Chapter 2 Installation ASDA-M
2-10 Revision December, 2014
Assume 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) × 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) × 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 ×（(N+1) × 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 = 27.2W. If it is
smaller than the built-in capacity of regenerative resistor, the built-in 60W
regenerative resistor will do. Generally speaking, when the need of the external load
inertia is not much, the built-in regenerative is enough. The diagram below describes
the actual operation. The smaller power of the regenerative resistor it has, the more
energy it accumulates and the higher temperature will be. When the temperature is
higher than a specific value, ALE05 occurs.
To Next Page
Loading...
Need help?
General
