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T
rms
I
=
rms
K
t
7.73 = 1.61 A
=
4.8
Power Requirements
The control must supply sufficient power for both the acceleration portion of the movement profile,
as well as for the duty-cycle requirements. The two aspects of power requirements include (1)
power to move the load, P
Power delivered to move the load is:
P
=
T(S
) (746)
del
m
63,025
Power dissipated in the motor is a function of the motor current. Thus, during acceleration, the
value depends on the acceleration current (I
current (I
). Therefore, the appropriate value is used for "I" in the following equation.
rms
P
= I
(R
)
2
diss
m
The sum of these P
and P
del
Example: Power required during the acceleration portion of the movement profile can be obtained
by substituting in equations 8 and 9:
13.75(2,000)
P
=
(746)=325.5 W
del
63025
P
= (2.86)
(4.5)(1.5)=55.2 W
2
diss
P = P
+ P
del
diss
= 325.5 + 55.2 = 380.7 W
Note: the factor of 1.5 in the P
"hot". This is assuming the winding is at 155° C.
Continuous power required for the duty cycle is:
P
= 7.73(2,000) (746) = 182.9 W
del
63025
2
P
= (1.61)
(4.5) (1.5) = 17.5 W
diss
P
= 182.9 + 17.5 = 200.4 W
In Summary
The control selected must be capable of delivering an acceleration current of 2.86 A, and a continu-
ous current of 1.61 A. The power requirement calls for peak power of 380.7 W and continuous
power of 200.4 W.
To aid in selecting, computer software programs are available to perform the iterative calculations
necessary to obtain the optimum motor and control.
Application Notes
, and (2) power losses dissipated in the motor, P
del
); and while running, it is a function of the rms
acc
determine total power requirements.
diss
calculation is a factor used to make the motor's winding resistance
diss
Page 16
(7)
.
diss
(8)
(9)
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