Delta ASDA-A3 Series User Manual page 76

ASDA-A3
Assume that the load inertia is N times the motor inertia, when the motor decelerates from
3,000 rpm to 0, the regenerative energy is (N+1) × Eo and the regenerative resistor needs to
consume (N+1) × Eo - Ec joules. Assume that the reciprocating motion cycle is T sec, then the
required power of regenerative resistor = 2 × ((N+1) × Eo - Ec) / T. The calculation is as follows:
Step
Set the capacity of the regenerative
1
resistor to the maximum.
2 Set the operation cycle (T).
3 Set the rotation speed (wr).
4 Set the load / motor inertia ratio (N).
Calculate the maximum regenerative
5
energy (Eo).
Find the regenerative energy that can be
6
absorbed by the capacitor (Ec).
Calculate the required capacity of the
7
regenerative resistor.
Example:
For the motor ECM-A3L-CY0604RS1 (400 W), the reciprocating motion cycle is T = 0.4 sec.
Its maximum rotation speed is 3000 rpm and the load inertia is 15 times of the motor inertia.
Servo drive
(kW)
0.4 ECM-A3L-CY0604RS1
Find the maximum regenerative energy: Eo = 0.74 joules (from the preceding table).
Find the regenerative energy that can be absorbed by the capacitor: Ec = 8.42 joules (from the
preceding table).
The required capacity of the regenerative resistor =
17.1 W
From the calculation above, the required power of the regenerative resistor is 17.1 W, which is
smaller than the specified capacity. In this case, the built-in 40 W regenerative resistor fulfills the
need. In general, the built-in regenerative resistor can meet the requirement when the external
load is not too great.
Item
Rotor inertia
Motor
J (× 10
0.15
Calculation and setting method
Set P1.053 to the maximum value.
Manual input.
Manual input or read the status with P0.002.
Manual input or read the status with P0.002.
2
Eo = J x wr /182
Refer to the preceding table.
2 × ((N+1) × Eo - Ec) / T
Regenerative energy
generated when the motor
-4 2
kg.m ) decelerates from 3000 rpm
to 0 without load Eo (joule)
0.74
2 × ((N + 1) × E - E )
c
0
T
Installation
Maximum
regenerative
energy of the
capacitance Ec
(joule)
8.42
2 × ((15 + 1) × 0.74 - 8.42)
= =
0.4
2
2-31