Actuators application notes - BEI Kimco Magnetics - #9

3.3 Calculation current available to produce torque:

12V/0.45Q = 27A

3.4 Calculation of current required to drive load:

56 oz. in./Kt = (56 oz. in.)(13.0 oz. in./A) = 4.3A

Example:

Bus Voltage - 36V DC Acceleration time - 2 seconds Load - 8" diameter aluminum disk,

0.300" thick Supporting chuck -5" diameter, V*

thick plastic, mounted directly to

motor shaft. Operating speed - 3,600 RPM Duty Cycle: <10%

Step 1 - Load Inertia Calculation

1.1 Aluminum disk:

Equation (9) shows that the iner­tia of a solid cylinder of length L, radius R, and density p may be determined as follows:

J = (0.0041 Sec²/in)R⁴Lp = (0.0041 Sec²/in)(4")⁴(0.300")(1.54 oz/in.³) = 0.4849 oz. in. Sec²

1.2 Plastic chuck:

J = (0.0041 Sec²/in)(2.5")⁴(0.5") (0.64 oz./in.³)= 0.051 oz. in. Sec²

1.3 Motorrotor: Tobeincludedlater.

Step 2 - Acceleration Calculation: Units of Radians/Sec² should always be used in acceleration rate calcula­tions, since "Radian" is a unitless number. The product of acceleration and inertia will drop out to units of torque.

Acceleration rate = operating speed/ acceleration time:

3,600 RPM/9.55=377 Rad/Sec, and Accel, rate = (377 Rad/ Sec) / (2 Sec) = 188 Rad/Sec²

Step 3 - Peak Torque Calculation: It was seen in Equation (4) that peak torque is the sum of the torque due to inertia, load, and friction. Since load (usually windage) and friction torque in applications such as disk certifiers are often negligible, the motor may be selected based on the product of iner­tia and acceleration alone:

T_p = (inertia) (acceleration rate) = (disk inertia + chuck inertia + rotor inertia) (acceleration) = (0.4849 + 0.051 + rotor, oz.in. Sec²) (188 Rad/Sec²)

Ignoring rotor inertia for now,

T = (0.5359 oz. in. Sec²) x (188 Rad/Sec²) = 100 oz. in.

Step 4 - Motor Selection: A motor with a peak torque rating higher than the value calculated in Step 3 should be selected. Again, referring back to Table 5, model DIN34-20 has a peak

torque rating of 120 oz. in. It also has a rotor inertia of 7.7 X 10³ oz. in. Sec² ; a value that is negligible compared to the total load inertia. (If the rotor iner­tia was comparable to or greater than the load inertia, it would be necessary to proceed with motor selection itera­tions, plugging the rotor inertia value in the inertia-times-acceleration-rate equation to insure a proper peak torque requirement.)

Duty cycle check: (100 oz. in.)20% = 20 oz. in., well within the contin­uous stall rating of 50 oz. in.

Step 5 - Winding Selection: The wind­ing should be selected in the same manner as shown in the single speed application example. In this case, the high peak torque requirement neces­sitates selection of a winding with a low resistance, so that enough current may be drawn to provide 100 oz. in. of torque. Winding A appears to be a suitable candidate.

Step 6 - Controller Selection: High acceleration rate applications require selection of a controller that has a peak current rating at least as high as the current dictated by motor winding parameters. The winding A selected in step 5 will result in a peak current magnitude of 100 oz. in./6.1 oz. in./A, or 16A. Again, considering motor efficiencies, a controller with a peak rating at least 20% higher than the calculation value, or about 20A, should be selected.

The high acceleration, single speed application also enables use of two-quadrant control.

The point-to-point positioning applica­tion is by far the most complex of all applications; not just from the stand­point of system integration and load compensation, but because of the dif­ficulty in estimating an average veloc­ity profile or duty cycle, in practice.

The remainder of this section de­scribes five basic power transmission mechanical configurations. Equations specific to each configuration are provided to enable determination of the motor performance required to effect a point-to-point move. Once determined, move speed, reflected torque, and reflected inertia may be used to calculate the peak torque, RMS torque, and speed of operation motor selection parameters discussed on page 5 and 6. The section will conclude with an example of a chemi­cal etching/plating process requiring point-to-point positioning capabilities.

It appears that model DIN34-26, wind­ing A, will provide the 56 oz. in. torque needed at 6,000 RPM operation.*

*/n practice, the winding resistance increases as the motor temperature increases. The PR losses increase; and at high speed-core losses and eddy current losses affect the efficiency of the motor. Again, the 20% safety margin utilized in motor sizing exercises helps ensure proper selection.

Step 4 - Controller Selection: The example shown above required a controller rated at least 70V DC output terminal voltage and 4.3A continuous output current. It should be noted that the voltage at the motor terminals is always lower than the input voltage to the controller due to losses in the elec­tronic components. In addition, brush-less motors are designed for efficien­cies around 80%. A controller with a rated current about 20% greater than the calculated value should ultimately be selected.

Another consideration in controller selection is two-quadrant vs. four-quadrant control. A two-quadrant unit provides positive torque control in two directions (clockwise and counter­clockwise). The four-quadrant unit provides positive and negative torque in both directions. Fixed speed opera­tion applications enable use of the less expensive two-quadrant technology.

In these applications the motor is selected to accelerate the load to a desired operating speed in a specified acceleration time. Once at speed, the motor needs to provide little torque to keep the load in motion. Input voltage, acceleration rate, load and motor iner­tia, and speed of operation are the pa­rameters to be considered, as shown in the following disk certifier example.

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