Mitsubishi Electric MR-J2S-*A, Инструкция по эксплуатации

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13 - 2
13. OPTIONS AND AUXILIARY EQUIPMENT
(2) Selection of the regenerative option
(a) Simple selection method
In horizontal motion applications, select the regenerative option as described below.
When the servo motor is run without load in the regenerative mode from the running speed to a
stop, the permissible duty is as indicated in section 5.1 of the separately available Servo Motor
Instruction Manual.
For the servo motor with a load, the permissible duty changes according to the inertia moment of
the load and can be calculated by the following formula.
Permissible
duty Permissible duty for servo motor with no load (value indication Section 5.1 in Servo Motor Instruction Manual)
(m 1)
ratedspeed
running speed
[times/min]
2
where m load inertia moment/servo motor inertia moment
From the permissible duty, find whether the regenerative option is required or not.
Permissible duty number of positioning times [times/min]
Select the regenerative option out of the combinations in (1) in this section.
(b) To make selection according to regenerative energy
Use the following method when regeneration occurs continuously in vertical motion applications or
when it is desired to make an in-depth selection of the regenerative option.
a. Regenerative energy calculation
Use the following table to calculate the regenerative energy.
M
Friction
torque
U
nb
al
an
ce
to
rq
ue
T
F
T
U
S
er
vo
m
ot
or
s
pe
ed
G
en
er
at
ed
to
rq
ue
Time
Up
No
tf(1 cycle)
Down
( )
( )
(Driving)
(Regenerative)
T
psd2
t2 t3 t4 t1
T
psa2
T
psd1
T
psa1
1)
2)
3)
4)
5)
6)
7)
8)
Formulas for calculating torque and energy in operation
Regenerative power Torque applied to servo motor [N m] Energy [J]
1)
T
1
(J
L
J
M
)
9.55 10
4
N
0
1
T
psa1
T
U
T
F
E
1
2
0.1047
N
0
T
1
T
psa1
2) T
2
T
U
T
F
E
2
0.1047 N
0
T
2
t
1
3)
T
3
(J
L
J
M
)
9.55 10
4
N
0
1
T
psd1
T
U
T
F
E
3
2
0.1047
N
0
T
3
T
psd1
4), 8) T
4
T
U
E
4
0 (No regeneration)
5)
T
5
(J
L
J
M
)
9.55 10
4
N
0
1
T
psa2
T
U
T
F
E
5
2
0.1047
N
0
T
5
T
psa2
6) T
6
T
U
T
F
E
6
0.1047 N
0
T
6
t
3
7)
(J
L
J
M
)
9.55 10
4
N
0
T
7
1
T
psd2
T
U
T
F
E
7
2
0.1047
N
0
T
7
T
psd2
From the calculation results in 1) to 8), find the absolute value (Es) of the sum total of negative
energies.