CHAPTER 4 DIRECT-CURRENT MOTORS - … · CHAPTER 4 DIRECT-CURRENT MOTORS Chapter Contributors...

CHAPTER 4

DIRECT-CURRENT MOTORS

Chapter Contributors

Andrew E. Miller

Earl F. Richards

Alan W. Yeadon

William H. Yeadon

4.1

This chapter covers methods of calculating performance for direct-current (dc)mechanically commutated motors. Section 4.1 discusses the electromagnetic circuitfor dc motors and series ac motors. Sections 4.2 and 4.3 establish some commongeometry and symbols and discuss commutation for dc motors. Section 4.4 presentspermanent-magnet direct-current (PMDC) calculation methods. Section 4.5 pre-sents series dc and universal ac/dc performance calculations. Sections 4.6 and 4.7 dis-cuss methods for calculating the performance of shunt- and compound-connected dcmotors. Finally, Secs. 4.8 and 4.9 discuss dc motor windings and automatic armaturewinding.

4.1 THEORY OF DC MOTORS*

4.1.1 DC Series Motors

A series motor operating on direct current has characteristics similar to those whenit is operated on ac current at power-system frequencies. However, it is best todescribe dc and ac operation separately so that comparisons can be made.

The general equivalent electrical circuit of the series dc motor and its physicalconstruction is shown in Figs. 4.1 and 4.2.The motor consists of a stator having a con-centrated field winding (Fig. 4.3) connected in series by way of a commutator to a

* Section contributed by Earl F. Richards.

wound armature (Fig. 4.4). One of the first things to be considered in the operationof the motor are the motor and generator action, which exist simultaneously in thearmature circuit of the motor. These two principles are (1) the instantaneous elec-tromotive force (emf), which is induced in the armature conductors when movingwith a velocity v within a magnetic field, and (2) the force produced on the conduc-tors as the result of carrying an electric current in this same magnetic field.

4.2 CHAPTER FOUR

FIGURE 4.1 Series motor diagram: ra = armature resis-tance measured at brushes, rf = main field resistance, and v =applied voltage.

FIGURE 4.2 Series motor.

It is known that the instantaneous force on a conductor of length carrying a cur-rent i in a magnetic field B is:

f = Bi sin θ N (4.1)

Or, in vector notation:

f = (i × B) (4.2)

where θ is the angle between the direction of the magnetic field and the direction ofcurrent flow in the conductor, B is in webers per square meter (teslas), i is in

DIRECT-CURRENT MOTORS 4.3

FIGURE 4.3 One pole of a series motor field winding.

FIGURE 4.4 Wound armature.

amperes, and is in meters. Motors, by design, have the armature conductors andmagnetic flux at quadrature to one another. Therefore, the force becomes:

f = Bi (4.3)

Assume a situation in Fig. 4.5 where a conductor of length is located in a mag-netic field and is free to move in the x direction perpendicular to the field. From thepreceding discussion, a force is produced on the conductor, causing it to move in thex direction. Then:

V − ir = e or Vi − i2r = ei (4.4)

So, differential electrical energy intothe conductor less differential i2r lossequals differential mechanical outputenergy:

dWelect = dWmech (4.5)

ei dt = f dx where f = Bi (4.6)

ei dt = Bi dx (4.7)

e = B = Bv (4.8)

This force causes movement in theconductor, which in turn causes a volt-age to be induced into that conductorwhich is opposite to the direction of theoriginal current. This is an importantconcept in the operation of motors in

general and one which is used to discuss the operation of the series motor. Thisinduced voltage is usually called a counter-emf (cemf) voltage because of its opposi-tion to the applied voltage.

This example also indicates that a reversible energy or power exchange is possi-ble (i.e., between the mechanical and electrical systems). Therefore, the samemachine may operate as a motor or a generator, depending on the flow of energy inthe armature.

In the motor mode, the field and armature of a dc series motor are supplied withthe same current by an applied voltage, and a magnetic field (flux) is produced in themagnetic circuit. Since the armature conductors (coils) are located in this field, eachof the conductors in the field experiences a force (torque) tending to make it move(rotate), and as we have just indicated, a countervoltage (cemf) is produced oppos-ing the applied voltage. Other than the cemf which the armature produces, we mustrecognize that the armature circuit (coils) also produces a magnetic field of its own.The armature, because of its commutator and brush construction, has a unidirec-tional current and therefore produces a fixed-direction magnetomotive force (mmf),measured in ampere-turns. This mmf is the product of the effective coil turns on thearmature and the current through those turns. It must be understood here that thearmature winding must be considered in developing those ampere-turns (i.e., theremay be parallel paths through the armature, with the possibility of the armature coilsbeing wound in series or parallel arrangements; hence, there will also be a division ofthe total armature current in each of the windings).This topic is addressed later.Thefollowing is a discussion of the action of the mmf produced by the armature.

dxdt

4.4 CHAPTER FOUR

FIGURE 4.5 Conductor moving in a magneticfield.

Armature Reaction. The conductors in the armature carry a current proportionalto the load. The magnetic field produced by this current reacts with the main fieldproduced by the same current flowing in the field coils. Figure 4.6 indicates two so-called belts of armature conductors (coil sides) under each pole face. Each of theconductors comprising these belts carries current in the same direction and henceproduces additive mmf. In addition, there are conductors which also carry unidirec-tional currents but are not under the pole arcs. The important consideration here isthe effect of the presence of magnetic material in the pole pieces, armature core, andarmature teeth. The flux paths through the armature are influenced by the reluc-tances of the paths. It is obvious that the reluctance of the flux paths under the polepieces is less than that of the paths adjacent to the brush area, which constitute amaterial of much greater reluctance, namely air.


FIGURE 4.6 Armature magnetic field.

Figure 4.6 indicates that the brushes are on the mechanical neutral axis (i.e.,halfway between the poles). The general direction of the armature mmf is along thebrush axis at quadrature to the main field. The armature conductors adjacent to thepoles produce flux densities in the air gap which are equal and opposite at the poletips. Keep in mind that the flux density produced by the armature is directly relatedto the armature current. (Also, a uniformly distributed flux density in the air gap isattributed to the main field and is directly related to this same armature current.)Now the net mmf (or flux in the air gap) is the result of both the main field mmf andthe armature field mmf. The resultant air gap flux is now increased at one pole tipand reduced at the other pole tip because of the armature reaction. The flux distor-tion in the air gap is illustrated in Fig. 4.7. In this figure, two poles and the armatureconductors beneath the poles have been unfolded to illustrate the distortion moreclearly. Ampere’s circuital law is useful here in determining the armature mmf. Theresulting air gap mmf is the result of the superposition of the field and armaturemmfs. MMF drops in the poles and armature iron are considered negligible com-pared to the air gap mmf. For reference, positive direction for the mmfs is assumedto be a flux out of a north pole. The armature mmf is shown as a linear relationship,but actually, because the armature slots are discrete, this relationship is actually

made up of small discrete stair-step transitions. However, it is shown as a smoothcurve here for easier analysis.

It is obvious that the air gap flux varies along the pole face. Another observationfrom the distortion mmf pattern is that harmonics are very present in the air gapmmf. Also, because of symmetry, only odd harmonics can exist in an analysis of theair gap mmf. It must also be remembered that in the case of a series ac motor, thevariation of flux distortion would be approximately the same, only pulsating.

The results of a Fourier analysis of the air gap flux harmonics for the case when(NfIf)/(NaIa) = 7/5 are given in Figs. 4.8 to 4.10.The figures show the fundamental, thefundamental plus third and fifth harmonics, and finally the fundamental and oddharmonics, up to and including the fifteenth harmonic. The magnitudes of the oddharmonics beginning with the fundamental are 8.94, 0.467, 1.41, 0.924, −0.028, −0.02,0.523, and 0.0094. That is, the series can be represented as follows:

φair gap = 8.94 cos θ + 0.467 cos 3θ + 1.41 cos 5θ + 0.924 cos 7θ

− 0.028 cos 9θ − 0.02 cos 11θ + 0.523 cos 13θ + 0.0094 cos 15θ + (4.9)

Of course, changing the pole arc or magnitudes of the armature and field mmfswill change the harmonic content. The change in flux density across the air gap pro-duces two effects: (1) a reduction in the total flux emanating from each pole, and (2)

4.6 CHAPTER FOUR

FIGURE 4.7 Flux distortion in the air gap.

a shift in the electrical neutral axis, lending to commutation problems due to the fluxdistortion.

The flux distortion which the armature produces has been called a cross-magnetizing armature reaction, and rightly so.The net effect is illustrated in Fig. 4.11,showing the resultant field where the armature cross field is at right angles to themain field. The result is a distortion of the net flux in the motor; the second effect,which was indicated previously, is a reduction in the total main field flux.This reduc-tion of flux is not too obvious; in fact, it would almost appear from Fig. 4.8 that thevector addition of these two mmfs would lead to an increase in flux which wouldoccur with the brushes on the mechanical neutral axis.


FIGURE 4.8 Fundamental frequency of air gap flux.

FIGURE 4.9 Air gap flux due to fundamental plus third and fifth harmonics.

This resulting decrease in flux can be attributed to magnetic saturation. Figure4.12 illustrates how this comes about when one operates the motor at the knee of thesaturation curve. The area ABC is proportional to the reduction in flux at the poletip with a decrease in flux density, while the area CDE is proportional to the increasein flux at the other pole tip. Note that the reduction in flux exceeds the increase in

4.8 CHAPTER FOUR

FIGURE 4.10 Air gap flux due to odd harmonics through fifteenth.

FIGURE 4.11 Field of armature with and without brushshift.

flux because of the saturation effect. This is sometimes called the demagnetizingeffect of armature reaction. This demagnetization effect can be on the order of about4 percent.

Both armature reaction distortion and flux can be reduced or eliminated by com-pensation windings in the pole face or by increasing the reluctance at the pole tips;the latter by either lamination design or a nonuniform air gap.


FIGURE 4.12 Armature reaction resulting in main pole flux reduction.

Because the purpose of the commutator and brushes is to change the current ina short-circuited coil from, say, a current of +I to −I in the length of time the coil isshort-circuited by the brushes, some arcing at the brushes is expected. One desires tokeep any voltage induced into the coil to a minimum in order to keep this arcing assmall as possible.

When the brushes are on the mechanical neutral position and the main fieldbecomes distorted, the coils being commutated will have an induced voltage fromthe distorted air gap flux. A shifting of the brushes to a new neutral position is sug-gested to keep the induced voltage to a minimum and keep the brush arcing low. Itshould be recognized that the amount of distortion of the field is a function of theampere-turns of the armature and hence is dependent on the motor load. One thenrealizes that the electrical neutral position is a function of the load and that shiftingbrushes to an electrically neutral position is therefore not a matter of shifting themto a unique point in space.

Brush shifting causes another effect—that is, a demagnetization effect on the airgap flux. This effect is in addition to the demagnetizing effect which occurs becauseof saturation. When the brushes are shifted, the pole axis of the armature is shifted.

The result is that the angle between the main field and the armature field is greaterthan 90°. This process is also illustrated in Fig. 4.11.

In summary, there are two processes which cause reduction of the air gap flux: (1)reduction due to cross-magnetization when the brushes are on the mechanical neu-tral position, which changes the flux distribution across the pole face and the net fluxbecause of saturation effects, and (2) demagnetization resulting from a brush shiftand change of the armature pole orientation with respect to the main field, resultingin the armature field mmf having a component in direct opposition to the main fieldmmf.

Reactance Voltage and Commutation. It was indicated previously that duringcommutation the current in the shorted coil(s) under a brush must reverse andchange direction. The self-inductance of the shorted winding by Lenz’s law inducesan emf in the shorted winding to oppose the change in coil current. This voltage issometimes called the reactance voltage. (Actually, it may be only a portion of thereactance voltage, as is shown later.) This voltage slows down the reversal of currentand tends to produce sparks or arcing as the trailing commutator bar leaves a brush.The reactance voltage hinders good commutation. The magnitude of this voltagedepends directly on the square of the number of coil turns, the current flowing, andthe armature velocity; it is inversely proportional to the reluctance of the magneticpath.

When shifting brushes to seek an electrical neutral position on a motor, the shift-ing is done in the opposite direction of armature rotation. It must be rememberedthat the reactance voltage is an e = L(di/dt) voltage and is simply due to the currentchange during commutation. It has nothing to do with induced voltage from air gapflux. Shifting the brushes really does not help the reactance voltage. Theoretically,there is no induced voltage on the neutral axis from the motor flux (in fact, dλ/dtshould be zero here since the coil sides are moving parallel to the flux). In order tocounteract the reactance voltage and induce a voltage opposite to the reactancevoltage, the brushes must be shifted further backward than the magnetic neutralaxis. At this point a dλ/dt voltage is induced into the commutated coil by the fieldflux, which counteracts the reactance voltage.The result is that two voltages in oppo-site polarities are induced into the shorted coil, thereby reducing brush arcing.

An important consideration is that there can be another component of the reac-tance voltage which can occur when two coil sides are in the same slot and both areundergoing commutation. There is then the following self-inductance term:

Reactance voltage = L11 + M12 (4.10)

where the coil being considered is the coil carrying current i1 and the coil in the sameslot undergoing commutation is carrying current i2. If there is more couplingbetween coils, more mutual terms can exist.

Brushes are a very important consideration, and contrary to normal electricalprinciples, one would assume that keeping the brush resistance low would assist inreducing arcing at the brush commutator bar interface. This is far from the truth. Infact, resistance commutation is now an accepted technology. Brushes normally havea graphite or carbon formulation and hence introduce, by their characteristics, resis-tance into the interface. If constant current density could be achieved at a brush forall loads and speeds, an ideal condition would exist for commutation. Figure 4.13illustrates what would occur if ideal conditions are assumed. The assumptions are asfollows:

di2dt

di1dt

4.10 CHAPTER FOUR


FIGURE 4.13 Linear commutation.

The brush width is equal to that of a commutator segment. The current density under a brush is constant. The reactance voltage and resistance are zero. The commutator bar insulation is small compared to the width of a commutator

segment.

Note that as the commutator passes beneath the brush, for the assumptionsgiven, the contact resistance is inversely proportional to the contact area. Thesequence in Fig. 4.13 illustrates how the current change occurs in an armature coilunder these ideal conditions. This change in surface contact changes the brush con-tact resistance; hence, with constant current density, the change in current flowchanges linearly according to contact resistance. Remember, the resistance isinversely proportional to the contact area. This ideal commutation has been termedlinear commutation because of the linear transfer of current in the commutated coil,as indicated in Fig. 4.13.

If the brush width is greater than that of one commutator segment, linear com-mutation still exists because the resistance still changes linearly; hence, the currentdensity remains constant. The commutation period is longer; hence, the rate ofchange of current di/dt is reduced, and hence the reactance voltage is reduced also.

All of the assumptions made in the previous analysis cannot occur, and nonlinearcommutation results. The armature coils have resistance, have self- and mutualinductance, and hence linear commutation cannot be attained.

If these nonlinearities do occur, then either undercommutation or overcommuta-tion occurs. These are illustrated in Fig. 4.14. Since brush heating is a product ofbrush resistance and the square of the current, linear commutation is preferred, as itminimizes brush heating. That is, the squared value of the three forms of currentcommutation in Fig. 4.13 is smallest for linear commutation.

4.12 CHAPTER FOUR

FIGURE 4.14 Overcommutation, undercommutation, andlinear commutation.

It is possible to develop a circuit model for the commutation process, as shown inFig. 4.15. The factors have previously been described, and they are restated here.Consider the subject coil of turns N1, and self-inductance L11, and the current i1

which is being commutated.

1. The voltage of self inductance (reactance voltage) of the commutated coil is

eL1 = L11 (4.11)

where i1 is the current being commutated and L11 is the self-inductance of the coilbeing commutated.

di1dt

2. The voltage of mutual inductance of an adjacent coil in the same slot carryingcurrent i2 and undergoing commutation and linked to the subject coil is

eM1 = M12 (4.12)

where M12 is the mutual inductance between adjacent coils.3. The voltage induced into the subject coil of turns N1 when cutting flux φa due to

the armature mmf is

ea = N1 (4.13)

Sometimes this is called commutating voltage; it opposes the reactance voltage.Thiscan be controlled by brush overshifting to make ea cancel the voltages eM1 and eL1.

dφadt

di2dt


FIGURE 4.15 Commutation voltages.

4. There are voltage drops due to coil resistance rc; Rc and Rc, the left and right con-tact resistances at the brush; and the brush resistance between commutator seg-ments rb.

The commutation process is not simple, and brush composition and interfacefilm, measurements, and conditions all can be made to assist in the commutationprocess. Maintaining a suitable interface film between the brush and the coppercommutator is extremely important. This interface is formed of copper oxide andfree particles of graphite film; it provides a general resistance commutation and sup-plies lubricant to reduce surface friction and heat between the commutator and thebrush.

Torque-Speed Characteristics of DC Series Motors. The electrical equivalentsteady-state circuit is the most appropriate method for analyzing the motor. Thisequivalent circuit is repeated in Fig. 4.16 for convenience.

4.14 CHAPTER FOUR

FIGURE 4.16 Electric circuit of series dc motor.

Writing Kirchhoff’s voltage law around the loop gives:

V = Ia(rf + ra) + Ecemf (4.14)

= Ia(rf + ra) + Eg (4.15)

The counter-emf Eg was addressed earlier when discussing the voltage inducedinto a conductor of length and moving with a velocity v perpendicular to a mag-netic field of flux density B. The counter-emf voltage Eg was expressed as e and givenby:

e = Bv (4.8)

It is necessary to modify this and express e in terms of the motor parameters.Beginning with Faraday’s law:

Eg = Ecemf = N V (4.16)

where N is the number of conductors per armature path when Z is the total numberof conductors (coil sides) on the armature.Then:

Z = number of slots × coils/slot × turns/coil × 2 (conductors/turn)a = number of parallel paths through the armatureP = number of polesn = armature speed, rpmφ = flux per pole, maxwells

Then:

Eg = Ecemf = × 10−8 = Keφn V (4.17)

where φ is in lines (maxwells) and

Ke = × 10−8 (4.18)

From the equivalent circuit, the developed power of the motor is EgIa; that is:

Pdev = KeφnIa W (4.19)

Pdev = hp (4.20)

Tdev = lbft (4.21)

Tdev = 7.04 φIa × 10−8 lbft (4.22)

Tdev = 22.5 φIa × 10−8 ozin (4.23)

The equivalent circuit voltage equation then becomes

V = Ia(rf + ra) + Keφn (4.24)

And the motor speed equation becomes

n = (4.25)

For no load speed:

nL ≈ (4.26)V

Keφ

V − Ia(rf + ra)Keφ

ZP

a

ZP60a

KeφnIa746

33,000

2πn

KeφnIa746

ZP60a

ZφnP

60a

∆φ∆t


For stalled torque:

λ = KeφIa,stalled (4.27)

= Keφ (4.28)

Efficiency of DC Series Motors. Earlier discussion has indicated that a number oflosses occur in the series motor. These losses can be summarized as follows:

1. Copper loss in the armature winding I2ara

2. Copper loss in the series field winding I2arf

3. Brush contact loss4. Friction (brush and bearing friction) and windage5. Core loss (hysteresis and eddy current)6. Stray load loss (losses in addition to those above)

A comment should be made on the brush contact loss, listed as number 3. Exper-imentation and Institute of Electrical and Electronic Engineers (IEEE) specifica-tions have suggested as an approximation that

Brush contact loss = brush voltage drop × armature brush current (4.29)

This expression was arrived at for carbon or graphite brushes by observing the brushvoltage drop as a function of current density. At high- and low-current steady-statedensities, the voltage falls off and approaches zero. However, in between these lim-its the voltage drop is approximately constant at about 1 V per brush. Therefore, fora pair of brushes this becomes 2 V. On small machines this voltage drop tends toincrease, since the tendency is for the voltage drop to increase for lower brush andcommutator temperatures.

Carbon and graphite materials have a resistivity many times that of copper andhave a negative temperature coefficient. This negative temperature coefficient isattributed to the rise in brush voltage drop in small machines.

Stray load loss, as the term suggests, is a function of motor load and changes inload. Changes in load produce changes in armature current and hence affect (1)magnetic saturation in the magnetic circuit, (2) armature reaction changes, and (3)eddy current loss changes.

Figure 4.17 presents a diagram displaying these losses. By definition:

Overall efficiency = = (4.30)

= (4.31)

Sometimes both electrical and mechanical efficiencies are of interest in order todetermine where improvements can be made.

Mechanical efficiency = (4.32)

Electrical efficiency = (4.33)electrical power output

electrical power output + electrical losses

useful mechanical outputmechanical output + rotational losses

useful mechanical output

total electrical input

outputoutput + losses

outputinput

Vra

4.16 CHAPTER FOUR

The conditions for maximum efficiency can be related to those losses which are con-sidered to be constant and those that vary with the motor load current. If the lossesare segregated as follows:

K1 = constant losses

K2 = losses which vary linearly with Ia

K3 = losses which vary as the square of Ia

Then the efficiency is

eff = = (4.34)

= (4.35)

Equating this expansion to zero gives the condition for maximum efficiency.

VIa[V − K2 − 2K3Ia] − [VIa − (K1 + K2Ia + K3I2a)]V = 0 (4.36)

−2K3I2a + K1 + K3I2

a = 0 (4.37)

K3I2a = K1 (4.38)

Thus, for maximum efficiency the constant losses must be equal to those that varyas the square of the armature current. This is typical for all different pieces of rota-tional electrical equipment.The constant losses usually are considered to be the corelosses, friction, and windage. Usually the brush loss is small.

4.1.2 AC Series Motors

One of the first considerations when considering ac operation of a series-woundmotor is: Does the motor develop a unidirectional torque when operated on ac sup-

VIa(V − K2 − 2K3Ia) − [VIa − (K1 + K2Ia + K3I2a)]V

(VIa)2

deffdIa

VIa − (K1 + K2Ia + K3I2a)

VIa

input − losses

input


FIGURE 4.17 Losses in a series dc motor.

ply voltages? We know the answer is yes. Since the armature and field are connectedin series, a current reversal in the field also produces a current reversal in the arma-ture. The torque is therefore in the same direction. As indicated previously, the con-struction must be an ac type with a completely laminated magnetic path. If it werenot laminated it would react as a solid-core inductor, with extremely high eddy cur-rent losses.

The basic phasor and electrical circuit diagram of a series ac motor is shown inFig. 4.18. The common current in both armature and field produce a motor fluxwhich is nearly in phase with the current, the small difference being the hysteresisand eddy current effects, which are not accounted for in the phasor diagram. Sevenvoltages are required to overcome the applied voltage. These are the voltages pro-duced by the leakage reactances and resistances of both armature and field, thetransformer voltage ETf, the commutator brush drop, and the counter-emf voltageEg. When the armature is rotating, the armature conductors, which are movingthrough an alternating field flux, generate an alternating voltage Eg which is in phasewith the flux, as indicated on the diagram.

4.18 CHAPTER FOUR

FIGURE 4.18 Series ac motor: (a) equivalent circuit and (b) phasor dia-gram.

(b)

(a)

This voltage is proportional to the magnitude of the flux and the rate at which itis cutting the flux (Faraday’s law). Since the flux is a sinusoidally varying quantity,the voltage will also vary sinusoidally at the same frequency as the flux wave. Thetransformer voltage ETf has been included and reflects the Faraday voltage inducedinto the field by the core flux. A similar voltage would be present in the armature,but it is small in comparison to ETf and has not been included in the phasor diagram.The phasor sum of these voltages then represents the applied voltage V and thebasic phasor diagram of the ac series motor. Here all voltages are root mean square(RMS) quantities. As previously mentioned, the effects of hysteresis and eddy cur-rents have not been taken into account, nor have the effects of commutation andbrush shifting. The power factor angle θ is also indicated on the diagram and repre-sents the power factor at the input motor terminals.

Several things are apparent from the phasor diagram:

1. At starting, the generated voltage Eg is zero. Hence, the starting current is lim-ited only by the impedances of the armature and field and the transformer volt-age ETf.

2. The motor draws a lagging power factor cos θ.3. As the motor speed increases at a constant line voltage, the motor current

decreases and the power factor improves, which is a good reason to run a motorat high armature velocities.

4. A reduction in frequency also improves the power factor by reducing the reac-tance drops in both armature and field.

At this point it should be obvious that dc and ac performance of series motorsmust differ due to reactance and transformer voltages and also due to ac losses(additional hysteresis and eddy current losses—that is, core losses—and possiblechanges in ac resistance as compared to dc resistance of the armature and field).Note that in the development of the phasor diagram leakage reactance voltages forboth the field and armature, as well as a transformer voltage in the main field, havebeen included. It is appropriate to develop some mathematical formulations andbasis for these voltages. In reality, the leakage reactance and transformer voltageresult from an application of Faraday’s law and are broken down into two parts toobtain each voltage. This Faraday voltage is expressed in instantaneous form andconsists of two parts, as follows.

ef = eTf + e (4.39)

where ef = Faraday voltageeTf = transformer voltagee = leakage reactance voltage

Then:

ef = (λM + λ) = Nf (φM + φ) (4.40)

where φM is the flux within the complete magnetic motor circuit and φ is the leakageflux.

Then:

λf = Nf + (4.41)

M

ddt

ddt


where is the ampere-turns of the main field and M and are the reluctances ofthe motor magnetic circuit and the leakage path, respectively.

Then:

λf = Nf i + i (4.42)

= i + i (4.43)

Then:

ef = + (4.44)

= Lf + L (4.45)

where Lf and L are the inductances associated with the two fluxes present in themain field winding.

Now, in the steady-state response Ef can be written as follows:

Ef = ETf + E = jωLfI + jωLI (4.46)

= jxfI + jxI (4.47)

where jx is the leakage reactance, and the voltage and current are expressed in RMSvalues.

If we assume that the applied voltage is sinusoidal, the Faraday voltage also willbe sinusoidal, and it is possible to express the transformer voltage ETf in terms ofother variables.

By Faraday’s law,

eTf = Nf (4.48)

φM = eTf dt (4.49)

if we represent

eTf = 2ETf sin (ωt + α) (4.50)

φM = cos (ωt + α) + φc (4.51)

where ETf is the rms value.The constant of integration φc is a transient quantity and disappears in the steady

state analysis. Hence:

φM = cos (ωt + α) (4.52)

φM,max = (4.53)2ETf

Nfω

2ETfNfω

2 ETfNfω

1Nf

dφMdt

didt

didt

didt

Nf2

didt

Nf2

M

Nf2

Nf2

M

Nf

NfM

4.20 CHAPTER FOUR

Or

ETf = = 4.44fNfφM,max (4.54)

This is the general transformer voltage relationship common in transformer theory.

Development of the Equivalent Circuit for a Universal Motor Considering BrushShift. Begin by considering a two-energy-source system (i.e., a main field and anarmature) with no brushes (Fig. 4.19), where:

va = voltage applied to armaturevf = voltage applied to fieldra = armature resistancerf = field resistanceia = armature currentif = field current

With no brushes, we have the following:

va = raia + pλa (4.55)

Vf = rfif + pλf (4.56)

2πfNfφM,max2


FIGURE 4.19 Consideration of universal motor brush shift.

where

p = d/dt (differential operator)

So, where

λa = aaia + afif (4.57)

λf = faia + ffif (4.58)

all s could be functions of θr.Then:

va = raia + iapaa + aapia + ifpaf + afpif (4.59)

vf = rfif + iapfa + fapia + ifpff + ffpif (4.60)

Now:

λaa = Naφa = ia = aaia and λaf = if = afif (4.61)

where

aa = af = (4.62)

With aa and the plot of aa versus θr (Fig. 4.20):

aa = + cos (2θr) (4.63)

See Figs. 4.21 and 4.22.

af = fa = −L cos (θr) where L = (4.64)

The self-inductance of the field is a constant witha round armature.

ff = constant = Lff ff = (4.65)Nf

2

f

NaNf

Raf

Lmax − Lmin2

Lmax + Lmin2

NaNfaf

N2a

aa

NaNfaf

N2a

aa

4.22 CHAPTER FOUR

FIGURE 4.20 Plot of aa.

FIGURE 4.21 Mutual terms between arma-ture and field.

FIGURE 4.22 Plot of af.

Using the expressions for aa , af , fa , and ff gives the following:

va = raia + ωr(Lmin − Lmax) sin (2θr) + + cos (2θr) pia

+ ωrL sin (θr)if − L cos (θr)pif (4.66)

And

vf = rf if + ωrL sin (θr)ia − L cos (θr)pia + Lffpif (4.67)

Now, add brushes at an angle θra. If the brushes were added with no shift, theresulting magnetic field of the armature would be at quadrature to the f axis. Thiswould fix θr at 90°. Since the brushes are shifted to θra , θr is fixed at 90° + θra. Also,shifting the brushes causes a voltage to be induced in the field winding from thearmature current, so fa ≠ 0. In fact, fa = constant = Lfa with θr = 90° + θra. Then:

va = raia − ωr(Lmin − Lmax) sin (2θra)ia

+ − cos (2θra)pia

+ ωrL cos (θra)if + L sin (θra)pif (4.68)

And

vf = rf if + Lfapia + Lffpif (4.69)

Now, defining the constants:

Laf1 = L cos (θra) Laa1 = (Lmin − Lmax) sin (2θra) (4.70)

Laf2 = L sin (θra) Laa2 = − cos (2θra) (4.71)

The equations in matrix form now are as follows:

= (4.72)

Or

vf = rf if + Lffpif − Lfapia (4.73)

va = ωrLaf1if + Laf2pif + raia + ωrLaa1ia + Laa2pia (4.74)

From the preceding, the equivalent circuit can be drawn (Fig. 4.23). From thisequivalent circuit, the transient response of the series motor can be calculated.

4.1.3 Permanent-Magnet DC Motors (Shunt PM Field Motors)

The general equivalent electrical circuit of a PMDC motor and its physical con-struction are shown in Figs. 4.24 and 4.25. The motor consists of a stator (Fig. 4.26)having permanent magnets attached to a soft steel housing and a commutator con-nected through brushes to a wound armature (Fig. 4.27). One of the first things to be

if

ia

rf + pLff −pLfa

ωrLaf1 + pLaf2 ra + ωrLaa1 + pLaa2

vf

va

Lmax − Lmin2

Lmax + Lmin2

Lmax − Lmin2

Lmax + Lmin2

Lmax − Lmin2

Lmax + Lmin2


considered in the operation of the motor are the motor and generator action, whichexist simultaneously in the armature circuit of the motor. These two principles are(1) the instantaneous emf, which is induced in the armature conductors when mov-ing with velocity v within a magnetic field, and (2) the force produced on the con-ductors as the result of carrying an electric current in this same magnetic field.

4.24 CHAPTER FOUR

FIGURE 4.23 Equivalent circuit of the series motor.

FIGURE 4.24 Permanent-magnet dc motor. ra = armatureresistance measured at brushes, e = motor counter-emf, and v = applied voltage.

It is known that the instantaneous force on a conductor of length carrying a cur-rent i in a magnetic field B is:

f = Bi sin θ N (4.1)

Or, in vector notation:

f = (i × B) (4.2)


FIGURE 4.25 Permanent-magnet dc motor.

FIGURE 4.26 Stator.

where θ is the angle between the direction of the magnetic field and the direction ofcurrent flow in the conductor, B is in webers per square meter (teslas), i is inamperes, and is in meters. Motors, by design, have the armature conductors andmagnetic flux at quadrature to one another. Therefore the force becomes:

f = Bi (4.3)

Assume a situation in Fig. 4.28 where a conductor of length is located in a mag-netic field and is free to move in the x direction perpendicular to the field. From thepreceding discussion, a force is produced on the conductor, causing it to move in thex direction. Then:

V − ir = e or Vi − i2r = ei (power) (4.4)

So, differential electrical energy into the conductor less differential i2r loss equalsdifferential mechanical output energy:

dWelect = dWmech (4.5)

ei dt = f dx where f = Bi (4.6)

ei dt = Bi dx (4.7)

e = B = Bv (4.8)

Thus, the force causes movement of the conductor, which in turn causes a voltageto be induced into that conductor which is opposite to the direction of the originalcurrent. This is an important concept in the operation of motors in general and onewhich is used to discuss the operation of the dc motor.This induced voltage is usuallycalled a counter-emf (cemf) voltage because of its opposition to the applied voltage.

dxdt

4.26 CHAPTER FOUR

FIGURE 4.27 Commutator connected to wound armature.

This example also indicates that areversible energy or power exchange ispossible (i.e., between the mechanicaland electrical systems). Therefore thesame machine may operate as a motoror a generator, depending on the flow ofenergy in the armature.

In the motor mode, the armature ofthe PMDC motor is supplied with a cur-rent by the applied voltage, and a mag-netic field (flux) is produced by thepermanent magnets. Since the armatureconductors (coils) are located in thisfield, each of the conductors in the fieldexperiences a force (torque) tending tomake it move (rotate), and as we havejust indicated, a countervoltage (cemf)is produced opposing the applied volt-age. Other than the cemf which the

armature produce, we must recognize that the armature circuit (coils) also producesa magnetic field of its own. The armature, because of its commutator and brush con-struction, has a unidirectional current and therefore produces a fixed-directed mmf,measured in ampere-turns. This mmf is the product of the effective coil turns on thearmature and the current through those turns. It must be understood here that thearmature winding must be considered in developing those ampere-turns (i.e., theremay be parallel paths through the armature, with the possibility of the armature coilsbeing wound in series or parallel arrangements; hence, there will also be a division ofthe total armature current in each of the windings).This topic is addressed later.Thefollowing is a discussion of the action of the mmf produced by the armature.

Armature Reaction. The conductors in the armature carry a current proportionalto the load.The magnetic field produced by this current reacts with the PM field. Fig-ure 4.29 indicates two so-called belts of armature conductors (coil sides) under eachpole face. Each of the conductors comprising these belts carries current in the samedirection and hence produces additive mmf. In addition, there are conductors whichalso carry unidirectional currents but are not under the pole arcs.The important con-sideration here is the effect of the presence of the PM material in the pole pieces,armature core, and armature teeth. The flux paths through the armature are influ-enced by the reluctances of the paths. It is obvious that the reluctance of the fluxpaths under the pole pieces is less than that of the paths adjacent to the brush area,which constitute a material of much greater reluctance, namely a large air gap.

Figure 4.29 indicates that the brushes are on the mechanical neutral axis (i.e.,halfway between the poles). The general direction of the armature mmf is along thebrush axis at quadrature to the main field. The armature conductors adjacent to thepoles produce flux densities in the air gap which are equal and opposite at the poletips. Keep in mind that the flux density produced by the armature is directly relatedto the armature current together with a uniformly distributed flux density in the airgap attributed to the PM field. Now the net mmf (or flux in the air gap) is the resultof both the PM mmf and the armature field mmf. The resultant air gap flux is nowincreased at one pole tip and reduced at the other pole tip because of the armaturereaction.The flux distortion in the air gap is illustrated in Fig. 4.30. In this figure, twopoles and the armature conductors beneath the poles have been unfolded to illus-


FIGURE 4.28 Conductor moving in a mag-netic field.

trate the distortion more clearly.Ampere’s circuital law is useful here in determiningthe armature mmf.The resulting air gap mmf is the result of the superposition of thePM and armature mmfs. MMF drops in the poles and armature iron are considerednegligible compared to the air gap mmf. For reference, positive direction for themmf is assumed to be a flux out of a north pole.The armature mmf is shown as a lin-ear relationship, but actually, because the armature slots are discrete, this relation-ship is actually made up of small discrete stair-step transitions. However, it is shownas a smooth curve for easier analysis.

4.28 CHAPTER FOUR

FIGURE 4.29 Armature magnetic field.

It is obvious that the air gap flux varies along the pole face. Another observationfrom the distortion mmf pattern is that harmonics are very present in the air gapmmf.Also, because of mmf symmetry, only odd harmonics can exist in an analysis ofthe air gap mmf.

= (4.75)

The results of a Fourier analysis of the air gap flux harmonics for this case aregiven in Figs. 4.31, 4.32, and 4.33. The figures show the fundamental, the fundamen-tal plus third and fifth harmonics, and finally the fundamental and odd harmonics, upto and including the fifteenth harmonic. The magnitudes of the odd harmonicsbeginning with the fundamental are 8.94, 0.467, 1.41, 0.924, −0.028, −0.02, 0.523, and0.0094. That is, the series can be represented as follows:

φair gap = 8.94 cos θ + 0.467 cos 3θ + 1.41 cos 5θ + 0.924 cos 7θ

− 0.028 cos 9θ − 0.02 cos 11θ + 0.523 cos 13θ + 0.0094 cos 15θ + (4.9)

Of course, changing the PM pole arc or magnitudes of the armature and PMmmfs will change the harmonic content.The change in flux density across the air gapproduces two effects: (1) a reduction in the total flux emanating from each pole, and(2) a shift in the electrical neutral axis, lending to commutation problems due to theflux distortion.

75

mmf PMmmf armature


FIGURE 4.30 Flux distortion in the air gap mmf.

FIGURE 4.31 Fundamental frequency of air gap flux.

The flux distortion which the armature produces has been called a cross-magnetizing armature reaction, and rightly so.The net effect is illustrated in Fig. 4.34,showing the resultant field where the armature cross field is at right angles to themain field. The result is a distortion of the net flux in the motor; the second effect,which was indicated previously, is a reduction in the total main field flux.This reduc-tion of flux is not too obvious; in fact, it would almost appear from Fig. 4.34 that thevector addition of these two mmfs would lead to an increase in flux which wouldoccur with the brushes on the mechanical neutral axis.

4.30 CHAPTER FOUR

FIGURE 4.32 Air gap flux due to fundamental plus third and fifth harmonics.

FIGURE 4.33 Air gap flux due to odd harmonics through fifteenth.

This resulting decrease in flux can be attributed to magnetic saturation. Figure4.35 illustrates how this comes about when one operates the motor at the knee of thesaturation curve. The area ABC is proportional to the reduction in flux at the poletip with a decrease in flux density, while the area CDE is proportional to the increasein flux at the other pole tip. It can be observed that the reduction in flux exceeds theincrease in flux because of the saturation effect.This is sometimes called the demag-netizing effect of armature reaction; it is a serious consideration when using PM as afield structure. This problem is discussed later.

Because the purpose of the commutator and brushes is to change the current ina short-circuited coil from, say, a current of +I to −I in the length of time the coil isshort-circuited by the brushes, some arcing at the brushes is expected. One desires tokeep any voltage induced into the coil to a minimum in order to keep this arcing assmall as possible.

When the brushes are on the mechanical neutral position and the main fieldbecomes distorted, the coils being commutated will have an induced voltage fromthe distorted air gap flux. A shifting of the brushes to a new neutral position is sug-gested to keep the induced voltage to a minimum and keep the brush arcing low. Itshould be recognized that the amount of distortion of the field is a function of theampere-turns of the armature and hence is dependent on the motor load. One thenrealizes that the electrical neutral position is a function of the load and that shiftingbrushes to an electrically neutral position is therefore not a matter of shifting themto a unique point in space.

Brush shifting causes another effect—that is, a demagnetization effect on the airgap flux. This effect is in addition to the demagnetizing effect which occurs because


FIGURE 4.34 Field of armature with and without brushshift.

of saturation. When the brushes are shifted, the pole axis of the armature is shifted.The result is that the angle between the main field and the armature field is ofgreater than 90 degrees. This process is also illustrated in Fig. 4.34.

In summary, there are two processes which cause reduction of the air gap flux: (1)reduction due to cross-magnetization when the brushes are on the mechanical neu-tral position, which changes the flux distribution across the pole face and the net fluxbecause of saturation effects, and (2) demagnetization resulting from a brush shiftand change of the armature pole orientation with respect to the PM field, resulting inthe armature field mmf having a component in direct opposition to the PM field mmf.

Reactance Voltage and Commutation. It was indicated previously that duringcommutation the current in the shorted coil(s) under a brush must reverse andchange direction. The self-inductance of the shorted winding by Lenz’s law inducesan emf in the shorted winding to oppose the change in coil current. This voltage issometimes called the reactance voltage. (Actually, it may be only a portion of the reac-tance voltage, as is shown later.) This voltage slows down the reversal of current andtends to produce sparks or arcing as the trailing commutator bar leaves a brush. Thereactance voltage hinders good commutation.The magnitude of this voltage dependsdirectly on the square of the number of coil turns, the current flowing, and the arma-ture velocity; it is inversely proportional to the reluctance of the magnetic path.

When shifting brushes to seek an electrical neutral position on a motor, the shift-ing is done in the opposite direction of armature rotation. It must be rememberedthat the reactance voltage is an e = L(di /dt) voltage and is simply due to the currentchange during commutation. It has nothing to do with induced voltage from air gapflux. Shifting the brushes really does not help the reactance voltage. Theoretically,there is no induced voltage on the neutral axis from the motor flux (in fact, dλ/dtshould be zero here since the coil sides are moving parallel to the flux). In order to

4.32 CHAPTER FOUR

FIGURE 4.35 Armature reaction resulting in main pole flux reduction.

counteract the reactance voltage and induce a voltage opposite to the reactancevoltage the brushes must be shifted further backward than the magnetic neutral axis.At this point a dλ/dt voltage is induced into the commutated coil by the PM fieldflux, which counteracts the reactance voltage.The result is that two voltages in oppo-site polarities are induced into the shorted coil thereby reducing brush arcing.

An important consideration is that there can be another component of the reac-tance voltage which can occur when two coil sides are in the same slot and both areundergoing commutation.There is then the following mutual inductance term whichmust be added to the self-inductance term:

Reactance voltage = L11 + M12 (4.10)

where the coil being considered is the coil carrying current i1 and the coil in the sameslot undergoing commutation is carrying current i2. If there is more couplingbetween coils, more mutual terms can exist.

Brushes are a very important consideration, and contrary to normal electricalprinciples, one would assume that keeping the brush resistance low would assist inreducing arcing at the brush commutator bar interface. This is far from the truth. Infact, resistance commutation is now an accepted technology. Brushes normally havea graphite or carbon formulation and hence introduce, by their characteristics, resis-tance into the interface. If constant current density could be achieved at a brush forall loads and speeds, an ideal condition would exist for commutation. Figure 4.36illustrates what would occur if ideal conditions are assumed. The assumptions are asfollows:

The brush width is equal to that of a commutator segment. The current density under a brush is constant. The reactance voltage and resistance are zero. The commutator bar insulation is small compared to the width of a commutator

segment.

Note that as the commutator passes beneath the brush, for the assumptionsgiven, the contact resistance is inversely proportional to the contact area. Thesequence in Fig. 4.36 illustrates how the current change occurs in an armature coilunder these ideal conditions. This change in surface contact changes the brush con-tact resistance; hence, with constant current density, the change in current flowchanges linearly according to contact resistance. Remember, the resistance isinversely proportional to the contact area. This ideal commutation has been termedlinear commutation because of the linear transfer of current in the commutated coil,as indicated in Fig. 4.36.

If the brush width is greater than that of one commutator segment, linear com-mutation still exists because the resistance still changes linearly; hence, the currentdensity remains constant. This has an advantage in that the commutation period islonger; hence, the rate of change of current di /dt is reduced, and hence the reactancevoltage is reduced also.

All of the assumptions made in the previous analysis cannot occur, and nonlinearcommutation results. Armature coils have resistance, have self- and mutual induc-tance and hence linear commutation cannot be attained.

If these nonlinearities do occur, then either undercommutation or overcommuta-tion occurs. These are illustrated in Fig. 4.37. Since brush heating is a product ofbrush resistance and the square of the current, linear commutation is preferred, as it

di2dt

di1dt


minimizes brush heating. That is, the squared value of the three forms of currentcommutation in Fig. 4.37 is smallest for linear commutation.

It is possible to develop a circuit model for the commutation process, as shown inFig. 4.38. The factors have previously been described, and they are restated here.Consider the subject coil of turns N1 and self-inductance L1, and the current ii whichis being commutated.

4.34 CHAPTER FOUR

FIGURE 4.36 Linear commutation.

1. The voltage of self inductance (reactance voltage) of the commutated coil is

eLa = L1 (4.76)

where i1 is the current being commutated and L1 is the self-inductance of the coilbeing commutated.

diadt


FIGURE 4.37 Overcommutation, undercommutation, andlinear commutation.

FIGURE 4.38 Coils under commutation.

2. The voltage of mutual inductance of an adjacent coil in the same slot carryingcurrent i2 and undergoing commutation and linked to the subject coil is

ema = M12 (4.77)

where M12 is the mutual inductance between adjacent coils.3. The voltage induced into the subject coil of turns N when cutting flux φa due to

the armature mmf is

ea = N1 (4.13)

Sometimes this is called commutating voltage; it opposes the reactance voltage.This can be controlled by brush overshifting to make ea cancel the voltages ema

and eLa.4. There are voltage drops due to coil resistance rc ; Rc and Rc , the left and right con-

tact resistances at the brush; and the brush resistance between commutator seg-ments rb.

The commutation process is not simple, and brush composition and interfacefilm, measurements, and conditions all can be made to assist in the commutationprocess. Maintaining a suitable interface film between the brush and the coppercommutator is extremely important. This interface is formed of copper oxide andfree particles of graphite; it provides a general resistance commutation and supplieslubricant to reduce surface friction and heat between the commutator and the brush.

We have previously discussed demagnetization which occurs at the leading tip ofa motor due to armature reaction.

The subsection on armature reaction shows that the mmf of this reaction couldbe closely represented as a triangular wave (Fig. 4.30). We are concerned in PMmotor design with the demagnetizing effect of this armature reaction. In addition tothe normal design demagnetization curve shown in Fig. 4.39, a set of intrinsic curvesis also indicated. These curves are extremely important because they represent lim-its on the amount of demagnetization which can be tolerated. The value of −Hrequired to remove the magnetization is given the value Hci. The PM of the motornormally does not span 180° electrical; therefore we are not faced with the totalampere-turns of demagnetization of the armature reaction, only that componentwhich exists at the pole tip. If we let β represent the angle subtended from the cen-ter of magnet to the magnet tip, a trigonometric relationship can be used to calculatethe demagnetization armature reaction at the magnet tip. If NaIa represents the max-imum value of a triangular wave, the demagnetization ampere-turns at the magnettip equals (β/90°)(NaIa). If this value is less than the design value of the radial mag-net length (i.e., lm Hci), loss of magnetization at the pole tip can occur.

One important consideration when using permanent magnets in machinery is thefear of demagnetization of the magnet. This must be given important consideration.The shape of the demagnetization curve of the permanent magnet is an importantproperty in this respect. For example, in Fig. 4.39, for the alnico magnet with an oper-ating point on a load line (see the next subsection, Permanent Magnets for DCMotors), only a small amount of demagnetizing ampere-turns per inch will com-pletely demagnetize the magnet, whereas the ceramic or rare earth magnets’ char-acteristics have much more latitude. However, there is a considerable difference inthe available flux densities among the alnico, ceramic, and rare earth magnets;

dφadt

dibdt

4.36 CHAPTER FOUR

hence, for required large flux densities the alnico magnets are superior. Demagneti-zation results primarily from armature reaction, which can be large because of largein-rush currents upon motor starting or possibly because of control applicationsinvolving dynamic braking.These currents cause a large demagnetizing effect acrossthe pole face, as previously described in the armature reaction discussion. Anotherimportant consideration when using magnets is the loss of magnetic properties asthe temperature of the magnet increases. Thus, the temperature range of motoroperation must be considered when employing magnets.


FIGURE 4.39 Typical demagnetization curves for three magnets.

Magnets for PMDC motors are selected by their characteristic demagnetizationcurves—that is, their residual flux density Br, their energy product BH, and theircoercive force Hc.

Ceramic magnets have a characteristic and a permeability very nearly equal tothose of air. That is, µ0 approaches 3.2 in English units. This is an important consid-eration because of the large effective reluctance of the magnetic path, and removalof the armature does not change the operating point of the magnet appreciably oraffect the performance of the magnet.

The air gap flux is not adjustable with permanent-magnet materials, of course.This means that changes in motor speed and torque must be made by changes inarmature current.

The performance of PMDC motors is very similar to that of wound-field DCmotors, and similar equations exist. This development is the next topic to beaddressed.

L1 + M12 + (rc + rb + 2Rc)ia + ea = 0 (4.78)

where Rc = brush contact resistancerc = coil resistancerb = brush resistance

dibdt

diadt

Permanent Magnets for DC Motors. A permanent magnet (PM), when magne-tized by an external source for use in a PMDC motor, has a remaining residual fluxdensity and responds to a normal hysteresis characteristic. The useful portion of amagnet in PM operation is in the second quadrant of the hysteresis loop and is usu-ally called the demagnetization curve of the magnet. It represents the relationshipbetween B and H of the magnet once it has been magnetized. Typical demagnetiza-tion curves for three PM materials are shown in Fig. 4.39. In this figure, Br is theresidual flux density and Hc is the coercive force. For proper utilization of a perma-nent magnet a user makes use of the energy product curve B H, shown in Fig. 4.39for the rare earth magnet only.

Figure 4.40 presents a series magnetic circuit in which a PM, an air gap, and mag-netic material have been introduced into a closed magnetic circuit. Ampere’s lawaround the circuit (Ohm’s law for magnetic circuits) is as follows:

Hmm + Hgg + Hss = 0 (4.79)

where Hm , Hg , and Hs are the magneticfield intensities of the PM, air gap, andsteel, respectively. The lengths of themagnetic paths in the PM, air gap, andsteel are respectively m , g , and s. Byneglecting any leakage flux, the flux is thesame throughout the series circuit. If weassume the cross-sectional areas of thePM, air gap, and steel are respectively Am,Ag, and As, then:

BmAm = BqAq = BSAS = core flux (4.80)

where Bm, Bg, and Bs are the respectiveflux densities. Since the relationshipbetween B and H in a material is the per-meability µ, we can state:

Hmm + g + s = 0 (4.81)

Or

Hmm + g + s = 0 (4.82)

Then:

Hm = − + Bm = − + Bm (4.83)

This is the equation of a straight line on the BH demagnetization curve. If µs >>µg , the expression reduces to:

≈ − (4.84)µgmAg

gAm

BmHm

sµsAs

gµgAg

Amm

sAmµsmAs

gAmµgmAg

BmAmµsAs

BmAmµgAg

Bsµs

Bgµg

4.38 CHAPTER FOUR

FIGURE 4.40 Permanent magnet with airgap introduced.

where

− (4.85)

is the slope of the line.Thus, the values of Bm and Hm can be found from the demagnetization curves and

the cross-sectional area of the magnet, the air gap, and the respective lengths of themagnetic paths. If flux fringing is neglected, usually Ag = Am, and the slope becomes:

= − (4.86)

This becomes an approximate expression for the so-called load line of the PM,and the intersection with the demagnetization curve becomes the operating point.

As previously indicated, the energy product B H, which has units of watt-seconds per cubic meter, is important in PM utilization. In a linear medium it can beshown rather easily that B H is an expression for energy density in a given volume.

W = (4.87)

= = = BmHm (4.88)

It may also be of interest to look at the energy density which is present whenoperating at the maximum energy product as it relates to the air gap and PM energydensity.

Looking at the expression (magnitudes only)

≈ (4.89)

If Bm = Bg flux densities in the PM and the air gap are equal. Then:

Bg = Hm = Hm if Ag = Am (4.90)

where Vm and Vg are the volumes of the PM and the air gap.

Bg = Hm = ⋅ Hm (4.91)

BgHgVg = BmHmVm (4.92)

This equation then indicates that the point of operation for maximum availableenergy occurs at the maximum energy product B H.

If designing for the maximum energy product BH, consider the magnet charac-teristic shown in Fig. 4.41, where Bd and Hd are corresponding values giving the max-imum energy product. Knowing the slope of the load line, the ratio Bd/Hd is known.If, as previously suggested, the area of the magnet Am and the area of the air gap Ag

are equal, then:

= −µ = known (slope) (4.93)

From this m can be obtained.

mµgg

BdHd

VmVg

BmHg

VmVg

BgHg

µgVmVg

µgmAg

gAm

µgmAg

gAm

BmHm

12

mm

φmAm

12

mφmAmm

energy in a volume

volume

µgmg

BmHm

µgmAg

gAm


It is appropriate at this point to introduce some examples with magnetic circuitsutilizing PMs.

Example 1. Suppose the following parameters are given for Fig. 4.42. Designthe magnet for the maximum energy product.

Use the demagnetization curve given in Fig. 4.43.

Assume: Am = As = Ag = 5 cm2 (4.94)

g = 0.5 cm s = 8 cm (4.95)

µs = µ0µr where µr = 3500 (4.96)

= (4π × 10−7)(3500) (4.97)

Operating temperature = 20°C (68°F)

From previous development:

Hd = − + if Am = As = Ag (4.98)

Hd = − + (4.99)

= −0.5 + (4.100)

= −(0.5 + 2.27 × 10−3) (4.101)

Note the magnitude of the second term in comparison to the first term in the paren-theses. Therefore, the approximation as previously indicated is appropriate.

Bdm 4π × 10−7

Bdm4π × 10−7

83500

Bdm

8(4π × 10−7)(3500)

0.54π × 10−7

Bdm

sµs

gµg

4.40 CHAPTER FOUR

FIGURE 4.41 Designing for maximum energy.

FIGURE 4.42 Figure for example PM circuitcalculations.

Hd ≈ = Bd (4.102)

From Fig. 4.43, at 20°C the maximum energy product is approximately 3.25, with Hd ≈ 130 kA/m and Bd ≈ 0.195 T.

Therefore:

m ≈ × 0.195 = 0.6 cm (4.103)

Example 2. Suppose we have a 48-V PMDC motor running at 1400 rpm with anarmature resistance of 1.5 Ω.The motor constant is 75 (V s)/Wb. If the power inputto the motor is 200 W and the rotational efficiency is 80 percent, calculate the mag-netic flux per pole.

Using the SI system

ω = 1400 × = 147 rad/s (4.104)2π60

3.98 × 105

130 × 103

3.98 × 105

m

0.5Bdm 4π × 10−7


FIGURE 4.43 Magnet performance curve.

The power developed therefore must be

200 + 200(1.0 − 0.8) = 240 W

The developed torque is then

Tdev = = 1.63 N m (4.105)

Using the equation developed in Example 3

ω = − T (4.106)

φ2p − φp + = 0 (4.107)

φ2p − φp + = 0 (4.108)

φ2p − 0.004φp + 2.9 × 10−6 (4.109)

φp = 3.04 × 10−3 Wb (4.110)

Note: This value of flux when divided by the magnetic area represents a value of fluxdensity and an operating point on the demagnetization curve of the PM. Anydemagnetization mmf due to armature reaction would lower the operating point onthe curve. Keeping the reluctance of the air gap high (i.e., large air gap) will reducethe armature reaction.

Torque-Speed Characteristics of PMDC Motors. The electrical equivalent steady-state circuit is the most appropriate method for analyzing the motor.This equivalentcircuit is repeated for convenience in Fig. 4.44, and the armature brush drop isincluded.

Writing Kirchhoff’s voltage law aroundthe loop gives:

V = Iara + Eg + Eb V (4.111)

VIa = I2ara + EgIa + IaEb W (4.112)

The counter-emf Eg was addressedearlier when discussing the voltageinduced into a conductor of length andmoving with a velocity v perpendicularto a magnetic field of flux density B. Thecounter-emf voltage Eg was expressed ase and given by:

e = Bv (4.8)

(1.5)(1.63)(75)2(147)

4875(147)

raTK2ω

VKω

Va(Kφp)2

VKφp

PDω

4.42 CHAPTER FOUR

FIGURE 4.44 Armature Circuit for the PMDCmotor.

It is necessary to modify this and express e in terms of the motor parameters.Beginning with Faraday’s law:

Eg = Ecemf = N V (4.16)

where N is the number of conductors per armature path when Z is the total numberof conductors (coil sides) on the armature.

Then:

Z = number of slots × coils/slot × turns/coils × 2 (conductors/turn)a = number of parallel paths through the armatureP = number of polesn = armature speed, rpmφ = flux, Wb

Then:

Eg = Ecemf = × 10−8 = Keφn V (4.17)

where φ is the flux per pole in lines (maxwells) and

Ke = × 10−8 (4.18)

From the equivalent circuit, the developed power of the motor is EgIa , that is:

Pdev = KeφnIa W (4.19)

Pdev = hp (4.20)

Tdev = lb ft (4.21)

Tdev = 7.04 φIa × 10−8 lb ft (4.22)

Tdev = 22.5 φIa × 10−8 oz in (4.23)

The equivalent circuit voltage equation (neglecting brush drop) then becomes

V = Iara + Keφn (4.113)

And the motor speed equation becomes

n = (4.114)

For no load speed:

nnl ≈ (4.115)V

Keφ

V − IaraKeφ

ZP

a

ZP60a

KeφnIa746

33,000

2πn

KeφnIa746

ZP60a

ZφnP

60a

∆φ∆t


For stalled torque:

Tstall = KeφIa,stall (4.116)

= Keφ (4.117)

Example 3. Given a permanent magnet motor whose armature resistance is 0.8Ω and that operates on a 24-v supply, assume the air gap flux is 0.002 Wb and themotor torque and voltage constant is 100. The motor is operating under a loadtorque of 1 N m (neglect brush drop).

V = Iara + Eg = Iara + Kφpω (4.118)

ω = − Ia but T = KφpIa (4.119)

Ia = (4.120)

Then:

ω = − T (4.121)

Where T = developed torque, N mω = motor speed, rad/secφp = flux per pole, WbK = motor torque and voltage constant

Then:

ω = −

= 120 − 20 = 100 rad/s (4.122)

Or

n = 100 = 955 rpm (4.123)

The stall torque can be found by setting ω = 0.Then:

Tstall = =

= 6 N m (4.124)

Efficiency of DC Motors Earlier discussion has indicated that a number of lossesoccur in PMDC motors. These losses can be summarized as follows:

1. Copper loss in the armature winding I2ara

2. Brush contact loss

24(100 × 0.002)

0.8V(Kφp)

r

602π

0.8(100 × 0.002)2

24(100)(0.002)

ra(Kφp)2

VKφp

TKφp

raKφp

VKφp

Vra

4.44 CHAPTER FOUR

3. Friction (brush and bearing friction) and windage4. Core loss (hysteresis and eddy current)5. Stray load loss (losses in addition to those above)

A comment should be made on the brush contact loss listed as number 2. Exper-imentation and IEEE specifications have suggested as an approximation that:

Brush contact loss = brush voltage drop × armature brush current (4.29)

This expression was arrived at for carbon or graphite brushes by observing thebrush voltage drop as a function of current density.At high- and low-current steady-state densities, the voltage falls off and approaches zero. However, in between theselimits the voltage drop is approximately constant at about 1 V per brush. Therefore,for a pair of brushes this becomes 2 V. On small machines this voltage drop tends toincrease, since the tendency is for the voltage drop to increase for lower brush andcommutator temperatures.

Carbon and graphite materials have a resistivity many times that of copper andhave a negative temperature coefficient. This negative temperature coefficient isattributed to the rise in brush voltage drop in small machines.

Stray load loss, as the term suggests, is a function of motor load and changes inload. Changes in load produce changes in armature current and hence affect (1)magnetic saturation in the magnetic circuit, (2) armature reaction changes, and (3)eddy current loss changes.

Figure 4.45 presents a display of these losses. By definition:

Overall efficiency = = (4.30)

= (4.31)

Sometimes both electrical and mechanical efficiencies are of interest in order todetermine where improvements can be made.

useful mechanical output

total electrical input

outputoutput + losses

outputinput


FIGURE 4.45 Efficiency of a permanent-magnet dc motor.

Mechanical efficiency = (4.32)

Electrical efficiency = (4.33)

The conditions for maximum efficiency can be related to those losses which are con-sidered to be constant and those that vary with the motor load current. If the lossesare segregated as follows:

K1 = constant losses

K2 = those losses which vary linearly with Ia

K3 = those losses which vary as the square of Ia

then the efficiency is

eff = = (4.34)

= (4.35)

Equating this expansion to zero gives the condition for maximum efficiency.

VIa[V − K2 − 2K3Ia] − [VIa − (K1 + K2Ia + K3I2a)]V = 0 (4.36)

−2K3I2a + K1 + K3I2

a = 0 (4.37)

K3I2a = K1 (4.38)

Thus, for maximum efficiency the constant losses must be equal to those that varyas the square of the armature current. This is typical for all different pieces of rota-tional electrical equipment.The constant losses usually are considered to be the corelosses, friction, and windage. Usually the brush loss is small.

4.2 LAMINATION, FIELD, AND HOUSING GEOMETRY*

4.2.1 Universal (Series-Wound) Motor Construction

Figures 4.46 and 4.47 show the general construction of a two-pole-wound field uni-versal motor. Figure 4.48 shows a larger four-pole series motor designed to run ondirect current.

The stator (field) and armature laminations are stacked, insulated, and woundwith magnet wire.The armature (Fig. 4.47) is wound in consecutive continuous loopsor coils. Each coil end is connected to a commutator bar. The field is wound with

VIa(V − K2 − 2K3Ia) − [VIa − (K1 + K2Ia − K3I2a)]V

(VIa)2

deffdIa

VIa − (K1 + K2Ia + K3I2a)

VIa

input − losses

input

electrical power outputelectrical power output + electrical losses

useful mechanical outputmechanical output + rotational losses

4.46 CHAPTER FOUR

* Sections 4.2 to 4.4 contributed by Alan W. Yeadon, Yeadon Engineering Services, PC.


FIGURE 4.46 Universal motor construction.

FIGURE 4.47 Universal motor.

magnet wire and the coils are connected in series such that they will provide oppo-site magnetic polarity. The field (Fig. 4.47) is connected in series with the armaturethrough the brush-commutator connection.

Armature Geometry. We start by analyzing the physical dimensions of the arma-ture lamination, which gives us the following:

Gross slot area. This tells us how much cross-sectional area is available toaccommodate the armature windings (before adding insulation and slot pegs).Net slot area. This is the net available area to accommodate the copper wireafter the slot insulators and pegs are subtracted from the gross slot area.Magnetic path length. For the armature, this is the distance that the flux willtravel and over which the mmf will be dropped.Magnetic path area. This is the cross-sectional area of the magnetic path as seenfrom the flux’s point of view. The smaller the path area, the higher the flux den-sity, and the more mmf will be dropped along the path.Weight of steel. This must be known to calculate the inertias of the armatureassembly.Inertia. This is used in determining start-up acceleration or load matching.Slot constants. This is used in estimating reactance. This is largely a function oflamination geometry.

Using trigonometry, we break the armature lamination up into several sections inorder to calculate the gross slot area. The radii used in the calculations are readilyavailable from most lamination drawings supplied by the manufacturer.

Using Fig. 4.49, determine the following dimensions:

4.48 CHAPTER FOUR

FIGURE 4.48 Four-pole dc series motor.

Ra = outside radius of the armature laminationRa1 = inside radius of the armature teethRs = radius of the shaft

Wast = width of the armature slot top in straight-line distanceRa2 = radius of the circle on which the arc segments of the bottoms of the slots

are centeredNat = number of armature teethLstk = length of the armature stackθpole = angle of pole arc


FIGURE 4.49 Armature lamination.

To find the tooth pitch angle θtp, between two consecutive tooth centers, indegrees:

θtp = (4.125)

To convert to radians:

θtpr = θtp = θtp (0.017453) (4.126)

Tooth pitch tp, the circumferential distance between two tooth centers, in inches, is

tp = Raθtpr (4.127)

Width of the armature slot along the circumference Wasc, in, is

Wasc = 2Ra sinr−1 (4.128)

Width of the tooth tip along the circumference Watt, in, is

Watt = tp − Wasc (4.129)

Area of the lamination consisting of the pie-shaped segment between tooth centersASEG1, in2, is

ASEG1 = (4.130)

Now that we know the area of the SEG1 section, we can use right triangles todetermine the areas of smaller sections and subtract them from the SEG1 section(Fig. 4.50). Due to symmetry, only half of the SEG1 section needs to be calculated.

For the triangle TR1, at the bottom of the slot, the angle is half of the tooth pitchangle:

θTR1 = (4.131)

The length of the base of TR1 is

LTR1 = (4.132)

The angle within TR1 that the arc of radius Ras sweeps is

θas = 90° −θTR1 (4.133)

The area of triangle TR1 is

ATR1 = (4.134)

The area of the slot SEG2 which is in TR1 is

ASEG2 = (4.135)R2

asθas(0.017453)

2

R2as

2 tan θTR1

Rastan θTR1

θtp2

R2aθtpr2

Wast2Ra

π180

360°Nat

4.50 CHAPTER FOUR

The area of the steel inside TR1 will be needed later:

Asteel,TR1 = ATR1 − ASEG2 (4.136)

To determine the net slot area, we need to know the length of the arc of SEG2, ininches

Larc,SEG2 = Rasθas(0.017453) (4.137)

The area of the square SQ1 at the joint of the tooth and the tooth tip is the squareof the radius Rat, since this arc segment is approximately 90°.


FIGURE 4.50 Armature lamination Slot.

ASQ1 = R2at (4.138)

Using the circle segment of Rat over θat (approximately 90°) as SEG3:

ASEG3 = (4.139)

Area of the steel in SQ1:

Asteel,SQ1 = ASQ1 − ASEG3 (4.140)

Length of the arc of SEG3, in: is

Larc,SEG3 = Ratθat(0.017453) (4.141)

The following calculations are to determine the area of the steel near the centerof the armature lamination. Length of the hypotenuse of TR1, in:

Lh,TR1 = (4.142)

Length of the hypotenuse of TR2, in:

Lh,TR2 = Ra2 − Lh,TR1 (4.143)

Length of the base of TR2, in:

LTR2 = Lh,TR2 cos θTR1 (4.144)

The length of the side of the armature slot with no arcs in it is between SQ1 andTR1. This length, in inches, is

Lsas = Ra1 − Rat − LTR1 − LTR2 (4.145)

TR3 is the large triangle that is made from the intersections of the slot center line,the extended side of the slot, and the inside of the tooth tip. Length of TR3, in:

LTR3 = LTR1 + Lsas + Rat (4.146)

Height of TR3, in:

HTR3 = LTR3 tan θTR1 (4.147)

Area of the large triangle TR3, in2:

ATR3 = (4.148)

Using the small triangle TR4 inside TR3, the length of the base of TR4, in, is halfthe linear width of the top of the armature slot.

LTR4 = (4.149)

Length of the hypotenuse of TR4, in:

Lh,TR4 = (4.150)LTR4

cos θas

Wast2

HTR3LTR32

LTR1cos θTR1

R2atθat(0.017453)

2

4.52 CHAPTER FOUR

Height of TR4, in:

HTR4 = (4.151)

Area of TR4, in2:

ATR4 = (4.152)

The gross slot area (GSA) is found by subtracting the area of TR4 and the steelfrom the large triangle TR3 and then doubling it. GSA, in2:

AGS,a = 2[ATR3 − (Asteel,SQ1 + Asteel,TR1 + ATR4)] (4.153)

The net slot area (NSA) is the area left after the armature slot insulation is put in.In order to determine the area that the insulation takes up, we need to know thelength around the slot.

Length of the inside of the armature tooth that is not included in any otherdimensions as of yet Liat, in:

Liat = HTR3 − Rat − Lh,TR4 (4.154)

Length of the armature slot insulation Lasi, in:

Lasi = 2(Lait + Larc,SEG2 + Larc,SEG3 + Lsas) (4.155)

Net slot area that the armature winding can occupy ANS,a , in2:

ANS,a = AGS,a − LasiTasi (4.156)

where Tasi is the thickness of the armature slot insulation.

Weight of Armature Steel. The stacking factor percentage is typically 92 to 95 per-cent, or:

Fstk = (4.157)

Effective magnetic armature stack length, Lstke, in:

Lstke = Lstk × Fstk (4.158)

Length of the hypotenuse of the large TR3 Lh,TR3, in:

Lh,TR3 = (4.159)

Distance from the center of the lamination to the top of the usable armature slotRast, in:

Rast = Lh,TR3 + Lh,TR2 − HTR4 (4.160)

Height of the armature slot top (where there is no wire, near the outside edge) Hast,in:

Hast = Ra − Rast (4.161)

LTR3cos θTR1

lamination thickness × number of laminations

actual stack length

LTR4HTR42

LTR4tan θas


Area of the armature slot top Aast, in2:

Aast = HastWast (4.162)

Cross-sectional area of the armature shaft Ashaft, in2:

Ashaft = πR2shaft (4.163)

Total area of the steel in the armature lamination Asteel,arm lam, in2:

Asteel,arm lam = Nat ASEG1 − AGSA + Aast + (4.164)

which takes into account the area of the shaft that is inside each large SEG1.Volumeof the steel in the armature stack Vas, in3:

Vas = LstkeAsteel,arm lam (4.165)

Weight of the steel in the armature stack Wtas, oz:

Wtas = 4.536 Vas (4.166)

where 4.536 is the density of steel in ounces per cubic inch.Volume of the armature shaft Vs, in3:

Vs = AshaftLshaft (4.167)

Weight of the steel armature shaft Wts, oz:

Wts = 4.536Vs (4.168)

4.2.2 Magnetic Circuits

Magnetic Paths in the Armature. The armature yoke is the area between the bot-tom of the slot and the shaft.The effective width of the armature yoke is larger for amagnetic shaft and smaller for a nonmagnetic one. For the initial calculations, useLstke = 1.0.

Width of the armature yoke (Figs. 4.51 and 4.52) for a magnetic shaft, in inches:

Way = Ra2 − Ras (4.169)

Width of the armature yoke for a nonmagnetic shaft, in inches:

Way = Ra2 − Ras − Rshaft (4.170)

Only one of the preceding formulas will be used for each design.Height of the armature tooth tip, which is the same as its magnetic path length, in

inches:

Lattm = Hatt (4.171)

The magnetic path length of the armature tooth is from the tooth tip to the yoke, ininches:

Latm = Ra − Ra2 + Ras − Hatt (4.172)

AshaftNat

4.54 CHAPTER FOUR

Area that the flux “sees” through the armature tooth path, the magnetic area ofthe armature tooth, in square inches:

Aatm = WatLstke (4.173)

The total magnetic area of armature teeth under a pole is the magnetic area ofthe armature tooth times the pole arc angle divided by the tooth pitch angle. Thisturns out to be the effective number of teeth under a pole times the magnetic areaof one tooth, in square inches:

AT,atm = Aatm (4.174)

where the ratio θpole/θtp must be rounded up to the next-higher integer. This round-ing up assumes that if any part of the tooth tip is over the pole, the flux will use thattooth as part of its magnetic path. Figure 4.53 shows a finite element analysis (FEA)of a universal motor with only the field coils turned on. Note that the flux lines uti-lize the teeth where there is any part of the tooth tip over the pole.

Magnetic area of the armature tooth tip, in square inches:

Aattm = WattLstke (4.175)

The total magnetic area of armature tooth tips under a pole is the effective num-ber of teeth under a pole times the effective number of tooth tips under a pole, insquare inches:

θpoleθtp


FIGURE 4.51 Armature magnetic paths and reactance dimensions.

AT,attm = Aattm (4.176)

Magnetic area of the armature yoke, in square inches:

Aaym = WayLstke (4.177)

The magnetic path length of the armature yoke (Fig. 4.54) depends on the shaftmaterial. For a nonmagnetic shaft, the path length in inches is

Laym = Rshaft + (4.178)Way

2π2

θpoleθtp

4.56 CHAPTER FOUR

FIGURE 4.52 Armature lamination slot.

And for a magnetic shaft

Laym = (4.179)

The magnetic path area of the shaft is accounted for in the armature yoke mag-netic path calculations.

4.2.3 Armature Conductors

In order to determine the resistance of the armature as seen by the brushes, the fol-lowing parameters need to be set:

Sac = number of teeth over which one coil is woundTpca = number of turns of wire per armature coil

CPS = number of coil sides in each armature slot

The length of one turn in the winding must be determined. There are three dif-ferent methods of determining the length of a turn.

Armature End-Turn Length—Method 1. This method uses the geometry of thelamination to estimate the average end-turn length. The accuracy of this method is

Rshaft + Way2

π2


FIGURE 4.53 Finite element analysis showing flux paths through stator teeth.

dependent on the capabilities of the winding machinery. Using geometry and thewidth of the armature yoke for a magnetic shaft, the mean end-turn length is foundby averaging the distances around the lamination at the radii of the top of the slotand the bottom of the slot. The mean end-turn length of one side of one turn, ininches, is

Lmeta = (4.180)

In order to understand the other methods of determining end turn length, theformulas for armature resistance are discussed first.

Sac(Rast)θtp(0.017453) + 2Way sin [(Sac/2) θtp]

2

4.58 CHAPTER FOUR

FIGURE 4.54 Armature lamination slot.

Armature Resistance. The mean turn length is the sum of the mean end-turnlength for each of two ends plus twice the stack length, in feet (because wire resis-tances are given in ohms per foot):

Lmta = (4.181)

Length of copper per armature coil, in feet:

Lac = Lmta × Tpca (4.182)

Resistance of one armature coil, in ohms:

Rac = LacρaFwsa (4.183)

where ρa is the armature wire resistance per foot and Fwsa is the armature windingstretch factor, usually 1.02 to 1.10.

The number of coils to be considered for armature resistance depends on thenumber of commutating coils, so the number of active coils for the armature is

Nca = Nat − 1 (4.184)

where Nat is odd, and

Nca = Nat − 2 (4.185)

where Nat is even.The net armature resistance (Fig. 4.55) is the combination of half the number of

active coils in parallel with half the number of active coils, in ohms:

Rat = (4.186)

Armature End-Turn Length—Method 2. This method assumes that you alreadyhave a motor and know Rat, Tpca, Fwsa, and ρa. From this, the Lmeta can be calculatedand then used for calculations for other motors with the same lamination. The fol-lowing four formulas determine the mean end-turn length for one end of one turn.

Rac = (4.187)

Lac = (4.188)

Lmta = (4.189)

Lmeta = (4.190)

Armature End-Turn Length—Method 3. Measure what looks like the average endturn on an existing motor.

Armature Slot Fill. The maximum armature slot fill depends on the windingequipment. A full slot could be anywhere from 40 to 55 percent. The slot fill is cal-culated by assuming that the wires are lying side by side as if they were square. Using

12Lmta − 2Lstk2

LacTpca

RacρaFwsa

2RatNca/2

(Nca/2)Rac2

2Lstk + 2Lmeta12


the outside diameter of the armature wire as Daw (this depends on insulation thick-ness), the percentage of armature slot fill is

Fas = × 100 (4.191)

Armature Copper Weight. Using pound-feet from the wire tables, the weight ofthe armature copper Wtac, is:

Wtac = LacNat (4.192)

Armature Inertia. The inertia of the shaft Js , oz in s2, is

Js = (0.0184)Rs4 Lshaft (4.193)

Assuming the steel is distributed over a solid cylinder, the inertia of the armaturelamination steel, Jas, oz in s2, is

Jas = − Js (4.194)

where the factor 772 is 2g, in s2.

LstkeLs

(Wtas) R2a

772

CPS

2lbft

(Daw)2TpcaCPS

ANSA

4.60 CHAPTER FOUR

FIGURE 4.55 Schematic of armature current paths.

Assuming that the copper in the end turns effectively distributes the total cop-per weight over a solid cylinder, the inertia of the copper in the armature Jac,oz in s2, is

Jac = − (0.02095)Rs4 Lstk (4.195)

The inertia of the commutator is split into the inertia of the copper bars and theinsulation on the inside (see Fig. 4.69 in Sec. 4.3.2). The commutator inertia Jc ,oz in s2, is

Jccu = (0.02095)Lcomm,copper 4

− 4

(4.196)

Jci = (0.0005086)Lcomm,ins 4

(4.197)

Jc = Jccu + Jci (4.198)

where Lcomm,copper = length of commutator copper measured parallel to shaftLcomm,ins = length of commutator insulation material

Dc = outside diameter of commutatordci = inside diameter of commutator copper

The total armature inertia Ja , oz in s2, is

Ja = Js + Jas + Jac + Jc (4.199)

Armature Balance. ISO Standard 1941 gives guidelines for acceptable balanceand vibration limits for electric machines based on their usage. In general, the lowerthe unbalance, the longer the motor life. Unbalance affects bearing wear by causingthe shaft to pound against the bearing, breaking down the surface and pumping out the lubricant. It affects brush life adversely by causing the brushes to instanta-neously lose contact with the commutator surface, resulting in electrical arcing,wear, and higher levels of EMI. If this occurs near a resonant speed, the conditionbecomes severe and the motor instantaneously ceases operation because the brushlifts completely away from the commutator, causing an open circuit.

Armatures should be aligned in the magnetic field such that there is a net mag-netic force pulling the armature toward the commutator end of the motor. Thisreduces end bounce and improves brush life.

4.2.4 Field (Stator)

The geometry of the stator lamination requires breaking it down into several smallareas in the same manner as used for the armature.

Using Figs. 4.56, 4.57, and 4.58 and the XY coordinate system as shown, this sub-section uses the given dimensions to determine the field slot area.

The following dimensions are needed from the lamination drawing. Some manu-facturers distribute drawings with only a few dimensions on them, but will give thesedetailed drawings if requested.

dci2

dci2

Dc2

(Wtac × 16)R2ast

772


4.62 CHAPTER FOUR

FIGURE 4.56 Example of a lamination drawing.

Lbpt = distance between opposite pole tipsXptc, Yptc = coordinates of pole tip center

Rpt = radius of pole tipXossc, Yossc = coordinates of center of arc of outside of stator slot

Ross = radius of outside of stator slotXssc, Yssc = coordinates of center of radius of stator slot center

Rss = radius of stator slotRosw = radius of outside of stator wallRisw = radius of inside of stator wall, which blends with arc of outside of sta-

tor slot

Depending on which stator dimension is given, the X coordinate for the center ofthe pole tip, in inches, is

Xptc = + Rpt (4.200)

Or, the distance between opposite pole tips, in inches, is

Lbpt = 2(Xptc − Rpt) (4.201)

Lbpt2


FIGURE 4.57 Stator (field) lamination.

The pole arc angle is twice the angle from the center of the lamination to the tipof the pole closest to the other pole. The pole arc angle, in degrees, is

θpole = 2 tan−1 (4.202)

Angle from the center of the stator to the center of the pole tip, in degrees:

θptc = tan−1 (4.203)

Distance from the center of the stator to the pole tip center, in inches:

Lptc = X2ptc +Y2

ptc (4.204)

Using stator triangle TRS1, which has the hypotenuse parallel to Lptc, the heightis

HTRS1 = Yssc (4.205)

YptcXptc

YptcLbpt/2

4.64 CHAPTER FOUR

FIGURE 4.58 Stator (field) slot area.

Angle between the height and the hypotenuse, in degrees:

θTRS1 = 90° − θptc (4.206)

For later use, the angle from the center of the stator to the center of the statorslot, in degrees, is

θssc = tan−1 (4.207)

The triangle TRS3 has the height of the difference of Y coordinates of the statorslot center and the outside of the stator slot center, in inches:

LTRM4 = Rss (4.208)

And the length, in inches:

LTRS3 = Xssc − Xossc (4.209)

Angle between the hypotenuse and the height of TRS3, in degrees:

θTRS3 = tan−1 (4.210)

Length of the hypotenuse of TRS3, in inches:

Lh,TRS3 = (4.211)

The small TRS2 has the leg from the center of the stator slot with its length, ininches:

LTRS2 = Rss (4.212)

Angle between the length of TRS2 and the hypotenuse, in inches:

θTRS2 = θTRS3 − θTRS1 (4.213)

Height of the triangle TRS2, in inches:

HTRS2 = LTRS2 tan θTRS2 (4.214)


Lh,TRS2 = (4.215)

Area of TRS2, in square inches:

ATRS2 = (4.216)

The arc segment SEG5 at the end of the stator slot traverses the angle, in degrees:

θSEG5 = 180° − θTRS2 (4.217)

Area of SEG5, in square inches:

LTRS2HTRS22

LTRS2cos θTRS2

HTRS3cos θTRS3

LTRS3HTRS3

YsscXssc


ASEG5 = θSEG5(0.017453) (4.218)

Length of the arc of SEG5, in inches:

Larc,SEG5 = θSEG5(0.017453)Rss (4.219)

Using θTRS4 = θptc, the height of the very large TRS4, in inches:

HTRS4 = Risw cos θptc (4.220)

Length of TRS4, in inches:

LTRS4 = Risw sin θptc (4.221)

Height of TRS5, in inches:

HTRS5 = HTRS4 − Xossc (4.222)


LTRS5 = LTRS4 − Yossc (4.223)

Angle between the height of TRS5 and the hypotenuse, in inches:

θTRS5 = tan−1 (4.224)

Angle of the stator triangle TRS6, in degrees, is

θTRS6 = θTRS5 − θptc (4.225)

since the angle of TRS4 is the same as the angle of the pole tip center.Length of TRS6, in inches:

LTRS6 = Risw − Rpt − Lptc (4.226)


HTRS6 = LTRS6 tan θTRS6 (4.227)


Lh,TRS6 = (4.228)


ATRS6 = (4.229)

Using triangle TRS7, the angle between the height and the hypotenuse, in degrees:

θTRS7 = 90° − θTRS6 (4.230)

Length of the hypotenuse of TR7, in inches:

Lh,TRS7 = Ross − Lh,TRS6 (4.231)

LTRS6HTRS62

LTRS6cos θTRS6

LTRS5HTRS5

R2ss

2

4.66 CHAPTER FOUR


HTRS7 = Lh,TRS7 cos θTRS7 (4.232)


LTRS7 = Lh,TRS7 sin θTRS7 (4.233)


ATRS7 = (4.234)

TRS8 shares a leg with TRS7, so the length of TRS8, in inches, is

LTRS8 = LTRS7 (4.235)


Lh,TRS8 = Lh,TRS3 − Lh,TRS2 (4.236)

The angle between the height of TRS8 and the hypotenuse is the same as the thirdangle of TRS2, in degrees:

θTRS8 = 90° − θTRS2 (4.237)


HTRS8 = Lh,TRS8 cos θTRS8 (4.238)


ATRS8 = (4.239)

The large SEG6 traverses the same angle of the large triangle which would be com-prised of TRS7 and TRS8. Angle for SEG6, in degrees:

θSEG6 = 180° − θTRS7 − θTRS8 (4.240)

Length of the arc of SEG6, in inches:

Larc,SEG6 = θSEG6(0.017453)Ross (4.241)

Area of SEG6, in square inches:

ASEG6 = θSEG6(0.017453) (4.242)

Gross slot area of the stator slot, in square inches:

AGS, f = ATRS2 + ASEG5 + ATRS6 + (ASEG6 − ATRS7 − ATRS8) (4.243)

Length of the stator slot insulation, in inches,

Lssi = HTRS6 + HTRS7 + HTRS8 + HTRS2 + Larc,SEG5 + Larc,SEG6 (4.244)

R2oss2

LTRS8HTRS82

LTRS7HTRS72


The net slot area available for the field winding on one pole is the GSA minus thearea of the insulation, in square inches:

ANS,f = AGS,f − LssiTssi (4.245)

where Tssi is the thickness of the stator slot insulation in inches.

Constants for Stator Reactance Calculations. As with the armature, the geometryof the stator slot must be known in order to calculate the reactance. Note that thefield slot can be turned to resemble the armature slot. Use Figs. 4.58 and 4.59 todetermine the following values which will be needed later.

4.68 CHAPTER FOUR

FIGURE 4.59 Dimensions for stator reactance.

The values of a and b, as similar to the armature, in inches:

a1 ≈ Ross − LTRS7 (4.246)

a2 ≈ 2Rss (4.247)

b = (4.248)

Since the stator slot has a different shape than the armature slot, the value d3 for thestator is the average of d1 and d2, in:

d2 = b (4.249)

d1 ≈ (4.250)Larc,SEG6

2

HTRS2 + HTRS8 + HTRS62

d3 = (4.251)

Length of the line across the wide opening of the slot e, in:

e = LTRS6 (4.252)

There is no c dimension as in the armature slot dimensions.Length of the interpolar air gap, which is the distance between the armature and

the stator yoke at their closest position, in inches:

Lgip = YST1 − RST1 − Ra (4.253)

where YST1 is the Y coordinate of the center of radius RST1.The points on the pole tip A, B, and C are used as the starting points of three arcs

which are drawn with the edge of the armature as the center. These arcs werederived from flux mapping using curvilinear squares, which will not be discussedhere. C is the point where line e intersects the pole. A is the point where the radiusof the pole tip would be parallel to the armature tooth tip. B is the point on the slotside of the pole tip which is halfway between A and C. The point B′ is the point thatthe average leakage flux would travel to on the stator yoke.

Arc length BB′ is the length of the arc of radius RBB′ (which is the distance fromA′ to B) shortened by the radius RST1.Arc length AC is the arc around the pole tip ofabout 90°.

RBB′ = X2ptc +(Yptc + Rpt −Ra)2 (4.254)

BB′ ≈ RBB′ − RST1 (4.255)

AC ≈ Rpt (4.256)

4.2.5 Magnetic Paths in the Field (Stator)

Using the quarter-stator drawing in Fig. 4.60, the portion of the stator is divided upinto seven magnetic paths. In order to determine the widths and lengths of thesepaths, the dimensions of four triangles need to be determined. All magnetic pathwidths are multiplied by 2 because this analysis is done on a per-pole basis and thedimensions being used are from a drawing of half of one pole. For the initial calcu-lations, use Lstke = 1.0.

Width of the stator yoke, in inches:

WSY = ROSW − RISW (4.257)

magnetic length of the stator yoke, in inches:

LSY = RISW + (90° − θssc)(0.017453) (4.258)

Magnetic area of the stator yoke, in square inches:

ASY = 2WSYLstke (4.259)

WSY2

π2

π2

d1 + d22


Angle of triangle TRM1 between the length and the hypotenuse, in degrees:

θTRM1 = θssc (4.260)

Width of stator path 1, in inches:

WS1 = − Xssc − Rss (4.261)

Length of triangle TRM1, in inches:

LTRM1 = WS1 + Rss (4.262)

Height of TRM1, in inches:

HTRM1 = (4.263)

Magnetic length of stator path 1, in inches:

LS1 = HTRM1 + Rss (4.264)

LTRM1tan θTRM1

Wstator2

4.70 CHAPTER FOUR

FIGURE 4.60 Stator magnetic path lengths and widths.

Magnetic area of stator path 1, in square inches:

AS1 = 2WS1Lstke (4.265)

The length of triangle TRM2 is the same as the width of the stator magnetic path2, in inches:

LTRM2 = WS2 = WS1 + Rss (4.266)

The height of TRM2 is related to the length of TRM3. Length of TRM3, in inches:

LTRM3 = HTRS1 − Rss (4.267)


HTRM2 = (4.268)

Length of the hypotenuse of TRM2, in inches:

Lh,TRM2 = L2TRM2+ H2

TRM2 (4.269)

Angle between the length and the hypotenuse of TRM2, in degrees:

θTRM2 = tan−1 (4.270)

The magnetic path length of stator section 2 is the radial length of the arc which hasthe radius of half the length of the hypotenuse, in inches:

LS2 = θTRM2(0.017453) (4.271)


AS2 = 2WS2Lstke (4.272)


HTRM3 = LTRM2 (4.273)


Lh,TRM3 = L2TRM3+ H2

TRM3 (4.274)

Angle between the hypotenuse and the height of TRM3, in degrees:

θTRM3 = tan−1 (4.275)

Width of stator section 3, in inches:

WS3 = Lh,TRM2 (4.276)

The magnetic path length of stator section 3 is the radial length of the arc which hasthe radius of half the length of the hypotenuse of TRM3, in inches:

LS3 = [θTRM3 − θTRM2](0.017453) (4.277)Lh,TRM3

2

LTRM3HTRM3

Lh,TRM22

HTRM2LTRM2

LTRM32


Magnetic area of stator section 3, in square inches:

AS3 = 2WS3Lstke (4.278)

Using the small TRM4 which is inside the stator slot, the length of TRM4, in inches:

LTRM4 = Rss (4.279)

Angle between the length of TRM4 and the hypotenuse, in degrees,

θTRM4 = 90° − θssc − θTRS1 (4.280)


Lh,TRM4 = (4.281)

Radius at which the width of stator path 4 is centered, in inches:

RS4 = X2ssc +Y2

ssc − Lh,TRM4 (4.282)

The width of magnetic stator path 4 is the length of the arc that is swung from thehypotenuse of TRM4 to the horizontal axis, in inches:

WS4 = RS4θssc(0.017453) (4.283)

The length of magnetic stator path 4 is the length of the arc with radius Rss that isswung from the vertical to the leg of TRM4, in inches:

LS4 = RssθTRS1(0.017453) (4.284)


AS4 = 2WS4Lstke (4.285)

In order to find the width of stator path 5, the length of the arc from the far sideof the pole tip to the horizontal must be determined, in inches:

Larc,S5 = [Lptc + Rpt]θptc(0.017453) (4.286)

The width of stator path 5 is the average of the arc length in Eq. (4.286) and thewidth of stator path 4, in inches:

WS5 = (4.287)

Magnetic length of stator path 5, in inches:

LS5 = RS4 − (Lptc + Rpt) (4.288)


AS5 = 2WS5Lstke (4.289)

Width of the magnetic path consisting of the actual pole, in inches:

WPM = Lptc (0.017453) (4.290)θpole

2

WS4 + Larc,S52

LTRM4cos θTRM4

4.72 CHAPTER FOUR

Magnetic length of the pole itself, in inches:

LPM = (Lptc + Rpt) − Rpole (4.291)

Magnetic area of the pole, in square inches:

APM = 2WPMLstke (4.292)

4.2.6 Field Conductors

In order to determine the resistance and slot fill of the field, the following parame-ters need to be determined by the designer:

Tpcf = turns of wire per field coilDfw = diameter of field wire

The length of one turn of the field winding must be determined. As with thearmature winding, there are three different methods of determining end-turn lengthfor the field.

Field End-Turn Length—Method 1. The mean end-turn length for the field is theaverage of the length of the shortest turn which is wound tightly across the pole andthe length of the turn on the outer edges of each slot.The shortest turn is the straightdistance across the pole, and the longest turn is the length of the arc swung across theouter edges of the slots. The field winding mean end-turn length, in inches, is

Lmetf = (4.293)

where π is 180° converted to radians. The end-turn length will vary slightly with slotfill, winding equipment, and the overhang of the field slot insulation relative to thestack.

In order to use method 2 of determining end-turn length, which is similar to thearmature method, you first need to know how the field resistance is determined.

Resistance of Field Windings. The field resistance is the total resistance of twofield coils in series.The mean turn length of one field coil, in feet (because wire resis-tances are given in ohms per foot), is

Lmtf = (4.294)

Length of copper per field coil, in feet:

Lfc = Lmtf × Tpcf (4.295)

Resistance of one field coil, in ohms:

Rfc = LfcρfFwsf (4.296)

where ρf is the field wire resistance per foot, found from the wire tables for theknown wire gauge, and Fwsf is the winding stretch factor, usually 1.02 to 1.10.

2Lstk + 2Lmetf12

2(HTRS1 − Rss) + πLTRS42


The net field resistance is the series combination of two field coils, in ohms:

Rf = 2Rfc (4.297)

Field End-Turn Length—Method 2. As with the armature, this method assumesthat you already have a motor and know Rf,Tpcf, Fwsf, and ρf (from wire tables). Fromthis, Lmetf can be calculated and then used for calculations for other motors with thesame lamination.

The following four formulas give the mean end-turn length for one end of onefield turn, in inches.

Rfc = (4.298)

Lfc = (4.299)

Lmtf = (4.300)

Lmetf = (4.301)

Field End-Turn Length—Method 3. Measure what looks like the average end turnon existing motor with the same lamination.

Field (Stator) Slot Fill. The maximum field slot fill depends on the windingequipment.A full slot could be anywhere from 80 to 100 percent.These percentagesare greater than for the armature because there is only one coil in the field slot andit is possible to use the slot insulation to extend the area of the slot. The field slot fillis calculated by assuming that the wires are lying side by side as if they were square.Using the outside diameter of the field wire as Dfw (found in the wire tables takinginto account insulation thickness), the percentage of winding slot fill is

Ffs = × 100 (4.302)

Field Copper Weight. Using pound-feet from the wire tables, the weight of thefield copper Wtfc, lb, is

Wtfc = 2 (4.303)

Turns Ratio. The field-to-armature turns ratio is the ratio of the total number offield turns to the total number of series-connected turns in one path of the armature.

ψfa = (4.304)

where P is the number of poles and Nca is the number of active coils in the armature.This ratio represents the mmf produced by the field in relation to the mmf pro-

duced by the armature. If the ratio is too high, the field produces strong mmf, whichinduces a higher voltage in the commutating coil. This high voltage causes arcingproblems during commutation. If the ratio is too low, the armature mmf weakens thepole too much and a loss of torque results.A good range for this ratio is 1.00 to 1.30.

PTpcf(Nca/4) 2Tpca

Lfc(lb/ft)

Fwsf

D2fwTpcf

ANS,f

12Lmtf − 2Lstk2

LfcTpcf

RfcρfFwsf

Rf2

4.74 CHAPTER FOUR

4.2.7 Magnet Circuit

This subsection derives a curve for the relationship of flux across the air gap versusthe related mmf drops.This curve tells you how many ampere-turns are necessary toproduce a specified air gap flux. Special consideration is taken to account for satu-ration and for the air gap.

The air gap is the area between the pole on the stator and the tooth tips on thearmature. The air gap is not uniform because of the spaces between the tooth tipsand because the pole is not perfectly round. Figure 4.61 shows that the air gap islarger at the pole tips than at the center.

First, take into account the longer airgap at the pole tips. On the manufac-turer’s drawing, the pole arc changes toa straight line at a specified angle θref,which is usually about 45°. When mea-suring from the center of the armature,at 90° − θref, the air gap starts to getlonger and the length continues toincrease out to the pole tip. (Longerrefers to the magnetic length.) The aver-age length of the air gap over this part ofthe arc is, in inches:

Lgap,tip = − Ra

(4.305)

The next formula gives an effectivemechanical air gap length by using aweighted average over half the pole arc.

The effective mechanical air gap length, in inches, is

Lgap,eff,mech = (4.306)

The magnetic length must account for the spaces between the armature teeth.Carter’s coefficient KC accounts for the flux that fringes from the pole to the sides ofthe armature tooth tips (Fig. 4.62). Carter’s coefficient can be determined by firstusing an available graph of the width of armature slot along the circumferencedivided by the effective mechanical air gap length versus σC, where σC is an arbitraryvariable name. Using σC:

KC = (4.307)

Another method for determining Carter’s coefficient (with no graph available) isby the following two formulas:

λ = tanr−1 − ln 1 + (4.308)

KC = (4.309)tp

tp − λWasc

W 2asc

4L2gap,eff,mech

Lgap,eff,mechWasc

Wasc2Lgap,eff,mech

2π

Watt + WascWatt + Wasc(1 − σC)

[(θpole/2) − (90° − θref)]Lgap,tip + (90° − θref)(Rpole − Ra)

θpole/2

Rpole + (Lptc − Rpt)2


FIGURE 4.61 Nonuniform air gap.

Effective magnetic air gap length, in inches:

Lgap = Lgap,eff,mechKC (4.310)

The magnetic width of the air gap is the length of the arc along the center of themagnetic air gap, in inches:

Wgap = Rpole − θpole(0.017453) (4.311)

Magnetic area of the air gap, in square inches:

Agap = WgapLstke (4.312)

Figure 4.63 shows the analogous electrical circuit for the magnetic circuit of a uni-versal motor. Each resistor is associated with an mmf drop, similar to an emf (orvoltage) drop in an electrical circuit. The resistor in the middle of the armature slotrepresents an mmf drop that occurs only at high saturation. The Trickey factor Ktr

accounts for this mmf drop and is discussed later.

Rpole − Ra2

4.76 CHAPTER FOUR

FIGURE 4.62 Fringing of flux at air gap.

Plot of Flux per Pole versus Ampere-Turns Excitation. Flux is required for anelectric motor to produce torque. In order to produce flux, mmf must be supplied.Each combination of armature and stator laminations will produce a differentamount of flux for a given supplied mmf. A curve of air gap flux (in kilolines φgap)versus mmf in ampere-turns of excitation can be determined by determining thesum of all mmf drops across the magnetic circuit (lamination set) for each of severalvalues of flux. In order to determine the mmf drop across a specific area, the fluxdensity must be determined. From the BH curve for the material being used, themmf drop in ampere-turns can be determined for the specific area.

For a given φgap, klines, the flux densities for the various paths are as follows: Bgap,kline/in2, in the air gap:

Bgap = (4.313)

Batt, kline/in2, in the armature tooth tips:

Batt = (4.314)φgapAT,attm

φgapAgap


FIGURE 4.63 Schematic of magnetic circuits.

Bat, kline/in2, in the armature teeth:

Bat = (4.315)

Bay, kline/in2, in the armature yoke:

Bay = (4.316)

For the following calculations, assume that the flux is in the return path in the sta-tor.

BSY, kline/in2, in the stator yoke:

BSY = (4.317)

BS1, kline/in2, in stator section 1:

BS1 = (4.318)


BS2 = (4.319)


BS3 = (4.320)


BS4 = (4.321)


BS5 = (4.322)

Bpole, kline/in2, in the pole:

Bpole = (4.323)

In order to calculate the total mmf drops in the magnetic circuit, you need to readthe magnetic field intensity H, (A turn)/in, from the BH curve for each section.Themmf drop for the section is the intensity times the length of the section.

Field intensity Hgap, (A turn)/in, in the air gap:

Hgap = (4.324)Bgap3.19

φgapAPM

φgapAS5

φgapAS4

φgapAS3

φgapAS2

φgapAS1

φgapASY

φgapAaym

φgapAT,atm

4.78 CHAPTER FOUR

MMF drop across the air gap, in ampere-turns:

gap = HgapLgap (4.325)

For the following equations, the value of H is read from the graph for each dif-ferent flux density B.

att = HattLattm (4.326)

Special consideration must be taken to account for the fact that when the fluxdensity reaches saturation in the armature teeth, the flux will set up on the sides ofthe teeth.At about 110 kline/in2, the armature teeth saturate, and any additional fluxwill exist along the sides of the teeth as well as in the teeth themselves (Fig. 4.64).The mmf drop due to the flux on the sides of the tooth can be considered by addingit to the mmf drop in the tooth itself. Also, as Bat becomes greater than 110,000line/in2, the initial calculation of Bat becomes Bat,apparent.

Bat > 110,000 line/in2 = Bat,apparent = Bata (4.327)


FIGURE 4.64 Flux in air beside teeth (Trickey factor).

The Trickey factor Ktr must be determined:

Ktr = − 1 (4.328)

The Trickey factor is the number of tooth widths in the air beside the teeth that theflux occupies at saturation. Note that it accounts for the air between the laminationsby dividing by the stacking factor.

A little extra effort is required to determine the field intensity H for a saturatedtooth. Using the BH curve for the material and the Bat,apparent which is over 110kline/in2, determine the apparent field intensity Htemp in the armature teeth. A newapparent flux density is Batal.

Bata = µHtemp + 3.19KtrHtemp (4.329)

where µ = Bata/Htemp

Therefore Bata = µHtemp

Htemp is altered until Batal = Bata. Bat then equals the new µHtemp.

Bat can also be derived graphically. See P. H.Trickey, vol. I-3-109 (1971). Using thisnew flux density, the new value for Hat can be determined from the curve for themmf drop in ampere-turns across the armature teeth for an apparent flux density>110 kline/in2.

The last mmf drop to calculate for the armature is that of the armature yoke ay,A turn:

ay = HayLaym (4.330)

Using the same method as before for the values of the mmf drops across the sec-tions of the stator:

sy = HSYLSY (4.331)

S1 = HS1LS1 (4.332)

S2 = HS2LS2 (4.333)

S3 = HS3LS3 (4.334)

S4 = HS4LS4 (4.335)

S5 = HS5LS5 (4.336)

pole = HpoleLpole (4.337)

Sum of the mmf drops, in ampere-turns:

total = gap + att + at + ay + SY + S1 + S2 + S3 + S4 + S5 + pole (4.338)

The value total represents the field ampere turns per pole necessary to providethe specified air gap flux φgap. This means that the product of the number of fieldturns Tpcf and the field current Iline must equal total. A curve of air gap flux versusampere-turns of excitation can be plotted by varying φgap. Remember that each curveis specific to the lamination set and that it does not account for any loss in mmf dueto armature reaction.

tpWatFstk

4.80 CHAPTER FOUR

Reactances. Reactances are largely a function of lamination geometry. In order tocalculate reactances, the slot dimensions and lengths must be known. The followingsubsections examine methods for calculating field and armature reactances.

Field Leakage Reactance. The first formula is from Puchstein (1961) for fieldleakage reactance, in ohms.

Xf,total = 2πfφcT2pcfLmtf × 10−8 (4.339)

where Lmtf is the mean turn length in inches of the field and φc is the number of linesper ampere-conductor per inch of mean turn length. A good value to use for φc is 6.

The next formula is from Trickey for field leakage reactance.

Xf/pole = (2πf(2Tpcf)2 × 10−8) (Ksf1 + Ksf2) + (4.340)

where tpcf = field turns per poleKsf1 = slot constant for part containing coilKsf2 = slot constant for pole tip fluxSfc = span of field coil, in

You also need to know the following in order to calculate armature reactances:

ψp = θpole/pole pitch = θpole/180°Nas = number of armature slots = Nat

Zs = armature conductors in a slot = Tpca(CPS)Za = total armature conductors = Tpca(CPS)Nat

f = line-current frequency, Hz

The span of the field coil is the arc length from the center of the field winding in theslot to the same position in the other slot, in inches.

Sfc = (Y2ssc +X2

ssc) 2 (4.341)

The slot constants Ksf1 and Ksf2 are shown here.

Ksf1 = + F (4.342)

Ksf2 = 1.5 (4.343)

The values for a1, a2, b, d1, d2, d3, e,AC, and BB that are used to calculate F are shownin Fig. 4.65.

F = 0.28 − 0.14 + + 0.08 (4.344)

Trickey’s total field reactance must be multiplied by the number of poles.

Xf,total = 2Xf/pole (4.345)

b/a22.87(a1/a2) + 0.008

a1a2

ACBB′

d3e

θssc + θptc2

π180

SfcP

3.19Lstke2P


Using either of the preceding two methods for field leakage reactance, find thetotal reactance.

Xtotal = Xar,T + Xf,total (4.346)

Note: Trickey’s method does not consider end-turn leakage.

Armature Reactances. In order to calculate reactances, the slot constants of thearmature slots must be determined. Figure 4.66 shows the dimensions which areneeded to calculate slot constants.You also must use the actual value for Lstke.Thesedimensions were calculated previously.

Armature Leakage Reactances Under a Pole, Ω

Xapf/path = (10.5) 2

DaLstkeψ3p × 10−8 (4.347)

For inductors in parallel:

Xapf,T = (4.348)

Armature Slot Leakage Reactance, Ω

Xas = Ksa (4.349)

where Ksa = slot constant for round-bottom armature slots. Use the following for-mulas.

20Zs2LstkefNasψpP2 × 108

Xapf/path2

Za2P

2fPLgap

4.82 CHAPTER FOUR

FIGURE 4.65 Dimensions for stator reactances.

F = 0.28 − 0.14 + + 0.08 (4.344)

Ksa = F + + (4.350)

where a1, a2, b, c, d, and e are dimensions of the armature slot found previously andshown in Fig. 4.66.

Armature End Leakage Reactance, Ω

Xae = (4.351)

Armature Interpolar Leakage Reactance, Ω

Xgip/path = 10.5 2

(Da + Lgip) (1 − ψ3p) × 10−8 (4.352)

Xgip,T = (4.353)

where Lgip = length of gap between armature OD and ID of stator yoke.

Xgip/path2

Za2P

fLstkeLgip

3.19f(2Lmeta)Z2aNas

P2108

2ce + a1

de

b/a22.87(a1/a2) + 0.08

a1a2


FIGURE 4.66 Armature magnetic paths and reactance dimensions.

Armature Tooth Tip Leakage Reactance, Ω

Xatt = log10 1 + (4.354)

Total Armature Reactance, Ω

Xar,T = Xapf,T + Xas + Xae + Xgip,T + Xatt (4.355)

4.3 COMMUTATION*

4.3.1 A Practical Approach to Commutation and Brush Selection

There are many factors that affect commutation.Among these are inductance, brushfriction, surface speed of the commutator, brush material, commutator finish, brushpressure, contact film, induced emf, current density, armature balance, and concen-tricity of the commutator to the shaft.

Perfect linear commutation, as shown in Fig. 4.67, would occur when the currentin the coil under commutation goes from a peak +I through 0 to a peak −I as thebrush position changes from full contact on one bar to complete shorting of one coilto full contact on the next bar. This condition would provide a relatively constantcurrent density in the brush.

π(tp − Wast)2Wast

14.7(Zs)2Lstke fNas(1 − ψp)P2108

4.84 CHAPTER FOUR

FIGURE 4.67 Overcommutation, undercommutation,and linear commutation.

Undercommutation (Fig. 4.67) occurs when the current stays at +IPK for most ofbrush travel and is abruptly changed to −IPK as the brush leaves the segment.

This effect is a result of the winding inductance trying to maintain the current inthe coil during commutation. The current density in the brush varies from low tovery high as the brush moves from segment 1 to segment 2.

Overcommutation (Fig. 4.67) occurs when the emf induced in the coil duringcommutation causes the current reversal to take place before the brush contacts theadjacent segment. In this case the current density in the brush is very high initiallyand drops to very low.

*Section contributed by Alan W. Yeadon and William H. Yeadon, Yeadon Engineering Services, PC.

In the cases of both undercommutation and overcommutation, the high brushcurrent densities cause excessive heating and wear.

Considering the list of things that affect commutation, it is obvious that perfectcommutation is impossible. However, there are some things to keep in mind whendesigning a motor which will tend to minimize commutation problems.

Brush pressure is important. Figure 4.68 shows relative brush and commutatorwear verses brush pressure. At low pressures, the wear is mostly electrical in naturebecause the poor contact results in high-resistance spots with localized heating andarcing. At high pressures, the wear is mostly mechanical and is due to the parts in effect grinding themselves away because of friction losses. For fractional-horsepower motors using commonly available brush and commutator materials, the“ideal” range is from 3 to 9 lb/in2. A good approach to this is to select 6 lb/in2 as anominal value and design the brush rigging and spring system to maintain as close to6 lb/in2 over the usable brush length as possible. One can use a constant force springor, if this is not practical, try to select a spring that starts out with a pressure of 7 to 8lb/in2 and has a pressure of 4 to 5 lb/in2 at the end of the usable brush length. In appli-cations where the brushes and commutators are submerged in a liquid, the pressureneeds to be increased to overcome hydroplaning.A good range is 10 to 16 lb/in2, with12 to 14 lb/in2 being preferred over the usable brush length.


FIGURE 4.68 Brush pressure versus wear.

The next thing to do is to ensure that the machine is commutating in the neutralzone. The neutral zone Wn is defined as the mechanical arc distance between thepole tips.

Wn = 1 − (4.356)

where θpole = pole angle, degreesDa = outside diameter of amatureP = number of poles

The magnetic neutral zone is located at some angle shifted from the mechanicalneutral position. The angle of shift is a function of the armature mmf. This angle canbe estimated from experimental data on similar motors.Try 15° to 20° to start.A bet-ter way is to make a finite element analysis and estimate the angle from the flux plotwith the motor at full load current. Next, consult a manufacturer’s catalog and select

πDaP

θpoleP360

a standard commutator that is about half the diameter at the armature laminationbeing utilized. Select a brush material that the manufacturer recommends for thetype of application in which the commutation system will be used. Design the brushthickness so that it will cover about one bar pitch of the commutator. Set the brushwidth to cover as much of the commutator bar as possible, taking into considerationarmature end play and the required mechanical and electrical clearances.

Calculate the commutation zone Wc. It is defined as the circumferential distancethe coil under commutation travels before current reversal is complete.

Wc = + + − Sac − (4.357)

where tbr = brush thicknesstcin = insulation thickness between barsDa = outside diameter of armatureP = number of polesa = number of parallel armature paths

Cb = number of commutator barsSac = number of teeth spanned by an armature coilNat = number of armature teethDc = finished outside diameter of commutator

Then calculate the commutation–to–neutral zone ratio ψcn.

ψcn = (4.358)

The result should be less than 1 for good commutation. According to Puchstein(1961), the optimum value is 0.667, but this seems practical only in larger machines.The precise calculation for this ratio is shown in Sec. 4.3.2.

Calculate the full load current of the motor.Then calculate the current density inthe brush.

γbr = (4.359)

This value should be less than the value recommended by the brush supplier for thetype of brush selected.

If the current density is not acceptable, select a brush material with a higher rating,or increase the brush thickness until it is acceptable. If the brush thickness is increased,the commutation–to–neutral zone ratio must be recalculated. If it is now greater than1, increase the commutator diameter until it is less than 1. Note that changing thebrush area will change the brush pressure. Because of this, it will be necessary torecheck the spring to ensure that the pressure remains in the preferred range.

Other tips on improving commutation include reducing the motor speed to assmall a value as possible and keeping the coil inductances low. When designing thebrush rigging, keep 0.002 to 0.003 in total side-to-side clearance and top-to-bottomclearance between the brush and its guide. Less clearance may cause binding prob-lems and may not allow the brush to remain in contact with the commutator. Try toextend the brush guide to within 0.020 in of the commutator surface if practical.Thiswill minimize brush movement, which will improve contact and reduce arcing. Bal-ance the armature to its ISO 1941 requirement. Keep the shaft straight and the bear-ings lined up and concentric to the field core.

IlinetbrWbr

WcWn

πDaCb

aP

CbNat

CbP

CbNat

(tbr − tcin)CbπDc

4.86 CHAPTER FOUR

Brush Dust and Slot Packing. As the commutator and brushes move relative toeach other, they wear, and fine particles of dust are released into the motor. Depend-ing on the brush materials used, this dust may contain copper and silver, but it ismostly carbon. In any event, the material is conductive.

As the particles are dislodged from the brushes and the commutator, they aremoved to the edge of the commutator slots and forced down into them. If the surfacespeed of the commutator is sufficient, the tangential forces will throw the particlesfrom the slots out into the surrounding area. This is the desired effect.

If the surface speed is too slow, or a contaminant, such as oil, is present, the parti-cles will be forced into the slots until the slots are packed with the material. Since thematerial is conductive, the effect of this packing is to short the bars together, creat-ing one or more shorted coils. This adversely affects motor performance and life. Toavoid this situation, keep oils away from the commutator by using slingers and anti-wetting agents. Run the motor at sufficient speed to throw the particles from theslots.The actual armature minimum speed required depends on the mass of the par-ticles and the commutator diameter, but an old rule of thumb is to keep the speedabove 1500 rpm.

Excessive dust buildup can cause other problems as well, such as shorting thebrush rigging to the end frames. This can be minimized by designing the brush rig-ging to trap the dust or by isolating the surrounding area with insulating materials.

4.3.2 Commutation System Design

A lot of effort goes into designing a motor that has good commutation. Good com-mutation saves on brush wear and reduces sparking. In order to calculate commuta-tion parameters, we need to know the following (refer to Figs. 4.69 and 4.70):


FIGURE 4.69 Commutator construction.

Cb = number of bars on the commutatorDc = outside diameter of commutatortbr = brush thickness (thickness which is tangential to the commutator)

Tcin = insulation thickness between commutator barsa = number of current paths in armatureP = number of poles

Nat = number of armature teethSac = number of teeth spanned by a coilDc = commutator diameter

Width of the commutation zone Wc and the neutral zone Wn:

Wc = + + − Sac − (4.360)

Wn = 1 − (4.361)

The figure of merit for the commutation–to–neutral zone ratio should be less than1.0 for good commutation.

ψcn = (4.362)

Flat commutators are commonly used where long brushes are required for longmotor life. Flat commutators allow the length of the brush to extend the motorlength instead of the motor diameter. Figure 4.71 shows a flat commutator and thedimensions used to calculate commutation parameters. The following dimensions(which can be substituted into the commutation parameters just given) need to bedetermined for the commutator and corresponding brush:

WcWn

πDaP

θpoleP360

πDaCb

aP

CbNat

CbP

CbNat

(tbr − Tcin)CbπDc

4.88 CHAPTER FOUR

FIGURE 4.70 Round commutator dimensions.

αbr,flat = angle which the brush encompassesDbr,path = outside diameter of the path the brush travels on the commutatordbr,path = inside diameter of the path the brush travels on the commutator

tcin = thickness of the insulation between commutator bars

The following formulas calculate the values to be used in the commutation zoneformula.

Geometric mean diameter of the brush path, to replace the diameter of the com-mutator, in inches:

DC = 0.707 2

+ 2 (4.363)

Radius of the brush path, in inches:

Rbr,path = (4.364)DC2

dbr,path2

Dbr,path2


FIGURE 4.71 Flat commutator dimensions.

The thickness of the brush for a flat commutator is the arc distance across thebrush at the radius of the brush path, in inches:

tbr = (αbr,flat)(Rbr,path) (4.365)

Commutation is the process of switching the direction of current in a coil. If thedirection of the coil were not switched, the motor would not turn. Figure 4.72 showsa current-carrying coil rotating in a magnetic field. The reaction of the flux from themagnets (or from the field poles in the universal motor) and the flux from the arma-ture coil sides causes the coil to rotate until a steady-state position is reached. Thedirection of current in the coil needs to be reversed in order to keep the coil turning.

π180

4.90 CHAPTER FOUR

FIGURE 4.72 Coil rotating in magnetic field.

Figure 4.73 shows the magnetic field resulting from the stator winding beingenergized.

Figure 4.74 shows a rotating armature with two coils. Coil 1 has end points oncommutator bars C1 and C1 which are about to be commutated by Brush 1.The cur-rent in the coils is in a steady state at this point. Figure 4.75 shows the brushes imme-diately before they will short out their respective coils. Figure 4.76 shows the coilsbeing short-circuited by the brushes. The current direction in the coils is in the pro-

cess of being switched. Note that the coil sides (which are into the paper) are nowmoving parallel to the magnetic flux lines shown in Fig. 4.73. This is beneficialbecause a short-circuited coil moving across a magnetic field will cause voltages tobe induced across the ends of the coil (which are the commutator bars). Figure 4.77shows the position of the brushes after the coil has been commutated.The current inthis coil is now in the opposite direction in a steady state.

Example. A universal motor with a C-shaped lamination is produced to run at240 V ac. The motor currently has the following attributes:

θpole = 100°Da = 1.238tbr = 0.250

Sac = 5 teethtcin = 0.015Dc = 0.89Cb = 24


FIGURE 4.73 Magnetic field with only stator energized.

4.92 CHAPTER FOUR

FIGURE 4.74 Armature coils with steady-state current.

FIGURE 4.75 Coils just about to be commutated.


FIGURE 4.77 Coils with new current direction in steady state.

FIGURE 4.76 Coils being commutated (shorted by brush).

For this motor, the widths of the commutation zone and the neutral zone are as fol-lows:

Wc = + + − (5) − (4.366)

Wn = 1 − (4.367)

Solving shows Wn = 0.864, Wc = 0.813, and ψcn = Wc/Wn = 0.94. These are acceptablevalues.

Now, change the motor to run at 115 V ac. To do this, double the wire sizes andhalve the turns to keep the slots full. Also, change the commutator to 12 bars with adiameter of 0.875. The neutral zone width remains the same, but the commutationzone becomes

Wc = + + − (5) − (4.368)

Wc = 0.656

ψcn = Wc

Wn = 0.759

Even better commutation!Reactance voltage of commutation, in volts:

Vrc = (4.369)

Where S = motor speed, rpmPa = number of armature paths

tcom = thickness of one commutator barTpca = turns per armature coil being commutatedKst = 1.0 if Sac is full pitchKst = 0.5 if Sac is less than a full pitchIL = line current

In ac operation there is a transformer voltage in the armature, in volts per coil:

Eta = 4.44φmaxfTpca × 10−8 (4.370)

where φmax is the air gap flux in peak lines at the given current.The transformer volt-age does not exist in dc operation.

Short-circuit voltage:

ES = 2Bv LstkTpca × 10−8 (4.371)

where

B ≈ (4.372)3.19ZaIap

(Lgip/2) 2P

1.33CbSPIL

a T2

pca (KstLstk + 0.2Lmeta) ∗ 10−8

tbr/tcom

π1.238

1222

1212

122

1212

(0.250 − 0.015)12

π0.875

π1.238

2100(2)

360

π1.238

2422

2412

242

2412

(0.250 − 0.015)24

π0.890

4.94 CHAPTER FOUR

Iap = (4.373)

and

v = S (4.374)

where v = velocity of armature surface, in/sPermissible commutation voltage:

Epc = E2ta + (ES + 2.5Vrc)2 (4.375)

Epc should be less than or equal to 5 to 7 V. This is difficult to achieve in small high-speed motors.

4.3.3 Flashover and Ring Fire

Under steady-state conditions, the field and armature fluxes and currents are rela-tively constant. Commutation voltages can be calculated using the previouslydescribed formulas, and a suitable commutation system can be devised. If, however,the motor is expected to see rapid load changes or voltage reversal, the steady-statecommutation system may no longer be suitable. Sudden load changes that cause themotor to rapidly accelerate or decelerate will cause transformer voltages to beinduced in the windings.

There are two components to these voltages. There is a voltage etae caused bychanging field distortion armature flux. It is largest near the pole tips and falls tozero under the pole. Since the commutated coil is nearest the pole tip, this voltageshows up at the brushes. If it is high enough, the voltage exceeds the insulationdielectric value between bars nearest the brush.This will result in arcing or flashoverbetween bars. This voltage has the value

etae = (4.376)

where Tpc = turns per coilP = poles

Pa = armature pathsdφa/dt = time rate of change of armature flux at the pole tips

The second is a voltage etap induced in the armature coils under a main pole. Thisresults from a change in armature mmf caused by rapidly changing armature currentas the load changes. Its value is

etap = (4.377)

where dφc /dt is the rate of change in flux in a coil under a pole with respect to time.These voltages are superimposed and seen at all bars. If these voltages are high

enough, the dielectric between all bars may break down, resulting in arcing orflashover between all bars. This phenomenon is sometimes referred to as ring fire. Itshould be pointed out here that the magnitude and polarity of these voltages isdependent on whether the load is increasing or decreasing and at what rate.

dφcdt

PPa

Tpc108

dφadt

PPa

Tpc108

πDa60

2IPa


Flashover and ring fire may be observed on starting some motors yet have verylittle effect on their life because it is intermittent and not seen under normal opera-tion.

It is obvious from the formulas that the best way to control flashover on motorswith variable loads is to limit Tpc. This usually means more armature slots and largercommutators with more bars. For a more detailed explanation of flashover, seePuchstein (1961) and Gray (1926).

4.96 CHAPTER FOUR

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