Handbook of Small Electric Motors MAZ Part 6 doc

Total or normal flux density B, T 1.1.2 The CGS System of Units The rationalized cgs system of units uses the magnetic units of gauss G and oersted Oe for flux density B and magnetizing

CHAPTER 1 BASIC MAGNETICS

Chapter Contributors

Mark A Juds Earl F Richards William H Yeadon

1.1

Electric motors convert electrical energy into mechanical energy by utilizing theproperties of electromagnetic energy conversion The different types of motorsoperate in different ways and have different methods of calculating the perfor-mance, but all utilize some arrangement of magnetic fields Understanding the con-cepts of electromagnetics and the systems of units that are employed is essential tounderstanding electric motor operation The first part of this chapter covers the con-cepts and units and shows how forces are developed Nonlinearity of magnetic mate-rials and uses of magnetic materials are explained Energy and coenergy conceptsare used to explain forces, motion, and activation Finally, this chapter explains howmotor torque is developed using these concepts

Although the rationalized mks system of units [Système International (SI) units] isused in this discussion, there are also at least four other systems of units—cgs, esu,emu, and gaussian] that are used when describing electromagnetic phenomena.These systems are briefly described as follows

MKS. Meter, kilogram, second

CGS. Centimeter, gram, second

* Sections 1.1 to 1.12 contributed by Mark A Juds, Eaton Corporation.

ESU CGS with e0=1, based on Coulomb’s law for electric poles.

EMU. CGS with µ0=1, based on Coulomb’s law for magnetic poles

Gaussian. CGS with electric quantities in esu and magnetic quantities in emu.The factor 4πappears in Maxwell’s equation

Rationalized mks. µ0=4π ×10−7H/m, based on the force between two wires

Rationalized cgs. µ0=1, based on Coulomb’s law for magnetic poles The factor

4πappears in Coulomb’s law

The rationalized mks and the rationalized cgs systems of units are the mostwidely used These systems are defined in more detail in the following subsections

1.1.1 The MKS System of Units

The rationalized mks system of units (also called SI units) uses the magnetic units

tesla (T) and amps per meter (A/m), for flux density B and magnetizing force H, respectively In this system, the flux density B is defined first (before H is defined),

and is based on the force between two current-carrying wires A distinction is made

between B and H in empty (free) space, and the treatment of magnetization is based

on amperian currents (equivalent surface currents)

Total or normal flux density B, T

1.1.2 The CGS System of Units

The rationalized cgs system of units uses the magnetic units of gauss (G) and oersted

(Oe) for flux density B and magnetizing force H, respectively In this system, the magnetizing force, or field intensity, H is defined first (before B is defined), and is

based on the force between two magnetic poles No distinction is made between B and H in empty (free) space, and the treatment of magnetization is based on mag- netic poles The unit emu is equivalent to an erg per oersted and is understood to

mean the electromagnetic unit of magnetic moment

Total or normal flux density B, G

Electron mass m e=9.109390 ×10−31kg

Plank’s constant h=6.6262 ×10−34J⋅s

Velocity of light c=2.997925 ×108m/s

Pi π =3.1415926536Acceleration of gravity g=9.807 m/s2

Time t 1.0 day =24 h

1.0 min =60 s

Velocity v 1.0 m/s =3.6 km/h

By convention, the magnetic field lines emerge from the north magnetic pole andenter through the south magnetic pole Two permanent magnets will attract or repeleach other in an effort to minimize the length of the magnetic field lines, which iswhy like poles repel and opposite poles attract Therefore, since the north magneticpole of a bar magnet points to the earth’s geographic north, the earth’s geographicnorth pole has a south magnetic polarity.

Hans Oersted discovered in 1820 that a compass needle is deflected by an tric current, and for the first time showed that electricity and magnetism are related.The magnetic field around a current-carrying wire can be examined by placing manytiny compass needles on a plane perpendicular to the axis of the wire This showsthat the magnetic field lines around a wire can be envisioned as circles centered onthe wire and lying in planes perpendicular to the wire

elec-The direction of the magnetic field around a wire can be determined by using the

right-hand rule, as follows (see Fig 1.1) The thumb of your right hand is pointed in

the direction of the current, where current is defined as the flow of positive chargefrom +to − The fingers of your right hand curl around the wire to point in the direc-tion of the magnetic field If the current is defined as the flow of negative chargefrom −to +, then the left-hand rule must be used

To summarize:

1 The north magnetic pole of a bar magnet will point to the earth’s geographic north.

2 Magnetic field lines emerge from the north magnetic pole of a permanent magnet.

3 Magnetic field lines encircle a current-carrying wire.

4 Magnetic field lines do not stop or end, but form closed curves or loops that

always encircle a current-carrying wire and/or pass through a permanent magnet

5 The right-hand rule is used with current flowing from positive to negative and

with the magnetic field lines emerging from the north magnetic pole

Figure 1.2 shows a magnetic circuit based on an electrical analogy In general, a

coil with N turns of wire and I amperes (amps) provides the magnetomotive force

NI that pushes the magnetic flux φthrough a region (or a material) with a cross

sec-tional area a and a magnetic flux path length l In the electrical analogy, a voltage V provides the electromotive force that pushes an electrical current I through a region.

The amount of magnetomotive force required per unit of magnetic flux is calledreluctance  The amount of voltage required per amp is called resistance R

NI

φ

FIGURE 1.1 Direction of flux (Courtesy of Eaton Corporation.)

FIGURE 1.2 Magnetic circuit with electrical analogy.

(Courtesy of Eaton Corporation.)

Visualization of Flux Linkage l Figure 1.3 shows 10 magnetic flux lines passing

through 4 turns of wire in a coil Each turn of the coil is linked to the 10 magnetic fluxlines, like links in a chain Therefore, the total flux linkage λis obtained by multiply-

ing the turns N by the magnetic flux φ,

as defined in Eq (1.19) In this case, themagnetic flux linkage λ =40 line turns,

where the units of N are turns and the

units of magnetic flux φare lines

1.2.2 Material Properties

B=magnetic flux density

H=magnetic field intensity

µ =magnetic permeability

The magnetic flux density B is

defined as the magnetic flux per unit of

cross-sectional area a The magnetic field intensity H is defined as the mag-

netomotive force per unit of magnetic

flux path l The magnetic permeability µ

of the material is defined as the ratio

between the magnetic flux density B and the magnetic field intensity The

permeabil-ity can also be obtained graphically from the magnetization curve shown in Fig 1.4 In

the electrical analogy, the current density J, the electric field intensity E, and the

elec-trical conductivity σare defined using ratios of similar physical parameters

FIGURE 1.3 Flux linkage visualization

(Cour-tesy of Eaton Corporation.)

= = = (1.29)The same can be done for the permeance , as shown in equation [1-30].

The inductance L is defined as the magnetic flux linkage λper amp I, as shown in

Eq (1.31) The inductance can be written in a form based on the material propertiesand the geometry (µ, a, and L), also shown in Eq (1.31) The B–H curve can be eas-

ily changed into a λ–I curve, as shown in Fig 1.5, and the inductance can then be

obtained graphically

1.2.4 Energy

W e=input electric energy

W f=stored magnetic field energy

W m=output mechanical energy

Wco=magnetic field coenergy

Energy Balance. All systems obey the first law of thermodynamics, which statesthat energy is conserved This means that energy is neither created nor destroyed.Therefore, an energy balance can be written for a general system stating that thechange in energy input to the system ∆W eequals the change in energy stored in thesystem ∆W fplus the change in energy output from the system ∆W m This energy bal-ance is illustrated in the following equation

Input Electric Energy We. The input electrical energy can be calculated by grating the coil voltage and current over time, as follows.

inte-W e=t

Substituting Faraday’s law, Eq (1.20), into Eq (1.33) shows that the input electrical

energy is equal to the product of the coil magnetizing current I and the flux linkage λ

W e=t

0I dt =λ

Stored Magnetic Field Energy. As can be seen in Fig 1.5, the flux linkage λis a

function of the magnetizing current I and depends on the material properties or the

magnetization curve The stored magnet field energy is calculated by integrating Eq.(1.34) over the magnetization curve By inspection of Eq (1.34), the area of integra-tion lies above the magnetization curve, as shown in Fig 1.6

The stored magnetic field energy can be calculated for linear materials by tuting Eq (1.31) into Eq (1.34) as follows Linear materials are characterized by a

substi-constant value of inductance L or permeability µ

1

2

1

2

1

2

1

2

Output Mechanical Energy Wm. The change in mechanical energy ∆W mis equal to

the product of the mechanical force F and the distance over which it acts ∆X.

where L and µare constant

A comparison of Eqs (1.37) and (1.45) shows that the stored magnetic field

energy W f and the magnetic coenergy Wcoare equal if the magnetic materials havelinear properties:

where L and µare constant

Electromechanical Energy Conservation (Mechanical Force, Torque). Themechanical forces and torques produced by electromagnetic actuators are derivedusing the energy balance from Eq (1.32) A graphical representation for the electro-mechanical energy conversion is shown in Fig 1.7 to help in visualizing the deriva-tion Figure 1.7 shows two operating states for an electromagnetic actuator State 1 is

characterized by coil current I1, flux linkage λ1, and flux path length l1 State 2 is

char-acterized by coil current I2, flux linkage λ2, and flux path length l2 The change in fluxpath length represents mechanical motion, which implies that there is a change in the

mechanical energy The change in the magnetization curve from flux path length l1to

l reflects a change in the inductance, as described in Eqs (1.30) and (1.31)

1

2

1

2

1

2

1

2

The change in electric energy and the change in stored magnetic field energy aredefined as follows.

The electric energy and the stored magnetic field energy for states 1 and 2 can beobtained by using the applicable regions above and below the magnetization curvedesignated as A, B, C, D, E, F, and G in Fig 1.7

W f 1=λ 1 0

T =d dθ12 (NI)2=d dθ12φ2 (1.75)

for constant L,µ

1.2.5 Application of the Force and Energy Equations to an Actuator

The purpose of this section is to show how the energy and force equations can beapplied to an actuator to determine the armature force The reluctance actuator

shown in Fig 1.8a will be used for this discussion The saturable iron regions of the actuator include the armature, which moves in the x direction, two stationary poles,

and a coil core The magnetic flux generally remains in the iron regions; however, itmust cross air gap 1 and air gap 2 to reach the armature Some of the magnetic fluxfinds alternative air paths which bypass the armature; these flux paths are called

leakage flux paths The air flux path shape and the reluctance equations are derived

from the reluctance definition =1/µa in Sec 1.3.

The first step in constructing an equivalent reluctance model is to identify eachiron flux path and each air flux path for the actuator The iron flux paths include thecore, pole 1, pole 2, and the armature, and the reluctances are designated as core,

P1,P2, and arm, respectively The air flux paths include gap 1, gap 2, leakage 1, andleakage 2, and the reluctances are designated as g1,g2,L1, and L2, respectively

By observing the expected paths of the magnetic flux (Fig 1.8a), an equivalent tance network can be assembled (Fig 1.8b), in which each reluctance is modeled as

reluc-an equivalent electrical resistor The coil is shown as reluc-an amp-turn NI source in series with the core reluctance (Fig 1.8b), and is modeled as an equivalent voltage source

directed toward the left (from −to +), according to the right-hand rule The magneticflux φis modeled as an equivalent electric current

Three magnetic flux loops (φ1,φ2,φ3) can be defined in the equivalent reluctance

network (Fig 1.8b) Three loop equations can be written in which all of the turn NI drops around each flux loop must sum to zero (magnetic Kirchoff’s law), as

amp-follows

Flux loop φ1

NI=0 = φ1(P1+P2+g1+g2+arm+L2) − φ2(L2) (1.76)Flux loop φ2

NI=0 = −φ1(L2) + φ2(L2+core) − φ3(core) −NI (1.77)Flux loop φ3

NI=0 = −φ2(core) + φ3(L1+core) +NI (1.78)The only unknowns in these equations are the magnetic fluxes (φ1,φ2,φ3), whichcan be determined by simultaneously solving the loop equations The resultingcore flux and the leakage flux in the second leakage path can be determined as fol-lows:

Magnetic flux in the core

Amp-turn drop across iron core

NIcore= φcorecore (1.84)Amp-turn drop across air gap 1

Amp-turn drop across air gap 2

Amp-turn drop across air leakage 1

Amp-turn drop across air leakage 2

The magnetic field intensity H and the magnetic flux density B for the nonlinear

iron regions can be determined as follows

Iron armature

Harm= Barm=aφ

ar 1 m

This completes the solution for the state of the magnetic field in the actuator The

x-direction force on the armature can now be determined by calculating the change

in the magnetic coenergy as a function of armature displacement in the x-direction.

The magnetic coenergy can be calculated for the entire actuator or for just the ing air gaps

work-Magnetic Coenergy Applied to the Entire Actuator. In general, the magneticcoenergy is calculated from Eq (1.44), which requires knowledge of the nonlinearmagnetic properties in terms of the flux linkage λand current I However, the mag-

netic properties for ferromagnetic materials are normally published in terms of the

magnetic flux density B and magnetic field intensity H, as B-H curves By

substitu-tion of the definisubstitu-tions for flux linkage λ =Nφ, magnetic flux density φ =Ba, and netic field intensity NI =H1 into Eq (1.44), the magnetic coenergy can be calculated from the B-H property characteristics, as follows:

prop-culated by integrating the area under the B-H curve, as follows The total magnetic

coenergy in the iron is the summation of the iron coenergies

Magnetic coenergy in the armature

Wco, arm=aarmlarmHarm0

Magnetic coenergy in the core

Wco, core=acorelcoreHcore

Total magnetic coenergy in iron

Wco, iron=Wco, apm+W co, P1+W co, P2+Wco, core (1.98)Air is not a ferromagnetic material, and by definition it has linear or con-stant magnetic properties Therefore, the magnetic coenergy in each of the airreluctances is identical to the stored magnetic field energy (∆W f= ∆Wco) The mag-netic coenergy in each of the air reluctances can be calculated from Eq (1.37),

as follows The total magnetic coenergy in the air is the summation of the air ergies

coen-Magnetic coenergy in air gap 1

Total magnetic coenergy in air

Wco, air=W co, g1+W co, g2+W co, L1+W co, L2 (1.103)The total magnetic coenergy for the entire actuator is the summation of the ironand the air coenergies, as follows

The armature force Farmcan be obtained by calculating the total actuator

mag-netic coenergy Wco, totat each of two armature positions, x1and x2 The resulting

armature force is the average force over the armature position change x =x1to x2,and in the direction of the armature position change

Total actuator magnetic coenergy at x =x1

Total actuator magnetic coenergy at x =x2

Average armature force over x =x1to x2

−

−

W

x1 co1

Magnetic Coenergy Applied to the Working Air Gaps. Since the armature force isproduced across the working air gaps (gap 1 and gap 2), the armature force can bedetermined by considering the coenergy change in the working gaps alone The totalmagnetic coenergy in the working air gaps is the summation of the air gap coener-gies from Eqs (1.99) and (1.100), as follows:

Magnetic coenergy of air gap 1

Total working air gap coenergy

Wco, gap=W co, g1+W co, g2=12 NI2

Average force

1.2.6 Summary of Magnetic Terminology

The analysis and design of electromagnetic devices can be accomplished by usingthe relations presented in Secs 1.2.1 to 1.2.4 The key equations from these sectionsare listed here

=1= µa l (1.30)Inductance

Wco=W f for constant L,µ (1.46)Mechanical energy

PROBABLE FLUX PATHS

Defining the permeance of the steel parts is very simple because the field is ally confined to the steel Therefore, the flux path is very well defined because it hasthe same geometry as the steel parts

gener-µ

µ

0

The flux path through air is complex Ingeneral, the magnetic flux in the air is per-pendicular to the steel surfaces and spreadsout into a wide area As an example, Fig 1.9shows five of the flux paths for a typical airgap between two pieces of steel The totalpermeance of the air gap is equal to the sum

of the permeances for the parallel flux paths.The permeance of each path can be calcu-lated based on the dimensions shown in Fig.1.10 and on Eq (1.30), as follows Path 1isthe direct face-to-face flux path Paths 2,3,

4, and 5are generally identified as fringingpaths H C Roters (1941) recommends that

the value for dimension h, as shown in Fig 1.10, should be equal to 90 percent of the smaller thickness of the two steel parts, h=0.9t However, it is easier to remember the slightly larger value of h=1.0t, and there is no significant loss in accuracy Two

examples of magnetic flux lines are shown in Fig 1.11 (iron filings on a U-shapedmagnet) and Fig 1.12 (finite element result flux-line plot) In these examples it iseasy to see the general flux path shapes shown in Fig 1.9

Path 1 The direct face-to-face air gap flux path has the same geometry as theperpendicular interface region between the two steel parts

Path 2 (Half Cylinder). The cross-sectional area of this flux path varies along the

length of the path Therefore, Eq (1.30) is modified as follows, where v2is the

FIGURE 1.9 Air gap permeance paths.

(Courtesy of Eaton Corporation.)

ume of flux path 2 Roters (1941) uses a graphical approximation to the mean pathlength, resulting in a permeance with a value of 2=0.26 µ0w, which is slightly larger

than that shown here

FIGURE 1.11 Magnetic flux lines illustrated by iron filings

on U-shaped magnet (Courtesy of H.C Roters and Eaton

Corporation.)

mean path length, resulting in a permeance with a value of 3=0.52 µ0w, which is

slightly larger than that shown here

l3= g+ r3=1.285g (average of inner and outer paths) (1.123)

Path 4 (Half Cylindrical Shell). The cross-sectional area of this flux path is

con-stant However, the magnetic flux path length increases as the radius r increases.

Therefore, Eq (1.30) is written in differential form and the permeance is calculated

FIGURE 1.12 Magnetic flux lines illustrated by a finite element

solution flux-line plot on a U-shaped magnet (Courtesy of the

Ansoft/DMAS finite element program and Eaton Corporation.)

by integrating over the radius as follows Roters (1941) uses the same procedure andshows the same results.

1, which is identical in shape to path 2 There is also a half cylindrical shell fluxpath (7) extending out of the page over path 6and into the page behind path 6,which is identical in shape to path 4 The values for flux paths 6and 7can beobtained from Eqs (1.121) and (1.131) by using the proper dimensions for the fluxpaths from Fig 1.10, as follows

Path 8 (Spherical Octant). There are also spherical flux paths on the corners, asshown in Fig 1.13 The permeance of these flux paths can be estimated using thesame technique demonstrated for evaluating flux paths 2through 5, as follows.The cross-sectional area of flux path 8varies along the path length Therefore,

Eq (1.30) is modified as follows, where v8is the volume of flux path 8 Roters (1941)uses a graphical approximation to the mean flux path length, resulting in a permeancewith a value of 8=0.308 µ0g, which is slightly smaller than that shown here.

1

8

increases Therefore, Eq (1.30) is written in differential form, and the permeance is

calculated by integrating over the radius as follows, where v9is the volume of fluxpath 9 Roters (1941) uses a graphical approximation to both the mean path lengthand the mean path area, resulting in a permeance with a value of 9=0.50µ0h, which

is slightly smaller than that shown here

Path 10 (Spherical Quadrant). The cross-sectional area of this flux path varies

along the path length Therefore, Eq (1.30) is modified as follows, where v10is thevolume of flux path 10 Roters (1941) uses a graphical approximation to the meanpath length, resulting in a permeance with a value of 10=0.077 µ0g, which is slightly

smaller than that shown here

ance is calculated by integrating over the radius as follows, where v9is the volume offlux path  Roters (1941) uses a graphical approximation to both the mean path

v10



l2 10

1

2

4

3

1

4

g

2

v10



l2 10

π

2

length and the mean path area, resulting in a permeance with a value of 9=0.25

µ0h, which is slightly smaller than that shown here.

1.3.1 Summary of Flux Path Permeance Equations

All of the flux path permeances are based on Eq (1.30) The relationships shown hereare for the special case of 90°and 180°angles between steel surfaces However, thetechniques that are demonstrated here can be applied to any geometry.The final forms

of the flux path permeances for Figs 1.9, 1.10, and 1.14 follow All of the magnetic fluxpaths represent fringing regions except for the direct face-to-face flux path (1).Direct face-to-face flux path (Figs 1.9 and 1.10)

dv11



l2 11

1

4

g

2

g

2

v11

l2 11

Quarter cylindrical shell (Figs 1.9 and 1.10)

5=2µ0 ln 1 +  (1.137)Spherical octant (Fig 1.13)

1.3.2 Leakage Flux Paths

The magnetic flux paths shown in Figs 1.9, 1.10, and 1.13 (the direct face-to-face fluxpath 1and the fringing flux paths 2,3,4, and 5) are based on the air gap geom-

etry These flux paths carry the magnetic flux across the working gaps g from the

magnet poles to the armature, as shown for the permanent-magnet reluctance ator in Fig 1.14 Also shown are the leakage flux paths L1,L2, and L3, which carrythe magnetic flux between the magnet poles, and prevent some of the magnetic flux

actu-from reaching the armature and the working gaps g The effect of each flux path is

described here

Direct Face-to-face Flux Path (1 ). This flux path is the highest-efficiency ducer of the force on the armature It also produces the majority of the force on thearmature The total magnetic flux through this path is limited by the saturation fluxfor the materials in the magnet poles and the armature

pro-Fringing Flux Paths (2 ,3 ,4 , and 5 ). The fringing flux paths increase the tem permeance, increase the total magnetic flux, and produce a lower-efficiencyforce on the armature than does the direct flux path Initial magnetic performanceestimates commonly use only the direct face-to-face flux path and ignore the fring-ing flux paths, in order to make the first calculations very easy and fast If the steel inthe magnet poles and armature is not saturated, then the fringing flux paths increasethe total magnetic flux and increase the armature force If the steel in the magnetpoles and armature is saturated, then adding the fringing flux paths does not changethe total magnetic flux, and since the fringing flux paths take magnetic flux awayfrom the direct flux path, the armature force is decreased

sys-Leakage Flux Paths (L1 ,L2 , and L3 ). These flux paths take magnetic flux awayfrom both the direct flux path and the fringing flux paths, and produce no force on the

h

g

w

π

armature Also, the magnetic flux carried by the leakage flux paths must be carried

by a portion of the magnet poles This causes the magnet pole material to reach thesaturation flux limit sooner than expected Therefore, the leakage flux paths cause

a large decrease in the armature force When a permanent magnet is placed nearthe working gap, the armature force is increased, because some of the leakage fluxbecomes fringing flux This essentially minimizes the leakage flux Conversely, theleakage flux paths are useful in permanent-magnet systems as a means of protect-ing the permanent magnet from large demagnetizing fields In this case, some ofthe demagnetizing flux bypasses the permanent magnet through the leakage fluxpaths

The leakage flux paths can be evaluated by using the following procedures

Path L1 (Half Cylindrical Shell). This flux path is identical in shape to path 4

The diameter of the internal half cylindrical shell d1is equal to 33 percent of the

per-manent-magnet length The radius of the external half cylindrical shell r1is equal to50% of the permanent-magnet length These permeance relationships are shownhere, based on Eqs (1.126) through (1.131) This flux path is valid only for alnico andearlier permanent magnets, which have effective poles at 70 percent of the magnetlength Ferrite and rare-earth permanent magnets have effective poles at 95 percent

of the magnet length; therefore, no magnetic flux is generated in this path and thepermeance is zero

FIGURE 1.14 Two-dimensional air flux paths around a permanent

mag-net reluctance actuator L1, L2, and L3are leakage flux paths. 2 ,  3 ,  4 ,

and  5are fringing flux paths (also shown in Fig 1.9) (Courtesy of Eaton

Corporation.)

Path L2 (Half Cylindrical Shell). This flux path is also identical in shape to path

4 The dimensions can be obtained by inspection from Fig 1.14, and the permeancerelationships shown here are based on Eqs (1.126) through (1.131)

a =D(h +g −r5) =Dh +g −  (1.176)

for a permanent magnet

If the permanent magnet is replaced with a coil, then this path has only half ofthis permeance The magnetic flux at the bottom of the flux path encircles zero coilamp-turns, and the magnetic flux at the top of the flux path encircles all of the coilamp-turns Therefore, a coil actuator contains only 50 percent of the magnetic flux inpath L3as compared to a permanent-magnet actuator.This corresponds to a 50 per-cent permeance value for the coil actuator

1

6

Total Permeance and Equivalent Reluctance Circuit. The total reluctance of thearmature air gaps (1,2,3,4, and 5) and the leakage flux paths (L1,L2, and

L3), as seen from the permanent magnet, can be calculated by drawing the equivalentreluctance network, as shown in Fig 1.15 The permeance of each armature air gap isthe sum of the air gap permeances

Armature air gap permeance

g=1+2+3+4+5 (1.179)Armature air gap reluctance

The reluctances of the leakage fluxpaths are listed here as the inverse of thepermeance values

FIGURE 1.15 Equivalent reluctance network

for the permanent magnet leakage shown in Fig.

1.14 (Courtesy of Eaton Corporation.)

When these equations are applied to a nonlinear system, the coenergy Wcomust beobtained by integrating each region over the local magnetization curve:

the coenergy Wcoand the stored field energy W fare equal and can be obtained from

Eq (1.37), where NI is the magnetizing force in the air gap,φis the magnetic fluxpassing across the air gap, and  is the air gap permeance

The total magnetizing force provided by the coil NIcoilis used to magnetize boththe steel part and the air gaps In general, the magnetizing force in the steel parts is

small, and the resulting magnetizing force in the air gap NI is nearly equal to the coil magnetizing force (NI≈NIcoil) However, when the steel parts become saturated, themagnetizing force in the steel parts becomes large and the magnetizing force in the

air gap becomes small (NI  NIcoil) The saturation of the steel parts produces a netic flux limiting condition in which the second form of Eqs (1.74) and (1.75) can

mag-be used effectively

1.4.1 General Air Gap Linear Equations

An actuator or motor generally has one or two magnetizing sources Some examplesare listed here

● A reluctance motor has one stationary coil and a moving armature that changes

the reluctance  or the permeance  of the air gap as it moves The mechanicalforce is generated by the change in reluctance

● A flux-transfer reluctance motor has two stationary coils and a moving armature

that changes the reluctance  or the permeance  of the air gap as it moves Themechanical force is generated by the change in reluctance Usually one of the sta-tionary coils provides a constant magnetizing force to produce a bias magneticfield which is modified by the second coil The coil that produces the constantmagnetizing force can be a permanent magnet

● A moving coil motor has one moving coil that changes the number of turns which

contribute to the total magnetizing force in the motor There may be a second coil(stationary) that provides a constant magnetizing force to increase the air gapstored magnetic field energy and the resulting force or torque The second coil can

be replaced by a permanent magnet The permanent magnet and coil can be

inter-changed to produce a moving magnet motor The mechanical force is generated by the change in the number of turns and is also called the Lorentz force.

The magnetizing force NI in Eqs (1.74) and (1.75) is expanded here to show the contribution from a second coil N a I a(either stationary or on a moving armature) and

the contribution from a bias field coil N f I f(stationary)

(NI)2=(N a I a)2+2(N a I a N f I f) +(N f I f)2 (1.188)Multiplying by the permeance gives the energy expression needed in Eqs (1.74)

and (1.75) Also, the definition of the self-inductance of the moving armature coil L a,

the self-inductance of the bias field coil L f, and the mutual inductance of both coils

L afcan be substituted into Eq (1.188) as follows:

force and torque equations It is assumed that the currents I a and I fare constant with

respect to the armature positions X and θ

Differentiation of each inductance term can be carried out in Eq (1.194) as

fol-lows It is assumed that field coil turns N fare constant with respect to the armature

positions X and θ

I2 = (N a I a)2 +N a I2 (1.196)

I a I f = N a N f I a I f +N f I f I a (1.197)

The terms in Eqs (1.196), (1.197), and (1.198) can be combined, or Eq (1.198) can

be differentiated directly to produce the following force and torque equations

2

pro-armature reaction force) The third term represents the Lorentz force on the

mov-ing coil due to the interaction with the magnetic flux produced by the bias field coil

1.4.2 Reluctance Actuators with Small Air Gaps and Saturation

When an air gap is very small, the permeance 1becomes very large in comparison

to the fringing flux paths 2through 10 Therefore, the total permeance totalcan beclosely approximated by the direct face-to-face permeance 1, Eq (1.30)

Reluctance Normal Force. The force for the actuator in Fig 1.16 can be written as

follows, based on Eqs (1.203) and (1.30) The motion of the armature X is defined to

be in the direction to close the air gap, and the initial length of the air gap is defined to

be g0.Also, N a=I a=0 because there is only one coil.The area of the pole face is a =T P D.

This is the permeance of the flux path between one pole and the armature Thetotal permeance of the flux path includes two air gap permeances in series There-fore, the total permeance Tis equal to one-half of the permeance for one air gap:

for small g and limited NI.

All of the variables in Eq (1.209) are constant except the armature position g.

Therefore, the armature reluctance force is proportional to the inverse square of thearmature position (or to the size of the working air gap) If the steel parts becomesaturated in some armature positions, then the air gap magnetizing force is reduced.Under this condition the air gap magnetizing force is a function of the armatureposition

2

When the steel parts become saturated the total magnetic flux φin the systemreaches a maximum limit, as shown in Fig 1.4 Equation (1.209) can be written as fol-lows by substituting Eqs (1.18), (1.21), and (1.30):

F T= (NI)2µ0 =  2

for small g

for small g and limited φand B

All of the variables in Eq (1.211) are constant Therefore, the armature tance force is constant regardless of the armature position when the system is satu-rated Due to symmetry, the force on each side of the armature or on each pole of themagnet frame is half of the total force, as shown here:

for a single pole, where g is small.

Maximum Possible Reluctance Normal Force. The reluctance normal force in Eq

(1.212) can be divided by the pole area a to obtain the normal magnetic pressure p

on the pole, as shown in Eq (1.213) The maximum normal magnetic pressure isdependent only on the saturation magnetic flux density (for small gaps) Steel typi-

FIGURE 1.16 Actuator with reluctance force produced normal to the armature bottom

sur-face, in the direction of motion X (Courtesy of Eaton Corporation.)

cally saturates at 1.60 T Therefore, the maximum possible normal magnetic pressurefor this steel is 150 lb/in2:

for a single pole, where g is small.

This relationship can be converted into the following simple design strategy If 150

lb of magnetic force is required, then at least 1.0 in2of steel is needed Conversely, if weare limited to 1.0 in2of steel, then the maximum possible magnetic force will be 150 lb

Reluctance Tangential Force. The force for the actuator in Fig 1.17 can be written

as follows, based on Eqs (1.203) and (1.30) The motion of the armature x is defined

to be in the direction to insert the armature into the stator cup Also, N a=I a=0because there is only one coil:

where C is small and NI is limited.

All of the variables in Eq (1.218) are constant Therefore, the armature tance force is constant regardless of the armature position If the steel parts becomesaturated in some armature positions, then the air gap magnetizing force is reduced.Under this condition, the air gap magnetizing force is a function of the armatureposition, and a new air gap magnetizing force must be calculated at each position

FIGURE 1.17 Actuator with reluctance force produced tangential to

the armature side surface, in the direction of motion x (Courtesy of

Eaton Corporation.)

When the steel parts become saturated, the total magnetic flux φin the systemreaches a maximum limit, as shown in Fig 1.4 Equation (1.218) can be rewritten asfollows by substituting Eqs (1.18) and (1.30):

F=  2

where C is small and φis limited

All of the variables in Eq (1.220) are constant except for the armature position

x Therefore, the armature reluctance force is inversely proportional to the square of

the armature position when the system is saturated Equation (1.221) is the flux

den-sity B limited form of Eq (1.220), which was obtained by substituting Eqs (1.21) and

(1.215) into Eq (1.220) Equation (1.221) shows that the force is independent of the

armature position x as long as the air gap flux density B is constant The force is

con-stant, and the air gap flux density is constant if the actuator steel parts are not rated:

where C is small and B is limited.

Pole Shaping. An example of the performance variations that are possible by ing the pole faces can be observed by comparing Eqs (1.209) and (1.218) and Figs.1.16 and 1.17 The normal force on the pole faces shown in Fig 1.16 and Eq (1.218) is

shap-inversely proportional to the square of the armature position, F∝1/g2 The tangential

force F on the pole faces shown in Fig 1.17 and Eq (1.209) is constant regardless of

the armature position A force characteristic between these two limiting conditionscan be achieved by using a cup-cone configuration Fig 1.16 is similar to the limitingconfiguration of a flat-face cone, and Fig 1.17 is similar to the limiting configuration

of a straight-sided cup Therefore, it is possible to shape the force characteristic as afunction of the armature position by modifying the pole face, as shown in Fig 1.18

Reluctance Tangential Torque. The torque for the actuator in Fig 1.19 can be ten as follows, based on Eqs (1.204) and (1.30) The motion of the armature θisdefined to be in the direction to align the armature vertically with the stator poles

writ-Also, N a=I a=0 because there is only one coil:

T T= (NI)2

where g is small and NI is limited.

All of the variables in Eq (1.227) are constant Therefore, the armature reluctancetorque is constant regardless of the armature position If the steel parts become satu-rated in some armature positions, then the air gap magnetizing force is reduced

FIGURE 1.18 Force curve performance variations due to modification of the pole shape.

(Courtesy of Eaton Corporation.)

FIGURE 1.19 Actuator with reluctance torque produced tangential to the armature end surface, in the direction of motion θ (Courtesy of Eaton Corporation.)

Under this condition the air gap magnetizing force is a function of the armature tion, and a new air gap magnetizing force must be calculated at each position.When the steel parts become saturated, the total magnetic flux φin the systemreaches a maximum limit, as shown in Fig 1.4 Equation (1.227) can be rewritten asfollows by substituting Eqs (1.18) and (1.30):

posi-T T= (NI)2µ0 = 2

where g is small and φis limited

All of the variables in Eq (1.229) are constant except for the armature position

θ Therefore, the armature reluctance torque is inversely proportional to the square

of the armature position when the system is saturated Again, as shown in the ous section, the torque produced at each end of the armature is one-half of the totaltorque:

where g is small and φis limited

Equation (1.231) is the flux density B limited form of Eq (1.230), which was

obtained by substituting Eqs (1.21) and (1.223) into Eq (1.230) Equation (1.231)shows that the torque is independent of the rotation angle θas long as the air gap

flux density B is constant The torque is constant, and the air gap flux density is

con-stant if the actuator steel parts are not saturated:

where g is small and B is limited.

Moving-Coil Actuator Lorentz Force.

The force for the moving-coil actuator inFig 1.20 can be written as follows, based

on Eqs (1.203) and (1.30).The motion of

the armature X is defined to be in the

direction to bring more of the movingcoil into the magnetic circuit Also, isconstant, because the size of the air gapdoes not change as the position of themoving coil changes Therefore, the rate

of change of the permeance with respect

to the armature position is zero

The length of wire on the movingcoil that is in the air gap permeancepath can be calculated as follows The

FIGURE 1.20 Moving-coil actuator producing

a Lorentz force in the direction of motion X.

(Courtesy of Eaton Corporation.)