AutomationDirect SureServo2 Manual - Halaman 15
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Halaman Cetak
Servo
Drive
Motor
(W)
SV2H-2B0N
SV2H-2B0B
15000
SV2H-2F0N
SV2H-2F0B
Assume that the load inertia is N times the motor inertia, and when motor decelerates from
3000 rpm to 0, the regenerative power is (N+1) × E
consume (N+1) × E
required power of regenerative resistor is Watts = 2× ((N+1) × E
The calculation is as follows:
Step
Set the wattage of the regenerative resistor to
1
the maximum�
2
Set the operation cycle (T)
3
Set the rotation speed in RPMs (wr)
4
Set the load / motor inertia ratio (N)
Calculate the maximum regenerative resistor
5
(E
)
0
Find the regenerative power that can be
6
absorbed by the capacitor (E
Calculate the required capacity (in Watts) of
7
the regenerative resistor
Example 1:
For the motor SV2L-204B (400W), the reciprocating motion cycle is T = 0.4 sec.
Its maximum speed is 3000 rpm and the load inertia is 15 times of the motor inertia.
Servo
Drive
Motor
(W)
400
SV2L-204x
Find the maximum regenerative power: E
Find the regenerative power that can be absorbed by the capacitor; E
table).
The required capacity of the regenerative resistor =
Inserting the values for the variables gives:
From the calculation above, the required power of regenerative resistance is 43.6 W, which is
slightly greater than the specified capacity. In this case, a built-in 40W regenerative resistor
does not quite fulfill/ah the need. In general, the built-in regenerative resistor can meet the
requirement when the external load is not too great. The leading "2" in the equation is for
the protection of the resistor sizing and is why P1.053 is double what the drive uses for regen
calculations.
SureServo2 User Manual – 1st Edition – 05/20/2021
Regenerative Power Generated when
Rotor Inertia
the Motor Decelerates from Rated
-4
J (x 10
kg·m²)
Speed to 0 RPM without Load
338
346�5
451
461�8
-E
joules. Assume that the reciprocate operation cycle is T sec, then the
0
c
What to Do
)
c
Regenerative Power Generated
Rotor Inertia
when the Motor Decelerates from
J (x 10-4kg.m²)
3000RPM to 0 without Load
0�15
= 0.74 joules (from the table).
0
2 x (15+1) x 0.74 – 6.24
0.4
Chapter 2: Installation
Max Regenerative Power of
E
( joule)
0
428�37
570�91
and the regenerative resistor needs to
0
-E
) / T.
0
c
Calculation and Setting Method
Set P1�053 to the maximum value
Manual input (T=seconds)
Manual input or read the status with P0�002
Code (09h)
Manual input or read the status with P0�002
E
= J * wr2/182
0
Note, the value 182 comes from the formula
1
182 =
below:
1
2 /60)
2 / × (
Refer to the table above
2 ×((N+1) × E
-E
) / T
0
c
Max Regenerative Power of
E
( joule)
0
0�74
= 6.24 joules (from the
c
2 x (N+1) x E
– E
0
c
T
= 43.6 W
the Capacitor Bank
E
( joule)
c
155�93
155�93
2
the Capacitance
E
( joule)
c
6�24
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