.2 W. If it is
smaller than the built-in capacity of regenerative resistor, the built-in 60W regenerative
To Next Page
Loading...
