Choosing a Motor
SECTION 9: CHOOSING A STEP MOTOR AND POWER SUPPLY VOLTAGE
The choice of a stepper motor and power supply voltage is entirely application dependent. Ideally the motor should deliver sufficient at the highest speed the application requires and no more.
Any torque capability in excess of what the application requires comes at the high cost of unnecessary motor heating. Excess torque capability beyond a reasonable safety margin will never be used but will exact the penalty of an oversized power supply, drive stress and motor temperature.
Learn to distinguish the difference between torque and power; high initial torque at low speed does not mean efficient motor utilization. Usually, power is the more important measure of a motor’s suitability to an application. To determine this, you must bias the motor’s operating point through power transmission gearing to operate the motor at its maximum power; normally just past its corner frequency.
The maximum shaft power sustainable with a drive running at 80VDC and 7A is around 250W, or one third of a horsepower. This is primarily achieved with double or triple stacked NEMA 34 motors.
NEMA 23 motors are physically too small to dissipate the resultant hear and NEMA 42 motors are too big to be properly impedance matched; if their current is less than a 7A drive’s limit then the voltage will generally be above the maximum voltage of 80VDC and vice versa.
The detent torque on a NEMA 42 motor is significantly higher than in smaller motors and is always a loss that must be subtracted from the potential available power output of the motor. In other words, the output power of a NEMA 42 motor drops more rapidly with speed than smaller motors. A NEMA 42 motor should be used only if high torque is required at low speed and it is not practical to gear down a smaller motor.
An efficient motor, defined as the smallest motor sufficient to meet the demands of the application, will run hot. Think of the motor as having fixed power conversion efficiency: Some percentage of the input power will be converted to heat and the rest will be converted to mechanical power. To get the maximum performance from the motor, the waste heat must be just under what the motor can tolerate. Usually this motor will be biased to operate just past the corner speed as well.
The place to start is to determine the load torque in oz/in, including the torque necessary to accelerate the load. The next step is to come up with the maximum speed the application has to operate at in full steps per second using the formula below. RPI is the revolutions per inch after the motor turns through the transmission, RPS is revolutions per second and PPS is the number of pulses per second from your step pulse source.
(DESIRED IPM * RPI) / 60 = RPS
RPS * 200 = PPS
Multiply the PPS value by the number of oz/in determined previously and divide the total by 4506. The answer will be how many watts mechanical are required from the motor to meet the load from the application. When picking a motor, choose one with 40% more than the calculated power. Below is an example of the equation completed for a load requiring 450 oz/in with a 3 TPI leadscrew and a desired IPM of 300.
(300 * 3) / 60 = 15
15 * 200 = 3000
(3000 * 450) / 4506 = 299 OZ/IN
299 * 1.4 = 419 OZ/IN
As you can see, you will want to use a motor with a rating of 419 oz/in for this application.