Schneider Altivar 61/71 Guía De Explotación página 42

Braking principle and calculation
2 Braking at constant speed
+
1
0
-
2
T
l
P
braki g
Using a drive
Using an asynchronous machine in quadrants II and IV makes the motor work as a generator and restores the electrical energy to the drive
DC bus via its inverter bridge.
The drive DC voltage cannot be returned to the line supply.
This is why the DC bus voltage increases when the motor is operating as a generator.
If the energy returned to the DC bus during braking exceeds the losses generated in the motor and the drive, then the DC bus voltage
increases.
To deal with this problem, it is necessary to increase the deceleration time or to use a braking unit.
The regenerated power depends on the inertia of the load and the deceleration ramp time.
The drive prevents itself from being locked out due to overvoltage by tuning the deceleration ramp time. To retain a short ramp time
(or follow the deceleration ramp) or to work with a driving load, it is necessary to use a braking option such as the braking unit.
Calculating the braking power
1) Calculating the braking time from the inertia
J ω ⋅
----------------- -
t =
b
T + T
b r
⋅
2π n
ω
--------------
=
60
ΣJ ⋅ ( )
n – n
1 2
--------------------------------- -
T =
b ⋅
9,55 t
b
T Motor braking torque
b
J Total inertia applied to the motor
Σ
n Motor speed ahead of gearbox
1
n Motor speed after gearbox
2
t Braking time
b
^
Peak braking power
P
b
–
Average braking power
P
b
during time t
b
1757084 11/2009
At constant speed, the braking power is constant throughout braking.
With fast dynamic braking (deceleration ramp < 2 s) the peak power lasts longer due
to the inertia of the load.
t
b
Example: Vertical downward movement, motor/generator test bench, gravity
t
conveyors, etc.
t
t
P
b
[Nm]
2
[kgm ]
[rpm]
[rpm]
[s]
[W]
[W]
⋅
T n
P ˆ
b 1
---------------- -
=
b
9,55
P ˆ
b
------
P
=
b
2
Motor
J = J + J
motor applied
J
applied
Machi e
Gearbox
1 2
1
i = J
machi e
2
J
machi e
=
2
i
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