Table of Contents

B. ESCAP IRONLESS ROTOR DC MICROMOTORS AND STEP MOTORS

1.0 INTRODUCTION 347

2.0 TABLE OF INTERNATIONAL UNITS AND CONVERSION FACTORS GERMANE TO MICROMOTORAPPLICATIONS 347

3.0 ADVANTAGES AND UNIQUE FEATURES OF AN IRONLESS ROTOR MOTOR 347

4.0 CONSTRUCTION OF THE ESCAP MICROMOTOR 349

4.14.24.34.44.5

General Construction and GeometryStator Housing AssemblyRotor AssemblyBrush Holder End-Cap Assemblyd.c. Tachometers

349350350350350

5.0 REVIEW OF PHYSICAL LAWS GOVERNING ESCAP MOTOR APPLICATIONS 351

5.15.25.3

Rectilinear MotionRotational MotionBasics of DC Circuits

351352355

6.0 BASIC MOTOR PHYSICS 356

6.16.26.36.4

Simplified Coil In a Magnetic FluxEscap Micromotor Energy Flow and ConversionDevelopment of the Equivalent CircuitFundamental Equations of the lronless Rotor Motor

356357357358

7.0 GEARBOXES APPLIED WITH ESCAP MOTORS 366

7.17.27.3

EfficiencyStall Rating of GearboxInertia Transfer

366366366

8.0 CONSIDERATIONS FOR THE CONTROL OF DC MICROMOTORS 367

8.18.28.38.48.58.68.7

General CommentsSummary of Feedback Techniques and ComponentsIronless Rotor Micromotor Model With Laplace Transform NotationOpen Loop Control ConsiderationsClosed Loop Control ConsiderationsServo Amplifiers and Control TechniquesAdditional Comments

367367369369370371376

345

9.0 TUTORIAL SELECTION OF MICROMOTOR APPLICATION PROBLEMS 3769.19.29.39.49.59.69.79.89.9

Two Problems In Angluar MotionDrive Motor Application ProblemDrive Motor Selection ProblemGearbox Application ProblemCase Study: The "DC 300" Cartridge DriveDynamic Braking of a DC MicromotorExample of Graphing Motor CharacteristicsConsiderations For Tachometer ApplicationsLight Chopper Drive Motor

376376379388381387388389390

10.0 REVIEW OF ESCAP STEP MOTORS 39010.110.210.310.410.510.610.710.810.910.10

DescriptionAdvantages and Unique FeaturesDetent TorqueHolding TorqueDynamic CharacteristicsDamping ConsiderationsResonance Accuracy Motor Drive TechniquesApplication ExampleReferences

380391392392394396396396397397401

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B. ESCAP IRONLESS ROTOR DC MICROMOTORS AND STEP MOTORS*

1.0 INTRODUCTIONThis chapter on dc micromotors is intended as a tutorial review of the major engineering considerations required for the properselection and application of ironless dc micromotors and tachometers. The text has been prepared for publication by Stock DriveProducts with the permission of Portescap U.S.

2.0 TABLE OF INTERNATIONAL UNITS AND CONVERSION FACTORS GERMANE TO MICROMOTOR APPLICATIONSThe International System of Units is used throughout this chapter, and Escap products are specified in this system of units.Consequently, the following tables have been prepared for the convenience the reader. A more complete treatment of theInternational System of Units and Conversion Factors is presented in the Designers Data Section of this handbook.

3.0 ADVANTAGES AND UNIQUE FEATURES OF AN IRONLESS ROTOR MOTOR

Absence of Rotor Poles no cogging (no preferred rotor position) very low torque and back EMF variation with rotor positionSmall Rotor Inertia rapid mechanical responseLow Rotor Inductance negligible electrical time constant low torque ripple low electrical noise minimum brush/commutator electro-erosionPrecious Metal Brush/Commutator System negligible voltage drop at the brushesNegligible Hysteresis and Eddy Current LossesVery Stable Magnetic Field very high over excitation to nonlinearity or demagnetization low stray fieldsHigh Power to Volume Ratio intense magnetic field narrow air gap high fill factor of copper conductorContinuous Copper Winding excellent dynamic balance

_____________* Prepared and Edited by: Peter J. Thornton. P.E.* Printed with the permission of Portescap U.S.* ESCAP Registered Trademark of Portescap.

347

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Low Internal Friction tow starting voltage low no-load currentHigh EfficiencyGood Compatibility with Simple Servo Drive ElectronicsLinear Characteristics

4.0 CONSTRUCTION OF THE ESCAP MICROMOTOR

4.1 General Construction and GeometryFigure 1 shows the major subassembly breakdown of the motor. Brushes, housing, rotor and magnet are clearly shown. Theprimary feature is the self supporting "basket" construction of the rotor. In this geometry (Figure 2), the rotor is a perfectlywound (one conductor placed next to its neighbor without gap) surface of a right circular cylinder bonded to an insulating discwhich supports the coil and commutator segments. The coil is wound in a unique skew configuration giving both electricalsymmetry and mechanical stability. Precious- metal brushes and commutator are used to provide a near zero voltage drop, lowfriction and long life (Figure 3). Multiwire brush construction is employed to provide parallel contact redundancy. Finally (Figure 4) the motor is assembled around a central permanent magnet and core which supports the bearings, outermagnetic return path and also the brush holder.

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4.2 Stator-Housing AssemblyThe stator consists of the central cylindrical permanent magnet, the core which supports the bearings, and the steel tube whichcompletes the outer magnetic return path. All three of these parts are held together by the motor front plate, or the mountingplate. The magnetic core is magnetized diametrically after it has been mounted in the magnetic system. Owing to the lowpermeability of the permanent magnet, the magnetic resistance is great enough to allow neglect of the weakening of the statorfield due to the rotor current.

4.3 Rotor AssemblyThe ironless rotor, in the form of a cylinder, rotates around the magnetic core. In Escape motors, the rotor is made in the form ofa skew wound coil and is fixed to a disc carrying the commutator. According to the motor dimensions, the commutator hasbetween 5 and 13 segments. The small commutator diameter results in low sliding-speed and the small friction gives highefficiency. The perfectly symmetrical construction of the skew wound coil makes for smooth and silent running. The crossing ofthe wires makes the self supporting rotor winding (armature) highly stable mechanically.

4.4 Brush Holder End-Cap AssemblyThe endcap includes the brush system. Owing to the use of precious metals for the brushes and the commutator segments, thelow contact resistance of the brushes is maintained throughout the life of the micromotor and ensures starting at very lowvoltages.

4.5 d.c. TachometersThe construction of the Escap tachometer is identical to that of the motor. However, some engineering differences in the brushcommutator system form the basic distinguishing features of the tachometer. The precious metals used for commutator andbrushes have been selected to combine maximum life time with low noise performance. Unlike tachometers with carbon brushes,the use of precious metal brushes allows operation at very low currents (

The actual value is largely determined by the precision of manufacture, primarily in coil winding.

typically generator output signal, 5segment commutator

a) value of the induced voltage(1 V) withactual ripple of 90 mV(9%)b) (scale 10 times larger than a) minimumtheoretical ripple, frequency per revolutionequals twice the number of commutatorsegments.

c) actual ripple value, twice perrevolutiond)signal for one complete revolution ofrotor

Figure 5 Tachometer Output Signal

5.0 REVIEW OF PHYSICAL LAWS GOVERNING ESCAP MOTOR APPLICATIONS

5.1 Rectilinear Motion5.1.1 Velocity, sometimes referred to as speed, is equal to distance (displacement) traveled per unit time (second).

Thus, v = dx dt

(v)velocity = (x)distance meters (t) time second

5.1.2 Acceleration is defined as the change in velocity per unit time. Therefore:

(a) acceleration = dv = dx dt dt v = Vo + at

5.1.3 Displacement is the distance traveled after time (t).Thus: (x) = v0t + 1 at + x0 2 and with zero initial velocity (when v0 = 0 and x0 = 0) x = 1 at 2

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5.1.4 Force is defined as a mass (m) multiplied by the acceleration (a) which it is experiencing.Thus: F = ma

Conversely, the mass of a body is represented by the force per unit acceleration.

5.1.5 The concept of inertia comes into play when describing the amount of force that is required to accelerate a body. A smallinertia requires less force to achieve a given acceleration than does a body with a larger inertia. Thus, load inertia is a criticalconsideration when making a motor application.

5.1.6 When a force is applied to a body which is free to rotate about some axis, the product of the magnitude of the force and itsforce arm is called the moment of the force about the axis. The force arm is defined as the perpendicular distance between the line of action of the force and the axis of rotation. It is this property of a moment of a force which produces torque and thus rotation of a body (load).

5.1.7 Work and energyWork is defined as the product of displacement and the component of force in the direction of displacement. Thus: W = FxThe kinetic energy of a moving body is expressed as:

Thus, the work done in accelerating a body equals the increase in its kinetic energy.

5.1.8 Mechanical power

Average power:

and instantaneous power:

and under constant force: P = Fv

5.2 Rotational Motion

5.2.1 Angular velocity

NOTE: 2pi radians = 1 rotation (360) By definition a radian is an angle whose arc length is equal to the radius of the arc. Also, "speed" should not be used to describe angular velocity. Rotational speed can only be expressed as RPM or RPS.

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5.2.2 Angular acceleration

Under acceleration angular velocity changes as follows: = o + t 5.2.3 Angular displacement after time (t) is defined as:

5.2.4 Handy equations of angular rotation physics with constant angular velocity ( = constant and o = 0) = t

alternate forms: w =

with constant acceleration ( = constant):

where = o + t

a =

with constant acceleration () and with zero initial velocity (o = 0):

angular acceleration of a load (J) with constant torque (M) is:

a =

5.2.5 Moment of inertia. The moment of Inertia (J) of a body referred to an axis of rotation is the product of the mass of thebody and the square of the distance between the center of mass and the axis of rotation. J = mr

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For a flywheel: (J) = mr and for a homogeneous disc:

5.2.6 Energy, work and powerRotational kinetic energy is expressed by:

And work (W) is defined as: dW = Md where M is the torque causing the displacement (). Under constant torque the work done over a given displacement is stated as: W = M (2 - 1)

Now

but is rate of doing work (Power)

and

Hence, power P = M (under constant torque)Net work = change in kinetic energy

and

M = J = J which is the rotational analogue of Newton's second law, F = ma in linear motion.

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5.2.7 MomentumThe law of conservation of momentum states that the total momentum of a system can only be changed by external forces actingon the system. In rectilinear motion the system momentum is given as the product of mass and velocity (mv). Whereas angular momentum is: L = J 5.3 Basics of DC Circuits 5.3.1 OHM'S law

R = resistanceE =voltageI = current

(ohms)(volts)(amperes)

variations:

E = lR and l =

5.3.2 Power P = power (watts)Power input to a network is computed as: P = El and the power (heat) dissipated in a resistance is given by: p = l R 5.3.3 Kirchhoff's rules Point rule: The algebraic sum of the currents toward any point of a network is zero. i = 0 loop rule: The algebraic sum of the voltage sources and the lR products in any loop of a network equals zero. E + iR = 0 5.3.4 Series and parallel resistance The equivalent resistance of resistors in series is the sum of the individual resistances. Thus: R = R1 + R2 + R3 + ... RN The equivalent resistance of resistors in parallel is computed as:

A special solution for only 2 parallel resistors is;

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6.0 BASIC MOTOR PHYSICS

6.1 Simplified Coil In a Magnetic Flux

Generator (Tacho)

Action: displacement (d)

Reaction: induced Emf (E)

Motor

Action: current (i)

Reaction: Force (F)

= 4500 GaussForce = f (, l, i)Torque = f (, l, i)Emf = f (, l, )Torque is current (i)dependentEmf is speed ()dependent

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l is a constant as determined by the winding length (l) and the magnetic flux density (). l = KT torque constant (motor) l = Kv voltage constant (Tacho)6.2 Escap Micromotor Energy Flow and Conversion

6.3 Development of the Equivalent Circuit

Vo = applied terminal voltage Lm = rotor inductance Rm = rotor resistance Ei = motor's induced back Emf i = circuit current flow

Summation of loop voltage drops yields:

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Since the Escap rotor inductance is 100 to 1,000 times less than that of conventional "iron-core" motors it can be neglected.

Therefore, the Lm term vanishes and the simplified voltage loop equation becomes:

Vo = Rm i + Ei

6.4 Fundamental Equations of the Ironless Rotor Motor

6.4.1 Motor Model

Since V = Rl + E we can substitute Kv for E since the back emf is equal to the angular velocity () multiplied by the motorconstant (K). Where Kv is the back emf constant and KM is the torque constant. V = Rl + Kv and M = KMl.

NOTE: Kv = KM = K This equality exists only when Kv and KM are expressed in metric units.

NOTE: Kv =

Therefore: V = Rl + K where K Kv = KM

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This same equation can be derived in another way as follows:

Air gap power in = air gap power out(from Figure 9) El = M Kl = M

Kl = M or l =

Power ln = dissipation + power out.(from Figure 10) Vl = lR + El Vl = lR + M Vl = lR + Kl V = lR + K

Alternate Forms:

6.4.2 Torque Speed

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The complete torque expression including mechanical losses is:

M = ML + Mf ML = M - Mf

Mf = Friction TorqueML = Load TorqueM = Motor Torque

NOTE: the origin of the torque speed characteristic (Figure 12) is shifted to (Mf, lNL).At no load current (lNL) the no load speed (w0) is:

at a particular load (ML) the corresponding speed () is:

In actual applications Mf

By definition: POUT = ML (Mechanical Power)

at peak power:

NOTE: Maximum mechanical power out cannot exceed 1/4 the stalled power input.

6.4.4 Efficiency

By definition: =

POUT = ML

Where lL = load current lS = stall current lNL = no load current

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Now PlN = VlL but V = lS R PlN = lSlL R

Solving for lL:

Substituting in efficiency equation:

6.4.5 Thermal Considerations

There is actually only one major criteria that should be taken into account when selecting an ironless rotor DC motor. The finalarmature temperature must not exceed its maximum rated value so that no separation in the winding occurs under highcentrifugal force. Given this consideration, it is evident that the ambient temperature (0), the two thermal resistances (Rth1 = rotor to case;Rth2 = case to ambient), and the average power dissipation (Pd), have to be very precisely evaluated in a given application. Pd = lR (Watts) = Pd(RTh1 + RTh2) (C) Where RTh1 & RTh2 are (C/Watt)The equation for the final armature (rotor) temperature is: f = 0 + f = 0 + (RTh1 + RTh2)Pd C The change in rotor resistance with temperature is expressed as: Rf = R[1 + (.004)(f - 20)]where "R" is that rotor resistance at 20C and .004 is the coefficient of thermal resistance for copper at 20C, in OHMS/C.

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Rf is the new resistance at temperature f.The maximum continuous average current is limited by the following thermal consideration: MAX 0 = = (RTH1 + RTH2) Rf lMAX

Where MAX is the maximum permissible rotor temperature (ie 100C for a standard motor). The increase in rotor temperature versus time with a constant dissipated power is:

where: t is elapsed time in seconds 1 is thermal time constant of coil (seconds). 2 is thermal time constant of tube (seconds).

6.4.6 Dynamic Performance

6.4.6.1 Starting under load conditions can be expressed as follows:

substituting:

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The solution for (0) = 0 is:

(see figure 14)

Where

but (mechanical time constant of motor from catalog)

NOTE: Speed of rotation (RPM):

6.4.6.2 Starting an unloaded motor can be stated as follows: Where: ML = 0 and JL = 0 = M and = 0

This means that after infinite time the unloaded motor (assume zero friction torque Mf) will attain the no-load angular velocitycorresponding to the power supply voltage. Thus:

and 0 = under these no load conditions.

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For an unloaded motor the initial acceleration would be:

Where: Md = motor starting torque JM = rotor inertia ld = current corresponding to the starting torqueIntegration of the function (t) with initial conditions of (t = 0) = 0:

(t = 0) = 0

+ c = 0 or c =

this new function is graphed in figure 15.

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7.0 GEARBOXES APPLIED WITH ESCAP MOTORS

where g = gear ratio

7.1 Efficiency: PlN = M

POUT = PlN

NOTE: Efficiency will change with temperature due to factors such as lubrication, gear mesh, etc.

7.2 Stall Rating of Gearbox: MSTALL = gM where MSTALL is motor stall torque

This torque (M) must never be permitted to exceed the maximum stall torque rating of the gearbox as stated In the catalog. Expressed as a motor current limit:

7.3 Inertia Transfer

Acceleration of the load JL is expressed as:

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Where:

(JLM) is the load inertia referred to the motor shaft.

8.0 CONSIDERATIONS FOR THE CONTROL OF DC MICROMOTORS

8.1 General Comments

Due to its inherent qualities the Escap micromotor is very suitable for a wide range of application types. The principal modes ofoperation are:

Open loop direct drive Closed loop velocity servo Closed loop position servo

8.2 Summary of Feedback Techniques and Components

8.2.1 Position feedback is required when critical positioning of the load is required such as in:

Welding robot head location Daisy wheel printer positioning Aircraft indicator drive

The most frequently used position feedback transducers (sensors) in use today are:

Potentiometers Synchros and resolvers Linear synchro Optical encoders Hall effect transistors Linear variation differential transformer Magnetic pickup

8.2.2 Velocity feedback is utilized when precise speed control of the motor is required or where a prescribed velocity profile isrequired to obtain controlled acceleration and deceleration: Examples of motor applications where velocity feedback might berequired are:

Daisy wheel printer ramping up and down of print wheel velocity to obtain minimum character positioning time. Tape drive applications requiring constant tape speed or maximum acceleration and deceleration of the tape. Carriage drive to obtain maximum accelerate/decelerate characteristics. Plotter pen drive where the pen is required to trace a prescribed path in a given time.

The most frequently used velocity feedback devices are:

Tachometers Optical encoders (velocity decoding)

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Motor back-emf sensing Phase locked quartz-based frequency comparator system

8.2.3 In summary, position control systems can be categorized into three primary functional areas:

Fixed displacement (point-to-point position) Translation/velocity profile A combination of both velocity and position

8.2.4 The most elementary components of a servo system are:

Motor Position transducer Velocity transducer Amplifier Controller (logic & comparator) Power source

These are arranged in block diagram form in figure 17.

The feedback transducers are either analog or digital sensors. Typically, tachometers and synchros are examples of analogfeedback devices whereas optical encoders and pulse counters are examples of digital devices.

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Controllers also fall into either the analog or digital category or a combination of both. The operational amplifier is the basicanalog element while the microprocessor is a digital control device.

8.3 Ironless Rotor Micromotor Model With Laplace Transform Notation

Figure 18 ironless Rotor Micromotor Model With Laplace Notation

VlN = input signaleemf = back emfK = motor constantl = motor currentS = Laplace operatorMf = motor friction torque

ML = load torqueJM = motor inertiaJL = load inertia = motor velocity = displacement

Note that with negligible inductance the classic Laplace term of reduces to .

This property is reflected in a negligible electrical time constant for ironless rotor motors which, in practice, never exceeds 1 %of the motor unloaded mechanical time constant.

8.4 Open Loop Control Considerations

The typical open loop operating mode for an Escap micromotor is a battery powered drive motor application. The high efficiency

and excellent regulation makes the micromotor well suited for these applications. We know from the torque speed

characteristics that a motors operating velocity is given by:

Thus, the regulation term defines how much the motor RPM will fall off with increasing torque. Proper choice of R and K iscritical especially in an open loop mode. Generally speaking, over a certain operating range, constant torque can be obtained by using a constant current power supply.This simply makes use of the fact that motor torque (M) = Kl. Within its current rating, a battery is well suited for use with amicromotor as a constant current source.

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Conversely, over a low torque operating range, a constant, motor output RPM can be maintained by using a constant voltage

power supply since

This mode of operation is only successful in low torque ( l lNL ) applications where the lR term is kept at a minimum.Obviously, as the torque demand increases the current must also increase and thus, the lR term increases and so does motorpower dissipation 1R.

It should be remembered that

Therefore for "good" regulation the power dissipation is greatly reduced for any given load torque. We again

see that the regulation is critical to the proper selection of a motor.

8.5 Closed Loop Control Considerations

8.5.1 ln an incremental motion control servo system the sensed position and/or velocity parameters, whether they be in digitalor analog form, are ultimately converted into control feedback signals and then into voltage and current form as drive power tothe motor. The motor thus executes this power command and the resulting motion is once again detected by the sensors. Thus,the feedback loop is "closed". This is illustrated in figure 17 in block diagram form.

8.5.2 Velocity profile

A key consideration in designing a servo system for Incremental motion is the velocity profile of the motor/load. The design goal is to optimize this profile, while minimizing some other parameter such as peak current or power dissipation. A common design problem may require an optimum velocity profile which will provide for minimum power dissipation (Pd) andminimum time (t) to perform a certain displacement (). Equations for this incremental motion are:

Torque:

Displacement:

Power Dissipation:

The simultaneous satisfaction of these equations while optimizing the desired conditions will result in a parabolic velocityprofile. Such a profile has an efficiency of 1.0. Other velocity profiles such as the trapezoid and triangle have efficiencies of 0.89 and0.75 respectively. In practice the trapezoidal velocity profile, being reasonably close to ideal, is generally used. Also, the ramp-up and ramp-downconstant acceleration is a simple matter to accomplish from a controls point of view. The trapezoidal velocity profile and itsassociated drive current profile are shown in figure 19. The drive circuit for the trapezoidal profile must provide a constantcurrent pulse. Both positive and negative.

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Figure 19 Trapezoidal Velocity Profile and Drive Current Profile

8.6 Servo Amplifiers and Control Techniques

8.6.1 General comments

To thoroughly cover the topic of servo control systems and their design would be an exhaustive exercise and certainly beyond thescope of this text. Whole chapters and even books have been written on individual areas of the total topic such as: damping,torsional resonance, amplifier design, etc. The intent of this text is to cover, in summary fashion, some of the salient issues inorder to provide the reader with an overview of servo amplifiers and control techniques. The reader is encouraged to studyfurther, his areas of interest. The references at the conclusion of this section represent an excellent library for further study. Amplifiers fall into two major categories: class A (linear) amplifiers, and switching amplifiers. Transistors may be used for eithertype but SCR's are often used in high power switching amplifiers. Simple speed control amplifiers are usually operated in a single quadrant and therefore are not bidirectional (no negativetorque). The true servo amplifier however, is capable of bidirectional drive and is able to operate in all four quadrants.

8.6.2 Linear amplifiers

The linear operating characteristics of a class A amplifier with no significant control lag within its designed bandwidth, make it theobvious choice for motor control applications. A single transistor speed control system is shown in figure 20.

A single transistor amplifier has a relatively low gain and therefore requires a large error signal before it will react. Higher gaincan easily be obtained by adding additional stages of amplification. Figure 21 gives an example of a speed control circuit havingtwo stages of current amplification and an integrated circuit (lC) incorporating a dual voltage amplifier.

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Transistor servo amplifiers are characterized by two primary output (drive) stages and consist of two basic circuitconfigurations. They are:

(a) The bridge or "H" consisting of four transistors and requiring a single dc power source; and,

(b) The "T" consisting of two complementary transistors and requiring two dc power sources.

See figure 22 for "H" and "T" circuits. The advantages and disadvantages of the bridge and "T" servo output stages are listed in the following tables.

Bridge "H" Stage

Advantages single power supply some voltage protection

Disadvantages difficult to drive in linear class difficult to obtain feedback

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"T" Stage Advantages easy to drive good feedback

Disadvantages requires 2 power supplies requires careful biasing to avoid dual conduction dead zone in characteristics

Both the "T" and "H" circuits may be operated as linear amplifiers or in the switching or "bang bang" mode. However, the "T" ismost often used as a linear amplifier while the "H" circuit seems to be more common in switching amplifiers.

8.6.3 Switching amplifiers

Switching amplifiers are the most versatile and perhaps most widely used servo drivers. There are three basic schemes used tocontrol power by switching amplifiers.

These methods are:

Pulse width modulation (PWM) Pulse frequency modulation (PFM) Silicon controlled rectifier (SCR)

The output waveforms of these three schemes are shown in figure 23.

All three of these control methods vary the power delivered to the motor by modulating the average power output over a giventime period. This can be observed in figure 23 where the average power is represented by the area under the curve during agiven period. The wider the pulse or the more frequent the pulse rate the greater the power. The generation of the pulses is performed by the controller logic and pulse generator circuits which in turn drive the outputamplifiers. Command signals and feedback signals of course control the pulse generation. Technically speaking, the SCR amplifier is a PWM amplifier but with a part sine wave output instead of a square wave. The SCRis also a lower frequency device and must be accompanied by a separate turn off circuit. SCR's are limited to high current andlow switching rates. These features generally make SCR controls less suitable for Escap micromotors.

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8.6.4 Selection of switching frequency

The major considerations in the selection of the switching frequency (fs) are:

fs must be above the servos dynamic response capability. Rule of thumb: fs > 10fBW where few is system bandwidth.

fs must be above the major system resonance points. fs>fRes.

The switching period must be greater than the transistor switching delay time (Td)

8.6.6 The ramp generator

One important element in a switching servo drive system is the ramp generator. This is the circuit responsible for generatingvarious sloping waveforms required for a particular velocity profile. Figure 24 shows a typical ramp generating circuit.

8.6.6 Phase locked loop servos

Phase locked loop servo (PLS) has become a very popular means of obtaining precise velocity control. This method exhibitsexcellent speed regulation. The principal of operation is simple. Basically the frequency of a feedback pulse is compared with a command frequency. The system adjusts itself until thefeedback is identical to the command at which time the system is said to be "phase locked". In this way the system outputvelocity is stabilized to the speed corresponding to the command.

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8.6.7 Back emf servo control

A back-emf servo utilizes the generated back-emf of the motor as a means of gaining velocity feedback information: The linearityand symmetry of the Escap mircromotor make it an excellent candidate for use in a back-emf servo. This scheme is low costand is well suited where low to medium performance is required. Please refer to the reference material for a detailed treatment ofthe subject.

8.7 Additional Comments

The subject of dc servo systems is vast. Thoroughly covering all aspects of the subject is beyond the scope of this text. Thereader is encouraged to conduct his own research to gain wider knowledge of the topic. Please consult the references at the endof this section for further reading on the following aspects of dc micromotor control:

Damping of ironless dc motors System instability Damping of ironless dc motors System instability System bandwidth Torsional resonance Circuit design Back-emf servos

9.0 TUTORIAL SELECTION OF MICROMOTOR APPLICATION PROBLEMS

9.1 Two Problems In Angular Motion

(A) It is required that a system reach a rotational speed of 2000 RPM in 20 MS. Initial velocity is zero. What is the angularacceleration required?

(B) It is required that a system have an angular displacement of 3 radians in 20 MS. Initial velocity is zero. What is the angularacceleration?

9.2 Drive Motor Application Problem

Customer has a "sniffer" drive problem. It is to be battery operated and the carbon brush motor he has selected consumesexcessive current. This motor has a 12 VDC rating, 14,000 RPM no load speed, a 5 ohm rotor, and 150 ma no load current. Theoperating conditions under load are a terminal voltage of 6 volts and 300 ma.

a) What is the load torque?

ML = K(l lNL) = 76.7 x 10-4(0.3 0.15) = 11.5 x 10-4Nm = 0.16 oz-in

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377

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b) What is the motor speed?

C) What is the efficiency? Pout = ML = 11.5 x 10-4 x 586.7 = 0.675 watt Pin = Vl - 6 x 0.3 = 1.8 watt = Pout/Pin = 37.5%

9.3 Drive Motor Selection Problem

This customer powers the above sniffer by Nickel-Cadmium rechargeable batteries of 1.25 - 1.30 volts per cell output. He wouldlike to use the least number of cells consistent with optimum efficiency and reasonable motor life. Small size is desirable.

a) What is your motor recommendation?

Try 22C11-216, Ra = 6.0 ohms @ 90F

l = lNL + lL = 0.119 +0.007 = 0.126A

V = 6.36 Volts (5 cells = 6.25 6.50 Volts)

b) What is the efficiency?

Pout = ML = 11.5 x 10-4 x 586.7 = 0.675 watt Pin = Vl = 6.36 + 0.126 = 0.801 watt

C) How does this efficiency compare to the motor's peak efficiency?

lNL = 0.007A

max = 84.4% (0.998 effective)

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d) What is temperature of rotor above case? Pa = lRa = (0.126) x 6 = 0.095 watt T = RP = 9 x 0.095 = 0.86C negligible

9.4 Gearbox Application Problem

A customer has a machine tool application where he needs exactly one revolution of a shaft in 1.0 1.5 seconds. He has a 24VDC supply available and load friction is 5 oz-in, inertia is 0.05 oz-in-s. Position of the output shaft will be sensed externally andon-off control applied to the motor.

a) What would your first try be using a B24?initially assume start-up and shut down times negligible relative to one second. Speed therefore is one revolutionin approximately one second or 60 RPM. Gear ratio is needed of approximately 100. Selecting the smallest 24VDC Motor, try the 26P11-210-100 with B24 ratio 128.

b) What would the motor and gearbox output speeds be under load?

Convert friction load to metric:MLf = 5 oz-in x 70.62 x 10-4Nm/oz.in = 353 x 10-4 Nm

Motor speed loaded:

output speed:

c) What time to speed (95%)?

Convert Inertia to metric:JL = 0.05 oz-in-s x 7.062 x 10-3 Kgm/oz-in-s = 3.53 x 10-4 Kgm

JM = 5.3 x 10-7 Kgm

3 = 0.057 Secd) Assuming equal times to start and stop, how long does it take from start to rest with a total angle of 1.0

revolution?

Ref: Figure 15For 3 = 2.05 = 2.05 x 5.6 x 0.019 = 0.22 Rad

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StartSlowStop

0.22 Rad5.84 Rad0.22 Rad

0.057 Sec1.043 Sec0.057 Sec

Total 6.28 Rad 1.16 Sec

9.5 Case Study: The "DC 300" Cartridge Drive

9.5.1 "DC 300" cartridge specifications

Excerpts from the "DC 300" cartridge specifications per American National Standard's ANSI X3.55-1977 and ANSI X3.56-1977necessary for our purposes are the following:

Friction force = FfThe worse case tangential force required at the outer driving radius (ra = .445 in) of the belt capstan to maintain a constantoperating speed.

Ff = 4.5 oz (1.25N)

Total inertia = mra2

The total equivalent inertial mass of all cartridge elements must not exceed m, given in linear units referred to the externalradius (ra) of the belt capstan.

m = .002 oz s2/in (.022 kg)

Inter block gap (lBG)The minimum length of the inter block gap should be:

lBG = 1.2 in (30.5 x 10-3)

Drive ratio:The ratio of the tape velocity, vt, to the tangential velocity of the external driving radius of the belt capstan, v, (which is alsoequal but opposite to the tangential velocity of the drive roller) shall be:

Vt/V = .76

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9.5.2 System requirements Density = 6400 bits/in (DC300XL) Tape read-write speed: vto The common speed for most applications is: Vto = 30 in/s therefore vo = 30/.76 = 39.5 in/s (lm/s) Tape rewind speed =Vtrw The common speed for most applications is: Vtrw = 90 in/s therefore: vrw = 90/.76 = 118 in/s (3m/s) Acceleration or deceleration time on read-write mode (ta; td): (lBG - .15 - .3)/Vto Where; .15 in., block integrity. Block integrity is a safety factor where a portion of the tape is DC erased before information block.

.3 in., distance between read head and write head

Acceleration or deceleration on rewind mode can vary from one system to another, however, it is greater than 75 ms. External drive roller radius: rc = .25 in (6.35. 10-3 m)

9.5.3 Model of Drive

withvo=39.5 in/s

M = .002 oz s2/inr = rc = .25 in.Ff = 4.5 oz.

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9.5.4 Solution

Because the greatest average power dissipation in the motor armature is when reading or writing on start-stop mode with thesmallest increment (or block) of information (128 bits = .02 in.), this will be used for an example of the worst case. The followingresults were obtained with a 23D21-216E Escap motor.

Torque

Tm1 = (Jm + mr2)vo/rta + FfrTm2 = Ffr

Tm3 = (Jm + mr2) vo/rtd + Ffr

Current

l1 = Tm1/kl2 = Tm2/kl3 = lm3/k

Voltage

U1 = Rl1 U2 = Rl1 + kvo/rU3 = Rl2 + kvo/rU4 = Rl3 + kvo/rU5 = Rl3

Dissipated power

Pd1 = Rl12

Pd2 = Rl22

Pd3 = Rl32

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Average dissipated power =

when

where:

l1 = A + B

l2 = B

l3 = A + B

Therefore,

in our case this will be: 2.32 W Optimal drive roller radius ro, which minimizes the average dissipate power, can be obtained by:

r optimum which minimizes

This results in the following optimal radius:

if Ff = 0, no friction force, "matching inertia"

ro = ( (m2/Jm2) + (ta2 + .5tato)Ff2/Jm2vo2)-1/4 ro= .15 inThe above results can be tabulated in the following curve:

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9.8 Dynamic Braking of a DC Micromotor

Many times customers not only ask for the acceleration characteristics but also need to know the dynamics of stopping. Thisaddresses the case of dynamic braking of a motor with both friction and inertial load. Gearhead parameters are included forcompleteness; however, a ratio and efficiency of unity may be used for cases involving direct motor output.

If we assume a switch closure across the motor terminals at time zero, the time to stop, ts, is calculated by the followingexpression:

where

m is the unloaded motor time constant (catalog value)

JL is the load inertia (Kg m2)

Jm is the motor inertia (Kg m2) is the gearbox efficiency (expressed as a decimal number less than unity)g is the gearbox ratio (expressed as a number greater than unity)WL is the load speed prior to braking (R/S)K is the motor constant (Nm/A)ML is the torque (Nm)R is the rotor resistance (ohms)

Note that the units of ts are the same as m.

Example as follows: A 22C11-210-5/B24.0-128 drives a 10 oz-in load at 55 rpm. Load inertia is 2 x 10-4 Kgm2. The time to stop is:

ts = 0.50 second

Due to the high deceleration torque developed by dynamic braking, care must be exercised when using a gearbox that theresulting reverse torque does not exceed the torque rating of the gearbox (both first stage and last stage ratings). Please refer toSection Two paragraph 5.0 of this handbook dealing with Minimization of Gear Train Inertia and the effects of "back driving"through a speed reducer.

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9.7 Example of Graphing Motor Characteristics

The graph of motor characteristics is based upon the following given information:

Motor: 26P11-216-35Continuous Current: 0.6 Amp.Max Desired rotor temp.: 85CAmbient temp.: 40CResistance at 20C: 9.4 Torque constant: 3.10 oz. in./Amp.Supply voltage: 12 vdcNo Load Current: 0.02 AThermal resistance: RTH1 = 5C/W RTH2 = 14C/W Calculate: RT = R20 [1 + 0.004 (T-20)] R85C = 1.26 X 9.4 = 11.8 PlN = l2R = 0.62 x 11.8 = 4.25 watts continuous T = 5C/W x 4.25 = 21C Case Temp TCASE MAX = 85 21 = 64C

9.71 MOTOR CHARACTERISTICS

26P11-216-35 12 VDC K = 3.10 oz in/A R20 = 9.4 lNL = 0.02A RTH1 5C/W RTH2 14C/W TA = 40C

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9.8 ConsideratIons for Tachometer Applications

The most significant considerations of a tachometer application are:

1. Voltage Constant (kv) expressed as volts per 1,000 RPM (V/K RPM). This defines the signal output of the tach Interms of Its mechanical input in RPM.

2. Linearity expressed as a percent. This defines the stability of the voltage constant over a given speed range and loadcondition. Between 500 and 5000 RPM Escap tachs have an unloaded linearity of 0.2% and with a 10 K load 0.7%. Because Escap uses precious metals in the brushes and commutator there is no requirement for a minimumcurrent flow to keep the brushes "clean". Therefore, Escap tachs may be used with high load impedances to takeadvantage of the 0.2% unloaded linearity.

3. Ripple is a result of the frequency of the induced emf. The ripple frequency is equal to twice the number ofcommutator segments (coils) multiplied by the Revolutions per Second (RPS). fR= 2 N (RPS)The ripple is modulated on the peak of the base output signal. For a 9 segment tach the peak-peak ripple isapproximately 1.52% of the peak output voltage.The effect of ripple can be minimized by choosing a high enough voltage constant and/ or operating RPM so as toimprove the signal to noise ratio. Filtered output can also be used but the result is to integrate the tach output thusreducing its response to quick changes in RPM.

4. A.C. Variation is a fluctuation in the base generator output signal. It is caused by non-symmetry in the coils. Escaptachs are characterized by their precise coil winding and thus have an absolute minimum of a.c. variation. Thefrequency of a.c. variation is equal to twice the RPS. fAC = 2 (RPS)

The amplitude of a.c. variation in Escap tachometers is approximately 3% of total ripple.

5. Thermal Stability. The temperature coefficient of Escap tachometers is - 0.02%/C due to the use of Al ccmagnets which are very stable and therefore will require little if any temperature compensation.

6. Inertia. The Escap tachometer is a low inertia device and is therefore well suited for certain applications whichcannot tolerate non-active load inertias, i.e.: sensing velocity characteristics of a very low mass load such as a printwheel.

Since the tachometer is primarily used as an analog sensor for measuring system speed (RPM) the designer usuallybegins by selecting a tachometer having an output signal (over entire system speed range) which is compatible withother system requirements and limitations such as: amplifier gain and saturation level; electrical noise; etc. There isno single "cook book" approach to tachometer selection. The designer must, as always, match his skills against theadvantages and disadvantages of all of the elements involved.

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9.9 Light Chopper Drive Motor

This application is for a motor to drive a slotted wheel which in turn interrupts (chops) a light beam at a frequency of 200 Hz. Thechopper wheel has only a single slot and an inertia of 0.2 gcm. Supply voltage available is 4 Vdc. Given: Vo = 4 Vdc Max

ML = 0.2 gcm (2 x 10-4 NM) Chopper freq. = 200 HzSolution:

= 2pif = 2pi x 200 =

Pin = ML = 2 x 10-4 x 1.257 x 103 = 0.25 Watts Propose to use 16 C11-210 Motor From catalog: K = 23 x 10-4 Nm/A R =7.5 lo = 0.015A

Now:

Vo = Rl + K Vo = (7.5 x 0.102) + 23 x 10-4 x 1257 = 3.66 volts 4 Volt supply is acceptable and motor choice checks out.

10.0 REVIEW OF ESCAP STEP MOTORS

10.1 Description

The Escap family of permanent magnet step motors are the result of a unique patented technology. The motors can be built with one phase per stack, with two or more phases per stack (each phase-covers a given angularsector) or with two or more phases imbricated in one single stack. Patents are protecting these different designs. The-followingdescribed motor has two phases arranged in one single stack. This new step motor design is based upon a homoheteropolarstructure. Figure 25 illustrates the motor's design in a simplified mechanical schematic.

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The motor is in the form of a thin axially magnetized disc made from somarium cobalt magnetic alloy. A special magnetizationprocess allows for a high number of magnetic poles and small step angles. The magnetic path is closed by the use of "C" shapedsilicon iron lamination cores. These cores are symmetrically arranged in the two stator halves. These lamination cores act as thestator "teeth" and are surrounded with a "bean" shaped coil for each electrical phase. The magnet is fixed to a shaft by virtue of two end-bells thus, forming the rotor assembly. It should be obvious that the inertiaof such a rotor assembly is very low an advantage of this motor. Figure 26 shows the construction of the Escap step motor.

The two stator halves forming the housing are precision molded of Ryton. This material has a very good modulus of elasticity, lowshrinkage, and excellent thermal stability. The four bean shaped coils are identical and are manufactured by standard windingmethods.

10.2 Advantages and Unique FeaturesThe advantages and unique features of the Escap P series motors can be summarized as follows: Very low rotor inertia (12 x 10-7 kgm2 for P532) High torque Low mass Low volume High power/mass ratio (150 W/kg-1 for P532) Excellent acceleration (140,000 rad/S2 for P532) High efficiency

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Low resonant frequency (250 Hz for P532)

Low system cost (motor/control package) Linear torque vs. ampere-turns characteristic Capable of very high step rates Low mechanical time constant No risk of demagnetization Compatible with simple low cost drive circuitry

10.3 Detent TorqueThe P series step motor does provide some holding torque with the windings do-energized. This torque is the result of aninteraction between the rotor poles and the stator poles. For this motor design the detent torque is a fourth harmonic of thefundamental sinusoid torque curve and is defined as

Tq = 2T4 Sin (4N 4)

During manufacture of the motor it is possible to increase or decrease the detent torque over a range of almost zero to about10% of the one-phase-on holding torque.

10.4 Holding TorqueThe P series motor is a two phase step motor. With one phase energized with a dc current, the rotor poles will align themselveswith the corresponding stator poles of the energized phase. A motor so aligned is in a position of stable equilibrium. If an externaltorque is applied to the motor shaft causing the rotor and stator poles to misalign, a counteracting torque is developed whichtends to restore the original condition of equilibrium. This restoring torque is called the static holding torque and its value varieswith rotor position. This torque is zero when the rotor and stator poles are aligned and increases with the angle of misalignmentup to some maximum value for the particular motor (static torque characteristics). With two phases energized the static holding torque is obtained by adding the torques of the two phases energized separately.Theoretically the two-phase-on scheme produces 2 times the torque developed with one-phase-on. In practice the actual statictorque is less because the individual torque curves are not sinusoidal. Under conditions of negligible detent torque the mathematical expression for Static holding torque (one-phase-on) is given by:

T = ni sin N

where N = number of pole pairs (25 for 100 steps per revolution) = torque per ampere-turn (a motor constant) ni = number of ampere-turns = total mechanical displacement (0 to 360)

The torque curves of Figures 27 and 28 present a graphic explanation of the various torques under one and two-phase-onoperation. A review of these curves yields the following helpful observations: (A) One-phase-on The stable equilibrium positions of the detent torque and the holding torque are the same. The stiffness of the stable equilibrium positions is increased by the detent torque. (B) Two-phases-on The stable equilibrium positions of the holding torque (phase 1 + 2 and 1 2) are unstable positions of the detent torque. The stiffness of the stable equilibrium positions is decreased by the detent torque.

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T1 DETENT TORQUE AMPLITUDET2 THEORETICAL MAXIMAL DYNAMIC TORQUE WITH ONE PHASE ENERGIZEDT3 THEORETICAL MAXIMAL DYNAMIC TORQUE WITH TWO PHASE ENERGIZEDT4 HOLDING TORQUE AMPLITUDE WITH ONE PHASE ENERGIZEDT5 HOLDING TORQUE AMPLITUDE WITH TWO PHASES ENERGIZED

A, B STABLE EQUILIBRIUM WITH ONE PHASE ENERGIZEDC, D STABLE EQUILIBRIUM WITH TWO PHASES ENERGIZED

A, C, B, D

SUCCESSIVE ROTOR POSITIONS WHEN HALF-STEPPING (8 STEP SWITCHINGSEQUENCE)

Figure 27 Descriptive Torque Curves

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T2 THEORETICAL MAXIMAL DYNAMIC TORQUE WITH ONE PHASE ENERGIZEDT4 HOLDING TORQUE AMPLITUDE WITH ONE PHASE ENERGIZEDL

TLA, B

A', B'

AMOUNT OF APPLIED LOADCOUNTERACTING TORQUE WHEN LOAD L IS APPLIEDSTABLE EQUILIBRIUM WITH ONE PHASE ENERGIZEDSTABLE EQUILIBRIUM WHEN LOAD L IS APPLIED

Figure 28 Descriptive Torque Curves

10.5 Dynamic CharacteristicsThe performance curves of the P312 and P532 step motors are shown in Figure 29. It should be noted that the start-stop torqueversus speed will be affected by both the load inertia and friction. However, the pull-out torque is usually affected by frictionalone. It should be realized however, that in some cases static friction and running friction are often different values. A good dynamic behavior can be claimed when this motor is compared to equivalent hybrid type 3.6 motors with a rare-earthcobalt magnet, or even to traditional hybrid type 1.8 motors. For the same number of steps per second, the P532 is rotatingtwice as fast as a 1.8 motor; the mechanical work is therefore equivalent, as long as torque is at least half the torque of thecompetition motor. It is effectively the case, even though the volume and the mass

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are respectively 27% and 42% lower. The rotor inertia is 5 times lower but, as a matter of fact, should be multiplied by 4 in orderto compare the inertia of both motors on a shaft which runs at the same speed. As far as mechanical power is concerned, theP532 motor is equivalent to a size 22 motor, 2" long, with 1.8 times more volume, 2.4 times more mass and 2.5 times moreequivalent inertia. The reason why the P532 has good performances with regard to its volume and its mass is essentially because of its low inertiaand low magnetic losses (due to a total silicon-iron magnetic circuit which weighs only 50g). At 1000 steps per second, the P532needs 0.4W and conventional 1.8 motors need 0.7W to 1.1W. At 2000 steps per second, this becomes 1.1W against 1.9 and2.9W. At 5000 steps per second, the P532 needs 4W, of which 1W is due to friction in the ball bearings viscous friction of themagnet in the air gap. Referred to a chopper drive, the Joule power in the coils represents about 6W. A total of 10W power lossescan so be figured out. This rough calculation gives an idea of the motor efficiency, since the available mechanical power is morethan 20W.

10.6 Damping Considerations

At low speed, the P532 and P312 have less losses than other step motors. Thus, the damping of the settling oscillations will needlonger time. However, the damping effect created by the short-circuit of the coils or simply by connecting the coils across a lowsource impedance is better, up to 3000 steps/s. At 1600 steps/s for example, an 80% power loss increase can be achieved byshort-circuit of only one coil or by connecting one phase across an adapted low resistance. Similar measurements with size 22 by2" long hybrid typemotors, shows that no damping effect occurs after 800 to 1500 steps/s under the same circumstances. The effect on damping by increasing the load inertia is to increase the settling time and overshoot amplitude. Likewise theeffect on damping by increasing the load friction is to decrease the settling time and overshoot amplitude. Friction sometimesimproves system performance.

10.7 Resonance

The mechanical resonance of a P532 motor with no load inertia is about 190 Hz. It is calculated by the following expression:

where: T is the holding torque, N is number of pole pairs J is total inertia

From this formula it can be observed that load inertia will reduce the primary resonant frequency. In applications where themotor must be operated near its primary resonance the addition of load inertia may provide a safer operating speed range.However, higher order harmonics may appear at higher speeds as inertia is increased. The addition of friction can sometimes beused to reduce the severity of resonance.

10.8 Accuracy

Step accuracy is defined as a non-cumulative error which represents the step to step error in one full revolution. Inertia andviscous friction do not affect step accuracy. Friction does however, create a dead band around the normal resting position of themotor. This is due to the fact that the rotor comes to rest in a position where the static torque matches the friction

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torque of the system. Thus, the rotor is offset from its ideal rest position by an angle where the static torque curve equals thefriction. This is called position accuracy and is not to be confused with step accuracy which is really a mechanical property ofthe motor. It should be obvious that the steeper the static torque curve the better will be the position accuracy. This is veryimportant in selecting the proper step motor.

10.9 Motor Drive Techniques

The performance of a step motor is greatly influenced by the type of drive circuitry utilized. An obvious advantage of any step motor is its compatibility with digital electronics. The motor making a fixed incrementaldisplacement (step) for each single pulse of energy supplied to it. Although step motors can be run closed loop, they have the cost saving advantage of being able to operate quite satisfactorilyin open loop mode. Provided, of course, that the response characteristics (torque, speed, etc.) of the motor are not exceeded.

The three types of driver configuration recommended for the Escap steppers are: resistance limited, (unipolar or

bipolar), or a bipolar chopper type.

The unipolar drive system is low cost and most commonly used in lower performance applications. Its disadvantage is due

to the fact that only one winding per phase is in use at any particular time.

The bipolar drive developes higher motor performance since both windings per phase are utilized. This requires either

series or parallel connected windings. Also the bipolar requires a dual polarity power supply or a transistor bridge for each motorphase. Bipolar driving yields a 2 increase in low speed torque for the same electrical input power as delivered by a unipolardrive. The bipolar chopper drive is known for its high performance and improved efficiency (due to the absence of externalresistance). This type of drive is best used at higher speeds. A chopper drive may cause audible noise due to the motor laminations vibrating at the chopper frequency. Single step motionis with a high acceleration due to the short current rise time of a chopper drive. The response therefore can be more oscillatoryespecially at speeds near the natural resonance of the motor.Half step techniques usually reduces resonant effects, and microstepping schemes will completely eliminate resonance and speedinstability problems.

10.10 Application Example

10.10.1 DescriptionThe application is a matrix dot printer carriage drive. The carriage is moving along a metallic bar; its weight is 5 oz. and thefriction is 4 oz. in. The motor has a 3.6 step angle and has to be able to drive the carriage at 1200 steps per second. The motionis transferred from the motor shaft to the carriage through a pulley and a cable. An acceleration time of 100 ms is allowed fromstandstill to 1200 steps per second. The supply voltage is specified below or equal to 24V.

10.10.2 Mechanical RequirementsWhen the P532 has to replace a 1.8 motor without any change in the electronics, especially the number of pulses per second, orin the carriage travel speed, one has to provide the system with a half diameter pulley. This sometimes creates a problem as faras the cable is concerned. In the present application, no pulley diameter is specified but we don't want the reflected inertia to behigher than 4 times the motor inertia. This will lead to the same kind of cable problems.

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The motor inertia is JM = 1.2 x 10-6 kg.m2

The load inertia should be less than 4 x 1.2 x 10-6 kg.m2 JL 4.8 x 10-6 kg.m2 If the pulley diameter is D = 0.45", then

x 5 x 28 x 10-3 = 4.6 x 10-6 kg.m2

The stress on each cable strand due to this radius of curvature is

d is the diameter of a single cable strand E = 2 x 1011 N/m2 (Young's modulus of elasticity) A good steel will tolerate = 1500 N/mm2; let's use = 700 N/mm2 for a more conservative calculation. Then

In addition, there will be a stress on the cable during the acceleration phase. It will be negligible, because there is no drasticacceleration requirement. Nevertheless if the P532 is used with a high acceleration rate like 105 steps/S2, there will be anadditional force on the cable:

F = m. = (5 x 28 x 10-3)

Using a 0.5 mm cable diameter, the steel section will be approximately 0.12 mm2. This means an additional stress of

Then

Finally, each strand diameter must be as low as 0.0015". In addition, the cable should be teflon coated. It may be easier to find a steel band in that thickness, instead of a cable. The following design may be used in that case

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10.10.3 Electrical RequirementsEach motor coil has 320 turns and R = 12 resistance. One should decide now if the coils will be connected in series, or inparallel. If a series connection is chosen, then the back emf will be (peak value)

= Torque per ampere-turns = 4.6 x 10-4 Nm n = Number of turns per coil = 320

Eemf = 22VIf a parallel connection is chosen, then the back emf will only be

The requirement made about the supply voltage assigns the coil connection to be parallel. Suggested driving technique: drive the P532 with constant voltage applied to an IC (SGS L 293), using a 2-phases-on scheme.This supply voltage calculation can be done as the following.

a) Required torque

= 0.032 Nm (4.56 oz. in.) Let's use a 1.5 factor to get a more conservative system. It becomes: Ts = 1.5 x 0.032 = 0.048 Nm (6.76 oz. in.)

b) The required number of ampere-turns; torque is related to the number of ampere-turns in the coils. With the influence of

the detent torque subtracted out, the running torque, which is times lower than the holding torque, is 0.048 Nm.

A running torque of 0.048 Nm consequently needs the holding torque to be 0.048 x 2 = 0.068 Nm (9.57 oz. in.)This torque can be reached if the number of ampere-turns in the 2 phases is 100 A-T.

C) Required supply voltage The voltage calculation is referred to

It is difficult to use this differential equation as it is. Our purpose is to get an idea of the supply voltage and the followingapproximation will give it close enough E = Rl + Eemf

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Normally is not negligible above 1000 steps/s. but it is hard to figure out.

With 100 A-T input power, the temperature raise is approximately 10C, the phase resistance the becomes R = 6 (1 + 0.004 x 10) 6.25 ohms then

The voltage drop in the transistor is about 1V; finally E = 14V The experiment shows that the actual value must be E = 15VIf a 10 series resistor is used to improve dynamic performances, then

An analog experiment indicates E = 19V

A further experiment result is the maximum running torque 0.042 Nm (5.92 oz.in.) slight less then expected. The differencerepresents the losses inside the motor which we did not take cared of.* The above calculation is only a guide and cannot replace experiment. An undesirable resonace frequency cannot especially bepredicted by the above formulas.

*If the motor is running at 400 steps/s with 15V supply voltage (400 steps/s is the frequency when the acceleration ramp starts),then the back emf is only 3.6V. It means that the current will raise to 1.6A. The chip cannot tolerate this current very long;consequently one should not leave this frequency for more than one step. If the motor is suddenly stopped, the current becomesvery high and must be switched off by using the inhibit function of the IC. When the series resistor is connected this situation isnot as critical.

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Physical Properties of Small DC Motors Using an Ironless Rotor: Dr. Erich JuckerPortescap

Reliability and Life of DC Motors: The New REE SystemDr. Marc HeyraudPortescap

Selecting Low Inertia d-c Servomotors for Incremental Motion: Dr. Marc HeyraudPortescap

Damping of D.C. Motors with Ironless Rotors: Dr. Erich JuckerPortescap

Inductance of Micromotors with lronless Rotors: Jean-Bernard KurethPortescap

DC Motors, Speed Controls, Servo Systems: An Engineering Handbook by Electro-Craft Corp.

Regulate motor-shaft speed better with an inactive bridge: James M. Pihl, Electronic Design, Feb. 15, 1978

Control the speed and phase of a dc motor by comparison against a control frequency: Mike Yakymyshyn, Electronic Design, 16Aug. 1977

Incremental Motion Control, DC Motors and Control Systems: Dr. Benjamin Kuo and Dr. Jacob Tol, SRL Publishing Co., 1978

Individual collections of application engineering data and instruction material by the following persons:

F. Prautzsch H. Dumas A. Ugnat V. Liengma P. Thornton

PortescapPortescapPortescapPortescapPortescap

A New Family of Multipolar P.M. Stepper Motors: Dr. Claude OudetPortescap

Various Step motor application and training notes prepared by: R. Welterlin and A. UgnatPortescap

Ironless Rotor D.C. Motors, A Consideration For Magnetic Tape Transport Design: A. UgnatPortescap

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IndexESCAP IRONLESS ROTOR DC MICROMOTORS AND STEP MOTORSINTRODUCTIONCONSTRUCTION OF THE ESCAP MICROMOTOR 349REVIEW OF PHYSICAL LAWS GOVERNING ESCAP MOTOR APPLICATIONSBASIC MOTOR PHYSICSGEARBOXES APPLIED WITH ESCAP MOTORS 366CONSIDERATIONS FOR THE CONTROL OF DC MICROMOTORSTUTORIAL SELECTION OF MICROMOTOR APPLICATION PROBLEMSREVIEW OF ESCAP STEP MOTORS