The Step Motor

8/8/2019 The Step Motor

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Chapter 9: Step motors 9.1

Figure 9.1: X-Y positioning table with step motors.

Chapter 9

The Step motor

9.1. INTRODUCTION.

Stepping motors can be looked upon as electric motors without commutators. Typically, all windings in

the motor are part of the stator, and the rotor is either a permanent magnet or, in the case of variable

reluctance motors, a toothed block of some magnetically soft material. All the commutations must be

handled externally by the motor controller, and typically, the motors and controllers are designed so that

the motor may be held in any fixed position as well as being rotated one way or the other. Most stepping

motors can be stepped at audio frequencies, allowing them to spin quite quickly, and with an appropriate

controller, they may be started and stopped "on a dime" at controlled orientations.

For some applications, there is a choice between using servomotors and stepping motors. Both types of

motors offer similar opportunities for precise positioning, but they differ in a number of ways.Servomotors require analog feedback control systems of some type. Typically, this involves a

potentiometer to provide feedback about the rotor position, and some mix of circuitry to drive a current

through the motor inversely proportional to the difference between the desired position and the current

position.

In making a choice between stepping motors and servomotors, a number of issues must be considered;

which of these will matter depends on the application. For example, the repeatability of positioning done

with a stepping motor depends on the geometry of the motor rotor, while the repeatability of positioning

done with a servomotor generally depends on the stability of the potentiometer and other analog

components in the feedback circuit.

Stepping motors can be used in simple open-loop control systems; these are generally adequate for

systems that operate at low accelerations with static loads, but closed loop control may be essential for

high accelerations, particularly if they involve variable loads. If a stepping motor in an open-loop control

system is overload, all knowledge of rotor position is lost and the system must be reinitialized;

servomotors are not subject to this problem.

Stepping motors are known in German as Schrittmotoren, in French as moteurs pas à pas, and in Spanish

as motor paso paso. The terms step-motor and stepper are also common.

This motor has a special place among the electrical motors by its applications (nearly always positioning)

and the fact that this motor must be treated in combination with his drivers and controlling technics.

The typical step motor applications are:

S positioning tables in three dimensions.

S robot arms

S positioning of heads for plotters, cutters, printers

S copy machines

S head positioning of disk-drives

Figure 9.1 shows a picture of an X-Y positioning table.

The drives in the X- and Y-directions are realized with

step motors and lead-screws.

The step motor as rotational motor is the most used

application but the step motor as linear motor is increasing. Stepper motors have the following benefits:

• Low cost

• Ruggedness

• Simplicity in construction

• High reliability


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Figure 9.2: Two pole PM-step motor.

1Figure 9.3: L is excited.

• No maintenance

• Wide acceptance

• No tweaking to stabilize

• No feedback components are needed

• They work in just about any environment

• Inherently more failsafe than servo motors.

There is virtually no conceivable failure within the stepper drive module that could cause the motor to

run away. Stepper motors are simple to drive and control in an open-loop configuration. They only

require four leads. They provide excellent torque at low speeds, up to 5 times the continuous torque of

a brush motor of the same frame size or double the torque of the equivalent brushless motor. This often

eliminates the need for a gearbox. A stepper-driven system is inherently stiff, with known limits to the

dynamic position error. Stepper motors have the following disadvantages:

• Resonance effects and relatively long settling times

• Rough performance at low speed unless a micro-step drive is used

• Liability to undetected position loss as a result of operating open-loop• They consume current regardless of load conditions and therefore tend to run hot

• Losses at speed are relatively high and cause excessive heating, and they are frequently noisy

(especially at high speeds).

• They can exhibit lag-lead oscillation, which is difficult to damp.

• There is a limit to their available size, and positioning accuracy relies on the mechanics (e.g.,

ballscrew accuracy).

Many of these drawbacks can be overcome by the use of a closed-

loop control scheme.

9.2. THE WORKING PRINCIPLE

The working principle is explained with the model of figure 9.2.

Figure 9.2 shows a step motor built with two magnetical circuits

A and B, making a right angle, provided with respectively coil

A1-A2 and B1-B2.

Each circuit has salient poles and two sup-windings called the

phase of the motor. Circuit A has the poles 1 and 3 with the

1 3 2 4 phases L and L . The poles 2 and 4 with the phases L and L

belonging to circuit B. When no phase is excited the rotor will

even so align to one of the poles. According to the flux

decreasing law, the rotor will take such a position that the

magnetical resistance for the (PM) induction field lines will be

as small as possible. The rotor will stand still between the poles1,3 or the poles 2,4. So there are 4 preferred positions for the

rotor. Which position the rotor chooses, is random.

The above mentioned positions are also the positions for the

situation of excited phases. This motor has 4 steps per

revolution.

1If phase L is excited the pole 1 will be a north pole and pole 3

is a south pole. The rotor gets the position as showed in figure

29.3. Is phase L excited identically the rotor makes a rotation

angle step of 90 clock wise. Figure 9.4 shows this situation.0

1 2 3 4Excite alternating the phases L ,L ,L and L the rotor will

rotate clockwise in a step wise manner. By increasing the

switching frequency, the rotor movement changes from step

wise to a continuous rotating (like a synchronous motor).


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2Figure 9.4: L is excited.

Figure 9.5: Full step mode.

Figure 9.6: Half step mode.

To get the other rotation direction you have to change the

sequence of the excited phases. For counter clockwise the

1 4 3 2 1sequence must be L , L , L , L and L .

They excite only one phase at the time is the full step mode.

Figure 9.5 shows schematically the full step mode. The (full)

step angle for this motor is 90 .0

Figure 9.6 gives the situation in which alternating 1 phase and

2 phase at the same time are excited. This use of the step

motor is called the half step mode.

The number of steps per revolution is enlarged two times. A

step angle of 45 is for most applications too large. More0

common step angles are 15 - 7,5 - 3,6 - 2 - 1,8 degrees.

Constructive decreasing of the step angle can be done by

taking more poles on the rotor or increasing the number of

phases.

9.3 STEPPING MOTOR TYPES.

Stepping motors come in two varieties, permanent magnet (PM) and variable reluctance (VR) (there are

also hybrid (H) motors, which are from the controller's point of view comparable with permanent magnet

motors ). To find out which type of motor you are dealing with, you can notice it when no power is

applied. Permanent magnet motors tend to "cog" as you twist the rotor with your fingers, while variable

reluctance motors almost spin freely (although they may cog slightly because of residual magnetization

in the rotor). You can also distinguish between the two varieties with an ohmmeter. Variable reluctance

motors usually have three (sometimes four) windings, with a common return, while permanent magnet

motors usually have two independent windings, with or without center taps. Center-tapped windings are

used in unipolar permanent magnet motors.

With an appropriate controller, most permanent magnet and hybrid motors can be run in half-steps, and

some controllers can handle smaller fractional steps or micro-steps.


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Figure 9.7: Construction of one stator phase.

Figure 9.8: Exploded view of a PM step motor.

For both permanent magnet and variable reluctance stepping motors, if just one winding of the

motor is energized, the rotor (under no load) will snap to a fixed angle and then hold that angle

until the torque exceeds the holding torque of the motor, at which point, the rotor will turn, trying

to hold at each successive equilibrium point.

9.3.1 The construction of the stator of a step motor.

Figure 9.7 shows the stator of a PM-step motor. A stator

core (iron) is provided with one stator phase. As

mentioned in paragraph 9.2 the PM-step motor needs

minimal 4 phases for rotating in two directions. This can

be realized with two coils making a right angle

magnetically to each other and using the current flow

into both direction. This is called a semi-four phase

system. The rotating direction is defined by the sequence

of the excitement of the coils in combination with the

polarity of the excitement. Using the coil for two

polarities very efficiently. The value of the step angle is defined by the number of poles of the rotor and

the number of phases of the stator. The stator phase can be equipped with some stator teeth as shown in

figure 9.7. Dividing the stator pole into some teeth increases the torque. The tooth pitch of the stator

must be equal to the pole pitch of the rotor, so that in the excited mode the teeth of the stator stand

opposite to the poles of the rotor. This construction of the stator can be used for all types of step motors.

9.3.2 The permanent magnet step motor.

The rotor is made of permanent magnetical material. As mentioned before the pitch of the stator must

fit to the pole pitch of the rotor. Allthe stator pole teeth together is one

position step of the rotor. Figure

9.7 shows the situation for an

excited phase. For a good

functioned motor the tooth of

different phases of a two phases

motor must be shifted half a pole

pitch of the rotor; for a three phase

motor the tooth must shifteda pole

pitch, etc. Figure 9.8 shows pictures

of a common applied PM-step mo-

tor. The stator exists of an upper

and lower iron shield with teeth and

a middle part also with teeth on both

sides. Between the iron shields and

the middle part cylindrical coils are

placed. The coils are enclosed by

the tooth of the iron parts. Through this construction the teeth are alternating north and south poles along

the circumference by an excited coil. This applies for both coils. If you excite each coil in two directions

you have a semi-four phase step motor. The rotor is a permanent magnet with an equal number of poles

as teeth of the stator circumference. The rotor poles will find an opposite magnetic pole of the stator as

a preference position. The teeth of the upper magnetic system and the lower system are half a pole pitch

shifted. The use of a permanent magnet gives the opportunity to build a compact motor with a relativelarge torque. The construction is easy to realize so the price is low.


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Figure 9.9: Five phase VR-motor.

Figure 9.10: Stack model of a VR step motor.

Figure 9.11: The hybrid step motor.

9.3.3 The variable reluctance step motor

The rotor of the VR step motor is made of iron and teeth around the circumference. The teeth are

comparable to the poles of the PM-rotor. The tooth pitch of the stator must be the same as the tooth pitch

of the rotor. The torque development is based of the

reluctance principle. According to this principle a

tooth of the rotor will align to the excited stator pole.

The more rotor teeth the more steps can be realized.

Unfortunately the VR step motor needs minimal 3

phases for rotating in 2 directions. The reason is that

the rotor teeth has not a magnetical pole of its own,

but is magnetized by the stator pole. The rotating

direction is defined by the sequence and not by the

polarity of the excitement of the phases.

The number of steps per revolution is the number of

teeth of the rotor multiplied with the number of phases. Figure 9.9 shows a cross-section of a five

phase VR step motor. The stator exists of 10 salient

poles, of which 2 poles facing to each other form one

phase. The construction of the stator teeth of

different phases is made in such a way that after exciting the two phases after each other the rotor teeth

will face the teeth of the stator phases. The rotor made one step. After exciting the 5 phases after each

other the rotor has rotate one rotor

tooth and made 5 steps. This type of

motor can realize many steps per

revolution (till 1000's of steps).

Another construction is build with so

called stacks. Each stack consists of an

rotor and a stator. One stator of a stack

is mostly one phase. The number of

stacks can be 3 til 24. Figure 9.10

shows a 3 phase VR step motor with 3

stacks. For a good working motor the

rotor or stator teeth must be a

tooth pitch shifted. In figure 9.10

the stator teeth are aligned; the

rotor teeth are shifted.

9.3.4 The hybrid step motor.

This type of step motor is a

combination of the PM- and VR-

step motor. So the favorable

properties of both are united. The

rotor of the H-step motor exists of

an axial magnetized permanent

magnet with iron end parts with

teeth on the poles. The teeth of the

end parts are mutually ½ tooth pith

shifted. The stator provides at least

4 phases with teeth on the poles.

The stator poles are surrounded at both end parts of the rotor. Making steps in one direction is realized


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Figure 9.12: References of the step motor.

Figure 9.13: The static torque of a PM-step motor.

by the sequence of the excitement of the phases like the PM-step motor. Figure 9.11 shows a cross

section of an H-step motor with 8 poles. The motor has 4 phases and 2 teeth per stator pole. The numbers

of the poles 1 and 5, 2 and 6, 3 and 7, 4 and 8 form the 4 phases.

9.4 THE CHARACTERISTICS OF THE MOTOR

For most electrical actuators the mechanical behavior is given by the torque speed characteristic. For step

motors you need three characteristics for the mechanical behavior:

ema. The static torque-angle characteristic describes the torque T in relation to the rotor rotation è.

Such a characteristic is also used for the synchronous motor. The characteristic gives the torque

during one step, when only one phase is excited with a constant (rated) current.

b. The torque frequency characteristic (pull-out curve) for the stationary situation. This

em step stepcharacteristic gives the torque T in relation to the step frequency f . The step frequency f

is the frequency of the steps of the motor. The pull out characteristic can be compared with the

torque speed characteristics of the other motors.

c. The starting and stopping curve (pull-in curve) gives the needed torque to accelerate anddecelerate in relation to the step frequency.

First the static torque characteristic is discussed, because it is the base of the other two curves.

9.5.1 The torque angle characteristic.

The torque angle curve will be given for a semi four phase

PM step motor. Figure 9.12 shows the references of the rotor

emrotation angle è and the motor torque T for an excited

load em phase A. The load torque T is opposite to the torque T .

em

If you measure the torque T as function of the rotation

angle è, you get the static torque curve of phase A. The

emtorque T is nearly acting like sin è, comparable to the

torque of the synchronous motor in relation to the load angle

load emä . For step motors with p pole pairs the torque T is

proportional with sin pè. Figure 9.13 gives of a p pole motor

the torque of phase A for -ð < pè < ð . Phase A means that+ +

sthe phase coil A is excited with a current I in the positive direction.


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Figure 9.14: Static torque of a VR step motor.

Figure 9.15: Error angle for two load torques.

(9.1)

sIf phase A is excited in a negative way with a current I it will de indicated by phase A. Suppose a no--

loaded motor must make a step from position pè= -½ð to pè=0. For this step a positive torque is needed.

emSo phase A must be excited during -½ð


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Figure 9.16: Torque for full and half step mode.

err sFigure 9.15 shows for two different load torques the è for nominal current I . As you can see the larger

the load torque, the larger the error angle. Equation (9.2) shows that the error angle is getting smaller

em,maxfor larger p and larger T .

9.6. MICRO STEPPING

The phase torques as function of the angle è of a two-phase PM stepper with p pole pairs are given by:

Phase: Torque: No-load position:

em,+A em,max noA T = -T sin(pè) pè = 0+

em,-A em,max noA T = -T sin(pè - ð) pè = ð -

em,+B em,max noB T = -T sin(pè - ½ð) pè = ½ð (9.3)+

em,-B em,max noB T = -T sin(pè - 1½ð) pè = 1½ð-

Figure 9.16 shows the

torques per phase for a

clockwise rotating motor. If

you excite only one phase at

any time you get the full

step mode. You get the half

step mode by exciting

alternating one and two

phases toge ther. For

example the sequence A,+

A B, B, B A, A, etc. is+ + + + - -

the half step mode for

rotation clockwise.

The total torque for the

phases A B is:+ +

em,+A+B em,+A em,+B em,maxT = T + T = -1,4T sin(pè - ¼ð) (9.4)

noThe no-load position is for pè = ¼ð. This torque is given in figure 9.17. The maximum torque is 1,4

times larger than of one phase. You can also write for the torque of the phases A and B:+ +

em,+A em s,A em,+B em s,BT = -K I sin(pè) and T = -K I sin(pè - ½ð) (9.5)

emK = a constant depending of motor parameters (comparable with the torque constant of a DC-motor).

s,A s,BI = the current of phase A I = the current of phase B

s,A s,BThe currents I and I can be made dependent on the variable â according:

s,A rated s,B ratedI = I cosâ and I = I sinâ (9.6)

em,+A em,+B em,+A+BThe torques T , T and T are than:

em,+A em ratedT = -K I sinpècosâ

em,+B em ratedT = K I cospèsinâ (9.7)

em,+A+B em ratedT = -K I sin(â-pè)

You can now create an arbitrarily step between two full steps by choosing the right value of â. This

principle is called micro stepping.

noThe no-load position is: â-pè = 0 (9.8)


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Figure 9.17: Circuit and current during switching on.

ser ter sFigure 9.18: Influence of R and U to i .

If you make the currents of the “half step” a factor ½2 smaller, the maximum torque will also decrease

with the same factor, so the torques of the “half step” and the full step are equal. So the angle â must be

¼ð.

Examples of values of â:

Full step mode:

no s,A rated s,B No-load positions pè = 0; â = 0 I = I ; I = 0

no s,A s,B rated No-load position pè = ½ð; â = ½ð I = 0; I = I

Half step mode:

nul s,A rated s,B No-load position pè = 0; â = 0 I = I ; I = 0

nul s,A rated s,B rated No-load position pè = ¼ð; â = ¼ð I = ½2I ; I = ½2I

nul s,A s,B rated No-load position pè = ½ð; â = ½ð I = 0; I = I

To create more micro steps between the full steps, you have to take more values of â in the right way.

For the right â you must divide 90 electrical degrees to the number of micro steps between two full steps

plus 1.For 2 extra step 90 degrees must be divided to 3, so â = 0 (full step), â = 30 (micro step), â = 600 0 0

(micro step), â = 90 (full step). The currents of the phases depend on the number of micro steps. Micro0

stepping increases the accuracy of the position possibilities and can compensate the error angle of loaded

motors. With smaller steps the motor rotates more quietly. For the same speed a higher frequency is

needed as for the full step mode. To hold the motor in a micro step position a continuous current in both

phases is necessary. The detent torque is in that case a great drawback. There are IC for micro stepping

available til about 400 micro steps.

9.8 SWITCHING ON AND OFF OF THE PHASES.

For a good functioning step motor the

step must made so fast as possible. So

sthe phase current I must reach the end

value as soon as possible. The current

sis defined by the resistance R , the

sinductance L and the terminal voltage

ter U . The influence of the switch

component is represented as a voltage

drop of the component. Mostly the

voltage drop is negligible. Figure 9.17

shows an easy switching circuit for

one phase and an extra serie resistor

ser R .ter After the switching on the voltage U the

scurrent i is defined by:

on s ser sThe constant ô = L /(R + R ) is called

the turn on time constant. Figure 9.17 shows

s ser onthe current i . For a high value of R the ô

swill be small and the current i increases fast.

(9.9)


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sFigure 9.19: The circuit and current i during turning on and off.

emFigure 9.20: The torque T around the no-load position.

ser If you want the same end value of the current for a high value R you need a higher value for the

on ter terminal voltage. Figure 9.18 shows a smaller ô for a higher voltage U . You must always try to reach

rated sthe rated value I for i .

rated on Normally the static torque curve is given for I . In practice you must try to make ô much smaller than

the necessary time needed for one step. The step time is mainly defined by the mechanical parameters

(inertia, friction, viscous friction).

sAfter the rotor reaches the end position of a step, the current i must decrease to zero as soon as possible.

Because you have to deal with an induction, this must be done in the right way otherwise voltage spikes

freewill occur across the switching element. Therefor a freewheel diode is necessary. An extra resistor R

saccelerates the fall of the current. Figure 9.19 shows that circuit. The current i is defined by:

off s ser sThe constant ô = L /(R + R + -

freeR ) is called the turn-off timeconstant. By increasing the value of

free sR the current i decreases faster.

This is limited by the fact that a too

freelarge value of R destroyed the

transistor. Just after turning off the

cevoltage U over the collector-

emitter junction of the transistor is

given by:

ce ter free rated U = U + R .I (9.11)

9.9 THE FREQUENCY TORQUE CHARACTERISTICS.

The step motor has two torque step frequency characteristics. The curves are based on the dependency

of the step frequency and the load torque. The pull-in curve gives the permitted frequency of a starting

and stopping stepper. The pull-out curve gives the permitted frequency for an already running motor.

This curve is comparable with the speed torque curve

of other motors.

9.9.1 The first starting step.

From a stand still situation the first step is defined by

viscthe inertia J, the viscous friction f , the load torque

load emT and the motor torque T of the stepper. During

the step there must be a balance between the different

torques. The rotor position is given by:

J = the inertia of the motor and load together.

(9.10)

(9.12)


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(9.14)

res resFigure 9.22: Step responses for f and 0,6f .

Figure 9.21: The rotor position during one step.

viscf = the viscous friction (very small).

shaft loadù = the angular speed of the shaft; T = the load torque (inc. friction).

emFigure 9.20 gives the static torque T . The torque is near the no-load position and can be approached

by:

em step no T = -K .(è-è ) (9.13)

Combination of (9.13) and (9.12) gives a solution for è as a function of the time:

resf = the mechanical resonance frequency of

the stepper with load.

A = integration constant depending on thestarting conditions of the step.

According to (9.14) the rotor will vibrate as

response on one step. At last the rotor reaches the

no-load position of the step. Figure 9.21 shows

the rotor position as reaction of one pulse. For

the course of the rotor position the same terms

are used known from the control theory. If you

want to make more steps, the frequency of those

steps must not be too high. It can happen that too

fast stepping, steps start in the maximum of the

overshoots and so steps are skipped. Figure

9.22 shows this effect with a too early generated

impulse.


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Figure 9.23 The pull-in curve.

Figure 9.24: Equivalent circuit for one phase.

A step frequency that equals the mechanical resonance frequency causes 'running out of order' of the

motor and can lead to bad resonance behavior. For this reason there is a limited step frequency in

relation with the load torque for the dynamic motor characteristic.

resFigure 9.22 shows that a step frequency of 0,6 f keeps the motor in to the right steps.

9.9.2. The pull-in curve.

As you have seen at the start (but also for stopping) the pulses may not follow too fast to each other. The

pull-in characteristic gives the step maximum frequency for a certain load that the motor will start to run.

Because there is no starting speed the inertia J has a great influence on the process. During the stepping

em emthe torque T is not constant (see the static torque) so you must calculate with an average value for T

visc acc. If you neglect the viscous friction f , only an acceleration torque T is necessary for the inertia J. -

accWith (9.12) the acceleration torque T is:

The average angle acceleration á will be:

From (9.14) and (9.15) you can deduce that an increasing load

loadtorque T the angle acceleration á will decrease. The motor

stepneeds more time to make a step. The step frequency f will

become smaller. At the end for a large load torque the motor

will not start any more. The other way around a no-load motor

has a very large disposable acceleration torque. In the no-load

step,nomode you find the highest starting step frequency (f ). Figure 9.23 is a sketch of the pull-in curve.

Some remarks for the curve:

em,max ave, max ave,maxS The holding torque T will be larger than T pointed in figure 9.23 , because T is

an average value.

S For a larger inertia (i.e. of the load) the curve will shift to the left.

accS If the viscous friction is not negligible, the torque T will become smaller. The step frequency

decreases.

9.9.2 The pull-out curve.

If the motor is already running there is no torque needed

for acceleration. The pull-out curve gives the step

frequency for a certain load torque for a running motor.It describes a stationary situation and therefore

comparable with the normal torque speed characteristics

for other motors. The pull-out curve for voltage supplied

PM stepper will be discussed. In the stationary mode the

equivalent circuit for one phase is like the Dc-motor. It is given in figure 9.24. You must calculate with

em s mut,saverage values. This applies to for the torque T the phase current I and the motional voltage U .

mut,sThe voltage U is proportional with the speed of the rotor and so proportional with the step frequency.

The no-load stepper needs only a torque for the friction, which requirs a relative small motor torque

em,ave s mut,sT . The phase current I has a low value so the value of U is high. A high speed is possible.

s mut,sThe needed current I for a loaded motor is much larger. For a voltage supplied stepper U must be

em

small, so the speed and the step frequency will decrease. In the stationary mode T will be equal to the

loadload torque T . Figure 9.25a shows the phases as a function of the rotation è in the full step mode.

(9.16)

(9.15)


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Figure 9.25: The torque for the no-load and loaded situation.

Figure 9.26: The pull-out curve.

Figure 9.27: Operation area of a step motor.

The turn on and off times are neglected.

When the phases are excited the maximum

parts of the static torque curves will be

followed.

sFigure 9.25b shows this for a value of I

ratedsmaller than the rated value I . By taking

the average value of each part of the torque

em,avethe total average torque T can be

em,avecalculated. For a loaded motor the T is

equal to the load torque. The phase current

sI is relatively high. Figure 9.25c is for a

nearly no-loaded motor. The average torque

is equal to the friction torque. The step

frequency of a no-loaded motor can be higher than a loaded motor, because the speed can be larger. In

the figures 9.25b and 9.25c it is impossible to see the different frequencies, because the figures are

srelated to the rotor positions. The maximum load torque is for the rated value of the current I , but for

step,lima frequency of zero. In the no-load mode you get the maximum step frequency f . Between the

step,limfrequency from 0 till f you can find a load torque by which the motor just can rotate. This is given

by the pull-out curve in figure 9.26.Figure 9.27 shows the pull-in and pull-out

curves together in one plot. The starting torque

is the same for both curves. The area under the

pull-in curve is called the start/stop area. The

area between the curves is called the "slew

rate" area. Suppose you must get from standing

still to a certain position as fast as possible,

than you have to cross the different kinds of

operation area of the motor. First, you have to

start the motor in the start/stop area. Next you

want to use an operation point on the pull-out

curve as long as possible. Such a point is only

reachable through the “slew rate" area. In this


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Figure 9.28:Resonance and unstabilities of a step motor.

area the step frequency must be carefully increased (not too fast). Doing this on the wrong way the motor

will stand still. There must be enough torque left for acceleration.

9.9.3 Resonances and unstabilities.

Because of not-linear effects (saturations, viscous

friction, turning on and off) the dynamic curves

have some dips and islands. In those areas the motor

is not functioning well and its behavior is not

predictable. Figure 9.28 shows those areas for a

certain step motor. By using higher or lower

frequencies those areas can be avoided. The

resonances are divided in three categories:

a. Low-frequency resonance. Those are of

mechanical nature lying around 100 Hz.The resonance is caused by the own

frequencies (or multiplies) of the

mechanical system. The frequencies are disturbing in the start/stop area.

b. Mid-range instabilities (500-1500Hz). The cause of those resonances are the electronic driver

circuits (choppers and bi-level).

c. Higher-range oscillation (2500-4000Hz). Those instabilities are not quite known yet . A

possible cause can be the behavior of the iron (magnetical nature).

The best solution against instabilities is to avoid the frequency. Another remedy is to reduce these

instabilities is to damp the movements of the step motors. Mechanical you can do that with oil-dampers

(only used for large motors). Electrical damping is realized with a 2-pole magnet on the shaft moving

in an extra magnetical field. So an electrical damp-force can be generated.

If the dips are really small you can try to skip those step frequency.

9.10 A COMPARE BETWEEN A PM- AND A VR-STEP MOTOR.

At the end some practical properties of the PM- and VR-step motor. The H-step motor can be seen as a

combination of both motors with the advantages of both types. The PM-step motor is implemented in

applications with limited space. Think of desk devices (i.e. printers). The De VR-step motor is not so

often used any more, but if it is used then in applications where a large torque required. In table 9.1 both

step motors are compared to each other.

Tabel 9.1

Type/properties PM VR

Holding torque smaller bigger

Detent torque yes no

Swing phenomenon for 1 step less more

Resonance phenomenon less more

Number of steps per revolution limited very large

Supply phases : 2 phases: 3


16/17


Figure Op.9.1 The torque of the phases A, B, C.

A general pronunciation about the price is difficult to give. The price depends on the construction and

materials of the motor, but also of the electronics of the supply and the controlling system. So each

application has its own price.

9.11. PROBLEMS.

9.11.1. Determine for the motors of figures 9.8, 9.9 en 9.11 the step angle.

9.11.2. On which manner (a mechanical) can the error angle be reduced? What can be the disadvantage

of it?

9.11.3. Draw the torque as a function of the position of a PM step motor for p=1 and 4 stator poles in

a not-excited mode. Use sinus forms. Give in the drawing also the own holding torque.

9.11.4. Why must the load torque be as low as possible for positioning problems?

9.11.5. Mention the positive and negative points of the bipolar control against the unipolar control.

em,ave9.11.6. Why are the maximum values of the torque T for the pull in and pull out curves equal to

each other?

9.11.7. A variable reluctance step motor consists of 5 stacks of one phase. The stator poles of the

have an even number of tooth as the accompanying rotor namely 24 tooth alike divided around

the circumference. What is the number of steps per revolution?

em9.11.8 The torque T of the phases A,B and C of

a 5-phase VR-step motor are given in

figure OP 9.1.

a. Calculate for full step the no-load

positions.

b. What are the no-load positions if you

excite every time 2 phases at ones? What

is the torque as a function of è?

c. What are the no-load positions if you

excite 3 phases at ones? What is now the

torque as a function of è?

d. What will you choose if you want in a simple way as much as possible steps per revolution,

without using micro-stepping?

9.11.9. A two phases (semi-four phases) PM-step motor has 8 pole pairs on the rotor. They want to

realize with the help of micro-stepping 4 extra steps between 2 full steps. The rated current is

nomI . Calculate the current levels and the no-load positions for the micro-steps.

9.11.10. For a two phases (semi-four phases) PM-step motor has a torque for phase A en B:

em,A s,A em,B s,BT = - 10i sin16 è mNm and ; T = - 10i sin(16 è-½ð) mNm. For full step the operation

s,A s,B ratedcurrents are : i = i = I = 0,5 A. They will create 2 extra steps with micro-stepping.

a. How many steps are made for one revolution (full steps mode)?

b. Calculate the no-load positions for the micro-steps; starting with the full step for è = 0.

loadc. Calculate the step error for full step mode if the load torque is T = 1 mNm.


17/17


Figure Op 9.2 A bipolar control circuit.

Figure OP 9.3: control circuit.

9.11.11. A three phases VR-step motor has a coil-resistance of 1 Ù and an average value for the

ratedinduction of 40 mH. The rate current I is 2A.

a. Design an unipolar control circuit with a turn time constant of 2ms and a turn off time constant

of 1 ms.

b. Estimate the power of the losses.

c. Calculate the supply voltage.

rreed. If this motor makes 600 steps per second. How many power is lost in the resistor R of the

free wheel circuit?

e. Calculated the maximum voltage over the collector-emitter junction of the transistor.

9.11.12 A PM-step motor has a coil-

resistance of 1 Ù and an average

value for the induction of 40 mH.

ter The terminal voltage U is 40 V.

ser The series resistor R has a value of

19 Ù. The controlling is bipolar , asyou can see in figure Op 9.2. The

switching transistors have their own

free wheel diode. All the switching

components are free of losses.

a. What is the value of the rated

ratedcurrent I , that has to be switched off ?

b. Calculate the turn off time constant. Approach e with (1 - t/ô)-t/ô

c. If this motor makes 600 steps per second. How many power is going back to the power supply?

d. Calculated the maximum voltage over the collector-emitter junction of the transistor.

9.11.13 For a two phases (semi-four phases) PM-step motor has a torque for phase A en B:

em,A em,BT = - sin4 è mNm and ; T = cos4 è mNm. They will create 1 extra step with micro-

stepping.

a. Calculated for three successive steps, starting for è = 0, the no-load position and the current

level for each phase.

load b. The load torque is T = 0,1 mNm. They realize half steps with and without micro-stepping.

Calculate for both situations the step error angle .

9.11.14 A variable reluctance step motor consists of 5 stacks of one phase. The stator poles of the

have an even number of tooth as the accompanying rotor namely 50 tooth alike divided around

the circumference.

a. What is the number of steps per revolution?

b. What happens with the pull-out curve if the inertia J of the system is decreasing?

9.11.15 Figure OP 9.3 shows the basic control circuit for one phase of a

PM-step motor. The rated current is 3,5 A. The coil has an

s freeinduction L of 0,2 mH and a resistance of 2 Ù and R = 20 Ù.

inThe turn on time constant ô is 0,04 ms.

ser a. Calculate the resistor R ?

ce b. What is the voltage U over the transistor junction, just after

turning off the transistor?

c. What is theoretical the maximum step frequency for one single

phase (Remember you have to turn on and off the current in the

coil)?

Documents

The Step Motor