Stepper motor and driver selection reference scheme
This welding machine requires the table to feed 13 times at the same distance. Previously, the ordinary motor was used and the position switch was used for positioning. It is now desirable to use stepper motor positioning instead of the travel switch and to simplify control. The parameters provided by the customer are as follows:
Workpiece quality: m1=500Kg; Workbench quality: m=1000Kg
Feeding distance: L=95mm/time; Feeding time: T≤2 seconds
Motor torque calculation
The load situation is shown below:
2. Calculation of inertia
Convert the moment of inertia of the system to the output shaft of the motor:
J fold = J2 + m*t 2 /(2π) 2 =0.03+2000*0.012/(2π) 2 =0.035(Kg·m 2 )
(J2 is the moment of inertia of the ball screw, which is approximately 0.03)
3. Motor selection
Due to the slow response of the stepper motor, it is initially decided to set the acceleration/deceleration time during the feed feed to 0.8 seconds, the constant speed operation time to 0.4 seconds, and the maximum speed to 8 revolutions per second (480 rpm). The speed map is as follows:
According to the above parameters, the motor can be rotated within 2 seconds:
0.8*8/2+0.4*8*0.8*8/2=9.6 (turn)
fulfil requirements.
Then the angular acceleration required for the motor to accelerate is:
w=8/0.8*2*3.14=62.8(1/sec)
Therefore, the output torque of the motor at 480 rpm is required to be at least:
T motor = J fold * w + T load = 0.035 * 62.8 + 5.2 = 7.398 (Nm)
Accordingly, 110BYG250D is recommended.
The drive is SH-21006C.
4. User feedback
The user has taken a little extra margin when selecting the model, and chose the 130BYG250D and SH-21006C, which are now in good condition.
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