1. Stepper Motor-1©1998-2003 by M. Zarrugh ISAT 303 Module II Electromechanical Actuators  The objectives of this module are to –understand the concept of actuation. –learn about types of actuators from the simple on/off switch to servo motor drives. –understand the physics and applications of stepper motors. –learn how to control output speed and position in a closed loop to create a desired motion profile. –learn how to select a stepper motor for a particular load level

2. Stepper Motor-2©1998-2003 by M. Zarrugh Actuators: What are They?  Actuators are devices which accept a small control signal to produce a large effect in the system output.  Actuator Types: electrical, mechanical (hydraulic or pneumatic)  Electrical: On/off switches and motors Actuator Plant or System Actual Output Desired Output Sensor Controller

3. Stepper Motor-3©1998-2003 by M. Zarrugh Mechanical Actuators: Hydraulic and Pneumatic Cylinders Cylinder 4-Way control valve Work being actuated (pushed) Piston Hydraulic cylinder: working fluid is liquid Pneumatic cylinder: working fluid is air Working fluid from and to pump Fluid flowing from the cylinder Fluid flowing to the cylinder

4. Stepper Motor-4©1998-2003 by M. Zarrugh Actuators: On/Off Switches  Electromechanical switches (relays): a low voltage control signal energizes a electromagnetic coil which closes a switch to close or open a high voltage (or current) circuit.  Programmable limit switches: to produce a regular sequence of on/off conditions.  Alarms/Annunciators: for monitoring process conditions and sounds an alarm if a pre-set condition is reached. Return spring Electro- magnet When electro-magnet is energized, contacts close and deliver power to the load +

5. Stepper Motor-5©1998-2003 by M. Zarrugh DriveProsConsTypical Use Motor Drives: Types, Pros and Cons Stepper Motor DC Motor AC (brushless) Lowest cost Compact size Digital control Limited power Small payload Low efficiency Easy to design Easy to maintain Quiet operation high speed Low maintenance Largest payload Complex implementation Motor adds inertia load Require bulky & costly inverters Cost the most Moderate conditions Low speed High power Inertial loads (high torque) Very high speed Hazardous use No maintenance

6. Stepper Motor-6©1998-2003 by M. Zarrugh Motor Drives: Steppers Fundamentals  Convert a series of input pulses (steps) into a proportional angular movement.  The motor shaft position is determined by the pulse count and its speed by the pulse rate.  The permanent magnet rotor is propelled by successively energizing peripheral electromagnets (stator poles or windings).  Motor must start and stop in the start & stop zone, but can operate in the slew zone. Start & Stop zone Torque Stepping Rate (speed) Starting and stopping limit “L-curve” Max. speed limit “U-curve” Slew zone N S

7. Stepper Motor-7©1998-2003 by M. Zarrugh Motor Drives: Steppers Fundamentals Example: A constant load torque of 70 oz.in is supplied by a stepper motor with following approximate torque-speed relations: L-curve: T (oz.in) = 200 - 0.050 V(steps/s) U-curve: T (oz.in) = 200 - 0.025 V(steps/s) T (oz.in) V (10 3 steps/s) 8 U 200 4 L Find: (a) Maximum allowable starting speed and running speed (b) Starting time if the motor runs 13 pulses at the max starting speed Solution: (a) The maximum starting speed V L is determined by equating the load torque to the supply torque given in the L-equation: 70 = 200 – 0.05 V L or V L = 2600 steps/s The maximum running speed V U is similarly determined. V U =5200 steps/s (b) During the starting period, the speed (or frequency) is 2600 steps/s; thus t = (no. of steps) / (steps/s) = 13 steps / 2600 steps/s = 5 ms. Load 70 VLVL VUVU

8. Stepper Motor-8©1998-2003 by M. Zarrugh Stepper Motor: Current Variations  The pole windings are wired into two separate sections called phases. The phases are sometimes divided further into two parts. Phases allow for many variations in current patterns: –Full-step current: both phases are always energized –Half-step current: the two phases are not always energized (most common)  Standard 2-phase 200-step stepper motor –50 teeth on each of the two rotor pole sections –stator has 8 poles each with 5 teeth –current in stator pole windings is sequenced to allow 1/4 tooth “effective” rotor rotation per step which results in the 200 steps.

9. Stepper Motor-9©1998-2003 by M. Zarrugh Stepper Motors: Drives  The drive is a separate section in the motor system. The drive responds to speed and position control commands by delivering appropriate electrical current to the motor to achieve the desired motion.  The translator section of the drive translates the step and direction signal into a set of pole switches to sequence the rotor movement. Translator Switch Set Stepper Drive Subsystem Direction Phase 1 Phase 2 Motor Step

10. Stepper Motor-10©1998-2003 by M. Zarrugh Stepper Motors: Performance Curves  Torque/speed curves are called motor performance curves.  These curves are the fundamental indicators of dynamic (motion and power) behavior. They specify combinations of torque and speed for safe operation.  At low speed, output torque depends on drive current setting.  At high speeds, output current depends on drive supply voltage. Torque Speed “U-curve” Input pulse Motor Current

11. Stepper Motor-11©1998-2003 by M. Zarrugh Stepper Motors: Performance Curves  The two windings halves in an 8-lead 200-step stepper motor can be connected either in series or in parallel.  The series connection doubles the number of winding turns and increases the output torque at low speeds, but reduces it at high speeds due to 4-fold increase in inductance. Series Parallel Series Parallel Torque Speed

12. Stepper Motor-12©1998-2003 by M. Zarrugh Motor Selection: Design Considerations  Motor torque/speed supply characteristics  Load torque/speed requirements  Load and motor inertia  Friction  Torque operating margin (>50%)  Duty cycle Torque Speed Motor Constant Load Load varies with speed Speed Time Duty Cycle Motor and Load Characteristics

13. Stepper Motor-13©1998-2003 by M. Zarrugh Motor Selection: Dynamic Considerations  Motor torque must overcome friction and inertia: Total torque = Friction torque + Acceleration torque T t = T f + T a  The friction torque, T f is a constant to be determined experimentally. Motor  The acceleration torque, T a depends on the moment of inertia J (oz-in 2 ) and angular acceleration  rad/s 2  : T a = J  Load m, J TaTa TfTf TtTt

14. Stepper Motor-14©1998-2003 by M. Zarrugh Motor Selection: Kinematic (Motion Profile) Considerations  In most positioning applications, a typical cycle of movement can be represented with a trapezoidal motion profile.  The cycle begins with a constant acceleration stroke, continues with a constant velocity stroke and ends with a constant deceleration segment.  Angular acceleration, speed and displacement are related by similar equations to linear motion. Speed,  Time, t t1t1 t3t3 t2t2 Accelerate  =  max /  t 1 Const. speed,  max Decelerate  =   max /  t 3 Linear Angular v = v 0 + at  =  0 +  t d =v o t + at 2 /2  =  o t +  t 2 /2 v 2 = v o 2 + 2ad  2 =  o 2 + 2 

15. Stepper Motor-15©1998-2003 by M. Zarrugh Motor Selection: Continuous Torque Requirement  At a given maximum operating angular speed, the motor torque must be at least 1.5 times the required continuous load torque.  The total required load torque T t is computed from the known friction torque and the time  t required to accelerate to  max. Motor Torque Speed,   max Continuous load torque TtTt >0.5 T t TmTm Given: T f = 12 oz-in  max = 10 rps = 10 rev/s  t = 0.2 s J tot = 50 oz-in 2 Find: Total load torque Minimum motor torque at  max Maximum Operating torque

16. Stepper Motor-16©1998-2003 by M. Zarrugh Stepper Motor: PROBLEM SOLUTION  =  max /  t = (10 rev/sec)/0.2 sec) = 50 rev/sec 2  = 50 rev/sec 2 (2  rad/rev) = 100  rad/sec 2 T T = T F + T  and T F was given as 12 oz-in. T  = J*  = 50 oz-in 2 (100  rad/sec 2 ) T  = 50 oz-in 2 (100  rad/sec 2 )(1/386 in/sec 2 )# T  = 40.7 oz-in T T = T F + T  = 12 + 40.7 = 52.7 oz-in = total load torque Minimun Motor Torque at  max =1.5 T T = 1.5(52.7) =79.1 oz-in. #conversion factor for converting gravitational units(in/s 2 ) to units of mass(oz-in 2 )  The pole windings are wired into two separate sections called phases. The phases are sometimes divided further into two parts. Phases allow for many variations in current patterns: –Full-step current: both phases are always energized –Half-step current: the two phases are not always energized (most common)  Standard 2-phase 200-step stepper motor –50 teeth on each of the two rotor pole sections –stator has 8 poles each with 5 teeth –current in stator pole windings is sequenced to allow 1/4 tooth “effective” rotor rotation per step which results

17. Stepper Motor-17©1998-2003 by M. Zarrugh Stepper Motor: APPLICATIONS  COMPUTER PERIPHERALS –FLOPPY DISC-POSITION MAGNETIC PICKUP –PRINTER-CARRIAGE DRIVE  BUSINESS MACHINES –CARD READER-POSITION CARDS –COPY MACHINE-PAPER FEED  PROCESS CONTROL –IN-PROCESS GAGING-PART POSITIONING –I.C. BONDING-CHIP POSITIONING  MACHINE TOOL –MULTI-AXIS MACHINES-X-Y-Z POSITIONING –GRINDING MACHINES-AUTO WHEEL DRESSING  The pole windings are wired into two separate sections called phases. The phases are sometimes divided further into two parts. Phases allow for many variations in current patterns: –Full-step current: both phases are always energized –Half-step current: the two phases are not always energized (most common)  Standard 2-phase 200-step stepper motor –50 teeth on each of the two rotor pole sections –stator has 8 poles each with 5 teeth –current in stator pole windings is sequenced to allow 1/4 tooth “effective” rotor rotation per step which results