Chapter 2 Installation ASDA-A2
Revision February, 2017 2-13
Servo Drive
(kW)
Motor
Rotor Inertia
J (× 10-
4kg.m2)
Regenerative power
from empty load
3000r/min to stop
Eo (joule)
The maximum
regenerative power of
capacitance
Ec (joule)
Medium
–High
Inertia
0.75
ECMA-L△1305
13.1 16.20 42.43
1.5
ECMA-L△1313
23.6 29.18 42.43
3.0
ECMA-L△1830
54.95 67.93 42.43
3.0
ECMA-J△1330
12.7 15.70 42.43
4.5
ECMA-L△1845
77.75 96.12 51.17
5.5
ECMA-L△1855
99.78 123.35 57.41
7.5
ECMA-L△1875
142.7 176.41 62.40
High
Inertia
1.0
ECMA-L△1308
17.1 84.56 42.43
1.5
ECMA-M△1309
11.18 55.29 57.41
Eo= J * 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:
Steps Item 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
To Next Page
Loading...
Need help?
General
