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Transcription:

Motor Info on the WWW Motorola Motors DC motor» http://www.freescale.com/files/microcontrollers/doc/train_ref_material /MOTORDCTUT.html Brushless DC motor» http://www.freescale.com/files/microcontrollers/doc/train_ref_material /MOTORBLDCTUT.html Stepper Motor» http://www.freescale.com/files/microcontrollers/doc/train_ref_material /MOTORSTEPTUT.html AC Induction ( squirrel cage ) Motors» http://www.freescale.com/files/microcontrollers/doc/train_ref_ material/motorblactut.html» http://www.sea.siemens.com/step/templates/lesson.mason?m otors:1:1:1

Question How would you make a DC motor move exactly 90 degrees and then stop?

Open-Loop Position Control PLC 4 V DC 4 V DC X0 Y X1 4 V DC motor 4 V DC

Closed-Loop Position Control +V r Ideally will drive motor such that output pot reading is the same as the input pot reading R i + Power Amp R i - R f +V r -V r DC Motor -V r

Open-Loop Position Control 1 V 0 V Stepper Motor Controller 50 On/Off steps from Pushbuttons, PLC, computer, or other controller 00 steps per revolution motor

Stepper Motors stepping motors move in response to a series of electrical pulses, one output "step" per input pulse open-loop control (without output position monitoring) is common three types of stepping motors are widely available permanent magnet, variable reluctance, and hybrid

Permanent Magnet (or PM) Stepping Motors rotor cross-section is gear shaped "teeth" of the gear form N/S poles of magnet Permanent magnet rotor Electro-magnet stator

Permanent Magnet (or PM) Stepping Motors an electrical circuit alternately switches the polarity of the stator poles as the polarity of a stator pole changes, the rotor will move to approach an equilibrium position equilibrium positions where N/S rotor poles align with the S/N stator poles

Figure 1.4 - Simple 1 step/rev hybrid motor South poles on this end of rotor North poles on other end of rotor

Selection of Stepper Motors steps per revolution (or degrees/step) actual output position assumed by the motor depends greatly on the static friction in the system maximum stepping torque cannot be exceeded or the motor will slip causes serious problems in open-loop control systems where the output is assumed to match the number of input pulses

Resolution Stepper motors rotate according to the number of input pulses Resolution: 500 steps/revolution or 0.7 degrees/step

Pulse Frequency

Wave Drive (full steps) Start one set of stator windings is energized, then the other 60 rotation 30 rotation 90 rotation

Two Phase Drive (full step) Start both stator windings are energized at the same time 60 rotation 30 rotation 90 rotation

Figure 1.7 - Half Stepping Start 15 rotation 30 rotation Phase 1 - ON Phase - OFF Phase 1 - ON Phase - ON Phase 1 - OFF Phase - ON

Half Step Mode Twice the resolution (steps/rev) from the same motor Much better smoothness at low speeds Less overshoot and ringing at end of each step Slight loss of torque can be improved with the "profiled current" method of Figure 1.10

"Micro" Step Mode Ratio the current in each of the two phases rotor will be proportionally attracted to the stator pole with the most current 100 to 500 times the resolution (steps/rev) from the same motor Very smooth at low speeds Much more complicated electronics

Stepper Motor Drive Circuits Stepping motor drive waveforms are typically generated from one of four circuits: resistance current limiting bilevel current limiting "chopper" current limiting, and variable voltage current limiting

Fig..4 Principle of R-L drive τ = L R τ = L R External resistance

Sample Problem Need to rotate the index table from stop to stop in 0.6 seconds What do you need to know to calculate the stepper motor torque required?

Torque Required τ = Jα

Knowns Thickness of table = 0.80 cm = 0.31 in Table diameter = 0 cm = 7.9 in Table material = aluminum ρ Al = 0.0065 kg/cm 3 = 0.096 lbm/in 3 Workpiece diameter = 3.0 cm = 1.18 in Workpiece height = 4.0 cm = 1.57 in Workpiece material = steel ρ St = 0.0078 kg/cm 3 = 0.8 lbm/in 3

From Physics & Dynamics J = 1 mr

Knowns Table mass =.66 kg = 5.83 lbm Table diameter = 0 cm = 7.9 in J table = 0.053 kg m = 18 lbm in Workpiece mass = 0.88 kg = 1.94 lbm Workpiece diameter = 3.0 cm = 1.18 in J WP = 0.00040 kg m = 1.35 lbm in

From Physics & Dynamics What is the moment of inertia of the workpiece with respect to center of the table? 1 J = mr + ma r a

From Physics & Dynamics a = total d d 0.085m table WP = = J = J + 8 J + table ( m a ) WP WP 3.36 in J total = 0.108 kg m = 368 lbm in

Acceleration-Velocity-Displacement Acceleration, A T Time, t Velocity, V T Time, t Displacement, S T Time, t

Acceleration-Velocity-Displacement? rad/s 0.60 sec α? rad/s 0.30 sec ω ω max T 45 θ θ max 0 T

From Calculus & Dynamics θ = 45 = π 4 rad = 1 ( ω )( 0.60 sec) max ω max ( ) = = α 0.30sec max α max = rad sec