MtMotor Sl tiS election - Philadelphia University · 2016-12-13 · 12/11/2016 8 DC Motors Permanent magnet (PMDC) • A permanent‐magnet dc motor is basically the same machine

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MSDM t S l tiMotor Selection

Dr. Ibrahim Al‐Naimi

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Motors Selection‐ Construction.

‐ Principle of operation.

‐ Equivalent circuit.

‐ Speed – torque characteristic equation/curve.

‐ Specifications.

‐ Power flow.

‐ Speed regulation‐ Speed regulation.

‐ Dynamic/mathematical Model.

‐ Advantages and disadvantages.

‐ Applications.

DC Motors• Construction:

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DC MotorsPrinciple of operation:

• The DC motor converts direct‐current electrical energy into rotational mechanical energy. It makes use of theinto rotational mechanical energy. It makes use of the principle that a wire carrying a current in a magnetic field experiences a force. The windings wrapped around a rotating armature carries current. The armature is the rotating member (rotor), and the field winding is the stationary winding (stator). The rotor has many closely spaced slots on its periphery. These slots carry the rotor windings. The rotor windings (armature windings) are powered by the supply voltage. An arrangement of commutation segments and brushes ensures the transfer of DC current to the rotating winding.

DC Motors

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DC MotorsEquivalent circuit:

DC Motors

• Separately excited and shunt.

• Permanent magnet.g

• Series.

• Compound.

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DC MotorsSeparately excited/shunt


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Speed control:

Adj ti th fi ld i t (F d b‐ Adjusting the field resistance (For speeds above the rated speed).

‐ Adjusting the terminal voltage (for speeds below rated speed.

‐ Inserting resistance in series with armature circuitInserting resistance in series with armature circuit (not used).


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DC MotorsPermanent magnet (PMDC)

• A permanent‐magnet dc (PM DC) motor is a dc motor whose poles are made of permanent magnets. p p gPermanent‐magnet dc motors offer a number of benefits compared with shunt dc motors in some applications. Since these motors do not require an external field circuit, they do not have the field circuit copper losses associated with shunt dc motors. Because no field i di i d h b ll hwindings are required, they can be smaller than

corresponding shunt dc motors. PMDC motors are especially common in smaller fractional and sub fractional horsepower sizes, where the expense and space of a separate field circuit cannot be justified.

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DC MotorsPermanent magnet (PMDC)

• A permanent‐magnet dc motor is basically the same machine as a shunt dc motor, except that the flux of a pPMDC motor is fixed. Therefore, it is not possible to control the speed of a PMDC motor by varying the field current or flux. The only methods of speed control available for a PMDC motor are armature voltage control and armature resistance control.

DC MotorsSeries:

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DC MotorsSeries:

DC MotorsSeries:• Series motor gives more torque per ampere than any other dc motor. It is therefore used in applicationsother dc motor. It is therefore used in applications requiring very high torques. Examples of such applications are the starter motors in cars, elevator motors, and tractor motors in locomotives.

• When the torque on this motor goes to zero, its speed goes to infinity. In practice, the torque can never go g y p q gentirely to zero because of the mechanical, core, and stray losses that must be overcome. However, if no other load is connected to the motor, it can turn fast enough to seriously damage itself.

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DC MotorsCompound:

DC MotorsCompound:

• Cumulatively compounded motor has a higher starting torque than a shunt motor (whose flux isstarting torque than a shunt motor (whose flux is constant) but a lower starting torque than a series motor (whose entire flux is proportional to armature current).

• In a sense, the cumulatively compounded dc motor combines the best features of both the shunt and the series motors. Like a series motor, it has extra torque for starting; like a shunt motor, it does not overspeed at no load.

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DC MotorsShunt, Series, and Compound

Shunt, Series, and Compound DC Motors

Shunt/Separately Excited Series Compound

‐ Relatively constant ‐ High starting torque. ‐ Speed regulation is speed from no load to rated load.‐ Speed regulation is approximately 5%.‐ Used in applications that required constant speed, and where high starting torque is not needed

‐ Used to drive hoist,electric locomotives.‐ Completely removing the load will cause the series motor to run away.

generally between 15% and 25%.‐ Has it greatest application with loads that require high starting torques and fairly constant speed or have pulsating loadstorque is not needed.

‐ Used to drive centrifugal pumps, fans, winding reels, conveyors, machine tool.

pulsating loads.‐ Used to drive shovels, metal stamping machines, reciprocating pumps, hoists, compressors.

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Brushless DC MotorDC Motor Disadvantages:Excessive sparking and brush wear.

• Small, fast dc motors are too small to use compensating windings and interpoles, so armature reaction and L dildt effects tend to produce sparking on their commutator brushes.

• In addition the high rotational speed of these• In addition, the high rotational speed of these motors causes increased brush wear and requires regular maintenance every few thousand hours.

Brushless DC Motor

DC Motor Disadvantages:• In some applications, the regular maintenance required by the brushes of these dc motors may be unacceptable. Consider for example a dc motor in an artificial heart‐regular maintenance would require opening the patient's chest.

• In other applications the sparks at the brushes• In other applications, the sparks at the brushes may create an explosion danger, or unacceptable RF noise.

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Brushless DC MotorDC Motor Disadvantages:• For all of these cases, there is a need for a small, fast dc motor that is highly reliable and has low noise and long life.

• Brushless motors combining a small motor much like a permanent magnetic stepper motor with a rotor position sensor and a solid‐state electronic switching circuit.

• They run from a dc power source but do not have commutators and brushes.

• The rotor is similar to that of a permanent magnet stepper motor, except that it is nonsalient.

Brushless DC MotorThe basic components of a brushless dc motor are1‐ A Permanent magnet rotor2‐ A stator with a three, four, or more phase winding3‐ A rotor position sensor4‐ An electronic circuit to control the phases of the rotor winding.

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Brushless DC Motor• A brushless dc motor functions by energizing one stator coil at a time with a constant dc voltage. When a coil is turned on it produces a statorWhen a coil is turned on, it produces a stator magnetic field Bs, and a Torque is produced on the rotor given by which tends to align the rotor with the stator magnetic field.

• At the time shown in the Figure below , the stator magnetic field Bs points to the left while the permanent magnet rotor magnetic field BR points up, producing a counterclockwise Torque on the rotor. As a result, the rotor will turn to the left.

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Brushless DC Motor

Brushless DC Motor• If coil a remained energized all of the time, the rotor would turn until the two magnetic fields are aligned, and then it would stop, just like a stepper motor. The key to the operation of a brushless dc motor is that it includes a position sensor, so that the control circuit will know when the rotor is almost aligned with the stator magnetic field.

• At that time coil a will be turned off and coil b will be d i h i iturned on, causing the rotor to again experience a

counterclockwise torque, and to continue rotating. This process continues indefinitely with the coils turned on in the order a, b, c, d, ‐ a, ‐ b, ‐c, ‐ d, etc., so that the motor turns continuously.

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Brushless DC Motor• The electronics of the control circuit can be used to control both the speed and direction of the motor The net effect of thisdirection of the motor. The net effect of this design is a motor that runs from a dc power source, with full control over both the speed and the direction of rotation.

B hl d t il bl l i• Brushless dc motors are available only in small sizes, up to 20 W or so, but they have many advantages in the size range over which they are available.

Brushless DC MotorBrushless DC Motor advantages:

• Relatively high efficiency.

• Long life and high reliability• Long life and high reliability.

• Little or no maintenance.

• Very little RF noise compared to a dc motor with brushes.

• Very high speeds are possible (greater than 50,000 / i )r/min).

Brushless DC Motor main disadvantage:

• Brushless dc motor is more expensive than a comparable brush dc motor.

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AC Motors

Construction:

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AC MotorsPrinciple of operation:

• The principle of operation of the induction motor is based on generating a rotating constant magnetic field.based on generating a rotating constant magnetic field.

• This rotating magnetic field interacts with a set of conductors arranged on the rotor, and short circuited at the ends with two rings.

• These are shown in the Figure, and as it looks like a squirrel cage it gives the motor its name (Squirrel cagesquirrel cage, it gives the motor its name (Squirrel cage induction motor, SCIM). This interaction between the magnetic field and the conductor induces a current in the bars.

AC MotorsPrinciple of operation:

• This induction action is what gives the motor its names, and makes it similar to the transformer action, in the fact that a voltage is induced into the rotor (sometimes called secondary) by current flowing in the stator (sometimes called the primary).

• A current flows in the conductors of the rotor, through the short circuiting rings at the end. This current in turn produces a magnetic field. It is the interaction between the stator magnetic field and the rotor magnetic field which induces the torque and causes rotation.

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AC MotorsRotating Magnetic Field:• The fundamental principle of AC machine operation is that if a three‐phase set of currents, each of equalthat if a three phase set of currents, each of equal magnitude and differing in phase by 120°,flows in a three‐phase winding, then it will produce a rotating magnetic field of constant magnitude. The three‐phase winding consists of three separate windings spaced 120 mechanical degrees apart around the surface of the machine.

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AC MotorsRotating Magnetic Field:

AC Motors

• Synchronous Motors

• Induction Motors (Asynchronous Motors)( y )– Squirrel Cage

– Wound Rotor

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AC MotorsSynchronous Motors• A synchronous motor has a 3‐ph winding stator produces rotating magnetic field. A dc current is applied to the rotor winding, which produces a rotor magnetic field.

• Permanent magnet Synchronous motor.

AC MotorsInduction Motors

• The distinguishing feature of an induction motor i th t d fi ld t i i d t this that no dc field current is required to run the machine.

• An induction motor has the same physical stator as a synchronous machine, with a different rotor construction.

• There are two different types of induction motor rotors which can be placed inside the stator. One is called a squirrel cage rotor, while the other is called a wound rotor.

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AC MotorsInduction Motors• A squirrel cage induction motor rotor consists of a series of conducting bars laid into slots carved in the face of theof conducting bars laid into slots carved in the face of the rotor and shorted at either end by large shorting rings.

• A wound rotor has a complete set of three‐phase windings that are mirror images of the windings on the stator. The three phases of the rotor windings are usually Y‐connected, and the ends of the three rotor wires areY connected, and the ends of the three rotor wires are tied to slip rings on the rotor's shaft. The rotor windings are shorted through brushes riding on the slip rings.

AC MotorsInduction Motors• Wound rotor induction motors have their rotor currents accessible at the stator brushes where they can beaccessible at the stator brushes, where they can be examined and where extra resistance can be inserted into the rotor circuit. It is possible to take advantage of this feature to modify the torque‐ speed characteristic of the motor.

• Wound rotor induction motors aremore expensive thanWound rotor induction motors are more expensive than squirrel cage induction motors, and they require much more maintenance because of the wear associated with their brushes and slip rings. As a result, wound‐rotor induction motors are rarely used.

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AC Motors

AC MotorsEquivalent Circuit of Induction Motor

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AC Motors

AC Motors

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AC Motors

AC MotorsInduction Motor Maximum (Pullout) Torque

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AC Motors• The torque‐speed characteristic for a wound‐rotor induction motor is shown in the following Figure. Notice on the figure that as the rotor resistance is increased, the pullout speed of the motor decreases (the slip increases), the starting torquethe motor decreases (the slip increases), the starting torque increase, and the maximum torque remains constant.

• It is possible to take advantage of this characteristic of wound rotor induction motors to start very heavy loads. If a resistance is inserted into the rotor circuit, the maximum torque can be adjusted to occur at starting conditions. Therefore, the maximum possible torque would be available to start heavy loads. On the other hand, once the load is turning, the extra resistance can be removed from the circuit, and the maximum torque will move up to near synchronous speed for regular operation.

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AC Motors• If a rotor is designed with high resistance, then the motor's starting torque is quite high, but the slip is also quite high at normal operating conditions.

• Recall that so the higher the slip the• Recall that , so the higher the slip, the smaller the fraction of air‐gap power actually converted to mechanical form, and thus the lower the motor's efficiency.

• A motor with high rotor resistance has a good starting torque but poor efficiency at normal operatingtorque but poor efficiency at normal operating conditions. On the other hand, a motor with low rotor resistance has a low starting torque and high starting current, but its efficiency at normal operating conditions is quite high.

AC Motors• An induction motor designer is forced to compromise between the conflicting requirements of high starting torque and good efficiency.

• One possible solution to this difficulty is to use a wound rotor• One possible solution to this difficulty is to use a wound‐rotor induction motor and insert extra resistance into the rotor during starting. The extra resistance could be completely removed for better efficiency during normal operation.

• Unfortunately, wound‐rotor motors are more expensive, need more maintenance, and require a more complexneed more maintenance, and require a more complex automatic control circuit than cage rotor motors. Also, it is sometimes important to completely seal a motor when it is placed in a hazardous or explosive environment, and this is easier to do with a completely self‐contained rotor.

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AC Motors• It would be nice to figure out some way to add extra rotor resistance at starting and to remove it during normal running without slip rings and without operator or control circuit interventioncircuit intervention.

• The following Figure illustrates the desired motor characteristic. The figure shows two wound‐rotor motor characteristics, one with high resistance and one with low resistance. At high slips, the desired motor should behave like the high‐resistance wound‐rotor motor curve; at low slips, it should behave like the low‐resistance wound‐rotor motor curve.

• Fortunately, it is possible to accomplish just this effect by properly taking advantage of leakage reactance in induction motor rotor design.

AC Motors

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AC Motors

Control Motor Characteristics by Squirrel Cage Rotor Design• The reactance X2 in an induction motor equivalent circuit

represents the referred form of the rotor's leakage reactancerepresents the referred form of the rotor s leakage reactance. Recall that leakage reactance is the reactance due to the rotor flux lines that do not also couple with the stator windings.

• In general, the farther away from the stator a rotor bar or part of a bar is, the greater its leakage reactance, since a smaller percentage of the bar's flux will reach the stator. Therefore, if h b f l d h f f hthe bars of a cage rotor are placed near the surface of the rotor, they will have only a small leakage flux and the reactance X2 will be small in the equivalent circuit. On the other hand, if the rotor bars are placed deeper into the rotor surface, there will be more leakage flux and the rotor reactance X2 will be larger.

AC Motors

Control Motor Characteristics by Squirrel Cage Rotor Design

• For example, the Figure below is a photograph of a rotor lamination showing the cross section of the bars in thelamination showing the cross section of the bars in the rotor. The rotor bars in the figure are quite large and are placed near the surface of the rotor. Such a design will have a low resistance (due to its large cross section) and a low leakage reactance and X2 (due to the bar's location near the stator). Because of the low rotor resistance, the )pullout torque will be quite near synchronous speed, and the motor will be quite efficient. Remember that:

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AC Motors


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AC Motors


• So, very little of the air‐gap power is lost in the rotor resistance However since R is small the motor's startingresistance. However, since R2 is small, the motor s starting torque will be small, and its starting current will be high.

• This type of design is called the National Electrical Manufacturers Association (NEMA) design class A. It is more or less a typical induction motor, and its characteristics are basically the same as those of a wound‐characteristics are basically the same as those of a woundrotor motor with no extra resistance inserted. Its torque‐speed characteristic is shown in the following Figure.

AC Motors

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AC Motors

Control Motor Characteristics by Squirrel Cage Rotor Design• The next figure, however, shows the cross section of an

induction motor rotor with small bars placed near the surfaceinduction motor rotor with small bars placed near the surface of the rotor. Since the cross‐sectional area of the bars is small , the rotor resistance is relatively high. Since the bars are located near the stator, the rotor leakage reactance is still small.

• This motor is very much like a wound rotor induction motor with extra resistance inserted into the rotor. Because of the l i hi h ll ilarge rotor resistance, this motor has a pullout torque occur ring at a high slip, and its starting torque is quite high. A cage motor with this type of rotor construction is called NEMA design class D. Its torque‐speed characteristic is also shown in the previous Figure

AC Motors


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AC Motors

Deep Bar and Double Cage Rotor Design

• Both of the previous rotor designs are essentially similar to a wound‐rotor motor with a set rotor resistance . How can a variable rotor resistance be produced to combine the high starting torque and low starting current of a class D design with the low normal operating slip and high efficiency p g p g ff yof a class A design?

AC MotorsDeep Bar and Double Cage Rotor Design• It is possible to produce a variable rotor resistance by the use

of deep rotor bars or double‐cage rotors. The basic concept is illustrated with a deep bar rotor in the following Figureis illustrated with a deep‐bar rotor in the following Figure. Figure a shows a current flowing through the upper part of a deep rotor bar. Since current flowing in that area is tightly coupled to the stator, the leakage inductance is small for this region.

• Figure b shows current flowing deeper in the bar. Here, the leakage inductance is higher. Since all parts of the rotor bar are in parallel electrically, the bar essentially represents a series of parallel electric circuits, the upper ones having a smaller inductance and the lower ones having a larger inductance (Figure c).

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AC MotorsDeep Bar and Double Cage Rotor Design• At low slip, the rotor's frequency is very small, and the reactances

of all the parallel paths through the bar are small compared to their resistances The impedances of all parts of the bar aretheir resistances. The impedances of all parts of the bar are approximately equal, so current flows through all parts of the bar equally. The resulting large cross‐sectional area makes the rotor resistance quite small, resulting in good efficiency at low slips.

• At high slip (starting conditions), the reactances are large compared to the resistances in the rotor bars, so all the current is forced to flow in the low reactance part of the bar near the statorforced to flow in the low‐reactance part of the bar near the stator. Since the effective cross section is lower, the rotor resistance is higher than before. With a high rotor resistance at starting conditions, the starting torque is relatively higher and the starting current is relatively lower than in a class A design. A typical torque‐speed characteristic for this construction is the design class B.

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AC MotorsDeep Bar and Double Cage Rotor Design• A cross‐sectional view of a double‐cage rotor is shown in next

Figure. It consists of a large, low‐resistance set of bars buried d l i h d ll hi h i f bdeeply in the rotor and a small, high‐resistance set of bars set at the rotor surface. It is similar to the deep bar rotor, except that the difference between low‐slip and high‐slip operation is even more exaggerated. At starting conditions, only the small bar is effective, and the rotor resistance is quite high. This high resistance results in a large starting torque. However, at normal operating speeds, both bars are effective, and the resistance is almost as low as in a deep‐bar rotor. Double‐cage rotors of this sort are used to produce NEMA class B and class C characteristics. Possible torque‐speed characteristics for a rotor of this design are designated design class B and design class C.

Design Class B Design Class C

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AC MotorsDeep Bar and Double Cage Rotor Design

• Double‐cage rotors have the disadvantage that they are more expensive than the other types of cage rotors, but p yp gthey are cheaper than wound‐rotor designs. They allow some of the best features possible with wound‐rotor motors (high starting torque with a low starting current and good efficiency at normal operating conditions) at a lower cost and without the need of maintaining slip rings d b hand brushes.

AC MotorsInduction Motor Design Classes• It is possible to produce a large variety of torque‐speed curves

by varying the rotor characteristics of induction motors. To help industry select appropriate motors for varying applications in the integral‐horsepower range, NEMA in the United States and the International Electrotechnical Commission (IEC) in Europehave defined a series of standard designs with different torque‐speed curves. These standard designs are referred to as design classes, and an individual motor may be referred to as a design , y gclass X motor. It is these NEMA and IEC design classes that were referred to earlier. The characteristic features of each standard design class are given below.

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AC Motors

Induction Motor Design Classes

‐ Design Class A

‐ Design Class B

‐ Design Class C

‐ Design Class D

‐ Design Class E and F

Design Class A Design Class B Design Class C Design Class D

‐Standard motor design‐Normal starting torque (More than 200% of full load torque).‐Full load slip < 5%.‐Maximum torque 200 –300% full load torque at

‐Normal starting torque with lower starting current and low slip.‐The same starting torque as class A with about 25% less current.The pullout torque >=

‐High starting torque with low starting currents and low slip (less than 5%) at full load‐The pullout torque is slightly lower than that

‐Very high starting torque (275% or more of the rated torque).‐Low starting current.‐High slip at full load.‐They are essentially ordinary class A induction300% full load torque at

low slip (less than 20%).‐High inrush current (500 to 800% of the rated current).‐Applications: Fans, pumps, lathes, and other machine tool.

‐The pullout torque >=200% of the rated load torque, but less than that of the class A design because of the increased rotor reactance.‐Rotor slip is low (less than 5%) at full load.‐Applications: Similar to those for design A, but

slightly lower than that for class A motors, while the starting torque is up to 250% of the full‐load torque.‐Built from double‐cage rotors (more expensive).‐Applications: They are used for high‐starting‐

ordinary class A induction motors, but with the rotor bars made smaller and with a higher‐resistance material.‐The high rotor resistance shifts the peak torque to a very low speed. It is even possible for the highest torque to occur at zero

design B is preferred because of its lower starting current. Design class B motors have largely replaced design class A motors in new installations.

torque loads, such as loaded pumps, compressors, and conveyors.

speed (100 % slip)‐Full load slip is typically 7 to 11%, but may go as high as 17% or more.‐Used in applications requiring the acceleration of extremely high‐inertia‐type loads, especially large flywheels used in punch presses or shears.

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AC Motors

• Lifts require high starting torques, and for this reason the standard squirrel cage motor is not suitable. The speed torque characteristic of a standard induction motor cantorque characteristic of a standard induction motor can be improved by using two cages, one with high rotor resistance and one with low rotor resistance. By adding the two speed torque characteristics of both, a high starting torque characteristic results. All lift motor today are double cage rotors. The use of variable voltage variable frequency drives might change the need for double cage rotors, as these drives provide good starting torque characteristics even from standard motors.

Single Phase Motors

• Three‐phase induction and synchronise motors are used in larger commercial and industrial settings However most homes and smallsettings. However, most homes and small businesses do not have three‐phase power available. For such locations, all motors must run from single‐phase power sources.

• There are two major types of single‐phase j yp g pmotors: – The universal motor

– Single phase induction motor.

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Single Phase Motors

• The major problem associated with the design of single‐phase induction motors is that, unlike three‐phase power sources, a single‐phase source does not produce a rotating magnetic field. Instead, the magnetic field produced by a single‐phase source remains stationary in position and pulses with time. Since there is no net rotating pulses with time. Since there is no net rotatingmagnetic field, conventional induction motors cannot function, and special designs are necessary.

The Universal Motor• The simplest approach to the design of a motor that will operate on a single‐phase ac power source is to take a dc machine and run it from an ac supply.

• The induced torque of a dc motor is given by:

• If the polarity of the voltage applied to a shunt or series dc motor is reversed, both the direction of the field flux and the direction of the armature current reverse, and ,the resulting induced torque continues in the same direction as before. Therefore, it should be possible to achieve a pulsating but unidirectional torque from a dc motor connected to an ac power supply.

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The Universal Motor• Such a design is practical only for the series dc motor, since the armature current and the field current in the machine must reverse at exactly the same time. For shunt dc motors, the very high field inductance tends to delay the reversal of the field current and thus to unacceptably reduce the average induced torque of the motor.

The Universal Motor• In order for a series dc motor to function effectively on ac, its field poles and stator frame must be completely laminated. If they were not completely laminated, their

l ld b h h l dcore losses would be enormous. When the poles and stator are laminated, this motor is often called a universal motor, since it can run from either an ac or a dc source.

• When the motor is running from an ac source, the commutation will be much poorer than it would be with a dc source The extra sparking at the brushes is caused bydc source. The extra sparking at the brushes is caused by transformer action inducing voltages in the coils undergoing commutation. These sparks significantly shorten brush life and can be a source of radio‐frequency interference in certain environments.

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The Universal Motor• A typical torque‐speed characteristic of a universal motor is shown

in the next Figure. It differs from the torque‐speed characteristic of the same machine operating from a dc voltage source for two reasons:

Th t d fi ld i di h it l t t 50– The armature and field windings have quite a large reactance at 50 or 60 Hz. A significant part of the input voltage is dropped across these reactances, and therefore EA is smaller for a given input voltage during ac operation than it is during dc operation. Since EA = Kφω, the motor is slower for a given armature current and induced torque on alternating current than it would be on direct current.

– In addition, the peak voltage of an ac system is times its rms value, 2p g yso magnetic saturation could occur near the peak current in the machine. This saturation could significantly lower the rms flux of the motor for a given current level, tending to reduce the machine's induced torque. Recall that a decrease in flux increases the speed of a dc machine, so this effect may partially offset the speed decrease caused by the first effect.

The Universal Motor

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The Universal Motor• The universal motor has the sharply drooping torque‐ speed characteristic of a dc series motor, so it is not suitable for constant‐speedso it is not suitable for constant‐speed applications. However, it is compact and gives more torque per ampere than any other single‐phase motor. It is therefore used where light weight and high torque are important. Typical

li i f hi lapplications for this motor are vacuum cleaners, drills, similar portable tools, and kitchen appliances.

The Universal Motor• As with dc series motors, the best way to control the speed of a universal motor is to vary its rms input voltage. The higher the rms input voltage, the greater the resulting speed of the motor.

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Single Phase Induction Motors

• Single‐phase induction motors suffer from a severe handicap. Since there is only one phase on the stator winding the magnetic field in a singlethe stator winding, the magnetic field in a single‐phase induction motor does not rotate. Instead, it pulses, getting first larger and then smaller, but always remaining in the same direction. Because there is no rotating stator magnetic field, a single‐phase induction motor has no starting torque.

Single Phase Induction Motors• This fact is easy to see from an examination of the motor

when its rotor is stationary. The stator flux of the machine first increases and then decreases, but it always points in the same direction. Since the stat or magnetic field does notsame direction. Since the stat or magnetic field does not rotate, there is no relative motion between the stator field and the bars of the rotor. Therefore, there is no induced voltage due to relative motion in the rotor, no rotor current flow due to relative motion, and no induced torque. Actually, a voltage is induced in the rotor bars by transformer action (dØ/dt) and since the bars are short circuited current flows(dØ/dt), and since the bars are short‐circuited, current flows in the rotor. However, this magnetic field is lined up with the stator magnetic field, and it produces no net torque on the rotor.

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• However, once the rotor begins to turn, an induced torque will be produced in it.

Th t b i th i hi h l i h• There are two basic theories which explain why a torque is produced in the rotor once it is turning:

1‐ Double‐revolving‐field theory.

2‐ Cross‐field theory

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Double‐revolving‐field theory• The double‐revolving‐field theory of single‐phase induction motors basically states that a stationaryinduction motors basically states that a stationary pulsating magnetic field can be resolved into two rotating magnetic fields, each of equal magnitude but rotating in opposite directions. The induction motor responds to each magnetic field separately, and the net torque in the machine will be the sum of the torques due q qto each of the two magnetic fields.


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Double‐revolving‐field theory• The torque‐speed characteristic of a three‐phase induction motor in response to its single rotatinginduction motor in response to its single rotating magnetic field is shown in the following . A single phase induction motor responds to each of the two magnetic fields present within it, so the net induced torque in the motor is the difference between the two torque‐speed curves as shown in the figure. Notice that there is no net gtorque at zero speed, so this motor has no starting torque.


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Single Phase Induction Motors• There are three techniques commonly used to start these motors,

and single‐phase induction motors are classified according to the methods used to produce their starting torque.

Th i h i diff i d i h f• These starting techniques differ in cost and in the amount of starting torque produced, and an engineer normally uses the least expensive technique that meets the torque requirements in any given application. The three major starting techniques are

1. Split‐phase windings

2. Capacitor‐type windingsp yp g

3. Shaded stator poles

• All three starting techniques are methods of making one of the two revolving magnetic fields in the motor stronger than the other and so giving the motor an initial nudge in one direction or the other.

Single Phase Induction MotorsSplit Phase Windings

• A split‐phase motor is a single‐phase induction motor with two stator windings, a main stator winding (M) and an auxiliary g g ( ) ystarting winding (A) as shown in the following Figure. These two windings are set 90 mechanical degrees apart along the stator of the motor, and the auxiliary winding is designed to be switched out of the circuit at some set speed by a centrifugal switch. The auxiliary winding is designed to have a higher resistance/reactance ratio than the main winding, so that the gcurrent in the auxiliary winding leads the current in the main winding. This higher R/X ratio is usually accomplished by using smaller wire for the auxiliary winding. Smaller wire is permissible in the auxiliary winding because it is used only for starting and therefore does not have to take full current continuously.

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Single Phase Induction MotorsSplit Phase Windings

• Since the current in the auxiliary winding leads the current in the main winding, the magnetic field BA peaks beforein the main winding, the magnetic field BA peaks before the main magnetic field BM . Since BA peaks first and then BM , there is a net counterclockwise rotation in the magnetic field. In other words, the auxiliary winding makes one of the oppositely rotating stator magnetic fields larger than the other one and provides a net starting torque for the motor.

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Split Phase Windings

• Split ‐phase motors have a moderate starting torque with a fairly low starting current. They are used for applications which do not require very high starting torques, such as fans, blowers, and centrifugal pumps. They are available for sizes in the fractional‐horsepower range and are quite p g qinexpensive.

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Split Phase Windings• In a split‐phase induction motor, the current in the auxiliary

windings always peaks before the current in the main windingwindings always peaks before the current in the main winding, and therefore the magnetic field from the auxiliary winding always peaks before the magnetic field from the main winding. The direction of rotation of the motor is determined by whether the space angle of the magnetic field from the auxiliary winding is 90° ahead or 90° behind the angle of the

i i di Si th t l b h d f 90° h dmain winding. Since that angle can be changed from 90° ahead to 90° behind just by switching the connections on the auxiliary winding, the direction of rotation of the motor can be reversed by switching the connections of the auxiliary winding while leaving the main winding's connections unchanged.


Capacitor single phase induction motor:‐ Capacitor‐start motor.

( )‐ Capacitor‐start‐and‐run (permanent split) motor.

‐ Two capacitor (capacitor‐ start, capacitor‐run) motor.

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Capacitor Start Motor:

• For some applications, the starting torque supplied by a split‐phase motor is insufficient to start the load on a motor's shaft. In those cases, capacitor‐start motors may be used. In a capacitor‐start motor, a capacitor is placed in series with the auxiliary winding of the motor.y g



• By proper selection of capacitor size, the magnetomotive force of the starting current in the auxiliary winding can beforce of the starting current in the auxiliary winding can be adjusted to be equal to the magnetomotive force of the current in the main winding, and the phase angle of the current in the auxiliary winding can be made to lead the current in the main winding by 90°. Since the two windings are physically separated by 90°, a 90° phase g p y y p y pdifference in current will yield a single uniform rotating stator magnetic field, and the motor will behave just as though it were starting from a three‐phase power source. In this case, the starting torque of the motor can be more than 300% of its rated value.

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• Capacitor‐start motors are more expensive than split‐phase motors and they are used in applications where aphase motors, and they are used in applications where a high starting torque is absolutely required. Typical applications for such motors are compressors, pumps, air conditioners, and other pieces of equipment that must start under a load.

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AC Vs. DC Motors• AC motors have become popular in many machine tools. AC

motors operate without brushes. They are more reliable, rugged in construction, and have less maintenance. AC motors are classified as single and 3‐phase. g p

• The velocity of the AC synchronous motor is controlled by the variable frequency supply.

• The popularity of alternating current motors (AC) is due to the following reasons.

‐Most of the power‐generating systems produce alternating currentcurrent.

‐ AC motors cost less than DC motors.

‐ Some AC motors do not use brushes and commutators. This eliminates many problems of maintenance and wear. It also eliminates the problem of dangerous sparking.

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AC Vs. DC Motors

• The AC motor is particularly well‐suited for constant‐speed applications. This is because its speed is determined by the frequency of the AC voltage applied to the motor terminals.frequency of the AC voltage applied to the motor terminals.

• The DC motor is better‐suited for applications that require variable speeds. An AC motors can also be made with variable speed characteristics but only within certain limits.

• For high torques and powers, AC induction motors are usually used, provided that accurate positioning is notusually used, provided that accurate positioning is not critical.

• AC motors are available in different sizes, shapes, and ratings for many different types of jobs.

Stepper Motor• Introduction

• Principle of operation

• Stepper Motor Types (Classifications)

• Stepper Motor Movement and Step Mode

• Stepping Angle Analysis

• Winding Types

• Characteristics• Characteristics

• Comparison

• Model and Dynamic Response (self reading)

• Motor Selection

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Stepper Motor

Introduction

• Stepper motor are electromechanical devices used primarily to convert information in digital form to mechanical motion.

• Stepper motors are categorised as doubly salient machines, which means that they have teeth of magnetically permeable material on both the stationary part (the ‘stator’) and the rotating part (the ‘rotor’)

Stepper MotorPrinciple of Operation

• A cross‐section of a small part of a stepping motor is shown schematically in the following figure Magnetic flux crossesschematically in the following figure. Magnetic flux crosses the small airgap between teeth on the two parts of the motor. According to the type of motor, the source of flux may be a permanent‐magnet or a current‐carrying winding or a combination of the two. However, the effect is the same: the teeth experience equal and opposite forces, p q ppwhich attempt to pull them together and minimise the airgap between them.

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Stepper Motor


• As the figure shows, the major component of these ( )forces, the normal force (n), is attempting to close

the airgap, but for electric motors the more useful force component is the smaller tangential force (t),which is attempting to move the teeth sideways with respect to each other. As soon as the flux ppassing between the teeth is removed, or diverted to other sets of teeth, the forces of attraction decrease to zero.

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• To produce a significant torque from a reasonable volume, both the stationary and rotating components must haveboth the stationary and rotating components must have large numbers of iron teeth, which must be able to carry a substantial magnetic flux

• The performance of the stepper motor depends on the strength of the magnetic field. High flux leads to high torque.torque.

Stepper Motor

Stepper Motor ClassificationV i bl R l (VR)• Variable Reluctance (VR)

• Permanent Magnet (PM)

• Hybrid

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Stepper MotorVariable Reluctance (VR)

• Single‐stack

• Multi‐stack

Stepper MotorSingle‐stack• The rotor is a nonmagnetized soft‐iron bar consists of several number of teethseveral number of teeth.

• The stator made of laminated iron and has number of windings (i.e. Phases).

• Each phase has number of salient poles. Each pole has number of teeth and being geometrically separated by a certain angle from the adjacent onecertain angle from the adjacent one.

• Examples of single‐stack VR stepper motor are shown in the following figures:

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3 phase, 6 poles, 4 rotor teeth.


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3 phase, 12 poles (teeth), 8 rotor teeth.


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Stepper Motor

The number of stator teeth is restricted by the numbers of phases and rotor teeth Each phase is distributed overphases and rotor teeth. Each phase is distributed over several stator teeth and, since there must be as many stator teeth directing flux towards the rotor as away from it, the number of stator teeth has to be an even multiple of the number of phases, e.g. in a three‐phase motor there can be 6, 12, 18, 24, . . . stator teeth. In addition, for , , , , , fsatisfactory stepping action, the number of stator teeth must be near (but not equal) to the number of rotor teeth.

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Stepper MotorMultiple‐stack• The multi‐stack variable‐reluctance stepping motor is divided along

its axial length into magnetically isolated sections (‘stacks’), each of hi h b it d b t i di (‘ h ’) I th twhich can be excited by a separate winding (‘phase’). In the cutaway

view of the following figure, for example, the motor has three stacks and three phases, but motors with up to seven stacks and phases have been manufactured.

• Each stack includes a stator, held in position by the outer casing of the motor and carrying the motor windings, and a rotating element. Th t l t f b i t d i l it hi h i t dThe rotor elements are fabricated as a single unit, which is supported at each end of the machine by bearings and includes a projecting shaft for the connection of external loads, as shown in the next figure. Both stator and rotor are constructed from electrical steel, which is usually laminated so that magnetic fields within the motor can change rapidly without causing excessive eddy current losses.

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Stepper MotorMultiple‐stack• The stator of each stack has a number of poles – the figureshows an example with four poles – and a part of the phase winding is wound around each pole to produce a radial magnetic field in the pole. Adjacent poles are wound in the opposite sense, so that the radial magnetic fields in adjacent poles are in opposite directions. The complete magnetic circuit for each stack is from one stator pole, across the airgap into the rotor through the rotor across the airgap into aninto the rotor, through the rotor, across the airgap into an adjacent pole, through this pole, returning to the original pole via a closing section, called the ‘back‐iron’. This magnetic circuit is repeated for each pair of poles, and therefore in the example of the figure there are four main flux paths.

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Stepper MotorPermanent Magnet (PM)

• The rotor in PM stepper motors is actually a permanent‐magnet. In most cases, the permanent magnet is in the g , p gshape of a disk surrounding the rotor shaft (No teeth). One arrangement is a magnetic disk which consists of north and south magnetic poles interlaced together (two poles). The number of poles on the magnetic disk varies from motor to motor. Some simple PM stepper motors has

l l h d k h l h honly two poles on the disk, while others may have many poles.

• The stator usually has two or more windings (phases), with each winding around a soft metallic core.

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Stepper MotorHybrid

• Hybrid stepper motors have two stacks of rotor teeth forming the two poles of a permanent magnet located g p p galong the rotor axis.

• The hybrid stepper motor consists of two pieces of soft iron, as well as an axially magnetized, round permanent magnet rotor. The term hybrid is derived from the fact that the motor is operated under the combined principles of the permanent magnet and variable reluctance stepper motors. The stator core structure of a hybrid motor is essentially the same as its VR counterpart.

Stepper MotorHybrid

• The hybrid stepping motor has a doubly salient structure (two stacks), but the magnetic circuit is excited by a ( ), g ycombination of windings and permanent magnet. Windings are placed on poles on the stator and a permanent magnet is mounted on the rotor. The main flux path for the magnet flux, shown in following Figure, lies from the magnet N‐pole, into a soft‐iron end‐cap, radially h h h d h h h hthrough the end‐cap, across the airgap, through the stator poles of section X, axially along the stator back‐iron, through the stator poles of section Y , across the airgap and back to the magnet S‐pole via the end‐cap.

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Stepper MotorHybrid• There are typically eight stator poles, as in the previous Figure, and each pole has between two and six teeth. The t t l l id d ith i di hi h dstator poles are also provided with windings, which are used to encourage or discourage the flow of magnet flux through certain poles according to the rotor position required. Two windings are provided and each winding (phase) is situated on four of the eight stator poles: winding A is placed on poles 1, 3, 5, 7 and winding B is on poles 2, 4, 6, 8. p , , , g p , , ,Successive poles of each phase are wound in the opposite sense, e.g. if winding A is excited by positive current the resultant magnetic field is directed radially outward in poles 3 and 7, but radially inward in poles 1 and 5. A similar scheme is used for phase B.

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Stepper MotorHybrid

• Both the stator poles and rotor end‐caps are toothed. For the motor illustrated in the previous Figure, each of the eight poles has two teeth, giving a total of 16 stator teeth, and the rotor has 18 teeth. Note that the stator teeth in sections X and Y are fully aligned, whereas the rotor teeth are completely misaligned between the two sections. If the magnet flux is concentrated in certain poles because of the winding excitation then the rotor tends to alignof the winding excitation then the rotor tends to align itself so that the airgap reluctance of the flux path is minimised.

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Stepper MotorStepper Motor Movement and Step Mode• Full step:

By applying full voltage to one or two phases of the motor, full stepping is achieved. The motor will advance by the stepping angle.

• Half step

By applying equal voltage to adjacent phases, half stepping can b hi dbe achieved.

• Micro step

By applying fractions of voltage to adjacent phases, micro‐stepping is achieved. Micro‐stepping can be applied in very small steps down to around 125th of the full step

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Stepper Motor

Stepper Motor Movement and Step Mode

Full stepFull step

• The following Figure shows a typical cross‐sectional view of the rotor and stator of a stepper motor. From this figure you can see that the stator (stationary winding) has eight poles, and the rotor h i l ( h)has six poles (teeth).

• For full step mode, the switching sequence is:

A, B, C, D, A, B, C, D, A...

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Stepper MotorStepper Motor Movement and Step Mode• Figure description: Movement of the stepper motor rotor as current is pulsed to the stator (a)motor rotor as current is pulsed to the stator. (a) Current is applied to the A and A’ windings, so the A winding is north, (b) Current is applied to B and B’ windings, so the B winding is north, (c) Current is applied to the C and C’ windings, so the C

d h (d) l d hwinding is north, (d) Current is applied to the D and D’ windings so the D winding is north. (e) Current is applied to the A and A’ windings, so the A’ winding is north.

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Stepper MotorStepper Motor Movement

Full Step

Degree 15 30 45 60 75 90 105 120 135

A 1 0 0 0 1 0 0 0 1

B 0 1 0 0 0 1 0 0 0

C 0 0 1 0 0 0 1 0 0

D 0 0 0 1 0 0 0 1 0

Stepper Motor

Stepper Motor Movement and Step Mode

Half stepHalf step• Another switching sequence for the stepper motor is called an half‐

step sequence. The switching diagram for the half‐step sequence is shown in the following figure. The main feature of this switching sequence is that you can double the resolution of the stepper motor by causing the rotor to move half the distance it does when the full‐step switching sequence is used. This means that a 200‐step motor,step switching sequence is used. This means that a 200 step motor, which has a resolution of 1.8°, will have a resolution of 400 steps and 0.9°. The half‐step switching sequence requires a special stepper motor controller, but it can be used with a standard hybrid motor. The way the controller gets the motor to reach the half‐step is to energize both phases at the same time with equal current.

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Stepper MotorStepper Motor Movement and Step Mode

Half stepI hi h fi h SW1 i d SW2 SW3 d SW4• In this sequence the first step has SW1 is on, and SW2,SW3 and SW4 are off. The sequence for the first step is the same as the full‐step sequence. The second step has SW1 and SW2 are on and all of the remaining switches are off. This configuration of switches causes the rotor to move an additional half‐step because it is acted upon by two equal magnetic forces and the rotor turns to the equilibrium position which is half a step angle The third step has SW2 is on andposition which is half a step angle. The third step has SW2 is on, and SW1, SW4 and SW3 are off, which is the same as step 2 of the full step sequence. The sequence continues for eight steps and then repeats. The main difference between this sequence and the full‐step sequence is that the energizing sequence for half step is A AB B BC C CD D DA.

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Stepper MotorStepper Motor Movement

Half Step

Degree 7.5 15 22.5 30 37.5 45 52.5 60 67.5

A 1 1 0 0 1 0 0 1 1

B 0 1 1 1 0 0 0 0 0

C 0 0 0 1 1 1 0 0 0

D 0 0 0 0 0 1 1 1 0


Micro step

• The full step and half step motors tend to be slightly jerky• The full‐step and half‐step motors tend to be slightly jerky in their operation as the motor moves from step to step. The amount of resolution is also limited by the number of physical poles that the rotor can have. The amount of resolution (number of steps) can be increased by manipulating the current that the controller sends to the p gmotor during each step. The current can be adjusted so that it looks similar to a sine wave. The following Figure shows the waveform for the current to each phase.

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Micro step


Micro step

• From the previous diagram you can see that the current• From the previous diagram you can see that the current sent to each of the four sets of windings is timed so that there is always a phase difference with each other. The fact that the current to each individual phase increases and decreases like a sine wave and that is always out of time with the other phase will allow the rotor to reach phundreds of intermediate steps. In fact it is possible for the controller to reach as many as 500 micro steps for a full‐step sequence, which will provide 100,000 steps for each revolution.

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Stepper MotorStepping Angle Analysis

(On board)

Stepper Motor

Type of winding• BifilarBifilar

• Unifilar

Type of drive• Unipolar

• Bipolar

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Unipolar Stepper Motor

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Bipolar Stepper Motor

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Stepper Motor• The torque of the stepper motor is proportional to the magnetic field intensity of the stator windings. It may be increased only by adding more windings or by increasing the current. A natural limit against any current increase is the danger of saturating the g y g giron core. Though this is of minimal importance. Much more important is the maximum temperature rise of the motor, due to the power loss in the stator windings. This shows one advantage of the bipolar circuit, which, compared to unipolar systems, has only half of the copper resistance because of the double cross section of the wire. The winding current may be increased by thesection of the wire. The winding current may be increased by the factor √2 and this produces a direct proportional affect on the torque. At their power loss limit bipolar motors thus deliver about 40 % more torque than unipolar motors built on the same frame. If a higher torque is not required, one may either reduce the motor size or the power loss.

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Stepper MotorUNIFILAR

• Unifilar, as the name implies, has only one winding per stator pole. Stepper motors with a unifilar winding will have 4 lead wires. The following wiring diagram illustrates a typical unifilar motor.

Stepper MotorBIFILAR

• Bifilar wound motors means that there are two identical sets of windings on each stator pole. This type of winding configuration simplifies operation in that transferring current from one coil to another one, wound in the opposite direction, will reverse the rotation of the motor shaft. Whereas, in a unifilar application, to change direction requires reversing the current in the same winding.

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Stepper MotorTorque Vs. Angle Characteristics (static torque curve)

• The torque vs. angle characteristics of a stepper t th l ti hi b t thmotor are the relationship between the

displacement of the rotor and the torque which applied to the rotor shaft when the stepper motor is energized at its rated voltage. An ideal stepper motor has a sinusoidal torque vs. displacement characteristic as shown in the following figure.

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Stepper Motor• Positions A and C represent stable equilibrium points when no external force or load is applied to the rotor shaft When you apply an external forcethe rotor shaft. When you apply an external force Ta to the motor shaft you in essence create an angular displacement, Ѳa. This angular displacement, Ѳa, is referred to as a lead or lag angle depending on whether the motor is actively accelerating or decelerating. When the rotor stops with an applied load it will come to rest at the position defined by this displacement angle.

Stepper Motor• The motor develops a torque in opposition to the applied external force in order to balance the load. As the load is increased the displacement angleAs the load is increased the displacement angle also increases until it reaches the maximum holding torque, Th, of the motor. Once Th is exceeded the motor enters an unstable region. In this region a torque in the opposite direction is created and the rotor jumps over the unstable point to the next stable point.

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Stepper MotorTorque vs. Speed Characteristics

• The torque vs. speed characteristics are the key to select the right motor and drive method for a specificthe right motor and drive method for a specific application. These characteristics are dependent upon (change with) the motor, excitation mode and type of driver or drive method. A typical “speed – torque curve” is shown in the following figure.

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Stepper Motor

Stepper Motor• The pull‐in curve describes the maximum constant start/stop rate that a fractionally loaded motor can achieve without loss of step. This curve is dependent onachieve without loss of step. This curve is dependent on the method of driving the motor and the load inertia. The effect of the latter is shown in the following Figure.

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Stepper Motor• The pull‐out curve describes the maximum stepping rate which a fractionally loaded motor can follow without losing steps, assuming sufficient time is allowed tolosing steps, assuming sufficient time is allowed to accelerate the motor by ramping the frequency of the command drive circuit. Within the start/stop region the motor can be started, stopped or forced to change direction of rotation following a sudden command change from the drive circuit. However, within the slew range the motor can only be accelerated or decelerated to the required speed and it cannot suddenly change direction.

Stepper Motor• Detent torque: The maximum torque that can be applied to the rotor of an unexcited motor without causing continuous rotation.continuous rotation.

• Holding torque: The maximum steady torque that can be externally applied to the rotor of an excited motor without causing continuous rotation.

• Maximum working torque: The maximum torque that can be obtained from the motor.can be obtained from the motor.

• Pull‐in torque: The maximum torque that can be applied to a motor when starting at the pull‐in rate.

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Stepper Motor• Pull‐in rate (speed): The maximum switching rate (speed) at which a fractionally loaded motor can start without losing steps.losing steps.

• Maximum pull‐in rate (speed): The maximum switching rate (speed) at which an unloaded motor can start without losing steps.

• Pull‐out rate (speed): The maximum switching rate (speed) which a fractionally loaded motor can follow(speed) which a fractionally loaded motor can follow without losing steps.

• Pull‐out torque: The maximum torque that can be applied to a motor when running at the pull‐out rate.

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Stepper Motor ComparisonCharacteristics VR PM Hybrid

Rotor Toothed soft‐iron nonmagnetizedrotors (low rotor

Magnetized rotor (moderate rotor inertia)

Two stacks of rotor teeth forming the two poles of a

inertia) permanent magnet (high rotor inertia)

Design Complexity Simple Moderately complex Complex

Cost Moderate Cheapest Most expensive

Step size Typical 15 15 to 7.5 degrees (typical 15)

3.6 to 0.9 degrees(typical 1.8)

Noise Noisy, specially With Quiet QuietNoise Noisy, specially Withsinusoidal exciting currents

Quiet Quiet

Detent Torque Zero Exist (cog when they are turned by hand)

Exist (cog when they are turned by hand)

Stepping Mode Normally run in full step only.

Full, half, and micro step modes.

Full, half, and micro step modes.

Stepper Motor Comparison

Characteristics VR PM Hybrid

Winding and drive Unifilar.Bipolar.

Unifilar and bifilar.Unipolar and bipolar.

Unifilar and bifilar.Unipolar and bipolar.Bipolar. Unipolar and bipolar. Unipolar and bipolar.

Torque (For the same motor size)

Low Moderate High

Speeds Low Moderate High

Application In applications requiring small step angle, high torque, and high speed in a g prestricted working space.

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Stepper Motor

Advantages1. The rotation angle of the motor is proportional to the input pulse.

2. The motor has full torque at standstill (if the windings are energized)

3. Precise positioning and repeatability of movement since good stepper motors have an accuracy of 3 – 5% of a step and this error is non‐cumulative from one step to the next.

4. Excellent response to starting/stopping/reversing.p g pp g g

5. Very reliable since there are no contact brushes in the motor. Therefore the life of the motor is simply dependant on the life of the bearing.

Stepper Motor

Advantages6. The motors response to digital input pulses provides open‐loop

l ki h i l d l l lcontrol, making the motor simpler and less costly to control.

7. It is possible to achieve very low speed synchronous rotation with a load that is directly coupled to the shaft.

8. A wide range of rotational speeds can be realized as the speed is proportional to the frequency of the input pulses.

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Stepper Motor

Applications

• Computer Peripherals (printer, plotter,...)p p (p , p , )

• Business Machines (card reader, copy machine,...)

• Process control (valve control, conveyors,...)

• Machine tool (CNC milling, turning,...)

• Robotics Robotics

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Motor Selection

• One of the most critical decisions the mechatronics engineering designer has to g g gmake is the selection of the type and size of motor required for the mechatronics system.

Motor Selection

A number of factors must be taken into consideration when selecting the type of motor to use:

• Positioning accuracy and resolution: In cases where accurate positioning is required, stepper motors offer an attractive option. In cases where this is not critical it is more convenient to use AC orthis is not critical it is more convenient to use AC or DC motors.

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Motor Selection

• Power and torque requirements: In sizing the motor, student should be able to meet both the power requirements as well as the torque requirements. The student has to have an appreciation of any out of balance loads and load related inertias. For high torques and powers, AC induction motors are usually used, provided that induction motors are usually used, provided thataccurate positioning is not critical.

Motor Selection

• Continuous duty cycle and intermittent duty cycle:In cases of continuous duty cycle no consideration needs to be given for the effect of acceleration, but in intermittent duty cycle applications with frequent starting and stopping, consideration has to be given to the effect of acceleration and deceleration.

• Speed range: Where large variations in speed are• Speed range: Where large variations in speed are required by the application, DC motors are ideal as they offer a wide speed range and good speed regulation.

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Motor Selection

Power/ Positioning Requirement forPower/ Torque

Positioning Accuracy

Requirement for Feedback

Stepper Motor Low Good No

Servomotors Low Good No (integrated)

AC squirrel cage High Poor Yesinduction motor

DC motor Low/high Poor Yes

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Stepper Vs. Servo• Stepper motors are operated open loop, while servo motors are

operated closed loop.

• Stepper motors are easily controlled with microprocessors howeverStepper motors are easily controlled with microprocessors, however logic and drive electronics are more complex.

• Stepper motors are brushless and brushes contribute several problems, e.g., wear, sparks, electrical transients.

• Servo motors have continuous displacement and can be accurately positioned, whereas stepper motor motion is incremental and its resolution is limited to the step sizeresolution is limited to the step size.

• Stepper motors can slip if overloaded and the error can go undetected. (A few stepper motors use closed‐loop control.)

• Feedback control with servo motors gives a much faster response time compared to stepper motors.

Motor Sizing and Selection Process

• The motor sizing and selection process is based on the calculation of torque and inertia imposed by the mechanical set up plus the speed and acceleration required by the application. The selected motor must be able to safely drive the mechanical set up by providing sufficient torque and velocity. Once the requirements have been established, it is easy the requirements have been established, it is easyto look either at the torque vs. speed curves or motor specifications and choose the right motor.

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The sizing process involves the following steps:

1. Establishment of motion objectivesj

2. Selection of mechanical components

3. Definition of a load (duty) cycle

4. Load calculation

5. Motor selection5. Motor selection

Motor Sizing and Selection ProcessEstablishment of motion objectivesA written outlining of the motion control application will help to establish the necessary parameters needed for the next steps.– Required positioning accuracy

– Required position repeatability

– Required velocity accuracy

– Resolution.

l l d f– Power supply voltage and frequency

– Linear or rotary application

– If linear application: Horizontal or vertical application

– Thermal considerations (Ambient temperature)

– What motor technologies are best suited for the application

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Motor Sizing and Selection ProcessSelection of mechanical components (drive Mechanism)

The engineer must decide which mechanical components are g prequired for the application. For instance, a linear application may require a leadscrew or a conveyor. For speed transmission a gear or a belt drive may be used.– Direct drive

– Special application or standard mechanical devices

– If linear application: Use of linear motor or leadscrew, conveyor...

– Reducer required: Gearbox, belt drive, etc.

– Check shaft dimensions: select couplings

– Check mechanical components for speed and acceleration limitations

Motor Sizing and Selection ProcessSelection of mechanical components (drive Mechanism)

Along with the type of drive mechanism the engineerAlong with the type of drive mechanism, the engineer must also determine the dimensions, mass and friction coefficient etc., that are required for the load calculation. The general items are explained below.

• Dimensions and mass (or density) of load

• Dimensions and mass (or density) of each part

• Friction coefficient of the sliding surface of each moving part

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Definition of a load duty cycle (Speed profile)The engineer must define the maximum velocity, maximum acceleration duty cycle time acceleration and decelerationacceleration, duty cycle time, acceleration and deceleration ramps, dwell time, etc., specific to the application.– Define critical move parameters such as velocity, acceleration rate

– Triangular, trapezoidal or other motion profile

– If linear application: Make sure the duty cycle does not exceed the travel range of linear motion device.

– Jerk Limitation required

– Positioning distance and positioning time

– Consideration of thrust load

– Does the load change during the duty cycle

– Holding brake applied during zero velocity

Motor Sizing and Selection ProcessLoad calculation• The load is defined by the torque that is required to drive the

mechanical set up. The amount of torque is determined by the i ti “ fl t d” f th h i l t t th t dinertia “reflected” from the mechanical set up to the motor and the acceleration at the motor shaft.– Calculate inertia of all moving components

– Determine inertia reflected to motor

– Determine velocity, acceleration at motor shaft

– Calculate acceleration torque at motor shaftCalculate acceleration torque at motor shaft

– Determine non‐inertial forces such as gravity, friction, pre‐load forces, external forces...

– Calculate constant torque at motor shaft

– Calculate total acceleration and RMS (continuous over duty cycle) torque at motor shaft

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Motor Selection

• The motor must be able to provide the torque p qrequired by the mechanical set up plus the torque inflicted by its own rotor. Each motor has its specific rotor inertia, which contributes to the torque of the entire motion system. When selecting a motor the engineer needs to recalculate the load torque forengineer needs to recalculate the load torque for each individual motor.

Motor Sizing and Selection Process• Decide the motor technology to use (DC brush, DC brushless, single phase or 3‐ph AC, stepper...)

S l t t /d i bi ti• Select a motor/drive combination

• Does motor support the required maximum velocity ? If no, select next motor/drive.

• Use rotor inertia to calculate system (motor plus mechanical components) acceleration (peak) andmechanical components) acceleration (peak) and RMS torque

• Does motor’s rated torque support the system’s RMS torque? If no, select next motor/drive.

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• Does motor’s intermittent torque support the system’s peak torque? If no, select next motor/drive.

D th t ’ f (t• Does the motor’s performance curve (torque over speed) support the torque and speed requirements? If no, select next motor/drive.

• If the ratio of load over rotor inertia exceeds a certain range (for servo motors and steppers 6:1) consider the use of a gearbox or increase the transmission ratio of the existing gearbox. Servo or stepper motors should not be operated over a ratio of 10:1.


• Large inertia ratios cause large overshooting and undershooting during starting and stopping, which can affect starting time and settling time. Depending on the conditions of usage, operation may be impossible.

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The basic motor selection criteria are:

• The motor’s rated speed must be equal to or exceed the application’s maximum speed

• The motor’s intermittent torque must be equal or exceed the load’s maximum (intermittent) torque.

• The motor’s rated torque must be equal to or exceed the load’s RMS torqueexceed the load s RMS torque.

• The ratio of load inertia to rotor inertia should be equal to or less than 6:1.

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Example (1): Conveyor Belt • A conveyor belt that moves objects is to be powered using a stepper motor. It has a gearing ratio of (r )and a diameter of the driving pulley ofratio of (rg)and a diameter of the driving pulley of (ds). The stepper motor has an inertia (Im), the gearbox has a high speed shaft inertia of (Ig1)and a low speed shaft inertia (Ig2). The conveyor belt pulleys have inertias of (Id) and (Is). The conveyor belt has a mass of (mC) and the objects on it have a total mass of (mL).

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Example (1): Conveyor Belt • The conveyor has to be advanced by a fixed distance (d) regularly in a specified period of time (T) In order to achieve this in the specified time(T). In order to achieve this in the specified time, the conveyor has to be accelerated at a constant acceleration until it covers half the distance d and then decelerated at a constant deceleration until it comes to rests within half the distance (d) as well. The values of acceleration and deceleration are equal.

Example (1): Conveyor Belt

• High positioning accuracy is required in this application, as the objects need to be positioned ready for the next operation (e.g., a manufactured items that need further processing, either automatic or manual). A stepper motor is ideal for this application, as it ll d iti iallows good positioning accuracy.

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Example (1): Conveyor Belt • The following are the parameters of the conveyor:

Example (1): Conveyor Belt

• The selection of the stepper motor is to be based on the table shown below. It shows 4 motor models with each motor’s available torque (holding torque)with each motor s available torque (holding torque) and its motor inertia.

2000 SM 30 4620 × 10‐6

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Example (2): Leadscrew Using Stepper Motor

Example (2): Leadscrew Using Stepper Motor• Specifications and Operating Conditions of the Drive Mechanism

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Example (2): Leadscrew Using Stepper Motor

Example (3): Leadscrew Using Servo Motor

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Example (3): Leadscrew Using Servo Motor

Example (4):

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Example (4):

Case of Study

Motor Sizing and Selection for Hoisting System• Introduction

• Output Mechanical Power• Output Mechanical Power

• Steady State Method (Power Only)

• Dynamic Method (Moment of Inertia and Starting Torque)

• Other Motor Parameters

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Motor Sizing and Selection for Hoisting System

Introduction

• When moving any load, two stages are encountered: Initial transient state and steady state. In the initial transient state, the speed of the load has to be increased up to the required speed. During the steady state, power has to be supplied to overcome the out of balance loads and keep masses moving at the rated speed. These two stages are shown diagrammatically in the Figure.


• During both of these stages, power needs to be supplied to overcome any friction and other related losses (e.g., air resistance). These two stages can be related to the typeair resistance). These two stages can be related to the type of energy which needs to be supplied to the moving mass:– Kinetic Energy

– Potential Energy

– Friction and other losses

• Based on the understanding of the three elementsBased on the understanding of the three elements above, the motor selection criteria can be developed.

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• Kinetic energy: During the initial acceleration phase, the motor needs to supply enough kinetic energy to the masses to accelerate them up to the required speed. Inmasses to accelerate them up to the required speed. In the case of lifts, these moving masses include the passenger load, the cabin mass, the counterweight, the ropes, the motor's rotor inertia, the sheave and gear inertias, and the hand‐wheel inertia (or the flywheel if one is fitted).

• In the case of an escalator or passenger conveyor, these masses would include in addition to the mass of the travelling passengers, the mass of the steps and the chain linking them.


• Potential energy: The motor has to be able to lift the out of balance load at the rated speed. In theory this energy is stored in these masses, and if the same masses comeis stored in these masses, and if the same masses come down again, then the energy should be returned to the supply.

• Frictional and other losses: In supplying the two elements of energy mentioned above, the motor needs to overcome the friction in the system (and the air resistance at high y ( gspeeds) and the losses in the gearbox.

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Motor Sizing and Selection for Hoisting SystemOutput Mechanical Power• When selecting a motor, it is important to note that the rating

of the motor (i.e., that on the nameplate) refers to net output f h Thi i l h d f hpower from the motor. This is equal to the product of the

output net torque (i.e., the output mechanical torque less the friction and windage losses) and the output rotational speed in radians per second. This output mechanical power will obviously be less than the input electrical power to the motor, due to the following losses (taking an AC motor as an example):– Stator copper losses.

– Core losses (Eddy current and hysteresis losses).

– Rotor copper losses.

– Friction and windage losses.


• Thus the efficiency of the motor is the division of the output mechanical power by the input electrical power.

• By using the input rated current voltage and power factor• By using the input rated current, voltage and power factor, and the output speed and torque, the efficiency of any motor can be calculated.

• Next, we examine the basic method of motor selection. Lift systems are used as an example for this. The same principles apply to any other hoisting system.principles apply to any other hoisting system.

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Steady State Method (Power Only)

• The first and most commonly used method is to check that the motor supplies sufficient power to move the outthat the motor supplies sufficient power to move the out of balance mass at the rated speed.

• The basic method can be used in most cases to select the size of the motor. The worst case is taken as that of lifting a fully loaded car in the up direction at the rated speed(assuming that the counterweight ratio is 50% or less).(assuming that the counterweight ratio is 50% or less). This is calculated in the following stages:


• The fully loaded passenger mass is reduced by the amount of counterweight. This represents the out of balance mass.

• This is multiplied by the acceleration of gravity, to determineThis is multiplied by the acceleration of gravity, to determine the force (N) needed to move this out of balance mass against gravity in the up direction.

• Multiplying this force by the rated speed gives the rate of flow of energy, or in other words the power (W). This represents the net output power of the system.

• This calculated net output power has to flow through the system, and thus a larger value of power has to be supplied to account for all losses in the shaft and gearbox. Thus, the net output power is divided by the shaft efficiency and the gearbox efficiency to provide the required motor power output rating.

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• The efficiency of the shaft and the gearbox (forward) can be taken as a combined figure of 85%, if no exact data is available. The forward efficiency of the gearbox is taken in this case because the motor is drivingmotor is driving.

• Notes:– The reverse efficiency is only taken if the braking calculation is being

carried out, which is outside the scope of this study. The forward and reverse efficiencies for worm gearboxes are different.

– The counterweight ratio specifies the mass of the counterweight. The counterweight is set to equal the sum of the mass of the car and acounterweight is set to equal the sum of the mass of the car and a percentage of the mass of the passenger. That ratio is equal to the counterweight ratio. The use of counterweighting allows the reduction of the power rating of motors, because the counterweight is used to compensate for the dead load, with the motor sized to deal with only the live loads.


• The formula for sizing the motor for a lift is as follows:

Where:– (P) is the rated passenger number in the car;

– (75) stands for 75 kg/passenger;

– (9.81) is the acceleration due to gravity m/s2;

( ) i h d d i /– (s) is the rated top speed in m/s;

– (CF) is the counterweight factor (a factor less than 1 and usually chosen to be less than 0.5);

– (η) is the total efficiency of the installation (taken around 85% if no other data available).

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• For a hydraulic lifts, the same formula can be used by replacing CF by ‐1.

• The counterweight ratio is important, and accounts for theThe counterweight ratio is important, and accounts for the fact that if the counterweight ratio is less than 50%, then the worst case scenario would be for a full car moving upwards. For example, if the counterweight ratio is only 40%, then when the car is full, only 40% weight of the passengers is compensated for by the weight of the counterweight, and the motor has to provide enough torque to lift the other 60%. Using a counterweight ratio of 40% is quite common. This is in recognition of the fact that the car rarely fills up to more than 80% of its rated load.


Example:

• A VVAC driven lift has a rated speed of 1.0 m/s, and a car load of 78 passengers If the overall system efficiency isload of 78 passengers. If the overall system efficiency is 68%, and the counterweight ratio of 45%, calculate the size of the motor.

Solution:

• Applying the above formula, gives:

• Thus, a 50 kW motor can be selected.

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• The efficiency of the system can also be determined based on the speed and whether it is geared or not. The Figure below shows how the system efficiency can vary against speed for a geared system.


Dynamic Method (Moment of Inertia and Starting Torque)• The basic power method only ensures that the motor can lift the out of balance masses at the rated speed It thuslift the out of balance masses at the rated speed. It thus only addresses the steady state situation (i.e., after all masses have started moving at the rated speed). However, the motor has to accelerate these masses up to the rated speed, and it has to do so at an acceptable value of acceleration.

• In this section we examine the moment of inertia method, as a checker for the basic method. It ensures that the motor is capable of supplying the required kinetic energy for translational and rotational masses at the required acceleration.

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Translational Masses• For all translational masses, the motor has to supply them with

the necessary kinetic energy to allow them to travel at thethe necessary kinetic energy to allow them to travel at the rated speed. The kinetic energy for a mass m moving at a velocity of v, is:

• In addition, the motor when lifting out of balance masses, has to provide the potential energy. The potential energy is expressed as:

where:– m is the mass of the object in kg;

– 9.81 m/s2 is the acceleration due to gravity;

– d is the distance the object is lifted.


• The translational masses in a lift system to be taken into consideration are:

C– Car mass.

– Passenger mass.

– Counterweight mass.

– Rope mass.

– Trailing cable mass (usually negligible in practice)g ( y g g p )

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Rotational Masses

• A mass m rotating around a certain point at a distance of r and at a speed v has a kinetic energy equal to:r, and at a speed v, has a kinetic energy equal to:

Where:– m is the mass of the object in kg;

– r is the radius of rotation in m;

– ω is the rotational speed in rad/s;


• By comparing this form to the translational form, it can be noticed that the linear speed corresponds to rotational speed in rad/s, the translational massm corresponds to aspeed in rad/s, the translational mass m corresponds to a rotational “mass” of (mr2) which is called the moment of inertia (I).

• The moment of inertia of the following items has to be included in the calculation: The rotor moment of inertia, the gearbox, the coupling, the brake drum, the g p ghandwheel or flywheel.

• The moment of inertia of all masses (translational and rotational) has to be referred to the motor shaft (also referred to as the high speed shaft, HSS).

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Derivation of the formulae• The formulae used to check that the motor is capable of

accelerating all the masses are derived in this sub‐section. They f h f ll i Firefer to the following Figure.


Derivation of the formulae• First, we need to discuss the method of referring the translational masses, to the motor shaft. This is necessary in order to calculate the rotational acceleration at the motor shaft, with all quantities referred to it.

• The translational masses are the mass of the car, the car‐load and the counterweight. Other masses which could be taken into consideration are the mass of the cables (ropes), th f th l t i l t ili bl d th fthe mass of the electrical trailing cable and the mass of any compensating cables if applicable. However, these masses are insignificant compared to the car, counterweight and passengers, so that they can be ignored.

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Derivation of the formulae

• To convert from translational quantities to rotational quantities referred to the motor, we need to find the qrelationship between the linear motion and the rotational motion. If the car travels by x metres, this will cause the sheave to rotate by:

Where ds is the diameter of the traction sheave in metres.


• Remember that each revolution is 2π radians. When this rotation is referred to the motor through the gearbox, it will be amplified by the reduction ratio of the gearbox, rg. Thus, the final rotation in radians will be:

Where:

ds is the diameter of the traction sheave in metres;

rg is the reduction ratio of the gearbox.

• Thus, the factor ds/2rg , (or its inverse 2rg/ds), can be used to convert between translational quantities at the sheave, and rotational quantities at the motor (or vice versa).

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Derivation of the formulae

• For inertia, translational masses are multiplied by the square of this factor to convert them from translational qmasses at the sheave, to rotational masses at the motor side. Thus, the inertia of the car, counterweight and passenger load, can be reflected at the motor side (high speed shaft) as follows:

Where,

Q is the rated load of the car;

C is the mass of the car;

C/W is the mass of the counterweight.


• Alternatively, if the sheave diameter and the gearbox reduction ratio are not known (or have not been selected yet), but the linear speed of the lift and the rotational speed of the motor have been decided, another ratio can be used, which is (note that n is divided by 60 to convert it to revolutions per second, and then multiplied by 2π toconvert the result to radians per second):

Where

n is the motor speed in rpm.

v is the linear lift speed in m/s.

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• In other words, these two ratios are identical:

• This calculation so far has ignored the efficiency of the whole installation. To these ratios should be added the efficiency of the whole installation (i.e., the efficiency of the shaft and the efficiency of the gearbox)the shaft and the efficiency of the gearbox).

• Thus, depending on which information is available, or how the components are selected, the relevant ratio could be used (note that both ratios are numerically identical). This is summarised in the following table.

• Where η is the total efficiency of the installation (i.e., the product of the gearbox efficiency and the shaft efficiency).

• 2 The inverse of these ratios would be used to convert from rotational acceleration or speed (at the motor shaft) to linear acceleration or speed (in the lift shaft).

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• Note that the forward gearbox efficiency is used in thisNote that the forward gearbox efficiency is used in this case. However, when the system is overhauling (Quadrant II, braked lowering) and the motor is acting as a generator, the power flow is from the shaft to the motor, and thus the reverse efficiency is used in that case.


Example (5):

• This example is based on a lift system. If no flywheel is fitted, calculate the initial linear acceleration in the two extreme cases:

(a) When the car is full of passengers, and travelling upwards.

(b) When the car is empty and travelling upwards.

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Example (5):

• Assume the lift has the following parameters:– The rated passenger load, Q=300 kg;

– The car mass, C=300 kg;

– The counterweight balance ratio is 50% (this means that the mass of the counterweight is selected as equal to the mass of the car plus 50% of the mass of the passengers);

– Sheave diameter, d, is 630 mm;

– Gear box reduction ratio, r, is 74:2;

– Rated torque of the motor is 25 5 Nm;– Rated torque of the motor is 25.5 Nm;

– Torque available for acceleration is twice the rated torque;

– Moment of inertia of the motor rotor and the brake drum 0.1 kg.m2;

– Lift speed is 1.2 m/s;

– Motor speed is 1350 rev/min.

– Total system efficiency of 55%.


Example (6):

• Assume a flywheel with inertia of 0.1 kg∙m2 is fitted to the lift in Example 5 Calculate the newfitted to the lift in Example 5. Calculate the new values of acceleration for the two extreme situations.

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Derivation of the formulae (continuo)

• The moment of inertia of the translational masses referred to the motor shaft can be expressed as follows:to the motor shaft, can be expressed as follows:

Where:– Q is the rated load of the car;

– C is the mass of the car;

– C/W is the mass of the counterweight;

– η is the efficiency of the installation.


• Alternatively, if the diameter of the sheave is not fixed yet, or the gear ratio is not fixed, it might be easier to use the motor rotational speed and the linear lift speed, asthe motor rotational speed and the linear lift speed, as follows:

Where:n is the motor speed in rev/min– n is the motor speed in rev/min;

– v is the linear lift speed in m/s.

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• Thus, the alternative form for the formula above becomes:

• The moment of inertia for rotational masses will include the motor moment of inertia, the brake drum and coupling moment of inertia, the gearbox moment of inertia, and the hand‐wheel (or flywheel) moment of inertia.

Where:– Jmot is the moment of inertia of the motor;

– Jgear is the moment of inertia of the gearbox;

– Jcoupling is the moment of inertia of the coupling and the brake drum;

– Jflywheel is the moment of inertia of the flywheel or hand‐wheel.

Motor Sizing and Selection for Hoisting System• The total moment of inertia referred to the motor shaft is the

sum of the moment of inertia of translational masses, and the moment of inertia of rotational masses:

• Then the out of balance torque is calculated. The out of balance mass depends on the counterweight ratio and the rated passenger load in the car.

Where:Where:– mo/b is the out of balance masses;

– P is the rated number of passengers in the car;

– 75 is the mass per passengers in kg;

– CF is the counterweight ratio.

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• From the out of balance m, the out of balance torque can be referred to the motor:

Where:– To/b is the out of balance torque;

– mo/b is the out of balance mass;

– 9.81 m/s2 is the acceleration due to gravity;

v is the rated speed of the lift in m/s;– v is the rated speed of the lift in m/s;

– n is the motor rotational speed in rpm;

– d is the diameter of the sheave in m;

– r is the reduction ratio of the gearbox;

– η is the efficiency of the installation.


• The value of the rotational acceleration can then be calculated by dividing the net torque by the total moment of inertia referred to the motor shaft:of inertia referred to the motor shaft:

Where:– α is the rotational acceleration in rad/s2;

J i th t t l t f i ti f d t th t h ft i– JTot is the total moment of inertia referred to the motor shaft in kg.m2;

– To/b is the out of balance torque in Nm;

– Tmax is the maximum permissible torque from the motor in Nm.

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Motor Sizing and Selection for Hoisting System• The maximum permissible torque is usually taken as 2 to 2.4

times the rated torque of the motor. This leads to:

Thi t ti l l ti b t d t li• This rotational acceleration can be converted to linear acceleration by using the conversion ratio(s):

• A value of acceleration of 0.8 ‐ 1.0 m/s2 is acceptable. If the value is more this is still acceptable if the drive is a variablevalue is more, this is still acceptable if the drive is a variable speed drive (i.e., ACVV, VVVF, DC SCR), because it drives the exact required voltage to achieve the required acceleration.

• If the drive is a two speed drive, then a flywheel might be needed to reduce the value of acceleration.


• If the acceleration is less than 0.6 m/s2, then the motor is not adequate, and a larger size motor with a higher torque needs to be selected, or the masses have to betorque needs to be selected, or the masses have to be reduced.

• To summarise the above derivation, all the formulae have been combined in one formula, shown here:

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Example (7):

• A lift system is designed to run at 1.75 m/s, with a car capacity of 28 passengers The car mass is 2000 kg andcapacity of 28 passengers. The car mass is 2000 kg, and the counterweight ratio is 50%. Select a motor which will run at a speed of 920 rpm from the table below.


Other Motor Parameters

• The calculations discussed so far in this Chapter have only concentrated on two parameters: Power rating of theconcentrated on two parameters: Power rating of the motor and torque rating (as well as the ratio of starting torque to rated torque).

• There are many other parameters that need to be taken into consideration when selecting the motor. Some of those are listed below:those are listed below:

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• Duty cycle: This parameter specifies how long the motor will be running for and how long it will be idle for. It is expressed as the ratio of on time to the total cycle time (e.g., 25% or 50%). It takes into consideration that some application run continuously without any speed change (e.g., fans, pump) and other applications will require frequent starting and stopping (e.g., lifts).

• Number of starts per hour: This is very specific to lifts and h i il li i h h ill dother similar applications where the system will start and

stop frequently. Manufacturers supply different tables depending on the number of starts per hour. Standard number of starts are usually 90, 120, 180, 240 starts per hour.


• Insulation class: This specifies the maximum temperature rating that can be supported by the insulation within the windings based on 20 000 hour life‐times. There arewindings based on 20 000 hour life times. There are generally four classes that can be selected, and these are specified by NEMA (National Electrical Manufacturers Association), which is the main trade association in the USA. It is estimated that for every 10 oC higher that a motor runs, it reduces its lifetime by a half.

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• Speed‐torque characteristics: For induction motors for example, different motor designs can produce different speed‐torque characteristics. NEMA (National Electrical Manufacturers Association), provides standard classifications (A, B, C, D and F) that describe typical relationships between speed and torque. Each classification is suitable for a certain application. For example, Class C describes a double cage high torque type of squirrel cage induction motor (SCIM) and is often usedof squirrel cage induction motor (SCIM) and is often used in elevator systems to provide high starting torque).

• Mean sound pressure: This specifies the (audible) sound level emitted by the motor measured at a distance of 1 m and is expressed in dB.


• Starting current: This specifies the starting current of the motor when started fully loaded direct on line as a ratio of the rated current.the rated current.

• Method of cooling: There are generally two methods of motor cooling. Natural ventilation just uses a propeller fitted on the motor shaft that allows the air to circulate (this method offers no cooling when the motor stops). Forced ventilation uses a dedicated fan with its own supply to cool the motor. This method has the advantage that is carries on cooling the motor even when the motor is idle.

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• Pole Changing Motors: When using induction motors and there is a need to run the motor at two different speeds, it is possible to select a motor that has two sets of windings each with different numbers of poles. For example, a motor can be selected with 4/16 poles (i.e., two sets of windings, one with 4 poles and one with 16 poles). At a frequency of 50 Hz, this motor will have two synchronous speeds: 1500 and 375 rpm respectively. This is often used in elevator applications where the high speed is used forin elevator applications where the high speed is used for most of the journey and the low speed is used for levelling. It is also used in industrial applications where the high speed is used for normal operation and the low speed is used for maintenance and inspection.

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