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Duty Cycle Torque
In addition to acceleration
torque, the motor must be able
to provide sufficient torque over
the entire duty cycle or
movement profile. This includes
a certain amount of constant
torque during the run phase,
and deceleration torque during
the stopping phase.
Running torque is equal to
friction torque (T
), in this case,
F
0.95 lb-in.
During the stopping phase,
deceleration torque is:
T
= -J
(a
) +T
dec
t
acc
F
=-(0.00052+.0.00313+0.0037) 1745.2 + 0.95
= -11.85 lb-in.
Now, the root-mean-squared (rms) value of torque required over the movement profile can be
calculated:
T
2
(t
)+T
2
(t
)+T
acc
acc
run
run
T rms=
t
+t
+t
+t
acc
run
dec
idle
=
(13.75)
(.12)+(.95)
(.12)+(11.85)
2
2
.12+.12+.12+.3
= 7.73 lb-in.
The motor, used for this example, can supply a continuous stall torque of 14 lb-in., which is
adequate for the application.
Control Requirements
Determining a suitable control is the next step. The control must be able to supply sufficient accel-
erating current (I
), as well as continuous current (I
acc
ments.
Required acceleration current that must be supplied to the motor is:
I
=
T
acc
acc
K
=
13.75 = 2.86 A
4.8
Current over the duty cycle, which the control must be able to supply to the motor is:
Application Notes
Figure 3. Example Application
Motor Parameters
Inertia = 0.0037 lb-in-s
Continuous Stall = 14 lb-in
Torque Constant (K
Resistance 4.5ohms
2
(t
)
dec
dec
(.12)
2
t
Page 15
Load = 200 lbs
2
Leadscrew
) = 4.8 lb-in/amp
T
Friction = 0.95 lb-in
(4)
) for the application's duty-cycle require-
rms
(6)
Inertia = 0.00313 lb-ins
2
Pitch = 5 revs/inch
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