DC Machines Lectures Notes


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Part I:   DC Machines

                                                           ELECTRICAL  MACHINES
















14.     Theraja Bl,  Theraja Ak  “ ELECTRICAL TECHNOLOGY”   .






CHAPTER 1 :                                          PRINCIPLE OF OPERATION   



     Right-hand screw rule (cross product operation in vector algebra)

              A1 X A2= A3    

A1, A2, and A3 are vectors. Rotate RH fingers from the direction of the first vector , A1, to the direction of the second vector, A2,  through the small angle (<180o) between them; the direction of A3 is then given by the extended RH thumb. In our applications, the magnetic field B is always the second vector.                         

Given a straight conductor lying in a magnetic field, and oriented perpendicular to the direction of the field. Let

B= magnetic flux density (also called magnetic induction), [T=tesla=weber/m2];

L=active length of conductor,[m].            u


Induced voltage (emf):          E=B.L.u.sinα    volts

u= speed of conductor perpendicular to its length[m/s];

α= small angle from u to B.

Direction of E  : RH screw rule from u to B .


Developed force :                  Fd= B.I.L   newtons [N]

I= current through conductor, [A].

Direction of Fd :  RH screw rule from   I  to  B.

-1.2 Single conductor


Consider  a straight conductor moving at a uniform speed   u  in a uniform magnetic field B,  and carrying a current  I.   Let

L=  active length(i.e. length of conductor segment immersed in the field);

Fm= applied mechanical force, [N].

u  &  B  give         E=B.L.u           and   I & B  give  Fd=B.I.L 

E.I= (B.L.u).I=(B.I.L). u=  Fd. u  -------------à     E.I= Fd.u= Pc = conversion power

Note: Because the speed  u  is constant, Fd and Fm  must be equal and opposite( otherwise there would be acceleration or deceleration).

E.I  is electrical power  , and Fd .u (Fm. u)  is  mechanical power.


                              I  in direction of  E;  Fd opposite to  u.        I  opposite to  E; Fd in   Direction of  u.     

                                                                 Generation action.                        Motor action 



                               V=E - I.R                                                                                   V=E +  I.R

                                                                                 Pin=V.I= (E+I.R). I

                        = Fd. u= Pc                                                                                    = E.I + I2.R= Pc+Pcu

Pcu= copper losses

                  Pout=V.I=  (E – I.R).I                                                                      Pout=             

                         =E.I – I2.R= Pc -   Pcu                                                                       =Fd. u=  Pc

                         = Pin   -    Pcu                                                                           Pin=  Pout    +     Pcu   

               Pin  -   Pout=    Pcu                                                                             Pin  -   Pout    =   Pcu   

-1.3  Wire loop


Consider now a wire loop rotating at a uniform speed in a magnetic field  B.  Conductors   a    and   b   are the active parts of the loop; the remaining parts are ‘ end- connections’ and ‘leads’.


                                                              Fda = Fdb   zero resultant force   Td= developed torque

                                             i  in direction of e ; Td  opposes rotation     i  opposes e; Td aids rotation

                                                                      Generator action.                                        Motor action.

KVL:   e= ea  +  eb = loop emf

-1.4 Slip rings

Slip  rings and brushes may be used to make  electrical contact with a rotating loop. Slip rings rotate with loop , while brushes are stationary to give sliding contact. A slip ring and a brush for each terminal.

The induced emf is alternating, and the developed torque oscillates (about the vertical position). Clearly, slip rings are not suitable for   dc  machines(they are used in  ac  machines).

-1.5 Commutator

A commutator is a conducting ring split into segments; each segment is electrically connected to one terminal of the loop. It is mounted on the shaft , but is electrically insulated from it. The brushes are stationary and make sliding contact with the segments.


A2 is always positive; A1 is always negative:  >>> et    is always positive , &  Td   is always  CW(when direct current supplied to brushes), but emf & current  within loop oscillate.  Although  et  and  Td  are now unidirectional ( i.e. remain in the same direction or sense), they do not represent steady  dc  operation because they fluctuate: each is maximum when the loop is in position  0  , and  zero when it is in position 2.


-1.6 Multiple loops

More uniform  dc  operation is achieved by using a number of loops displaced from each other in space. The emf’s add (series connection), and the developed torques aid each other. As the number of loops is increased, ideal  dc  operation is approached.


                                                                                       2 loops                               4 loops

-1.7  Magnetic circuit 

The magnetic field may be obtained by means of permanent magnets (PM), or, more commonly, by means of electromagnets (field coils with iron cores).


   Permanent magnet                                electromagnet                         practical construction

(with soft iron extension)

The value of the resulting flux is determined by the mmf (magnetic motive force) ( of the PM or electromagnet) and the magnetic reluctance in the path of the flux. Iron has very high magnetic permeability, so that it is the air gap in the path of the flux that limits its value. The air gap must therefore be made short to increase the effective flux. The air gap cannot be avoided completely(why?) .


Air core                                                     cylindrical iron core                  slotted iron core(armature)

-1.8 Multiple poles

A  dc  machine can have 2,4,6,8,…….   Poles(an even number-why?)

2p=number of poles; (i.e. p=number of pole pairs)

One revolution=360 mechanical degrees;

One pole pair= 360 electrical degrees;

Electrical angle=p   X   mechanical angle;

Pole pitch =180o electrical= 360o mech/2p  .

Each armature loop is placed over one pole pitch. Electrically, every thing repeats after two pole pitches.


-1.9 loop emf 

D= diameter of the armature[m];

L=active length of armature[m];

n= rotational speed [rps=revolution per second];

u= speed= πDn [m/s];

ωr=angular speed= 2 πn  [rad/s].

Ap=armature surface area corresponding to one pole pitch= πDL/2p  [m2].

φ= flux per pole [wb=weber]; total magnetic flux through pole face: same for all poles;

B = air gap flux density [T=tesla=wb/ m2]; normal field at armature surface;

Bav = average air gap flux density [T].

Φ=∫Ap B. dA    ;   Bav= φ/Ap

The actual air gap flux density B is distributed around the periphery of the armature as shown below. The average gap flux density Bav  is constant over a pole pitch. Consider side “ a” of the wire loop; the instantaneous emf is

ea= B.L.u       (ea (t) has the same waveshape as B)

The average emf induced in side  “a” is 

Ea= Bav. L.u     ( constant over alternate pole pitches)





Air gap







Average commutated emf

Wire loop :    Eloop= Ea + Eb =2 Bav. L. u=2 π.D. L.n . Bav= (2 π.D. L.n)(φ/Ap)        

                                                                   = 4p.n.φ=(2P. ωr. φ)/ π                        (=∆φ/∆t=2 φ/(1/2pn))


Conductor   (one side of loop):

                       Econ=1/2 . Eloop= Bav. L. u= π.D. L.n . Bav=2p.n.φ=(P. ωr. φ)/ π    


Coil        (N=number of turns=number of loops in series):

           Ecoil= N x Eloop=2 N. Bav.L.u= 2 π.D. L.N.n . Bav=4p.N.n. φ=(2P. N.ωr. φ)/ π       (=N∆φ/∆t)     


-1.10 loop torque      

The instantaneous force on side “a” is

  fa= B. Ia.L

The normal field  B  varies over a pole pitch, and hence  fa also varies. The average force on side “a”   is

Fa=  Bav. Ia.L

Bav , and hence  Fa , are constant over a pole pitch corresponding to a given pole. Moreover, because of commutation,  Ia reverses over the next pole pitch so that  Fa remains in the same sense around the axis of the armature. The resultant force due to all armature conductors is zero , but there is developed torque because all forces act in the same sense around the axis.

Average developed torque

Wire loop        Tloop= Fa. D/2  + Fb.D/2= D/2(Bav. Ia.L   + Bav. Ib.L  )= D.L.I.Bav   [Nm]

Where   Ia=Ib ,   >>>>   Tloop=D.L.I(φ/Ap)=2p.I. φ/π


Coil  ( N turns):  Tcoil= N X Tloop =D.L.N.I. Bav= 2p.N.I. φ/ π

For constant rotational speed  n , and assuming no friction, the developed torque is equal to the applied mechanical torque.


-1.11 Conversion power

A  dc machine may run as a generator or as a motor. In each of the two modes of operation, there is induced emf and developed torque( given by the equations of section 1.9 and 1.10). What determines the mode (generator or motor) is the directional relationship between  E, and I , and between  Td   and  n:


Generator mode:  I  with  E,    Td  opposite  n ;

Motor mode       :  I  opposite  E,      Td  with  n.

Recall the coil equations of section  1.9  and 1.10 ;  E=Ecoil  and    Td=Tcoil,  we can write

                       E.I= (4p.N.n.φ). I = 4p.N.ωr. φ.I/2π= (2p.N.I. φ/ π) .ωr = Td. ωr

This represents electromechanical energy conversion. The conversion power is defined as

                       Pc=  E.I  =  Td. ωr      {  E.I  on the electrical side----- Td. ωr  on the mechanical side

Generator action   


R=resistance of the coil

Tm=mechanical drive torque


Mech  i/p:   Pin= ωr. Td =Pc

Elect o/p:    Pout= V.I= (E – I.R).I= E.I – I2. R=  Pc  -  Pcu

               >>>> Pout=   Pin  - Pcu

Motor action





Elect i/p:  Pin= V.I=(E + I.R).I = E.I + I2.R  = Pc + Pcu

Mech  o/p:  Pout = ωr. Td=  Pc

         >>>>> Pout = :  Pin -  Pcu

More generally,                         Pout = :  Pin  - losses

Where the losses include( in addition to copper losses) mechanical losses(friction and windage), and iron losses(hysteresis and eddy current). 

Example  1:

Given that the air gap field is distributed sinusoidally with a maximum flux density of   Bm.  Show that      φ= Bm.D.L/p            and     Bav= 2. Bm/π

Solution:      Bav=1/ π .sinѳ = [-cosѳ   =

                       φ=Bav. Ap=     .

Tutorial 1:

Given that the air gap field is distributed as shown in fig. 1  over a pole pitch  yp. Show that

φ =  Bm. D.L/p               and         Bav=α.Bm         where α =            





                                       Fig. 1                            

Example 2:

The armature of a   dc  machine is  80 cm long and has a diameter of 50 cm. The maximum air gap flux density is 1.5 T. The pole arc covers 70% of the pole pitch. The armature speed is 500 rpm.

(a)       If  the machine has 2 poles, find the flux per pole and the average air gap flux density when the field distribution is (i) sinusoidal  ,(ii) as in fig.1 of T.1.

(b)      Repeat part  a for a six pole machine.

For all cases of parts  a and b, find the average emf, developed torque, and conversion power for a full pitch wire loop on the armature & carrying 9 A. current.


-a(i)  Sinusoidal field distribution

Bav=  =  =0.955 T

φ = Bav. Ap = 0.955 X  =0.6  Wb.

Eloop = 4p.n.φ= 4 X 1 X  =20  Volts

Tloop =  .I.φ=  2 x9 x 0.6/π =3.437  N.m

Pc= ωr. Tloop =2π X  = 180 Watts

-a(ii) field distribution as in fig.1

Bav= α. Bm= . Bm= 0.7 X 1.5= 1.05 T

φ = . Bm.D.L/p=  0.66  Wb.

Eloop=4.p.n. φ=22 Volts

Tloop=  .I.φ=   3.78  N.m

Pc= ωr. Tloop = 198 Watts

-b(i) when 2p=6

φ= 0.2 Wb.

Eloop=20  Volts

Tloop=3.437  N.m

Pc= 180  Watts


φ= 0.22   Wb.

Eloop=22  Volts

Tloop=3.78   N.m

Pc= 198  Watts


Tutorial 2:

The armature of a 4-pole  dc  machine is rotating at 840 rpm. The armature length and diameter are 40 cm  and  30 cm respectively. The flux per pole is 65 mWb. For a 5-turn full-pitched armature coil:

-a. Find the average emf induced in the coil.

-b. What current must flow through the coil if it is to develop a torque of 8.0 N.m? what is the resulting conversion power?

Answer(36.4 V, 7.25 A.; 264 W.)


CHAPTER 2 :                                          construction

Electrical  machines are essentially electrical and magnetic circuits coupled to each other to develop  emf’s and torques. Actual machines can vary greatly in details.

2.1 Materials    

a. Iron : high magnetic permeability >> minimize reluctance of magnetic circuit>> high working fields.

High grade steel (e.g. silicon steels) for   magnetic cores. Lower grade steels(e.g. cast iron , cast steel, mild steel) for constructional parts. Relative permeability of the linear part of the magnetization (B-H) curve is of the order of 1000, and is higher for higher grade steels.

Saturation limits maximum flux density in iron to  2 T or less; the resulting average gap flux density is then around 0.8  T   for practical industrial machines.

b. copper:  high electric conductivity>>> minimize resistance of electric circuits>> high working currents. Aluminum is seldom used because of space limitations; (why is silver not used either as long as its resistivity is the lowest?)

c. Air: air gap in path of flux necessary to allow motion.

d. Insulating materials: they are necessary to insulate conductors from each other and from adjacent iron parts (which are also conducting).

The economic factor:  machines are generally designed to yield the required performance at minimum cost.



materials + cost of manufacture.

Required performance: described in terms of operating voltage, current, power, torque, speed , efficiency, weight, volume, reliability, temperature rise, function, noise, pollution , etc.

2.2  Losses

Power losses in  dc  machines include :  copper losses (I2.R losses in conductors); iron losses( hysteresis & eddy currents); friction and windage. All losses are undesirable because they represent wasted power, and cause the temperature to rise.

Materials retain their desired properties( electrical, magnetic, and mechanical) within specific temperature limits. Temperature rise is most critical to insulating materials: their insulation capability deteriorates at sufficiently high temperatures( around 100 o C, depending on the particular insulator), and ultimately breaks down causing short circuits and possibly total machine failure. To limit temperature rise, the machine design must minimize losses , and provide for efficient operation.




In general , heat is generated through the volume of the conductor or iron core, but can be dissipated only through its surface. To improve cooling of such parts, their surfaces must be made larger, which means that machines become bigger.

Changing magnetic fields induce emf’s not only in copper conductors( which is desirable), but also in iron parts. As iron has a relatively small resistivity, currents will circulate in it. These  ‘eddy currents ‘ distort the field distribution and, more important, yield  I2.R  losses in the iron. To limit these losses, and the resulting temperature rise, some cores are made of steel laminations( instead of solid steel); these are punched sheets( or stampings) around 0.5-1.0 mm thick with insulated facets(  paper or suitable coating) stacked and bolted together. The laminations are stacked in the direction of the induced  emf so that the insulated surfaces will be in the way of circulating currents.

Stacking factor =    0.9-0.97


MAIN PARTS OF A  DC  MACHINE(see figure on page previous page)

2.3 Stator (field)

Yoke :  lower grade steel; supports poles; return path for flux; encloses machine.

Poles: core and shoes ; high grade steel. Pole core may or may not be laminated. Pole shoes are usually laminated( flux fluctuation due to rotating armature slots). Pole shoes (i) reduce reluctance of the air gap( by increasing its area); (ii) improve flux density distribution in the air gap; and (iii) provide mechanical support for the field coils.

Field coils: provide the mmf for the main working flux. There may be more than one set of coils. i.e.  2 or more coils on each pole. The coils of one set are identical




4 Rotor  (armature)

The armature is made of high-grade steel laminations punched and stacked together to form a cylinder or drum. It is mounted on the shaft (directly for small machines, and by means of a spider for larger machines). It may have radial and axial ventilation ducts for

Cooling. The armature windings are placed in slots around the armature periphery. The conductors are held in place by wedges, or by bands wrapped around the armature. The slots are sometimes skewed to reduce noise.


2.5 Air gap

The air gap is the space between stator and rotor needed to allow relative motion between them (i.e it is mechanically necessary). Magnetically , it  introduces a high reluctance in the path of the working flux ( most of the field mmf is consumed in the air gap) ; it is therefore made as short as possible, typically 0.5-5.0 mm. The armature surface and the pole shoes facing   it require precise machining to avoid asymmetry and vibration.

2.6 Commutator

The function of the commutator is to interface between the alternating currents and emf’s  in the rotating armature coils on the one side , and the direct current and voltage at the machine terminals. It is made up of copper segments insulated from each other and mounted on the shaft (i.e it rotates with the rotor) in a cylindrical form. V-rings hold the bars in place. The leads of each armature coil are connected to risers ( each of the two leads of a coil goes to a different commutator segment).

Brushes are carbon or graphite blocks mounted in  stationary holders with spring pressure to maintain  good electrical contact  with the rotating commutator segments. The brushes wear out with time and must be replaced regularly.

The commutator is a critical part of the machine:  from  brushes,   carbon fillings and dirt accumulate causing current leakage between segments. Intermittent contact between brushes and segments often leads to sparking, which may become quite serious.

2.7 Mechanical items

Shaft; spider; bearings (with grease pack or lubrication system).

Bolts; clamps, spacers, brackets.

Enclosure (frame); ventilation windows, lifting eye; base .

Terminal box; shaft mounted fan; brush gear; etc.





3.5.1 Lap Winding

In  lap windings , the two coil ends are connected to adjacent commutator segments; i.e.  if the lead from side   a  is connected to commutator segment   x, then the lead from side   b   is connected to segment   x±1 . Thus 


The winding is continued until all coils are traversed up to the last coil C,  which then closes on the first coil. A lap winding can be made to fit any number of coils and poles,  C  and  2p. The choice of progressive or retrogressive windings has no significant effect.


                   Progressive                                   Retrogressive

         YC=1                                                yC= -1



3.5.2  Wave winding

In wave windings, the two coil ends are connected to commutator segments that are approximately  two pole pitches apart ( where a pole pitch is now measured in number of segments  or coils,  i.e.  C/2p) . The  commutator pitch is then

yC= (C ± 1)/P    >>>>  C=p.yC ±1

After traversing all pole pairs, the winding should return to a commutator segment adjacent to the initial one, either the one just after it ( progressive ), or the one just before it (retrogressive).


The winding is continued until all coils are traversed with the last coil closing on the first coil.  Unlike the lap winding, the wave winding cannot be made to fit just any number of coils and poles: in the above equation, the values of  C  and  p must be such that   yC  turns out to be an integer. ( If a wave winding is to be placed on an unsuitable armature, some coil positions must be left out; for mechanical balance, dummy coils are placed in these positions ( dummy coils are not connected electrically).



3.6 Parallel paths

From the previous section, it can be seen that the armature coils form a closed winding that is tapped by the brushes at certain points.  In effect, the brushes divides the winding into a number of parallel paths each of which is composed of a number of coils in series. The terminal current is divided among the parallel paths. At any moment during operation, there are  short-circuited coils; these are not included in the paths. The circuit diagrams for the lap and wave windings are shown below. In each case, the  terminal voltage is equal to the voltage of the individual parallel paths.

From the sequence diagram in page 29  for the lap wdg., it can be seen  that all the brushes are necessary: if any brush is removed, there will be opposing emf’s in series (which would cancel out). Thus the number of brushes is equal to the number of poles, and hence the number of parallel paths is also equal to the number of poles:

Lap wdg:  2a=2p, >>>     a=p  , 2a=number of parallel paths;  a=number of pairs of parallel  paths





From the sequence diagram in page 36 for the wave wdg, on the other hand, it can be seen that only two brushes are really necessary; alternate  brushes are connected to each other internally through short-circuited coils. Thus 2 brushes may be disconnected and removed without affecting induced emf’s and currents. In other words, the brushes divide the wdg into two parallel paths only:

Wave wdg:  2a=2,  >>>>> a=1.

The extra brushes may be removed, but in practice ,they are sometimes kept to obtain better current distribution over the commutator.



                                                                               General symbolic representation       


3.7 Comparison of lap and wave windings

A lap wdg has 2p parallel paths and must have 2p brushes (or brush groups). A wave wdg has two parallel paths and requires only two brushes (or brush groups), but it can have 2p brushes.

A lap wdg can be made to fit any number of coils and poles. A wave wdg can be fitted only if the number of coils and poles yield an integer commutator pitch  yC (=(C  1)/p).

A brush on a lap wdg generally short-circuits one coil during commutation. In a wave wdg having only two brushes, each brush short-circuits   p  coils in series; in a wave wdg having   2p   brushes, a coil is short-circuited by two brushes in series.

Assume now that a given wdg can be connected in either lap or wave. The coil emf and current are the same in both cases.  The lap-connected wdg will have a high terminal current and a low terminal voltage, while the wave-connected wdg will have a low terminal current and a high terminal voltage (for 2p 4). The two wdgs will have the same power.

Conversely, assume a machine is to have a specified terminal voltage and a specified terminal current. If it is connected in lap, it will have a low  current per coil, and a high voltage per coil; therefore each coil will have many turns of small cross-section, and a higher level of insulation is needed. If it is connected in wave, it will have a high current per coil and a low voltage per coil; therefore the coils will have relatively few turns and large sections, and a relatively lower level of insulation is needed. In this respect, wave wdgs are better than lap wdgs because they allow for better cooling of wdgs, and because they have higher space factors ( space factor= copper area/total slot area). In general, wave wdgs are used in almost all machines up to         50 KW, and in all high voltage machines. Lap wdgs are used mostly in large machines having low voltage/high current ratings.

3.8 Equalizers 

In lap wdgs , each path is associated with a pair of poles. Thus if there is asymmetry in the magnetic circuit(due to wear of bearings, or eccentricity of shaft), the emf’s induced in the paths will not all exactly equal to each other. Unequal emf’s connected in parallel give rise to circulating currents that can be quite large ,  causing heavy and unnecessary heating.

To reduce this effect  , points that should have the same voltage are connected to each other by means of equalizers (or equalizing connections,  or equalizing rings).  Such points are located two pole pitches apart. Each equalizing ring is connected to  p  points  in  alternate  paths.

In the wave wdg, there are only  two paths through the armature. The coils of each path are distributed uniformly around the armature, and hence cover all pole pairs.  Any asymmetry in

 the magnetic circuit will have the same effect on both paths, so that the two induced  emf’s will be identical. Thus wave wdgs do not require equalizers, which is another advantage of wave  wdgs.


                       R             IC               R       V                  At no load  I=0 and  IC=


                   E1                                      E2




3.9 Multiplex windings

The lap and wave wdgs described so far are called simple or simplex wdgs. Duplex wdgs are composed of two simplex wdgs interleaved around the armature, and connected so as to have twice as many parallel paths:

Duplex lap  wdg :  2a=4p                                                     Duplex wave wdg:  2a=4

                                  a=2p                                                                                           a=2

similarly, there may be triplex wdgs, and so on.  Such multiplex wdgs  are  rarely used.


3.10 Armature calculations

Consider a symmetrical wdg composed of  2a  identical parallel paths. It has

C/2a   coils/path,                                                                                                                                          Ip        

C/a   coil sides/paths,                                                                                                                               Ip                                                                                       VA

Z/2a =NC/a  conductor/path                                  Rp

Let                                                                             Ep

Ep=induced emf in each path;

Ip=current flowing through each path;                                        brush                             IA

Rp=resistance of each path.                                                                 Rpp                          VA

Also let                                                                                                  EA                                                       

VA=armature terminal voltage;

IA=armature terminal current;                                                Thevenin equivalent  for armature wdg

EA=armature emf= Thevenin equivalent  emf  for armature winding;

Rpp=Thevenin equivalent resistance for armature winding;

RA=total effective armature resistance.

Applying  KVL:

Generator:     VA=EA  -  IA.RA

Motor       :     VA=EA  +  IA.RA   

3.10.1 Armature resistance

Single loop :  Rloop=

Where    = resistivity of copper at the working temperature;     IA                                         IA

               lm=mean length of a single loop;                                   RA                                            RA

               A=cross-sectional area of the conductor.                                             VA                             VA      



coil   Rcoil=N x Rloop=                                                         EA                                          EA

Path   Rp=  X Rcoil= =

Winding : Rpp=  Rp= =                                                  Generator                            Motor

                                                                      Common symbolic representation of armature winding

Effective armature resistance:   RA=Rpp  + Rbrushes + Rcontact;

Rbrushes  is the resistance of the carbon brushes. Rcontact is the resistance of the brush/commutator contact surface; it is nonlinear, and is usually taken as equivalent to a constant volt drop(for example  1 volt).

3.10.2  Armature emf

The average coil emf derived in section 1.9 is the same for all armature coils. Thus, for each path

Ep=  X Ecoil=  n.φ

But all paths are in parallel with each other, so that the armature emf is equal to the individual path emf’s:

EA= Ep=  X Ecoil=  n.φ=Ke.n.φ                                              (Ke= =  )

An alternative expression can be obtained in terms of ωr   ( =2π.n) :

EA= ( Ke/2π) ωr.φ= K. ωr                                                           (K=Ke/2π= = )


3.10.3 Torque

According to  KCL, the terminal current  IA  is the sum of all path currents:

 IA=2a X Ip                          >>>>>>    Ip=IA/2a

A path is composed of coils in series, so that the path current  Ip flows  through the individual coils:


The average coil torque was derived in section 1.19 :

Tcoil=  N.Icoil. φ=  N.Ip. φ=  .IA

The total armature torque is the sum of all coil torques acting in the same direction and aiding each other:

T=C X Tcoil=  .IA.φ=K. IA                                                 ( K==  as before )

3.10.4  Conversion power

The conversion power corresponding to electromechanical energy conversion in the machine is given by:

On the electrical side : PC= EA.IA

On the mechanical side : : PC= ωr.T

Note that      PC= EA.IA=(k. K. IA.φ ). ωr=  ωr.T

Using  Pin  to denote input power to the armature, and  Pout  to denote output power from the armature, we have:

Generator  :  Pin= ωr.T= PC    ,                   Pout=VA. IA=EA.IA  -   IA2.RA=  PC  -   PCu

Motor          :  Pin=VA.IA= EA.IA  +  IA2.RA=  PC  +   PCu        ,    Pout= ωr.T= PC   

PCu  is the total copper loss(or ohmic loss) in the armature.

Example 3 :

A 6-pole machine has 53 slots with  8 conductors/slot. The flux per pole is 50 mWb, and the speed is 420 rpm. Calculate:

-1. The number of turns per coil;

2. Ecoil, EA ;

3. Tcoil,  T;

4. conversion power , If the winding is  (a) simple lap winding;

                                                                      (b) simple wave winding.

Assume armature current to be  IA=50 A.

Solution: (a) simple lap winding

2a=2p=6,    C=53,   Z= 53 x 8=424  conductors,   2C=Z/N=2  X 53=106  > N= 424/106=4 turns/coil 

Ecoil=N x Eloop=4 X 4p.n.φ=16 X 3X (420/60) X 50 X 10-3=16.8  V.

EA=Ep=(C/2a).Ecoil= 53 X 16.8/6=148.4  V.

Or  EA=   n.φ= 148.4  V.

Tcoil=N  X  Tloop= 4  X  .IA.φ= 3.183  N.m

T=C. Tcoil=53 X 3.183=168.7  N.m

Pc=EA.IA=148.4 X 50=7420  W.

.(b)Simple wave winding

 2a=2,  Ecoil=16.8  V.

 EA=Ep=53 X16.8/2=445.2  V.

       Tcoil=N.Tloop=  .IA.φ=9.549  N.m

       T=C.Tcoil=506.11  N.m

       Pc= EA.IA=22260  W.

Tutorial 3 :

A 6-pole , 1500 rpm  dc  machine is lap wound with  732  active conductors ,each carrying  20 A.  The flux per pole is  30  mWb. (a) find  IA , EA ,  T  and Pc.

.(b)Repeat part (a) if the machine is reconnected in simple wave.


Example 4:

A 10-pole simple lap wound generator is rated at  110 V.,  600 A.,   and 750 rpm. It has a winding resistance of  7.2  mΩ, and is wound in  163  slots  with 4  coil sides/slot  and  2  turns/coil. Assume a brush  voltage drop of 1.5 V. (a) find the rated load power,  (b) find the resistance ,  emf,  and terminal voltage per coil and per turn,  and (c) find the developed torque and flux per pole.


Pout=VA.IA=110 x 600=66 KW


RP=RA  x 10=72 mΩ

Coil sides= S  X  coil sides/slot

         2C=163  X  4= 652  coil sides

C= 326  coils

Coils/path= 32.6

Rcoil= = 72/32.6=2.2  

EA= VA  + IA.RA  + Vbrushes= 110  +600  X 0.0072  + 2  x1.5=117.32  V.= Ep

Ecoil= =117.32/32.6=3.55  V.= 4pNnφ>>>>> φ=7.1 mWb.

Vcoil=Ecoi   - IP. Rcoil=3.417 V.

Vturn=Eturn -   IP. Rturn=1.78   -60 X 1.1 X 10-3=1.714   V.

T=K.IA.φ=  .IA.φ=5 x326 X2 X600  X 7.1 X10-3/5π=884    N.m


Tutorial 4:

Repeat  the same requirement in Ex. 4, if the flux and speed are kept the same , for wave winding.


CHAPTER  4                                             THE MAIN FIELD

The  operation  of  dc  machines  is  based  on the interaction between the armature conductors and the air gap field, which results in induced emf and developed torque. The main field is the field produced by the field coils (or PM’s) on the stator.

To be able to study the main field  by itself , we shall assume there is no armature current, i.e. the machine is operating at no load.

4.1 Main field distribution

If the field coils act alone( i.e. no current in armature conductors), the flux in  the  dc  machine  will have the general pattern  ,such that  most of the flux in the pole cores crosses the air gap to link the armature windings; this is the useful flux per pole  φ. However, some of the flux lines complete their paths without linking the armature wdg;  this is the leakage flux .

The useful flux  φ is produced by the mmf of the field coils. The  mmf per pole  Mf  is the sum of the mmf’s of all coils placed on one pole ( which may be one, or two, or more). Thus

Mf= f.IF  ampere-turn/pole

If you follow the lines of the useful flux, you will find that the path of the useful flux is composed of the following parts in series :  stator yoke  , pole core,  pole shoes, air-gap,  armature teeth, and armature  core.  As  a series circuit, the over-all reluctance is dominated by the highest reluctances  in the path, which are (a) the air-gap, and (b) armature teeth(when saturated).




CHAPTER 6                                               Commutation

The commutator is a characteristic feature of  dc  machines. Its purpose is to match the alternating currents and voltages of the armature  coils to the direct current and voltage of the brushes as already  explained in chapters 1 and 3. However, the commutation process is quite complicated,  and gives rise to secondary effects that place limits on the over-all performance of the machine.

6.1 The process of commutation 

Fig. 6.1 shows a general arm coil  C  moving to the right as it rotates with the armature; it is connected to commutator bars  a   and  b   which move with it. (a) when the coil sides are under the poles,  the coil is part of a certain armature path and carries a path current  Ia 

Ia  =

-(b) As the coil sides approach the q-axis (or brush axis), there will be an instant   t1  at which the brush contacts   bars   a   and   b  simultaneously; thus starts the short circuit of the coil by the brush. (c) The coil continues to be short-circuited by the brush;  it is said to be ‘undergoing commutation’. (d) As  coil sides move away from the q-axis , there will be an instant  t2   at which  bar   b   breaks  contact  with the brush so that the short circuit ends. (e) Coil sides move under poles, and  the coil is now part of a different path;  the coil current is  Ia  again, but in a direction opposite to the original one.

Clearly, then,  the coil is short-circuited for an interval  Tc

  Tc = t2  -  t1

During this interval, the coil current changes from  Ia   to    -Ia  ;  i.e.  it reverses or

‘ commutates’.   As shown in fig. 6.2,  the change in current must follow some time-curve from the point (t1, Ia)  to the point (t2, -Ia).  Depending  on various conditions that will be explained in later sections, we may have linear commutation  (curve 1),  over-commutation (curve 2), or under-commutation (curve 3).

To calculate the SC interval (or commutation interval), let    uc  denote the speed of the bars; thus

uc= 2πrcn

where  rc  is the radius at the commutator surface. From fig. 6.3 , it is seen that the leading edge of bar   a  moves  from  x1  at  t1   to   x2  at   t2;


uc =


Tc=  (                                  (yo=

As expected , the length of the  SC  interval , Tc,  is determined by the speed of rotation n ,  the relative dimensions  of  bars and  brush, and the number  of commutator bars.


 Fig. 6.3 aid to calculate  Tc

Fig. 6.1 The process of commutation>>


                                                                                                                  Fig. 6.2  Reversal of current in coil undergoing commutation

6.2 Equivalent circuit of commutating coil

During  commutating , the coil  SC  current  is   circulates in a path  composed of :  the coil itself,  risers,  bars, contact surfaces, and brush(see fig. 6.4).  A simplified  equivalent  circuit is shown in fig. 6.5 with  :

Rc= coil resistance;

Ec= rotational  emf in coil= 2N(B.L.u)

Lc=self inductance of coil;

r1=contact resistance between brush and trailing bar;

r2=contact resistance between brush and leading bar.

The circuit of fig. 6.5 involves the following simplifications:

1.        the resistance of riser, bar , and brush is negligible w.r.t contact resistance;

2.        mutual inductance with adjacent coils is neglected;

3.        brush assumed to short circuit one coil at a time.

Note that lower-case symbols are used for quantities that are time varying during   Tc ; these are  ec , is , i1 , i2,  r1 , and  r2 ; indeed, is and possibly  ec reverse during  Tc .  Also  note that from  KCL

i 1=Ia  -  is    and i2= Ia +  is


so that

i1   +  i2 =2Ia

as expected .

The terminal voltage of the coil  vc  is given  by

Vc=ec  -  isRc  -  Lc(dis/dt)

The rotational  emf  ec is small because field is small around the q-axis( see ,for example , fig. 5.5).  Lc (dis/dt) is called the reactance voltage ; it is induced by the change in  is.


Fig. 6.4 current distribution in brush and       Fig. 6.5 Equivalent circuit for brush

              Commutating coil.                                    And commutating  coil.


Figure 6.6,     Fig. 6.7  ,  &           Fig. 6.8



6.3 Linear commutation (resistance commutation)

In small machines, the coil voltage  vc is  smaller than  the contact drops  i1r1 and i2r2 .  If  we assume that  vc is negligibly  small, then  the equivalent  circuit  of fig. 6.5  reduces to that of fig. 6.6 . In this case , r1 and  r2  are in parallel  so that by current division


If we further assume that  r1  and  r2  are linear resistances (which in fact they are not) , then


Where the contact areas  A1  and  A2  are defined in fig.  6.7 .  Thus


i.e. the current division between bars is in direct proportion  to their respective  contact areas.  If in the above expression we substitute for i1  and  i2 in terms of  Ia  and  is (see section 6.2) , and rearrange ,  we get

is=                     ( derive this equation ?)

as the commutator slides against the brush at constant speed,  A1  increases linearly with time , while  A2 decreases  linearly with time.  Thus  is   varies  linearly  from  Ia    at   t1   to  -Ia   at   t2   ,  and  we  have linear  commutation  as in curve  1 of  fig. 6.2  . Linear  commutation is  also  called  resistance  commutation  because  the current variation  is controlled  by the contact  resistances   r1   and   r2 ( see first equation in this section).

Note that we  derived linear commutation as an approximation  based  on two assumptions : negligible  vc  and linear contact resistances. These assumptions do not generally hold in practice so that we seldom  have linear  commutation.  Commutation approaches linearity in small machines where these assumptions are approximately true.




6.4  Reactance voltage 

Reactance voltage is the voltage induced in the coil due to the time variation of  is ; it appears across   Lc  in the equivalent circuit of fig. 6.5  , and is equal  to  Lc (dis/dt).  Reactance  voltage has a great effect on commutation process,  so that linear commutation ( which is based on neglecting reactance voltage) is usually to be achieved in practice. The role of reactance voltage in the commutation process may be described qualitatively  as  follows:

The reactance voltage is induced by the change of coil current from  Ia   to   -Ia .  According  to Lenz’s law, the reactance voltage will be induced in such a way as to oppose what is causing  it ,  i.e.  it opposes the change in current.  Therefore  the reactance voltage retards or delays the change in current.

Due to RV, then, the current tends to follow a curve above that of linear commutation, for example curve 3 in fig. 6.2;  the greater the coil inductance  Lc , the higher the curve.

If current reversal is not complete ( i.e. current has not reached  -Ia) when bar   b   breaks contact with the brush  at  t2 ,  the curve will be as shown in fig. 6.8.  This  results in sparking which is explained as follows:

At  t2  the coil  current  attempts  to  jump  to  -Ia  almost instantaneously. This results in very high  RV (why?), which causes breakdown in the air. The arc provides  a path between  brush  and bar   b    through which current flows to complete its reversal  to  -Ia .

Sparking is harmful because it causes heating and hence wear of both brush and commutator  bars.  It  becomes more severe as load increases ( as the armature current  IA increases,  so does the path current  Ia).

Treatment of sparking   

In some small machines, the resistive contact drop is much greater  than the  RV  so that sparking is limited by the effect of resistance commutation,  i.e. commutation approaches the linear case.

In larger machines, some additional means must be found to limit sparking, i.e. to counter the effect of  RV   which is the prime cause of sparking as explained above. Modern machines use interpoles , while older machines (and some small machines) use brush shift;  these two methods  are explained in the following sections:

6.5 Interpoles  ( commutating poles, compoles)

Nearly all integral horsepower machines have interpoles(IP).   IP are narrow poles with large air-gap placed between main poles as in fig. 6.9. Their coils are connected in series with the armature so that the  IP   field is proportional to arm  current  IA (the large air gap prevents saturation in the iron). The  IP  field  acts on commutating coils at the q-axis.

The  IP  mmf  Mi is given by


Where  Ni   is the number of turns in each   IP  coil. The number of turns   Ni  is chosen to make the  IP  mmf  some  25%  grater than  Mam  , the cross-magnetizing armature mmf  at the q-axis(see section  5.1) ; thus

Mi=1.25 Mam      so  that      Ni =1.25 (NC/4pa)

In this way,  Mi   is  made to serve two purposes: (1) the additional  25% neutralizes the commutating coil flux (which induces the RV). This is clear in fig 6.10  a,b, and  c: the IP  field not only reduces the q-axis field to zero,  but drives additional flux in the negative direction to neutralize  RV. Figs. 6.10   d   and   e   show  the resultant field in interpole machines, without and with compensating windings; compare them with figs 5.5 and 5.10 ( NB  IPs treat arm reaction in the q-axis, while compensating wdgs treat arm reaction under the poles).

As the machine is loaded, the armature current  IA increases so that armature reaction  and   RV  increase;  but the  IP    field  is  also  α  to  IA,   and will increase  automatically to neutralize armature reaction (in the q-axis) and   RV .  IPs will continue to do their job properly for either  mode of operation, motor or generator, and  for  either direction of rotation, forward or reverse.

Fig. 6.11  shows the general connection of  a   dc  machine. Not all  windings shown are present in all machines.  IP  or commutating  wdgs are found on integral horsepower machines (rated power  >  one  hp);  compensating  wdgs  are found on large machines and  on some special  machines;  many machines have only one main field wdg,  shunt or series;  compound machines have both. The  terminals of main field wdgs (shunt and series ) are usually brought out to the terminal box to allow user manipulation;  the terminals of compensating and commutating wdgs are not brought out to the terminal box because they are permanently connected in series with the armature.












6.6 Brush shift

A second method for improving commutation to limit sparking is to shift the brushes from the q-axis  .  The  principle  is as follows:

Recall figs. 5.4 and 5.5 which show how the magnetic neutral axis (mna) moves away from the q-axis due to arm reaction . if now the brushes are shifted in the same direction ,  they will be in a region where the arm field opposes the main field. At certain   location the  two fields cancel out;  placing the brushes at this location eliminates the rotational  emf   ec (see fig.6.5).  This is not enough  because there still is the  RV . To neutralize  RV , the brushes are shifted a little further in the same direction;  the sides of commutating coil will then be subjected to  a small (but nonzero) field that opposes the coil flux  which induces the  RV. If the opposing fluxes can be made equal, the  RV  is eliminated.

As a method for improving commutation, brush shift is not as good as  IPs because it has the following disadvantages:

-1. As  the load  on the machine changes, the arm current IA changes so that  AR  and the  mna shift also change. For correct operation, the brush shift must be changed accordingly, which is impractical. In practice, the brushes are placed in a position that gives minimum sparking at rated load, so that there may be considerable sparking at other loads .

2. From fig. 5.4, it is seen that the mna shifts in the direction of rotation for generator operation, and in the direction opposite to rotation for motor operation. Thus brush shift (which is in the same direction as the mna shift) cannot be used with motors intended to run in both directions, or with machines intended for variable mode of operation: if brush shift is correct for one case , it is incorrect for the other!

3. Brush shift causes demagnetizing armature reaction(see fig. 5.12).

Because of these disadvantages , brush shift is used only in small fractional horsepower machines(rated power less than one hp) where it is not economical to use  IPs. Brush shift was also used in old machines before the invention of IPs.  


CHAPTER 7                                    POWER CONVERSION AND LOSSES

The input power to the  dc  machine  undergoes electromechanical conversion to produce the output power; the process yields a number of losses that appear as heat which has harmful effects on the performance of the machine.

7.1 Power balance

Most of the input power supplied to a  dc machine is converted into useful output power; the reminder of the input power is loss and heat. The principle of conversion of energy requires total power balance

Pin=Pout   +   LOSSES

It is sometimes useful to think of power as ‘flowing’ through the machine. Power flow is divided into two stages, the border line being the actual electromechanical energy conversion process.


It is also called the internal power because it is defined within the machine;  in contrast, Pin and Pout  are external powers that can be measured.  EA  and   Td  are internal quantities that cannot be measured directly.

Pin= Pc  +  LOSS1                 and         Pc=Pout  + LOSS2

The total loss is made up of  2 parts: LOSS1    occurs before conversion, and LOSS2  occurs after conversion. Clearly


Motor operation

The  input power is electrical, and the output power is mechanical. Part of the input power is lost as electrical(copper) losses in the windings, and the remainder  is available for electromechanical energy conversion;  part of the converted power is lost supplying the losses due to rotation,  and the remainder is available as a mechanical output power to drive the load. Note that

Pmech< Pc    >>>>>  ωrTL< ωrTd     >>>>>  TLd


That is, the shaft torque available at the load, TL , is less than the developed torque  Td ; the difference is needed to overcome opposing torques within the motor (such as bearing friction).

Generator operation

The input power is mechanical, and the out put  power is electrical. Part of the input is lost as rotational losses, and the remainder is available for electromechanical energy conversion; part of the converted power  Pc  is lost as electrical (copper) losses in the windings, and the remainder is available  as electrical output power to supply the load. Note that

Pmech> Pc  >>>> ωrTpmrTd   >>>> Tpm>Td

That is,  the shaft torque produced by the prime mover,  Tpm , is greater than the developed torque Td ; the difference is the torque needed to overcome friction and other opposing torques.

7.2 Losses

The losses of a  dc  machine are of various types and occur in different parts of the machine. Although different losses are produced differently,  they all appear as heat, i.e. they represent conversion to unless thermal energy. The heat generated by the losses has two major effects:

(i)                    Losses raise the temperature inside the machine, and thus affect the performance and life of the materials of the machine,  particularly insulation. Therefore losses determine the upper limits on machine rating.

(ii)                  Losses  are a waste of energy, and energy costs money; therefore losses result in a waste of money( in the operating cost of the machine).

Losses cannot be eliminated, but they can be reduced by proper design;  the design must also provide for ventilation to disperse the heat generated. Thus losses have a significant effect on the initial cost of the machine.

The cost of wasted energy in item(ii) above is important with industrial motors where the powers involved are quite high ; it is not important with small control motors where the powers involved are very small. However, the temperature rise in item (i) is important for all motors.


Electrical losses

Electrical losses are also called copper losses, winding losses, I2R losses, and ohmic losses. copper  losses occur in all windings due to flow of current through them; they are

LOSSarm=IA2RA=armature circuit copper loss

LOSSser=IS2RS=series field winding copper loss

LOSSsh=If2Rf=shunt field winding copper loss

In computing LOSSarm , RA includes the resistances of commutating and compensating windings(if present). The series field current  IS may or may not be equal to the armature current IA .

The copper loss in a given wdg is proportional to the square of the current in that wdg;  if the current is doubled, the copper loss increases four times.   LOSSarm  and LOSSser depend on armature current,  and hence they depend on the load on the machine ( IA  increases  with load).  LOSSsh depends on the terminal voltage, and varies with its square.

The above expressions can be used to calculate  copper losses using measured values of winding resistances. The wdg resistance must be at the correct wdg temperature ; if the temperature at which the loss is required is not known,  it is assumed to be  75 oC . if the wdg resistance is known (say by measurement) at a temperature  T1 , it can be found at a different temperature T2  from


The brush contact loss is also an electrical  loss. Since the brush contact drop  Vb  is approximately  constant over a wide range of armature currents,  the loss is proportional to the armature current itself( and not its square as in wdg losses):

Losscontact= IAVb

Magnetic losses

Magnetic losses are also called iron losses or core losses. They result from hysteresis and eddy currents in cores subjected to varying magnetization, i.e. mainly in the armature teeth and core, but also in the pole shoes (due to armature slotting).

Iron losses are distributed in the cores in complicated patterns,  so that there are no simple formulae that give their values accurately. It is known, however, that iron losses depend on the magnetization level(flux density) in the cores, and on the frequency with which it alternates,  f=pn.

For the hysteresis loss, we have

LOSShystα fBxmax

Where the constant of proportionality is determined by the volume of the core and its magnetic characteristic(hysteresis loop). The steinmetz exponent  x  depends on the type of iron used, and ranges from 1.5  to  2.5 (usually around  2); it is an empirical constant (obtained from experience ant testing, not from electromagnetic theory). For the eddy current loss, we have

LOSSeddyα f2B2max(lamination thickness)2

Where the constant of proportionality is determined by the volume of the core and its electrical characteristics(resistivity). Clearly, thin laminations reduce eddy current losses. The  armature is always laminated , and the pole shoes are usually laminated. If a motor is to be driven from a modern solid-state controlled rectifier, all cores must be laminated(including poles and yoke).

Mechanical losses

Mechanical losses arise from friction and windage(friction with air) during rotation. They depend on the speed of rotation, each type of mechanical  loss being proportional to some power   of  n.   Bearing friction loss depends on the type of bearing used and on the viscosity of the lubricant;  improper lubrication (too little or too much) increase the loss.

Brush friction loss is proportional to the area of contact and to the brush pressure;  it also depends on the brush and commutator materials, their state of polish, and the temperature at the contact surface; it is often the largest friction loss. Windage losses arise from moving the air around the armature(air friction); they depend on the shape of the rotating surface (smooth or rough). Ventilation loss is an additional windage loss due to fans and vent  ducts used to cool the machine.

Stray load loss

Stray load losses are additional losses that occur in the machine when loaded , and cannot be included with the conventional losses listed above. They include:

-additional core loss resulting from armature reaction distortion;

-copper loss due to short circuit current during commutation(in commutating coils, commutator segments, and brushes);

-non uniform current distribution in large armature conductors.

Stray load losses are small and difficult to calculate. They may be neglected for small machines, and are usually assumed 1% of output for large machines.

7.3 Classification of losses

Table 7.1 Classification of losses in  dc  machines.




With load


Armature circuit copper loss







α IA2


Series field copper loss




α IA2


Shunt field copper loss




α Vt2

Brush contact loss











α Ia



Hysteresis loss







α fBxmax


Eddy current loss











α f2B2max



Friction loss





α power of n

Windage loss





α power of n

Stray load loss







Table 7.2 Typical  values of  dc  machine losses for industrial motors in the range 1-100 KW; lower percentage losses are for the higher rated motors.



Percent of rated power

Armature cct electrical loss

Shunt field electrical loss

Rotational losses






Table 7.3 Typical efficiencies of industrial motors.

Rated power  KW












Constant and variable losses

Constant losses are losses that do not change as the load on the machine changes; they are independent of armature current, and include mechanical losses, core losses, and shunt field winding loss. Variable losses are  losses that increase as the load on the machine increases; they are electrical losses including armature circuit copper loss , series field loss, and brush contact loss. Copper losses increase with  I2A , while brush contact loss increase with  IA   itself. We may therefore write

LOSSEStotal=Ko + K1 IA +K2I2A 

The first term on the RHS represents constant losses, while the second and third terms represent variable losses. At full load, constant losses are 4-20%, and variable losses are 3-6 %; see table 7.2.


Stray load losses are indeterminate functions of armature current and speed. They complicate classification, but are small enough to be neglected in most cases.

7.4 Measurement of losses

There are a number of practical tests to measure  the various machine losses. In most cases, a given test yields the sum of two or more losses together; sometimes the component losses can be separated by further testing.

In testing, it is quite easy to measure electrical quantities (resistance, voltage, and current) and speed, somewhat difficult to measure torque, and quite difficult to measure magnetic quantities (flux and flux density). Powers are determined, on the electrical side, by the product of voltage and current, and on the mechanical side by the product of torque and angular speed.

In a load test, the machine is loaded at a given speed and field excitation(i.e. field current); the input and output powers are measured. The total loss at that speed , excitation, and load can be obtained from

LOSSEStotal=Pin  -  Pout

The total loss can be separated into electrical and rotational losses by calculating  I2R  products in the various windings using wdg currents measured during the load test and (hot) wdg resistances measured previously. With the electrical losses thus calculated, the rotational losses are obtained from

LOSSrotational=LOSSEStotal  - LOSSelec

Load tests for large  machines are impractical in test labs: they require very large loads, and waste large amounts of energy. There are other tests that yield the losses individually.

In a no load test, the machine is driven by a suitable prime mover (possibly another machine) with its terminals open circuited( i.e. it operates as unloaded generator). The input power to the test machine is measured mechanically(torque and speed), or electrically by measuring the input power to the drive motor and subtracting its losses(which must therefore be known). If the test machine is unexcited then  Pin=LOSSmech. If the test machine is then excited but left unloaded we get  Pin=LOSSrot ; the core loss is obtained from 

LOSScore=LOSSrot  -LOSSmech

(Exercise: suggest a test for separating the brush friction loss).

If no suitable drive (prime mover) is available, rotational losses may be obtained by running the machine as a motor with no external load; this is the running light test, or Swinburne test.  

The  input power will mainly go to rotational losses, but there will also be  a little copper loss (why?) . The copper loss may be computed and subtracted from the input power to yield rotational losses. In the running light test, rotational losses cannot be separated into mechanical and core losses (why?).

As seen from the above tests, it is always possible to determine copper losses from the measured  values of wdg currents during the tests, and previously measured wdg resistances. Winding resistances are measured by standard methods (voltmeter-ammeter, Wheatstone bridge, etc.); the wdg temperatures must be monitored at the time of resistance measurement (why?), or the measurement is made with the machine hot (for example directly after a load test).

The no load and running light tests determine machine losses without loading it. Other tests have been devised to operate the machine at full load conditions without requiring an external load.

An example of such tests is the opposition test (Kapp-Hopkinson test ) which requires two identical machines. The machines are coupled mechanically, and connected in parallel with each other to the mains. By increasing  the excitation for one machine and decreasing it for the other, the first will operate as a generator, and the second as a motor: the generator supplies the motor electrically, while the motor drives the generator mechanically; the power input from the mains supplies the losses to keep the system running. By suitable adjustments of the field rheostats , the armature currents can set to their rated values. With the two machines running at or near rating (current , voltage, and speed), they will develop full load losses, yet the disadvantages of a load test have been avoided: no external load is needed, and the mains supplies only machine losses, not their full power.









In a heat run (or temperature-rise test), the machine is run at  full load to develop full-load losses; the test takes 20-60 minutes until the temperature reaches steady  state corresponding to its rated operating value. The no load and running light tests cannot replace the load test in a heat run because they do not generate all machine losses simultaneously. An opposition test, on the other hand, can be used in a heat run.

7.5 Efficiency

The efficiency of a machine is defined by


The  first expression is general, the second is suitable for generators (Pout measured electrically), and the third is suitable for motors (Pin measured electrically). The efficiency is a fraction less than  unity, and is usually expressed in percent. Table 7.3 lists typical values of  dc machine efficiencies.

The general definition of efficiency given above can be analyzed into two component  efficiencies corresponding to the two stages of power flow:


For  a motor, this becomes


and for a generator, it becomes


maximum efficiency

we have already divided losses into constant and variable ; the efficiency was then expressed as in the eq. given in page 74. Now consider motor operation: neglecting the small shunt field current, the input power is


Using     η= = 1 -

η=1 - +K1 +K2 IA)

This gives efficiency as a function of armature current  IA ;  to locate the point of maximum efficiency , we differentiate the above eq. w.r.t  IA  ,and to find the maximum efficiency point , we equate the differential to zero:


And equate to zero , we get


Thus maximum efficiency occurs when the copper losses K2    equal the constant losses 

( or , as an approximation, when variable losses equal constant losses- i.e. assuming the brush contact loss K1IA  is small). Industrial machines are usually designed to have maximum efficiency for  IA between half and full load values( because the machine operates at less than full load most of the time); the exact choice is not critical because the efficiency curve is flat around the maximum value.

For industrial motors, traction motors, and other power-application motors, efficiency is quite important, but it is only one of a number of factors that determine how good a machine is ; the other factors include power/weight ratio, power/cost ratio, reliability, maintenance requirements, vibration and noise, etc. For small control motors, efficiency is of little  importance; the main factors of interest include accuracy , cost, size, speed of response, reliability, noise, weight, interference, etc.




Chapter 9                                    MOTOR OPERATION

In motor operation, a dc machine is supplied electrically, and drives a mechanical load.

9.1 Governing equations

In motor operation, we are interested in the output torque and shaft speed, and their influence on the current drawn by the motor. From previous chapters, the emf and torque equations for a  dc machine are




φ in these equations is the resultant useful flux per pole, i.e. it includes any demagnetization due to armature reaction. From  KVL, we have

EA=V  -  (ΣIR  +Vb)

V  is the voltage applied to the motor terminals,  and  ΣIR is the total series resistive drop( arm wdg, commutating wdg, compensating wdg, series field wdg, and any additional series resistance). To simplify our study, we shall approximate this equation to

EA=V  -  IAR

i.e the brush voltage drop is ignored, and  R  includes all resistances in the path of the armature current. Dividing the above eqn. by Ke.φ , we get


This equation tells us that speed is determined primarily by the applied voltage  V  and the flux  φ , with some reduction due to the series voltage drop IAR (which depends on current, and hence on load torque). The above eqn.s allow us to understand motor operation.

Load :  torque and current

The developed torque  Td   is  slightly  greater than the load torque  TL due to rotational losses:

Td=  TL  + Trot loss

The greater the load on the motor , the greater the current it draws from the supply. The load torque determines the current of the motor.

Actually, the above eqn. holds only under steady-state conditions, i.e. when the speed is constant. Under transient (or dynamic) conditions, the two sides of the equation are not equal, and the difference between them produces an acceleration

J  =Td –(TL  +Trot loss)

Where  J  is the moment of inertia of the rotating parts (rotor, shaft, and load);    is the angular acceleration. The above eqn. is a development from  Newton’s law  F=ma (i.e. it relates to the mechanics of the system).

Suppose that the motor is running at some constant speed so that     Now suppose the load on the motor suddenly increases: the motor will slow down according to the dynamic eqn. above. But this causes the induced  emf  to decrease. The resulting increase in the difference between  V  and EA must be balanced by an increase in the armature current  IA. The increase in current increases the developed torque  Td,  and the initial increase in load torque is thus met. The motor now operates at steady state again, but at a reduced speed.

Note that if the series resistance  R  is small, then only a slight change in speed is sufficient to cause large changes in armature current (and hence in developed torque).

After a disturbance (sudden change in load), the time it takes the motor to settle at a new speed is called the response time. It is determined by the electrical time constant of the motor and the mechanical time constants of the motor and connected load. In certain applications, particularly automatic control systems, the response must be quick, and the motor is designed to have low inductance and low inertia.

The series resistance in the armature circuit is small, so that the induced  emf  EA is approximately equal to the applied voltage  V, hence for a given value of flux φ , the speed is determined primarily by the applied voltage  V.

The flux  φ  is determined primarily by the main field  mmf (i.e. by field current ), and may be controlled by field resistors. The torque is directly proportional to flux  ,but the speed is inversely proportional to it. Thus an increase in flux tends to decrease speed, and a decrease in flux tends to increase speed. Clearly, then, armature reaction tends to increase speed, while the series field in cumulative-compound motors tends to decrease speed.

9.2 Definitions  

We shall need the terms and concepts defined below in our description of motor operation and the factors that affect it.

Mechanical characteristics 

The  mechanical characteristic of a  dc motor is the curve relating the motor’s two output variables, torque and speed; the curve shows how speed changes with load.

We can rewrite the eqn. of speed as follows:


for constant  V  and  φ  , the above eqn. represents a straight line with negative slope, fig. 9.2. The first term on the RHS gives the vertical intercept(no-load speed), and the coefficient of  Td  in the second term gives the slope. The load torque  TL  is a little less than the developed torque  Td , so that the relationship curves below the straight line. The shape of the curve may be further modified due to changes in the flux  φ (which affects both slope and intercept) as the motor load changes; the flux changes with load when there is a series field, and when the demagnetizing effect of armature reaction is not negligible.

When the motor is driving a mechanical load, the torque and speed are found from the motor mechanical characteristic and the load torque-speed characteristic, fig.9.2; that is , the operating point (T1, n1)  is found graphically. Fig. 9.3  shows some typical characteristics of mechanical loads.


The operating point may or may not be stable. In fig. 9.4a it is stable: if the speed suddenly increases from  n  to  n’ , the load torque     will be greater than the motor torque    , causing deceleration back to the operating point (n,T) ; the operating point will also be restored to (n,T) for a sudden decrease in speed (try it).

In fig. 9.4 b, the operating point is unstable : if the speed suddenly increase from  n  to n’ , the motor torque    will be greater than the load torque    , causing acceleration and further increase in speed away from the operating point (n,T) ; a sudden decrease in speed may  result in stall(zero speed). Clearly, then, the stability of the operating point depends on the relative shapes of the motor and load torque-speed characteristics.

Speed control

n=    indicates that the speed may be controlled by means of the applied voltage,  main flux, and the series resistance;  these parameters may be adjusted manually or automatically. Although the armature current  IA  appears in the equation, and hence affects speed, it is not a proper controlling parameter because it cannot be adjusted as desired, but is determined by the mechanical load.

Now a given mechanical characteristic corresponds to a particular setting of the applied voltage , field control resistor, and series resistance. If any of the settings is changed, operation shifts to another curve. Therefore the operating point may be moved from one curve to another by changing the setting of one or more of the control parameters, fig. 9.5. The speed may be kept approximately constant by automatic regulators that sense the shaft speed and adjust one of the control parameters to keep it at the set value.


Speed regulation

The speed regulation of a motor at a given load is defined by


It is a figure of merit that indicates how constant the shaft speed is with load. For many applications, a good drive motor is one which maintains its speed constant over a wide range of loads. The speed regulation of motors equipped with automatic speed control is almost zero.

A low value of speed regulation is not always desirable. There are applications that require the motor to change its speed with load,  for example to keep the torque or output power constant. A main feature of the dc motor is that its operation can be tailored to suit any type of load requirements.


9.3 Constant-flux motors  (permanent-magnet; separately-excited; shunt)

The difference between shunt and separately excited motors is that the field of a shunt motor is fed from the same source as the armature, while the field of a separately excited motor is fed from a different source , possibly at a different voltage. In both cases, constant field voltage and resistance result in constant field current( If does not change with load), and hence constant main field flux. Permanent magnet motors also operate with a constant main field.

If the demagnetizing effect of armature reaction is neglected, the developed torque Td will be directly proportional to the armature current  IA , so that the two variables are related by the straight line shown dotted in fig. 9.6 .  Armature reaction may reduce the flux  φ , and hence reduce  Td , so that the actual relationship between  Td   and   IA  is curved slightly below the straight line. The load torque  TL  is less than the developed torque due to rotational losses, so that the TL  curve is slightly below the  Td   curve, fig.9.6. The relationship between torque and current is sometimes called the torque characteristic of the motor.

For constant-flux motors, the mechanical characteristic is a straight line with a slight negative slope, fig.9.7. Armature reaction may reduce the useful flux and hence increase the speed, so that the mechanical characteristic curves slightly above the straight line. This upward curvature may lead to instability, it is avoided by designing the motor to have no demagnetizing armature reaction (by the use of interpoles ); and  by adding a weak series field to compensate for the reduction in flux (stabilized shunt motor).


The reduction in speed with load is very small for constant-flux motors. The mechanical characteristic is said to be hard, and the motors operate in an essentially constant speed mode.

9.4 Series motor

The main field flux of the series motor changes with load current according to the OCC; therefore the series motor is characterized by variable flux, as opposed to the constant flux motors. At light loads, operation is on the linear part of the OCC, so that

φαIA    and   Tdα

Thus the torque characteristic follows a parabola at light loads, fig. 9.8. At heavy loads, the machine will be saturated so that the flux is almost constant, and operation approaches that of constant-flux motors

φ  constant          >>>>>    TdαIA

The torque characteristic approaches a straight line at heavy loads, fig.9.8.

Applying the same reasoning  to the mechanical characteristic , we see that

At light loads  :    n

And  at heavy loads :  n      (similar to shunt)

The mechanical characteristic will then have the general shape shown in fig.9.9 . The change of speed with load is quite large; the mechanical characteristic is said to be soft, and the series motor operates in a variable speed mode. The motor has a high starting torque, but the torque quickly decreases as speed goes up. At no-load the speed becomes so high that it can damage the motor; therefore series motors are never run unloaded, and are always rigidly coupled to their loads(i.e. belts are never used).


9.5 Compound motors

A compound motor has both shunt and series fields. For cumulative compounding, the motor characteristics will move from shunt c/s  in the direction of series c/s as load increases(i.e. as the series field becomes stronger); see fig.s. 9.10, 9.11. The actual shape of the mechanical c/s is determined by the degree of compounding, i.e. by the ratio  , fig. 9.11. Differential compound motors have rising mechanical c/s because of the reduction in main field flux with load.



 Armature voltage control

In PM and separately-excited motors, the voltage applied to the motor can be varied with the field remaining constant. Different voltages then give different intercepts(different no-load speeds), and we get a family of parallel (i.e. same slope) mechanical c/s as in fig. 9.12.

Similar downward shifts occur for the series motor, fig.9.13. The simplest method of obtaining variable  dc voltage is to use a voltage divider, but this method is impractical and uneconomical; it is used only for testing.

In modern applications, variable  dc voltage for the armature is often obtained from a solid-state controlled rectifier, with the field fed from an uncontrolled rectifier, fig.9.14.  The firing angle of the controlled rectifier may be changed manually, but in practice it is adjusted automatically using a speed signal or armature current signal (i.e. load), or both for optimum control.

In road vehicles, the supply is itself   dc, and hence needs no rectification. Voltage control is often obtained by an electronic chopper circuit. Choppers may use pulse-width modulation  PWM  at constant frequency, or pulse-frequency modulation  PFM  with constant pulse width.

Another effective method for obtaining smooth voltage control is the Ward-Leonard system. The  dc  motor is fed from a  dc  generator driven by some prime-mover (eg  ac motor or Diesel engine). By  varying the field excitation of the generator , the armature voltage of the motor varied (and can be even reserved). The motor field is fed from an exciter (small  dc generator) or rectifier at constant voltage. The Ward-Leonard system is generally more expensive than a solid-state drive, but has compensating advantages for certain applications.

Armature resistance control

For a given load torque, and hence given current, placing an external resistance in series with the armature, reduces the emf and hence speed. The increasing value of resistance increases the slope of the mechanical c/s , but the intercept remains unchanged. Armature resistance control may also be used with series motors ; at heavy loads the machine is saturated and operation approaches that of constant-flux motors with slope increasing as resistance is increased. Armature resistance control is inexpensive and simple to use with small motors, but it is impractical and wastes energy with large motors.

Field control  

This method of speed control may be used with shunt and separately excited motors. If the field circuit resistance is increased, the field current, and hence the main field, will be reduced, and the speed will increase . The higher the field resistance, the higher the intercept and the greater the slope(i.e. the c/s becomes softer).

The flux cannot be reduced indefinitely because the speed becomes too high and may damage the motor. Moreover, if the main field becomes too weak, the demagnetizing effect of armature reaction becomes prominent (relatively large) which may lead to instability.

9.7 Starting

At  the moment the motor is switched on, it is at standstill, so that there is no induced  emf. The entire line voltage is applied across the armature resistance ,since we have

V= Istart.RA     and    Istart=  

The starting current is therefore very high , especially for large motors which have very small armature circuit resistance. The starting current may be more than  20  times rated value, and would damage the motor unless some means is found to limit it.

Once the motor starts to rotate , the internal  emf  begins to build up and thus reduce the current in accordance to   IA=

Note also that the rate at  which the  emf  builds up depends on the rate at which the motor accelerates from standstill which, in turn, depends on the starting torque;

That is , a high starting torque is desirable for rapid initial acceleration(and hence rapid build-up of  emf, and hence rapid reduction of the high starting current).

Direct on-line starting(DOL)

DOL  starting means simply connecting the motor to the supply through a switch. This method can be used only with small motors where (a) the armature resistance is high enough to limit the starting current , and (b) the rotor inertia is small enough to allow rapid acceleration( and hence rapid build-up of  emf  leading  to rapid reduction of current).

Variable voltage starting

Motors supplied from Ward-Leonard sets or controlled rectifiers can be started by raising the supply voltage gradually from zero. The low initial voltage results in a reduced starting current.

Resistance starting

This is the most common method of starting  dc  motors. A specially designed variable resistor is connected in series with the armature, fig. 9.21. When the moving contact is moved from the  OFF  position to the  START  position, all sections of the starting resistor are in the circuit so that the starting current is limited to


The value of  Rstart is chosen to limit the starting current to a safe value, usually 1.5-2.5 times rated current. Although the starting current is still greater than the rated current, it is considered safe because it flows only for a short time.  Moreover , a relatively high current is needed to obtain a high torque for rapid acceleration.

As the motor builds up speed(and hence emf), the starting resistance is cut out section by section until it is totally out of the circuit. During this process, the starting current  and the induced  emf  follow stepped curves of the forms shown in fig.9.22. In principle, a given section is cut out when the current  has fallen to some minimum value, say rated current; upon cutting out section, the current will jump up again to a value limited by the sections remaining in the circuit(maximum safe value).

During starting, full voltage must be applied to the shunt field winding to make the flux maximum; this maximizes starting torque and  emf, and prevents overspeed (the high starting current may cause severe armature reaction, i.e. reduce the flux). For this reason , the starting resistor is connected in the armature circuit and not in the line, and the field control resistor is shorted out during starting.



Fig. 9.22                                                                                    Fig. 9.21 Resistance Starting.


Grading of starting resistance of shunt motors      

Either lower value of current may be fixed or the number of starter steps may be fixed.


Resistance in the circuit on successive studs from geometrical progression , having a common ratio equal to                                                                                 R1





For 4-studs starter:

                                                                                                                 1         2    3          4

                                                                                                         +                                     A                                                                                                                           


                                             When arm  A  goes to point 1   IA=Imax=     where  R1=RA + starter resistance.

As the motor speeds up , its  emf grows and hence decreases  IA  as shown by the curve in fig.9.22. When current has fallen to predetermined value  Imin, arm  A  is moved to stud No. 2 . let the value of back emf be  Eb1  at the time of leaving stud No.1. Then

Imin=                                 ………..(1)

When  A  touches stud No.2, then due to diminution of circuit resistance, the current again jumps up to its previous value   Imax. since speed had no time to change, the back  emf  remains the same as initially.

Imax=                              …………..(2)

From (1), and (2), we get  =          …………(3)

When   A  is held on stud No.2 for some time, then speed and hence back  emf  increases  to a value  Eb2, thereby decreasing  the current to previous value   Imin , SO THAT

Imin=                            …………..(4)

Similarly, on first making contact with stud No.3, the current is

Imax=                           …………….(5)

From  (4), and (5),  we get             ……….(6)

When   A  is held on  No.3  for some time, the speed and hence  back  emf  increases  to a new value  Eb3 ,  thereby decreasing the armature current to a value  Imax such that

Imin=                        ……………….(7)

On making contact with stud No. 4 , current jumps to  Imax       given by

 Imax=                      ………………..(8)

From (7), and (8), we get                     ………(9)

From (3),(6),and (9), it is seen that

    =   =  =  =k (say)      ……..(10)

Obviously,  R3=kRa ,  R2=kR3= k2Ra


In general , if  i  is  the number of live studs and therefore (i-1)  the number of sections in the starter resistance, then

R1=ki-1Ra    or    ki-1

Or  (  )i-1=  ki-1

 or           ki-1= =




since  R1=    and   Ra  are usually known and  k is known from the given values of maximum  and minimum currents(determined by the load against which motor has to start ), the value of  i    can be found and hence the value of different starter sections.

9.8 Braking

When the electric supply to the motor is switched off, the rotation does not stop immediately, but continues until the kinetic  energy of the rotating parts (rotor and load) is dissipated. But in many applications, such as electric trains, vehicles, cranes, and lifts, the motor must be stopped quickly, and hence  some form of braking is required at switch-off.

External braking 

Quick braking can be achieved by an external friction brake mounted on the shaft and operated by a solenoid (electromagnet). At the instant the supply is switched off, the brake is applied to stop rotation. In effect, the kinetic energy of the rotating parts is dissipated quickly as heat in the brake pads.

The eddy-current brake  is another type of external brake. It is made up of a conducting disc mounted on the shaft , and a set of stationary coils adjacent to the disc. At the instant the supply to the motor is switched off, the brake coils are energized to induce  eddy currents in the rotating disc. The field of the coils and the currents of the disc produce a torque that opposes rotation(generator action)  and hence slows the shaft rapidly. In effect, the kinetic energy of the rotating parts is dissipated as heat in the disc of the brake.

Electric braking

Instead of using an external brake, it is sometimes possible to use the properties of the dc machine  itself to achieve quick braking, or to assist in braking.

In dynamic braking  (or rheostatic   braking), a resistor (possibly the starting resistor itself) is connected across the armature terminals at the instant it is disconnected from the supply. With the shunt field still excited, the machine acts as a generator loaded by the resistor; the armature current reverses, and the developed torque now opposes rotation. In effect, the kinetic energy is dissipated as heat mainly in the resistor, but also in the armature wdg. During braking, it is preferable to energize the field from the line and not from the armature; otherwise, braking action stops when the speed falls below the critical value.

A  dc   motor is said to be regenerating when its  emf exceeds the applied voltage so that the armature current reverses and the machine becomes a generator that returns electrical   power to the supply; the source of the power is the kinetic energy of the rotating parts, and hence regeneration slows the motor down. Regenerative  braking uses this principle to aid in stopping the motor or in slowing it down; regeneration is achieved by strengthening the field or by reducing the applied voltage. The main advantage of regenerative braking is the saving of energy, which is returned to the supply and not dumped as heat as in the other methods of braking. It is often used in electric trains to exploit downhill runs, and in cranes to exploit the descending part of the duty cycle.  Regenerative braking can be used only if the electric supply is capable of accepting electrical energy from the motor (eg chargeable batteries or  dc mains); standard controlled rectifiers cannot accept electrical power from the motor unless they are modified for the purpose (eg  by the inclusion of inverters). In regenerative braking, the braking action stops when the speed becomes low enough to reduce the  emf  below the terminal voltage.

A strong braking effect down to zero speed is obtained by plugging(or counter-current braking). The supply connections to the armature are reversed , so that the supply and armature  emf act as series sources aiding each other to circulate  a heavy counter-current


The machine operates in the generating mode with a heavy current, and hence with a strong braking torque. A series limiting resistor is inserted in the circuit to avoid damaging currents; if the starting resistor is used , the plugging current will be twice starting current(i.e. up to 5 times rated current). During plugging, the kinetic energy of the rotating parts plus heavy power from the supply (VIA) are dissipated in the armature winding and the limiting resistor. The supply must be disconnected from the motor at the instant the speed reaches zero, otherwise the motor will run in the reverse direction. Plugging involves such heavy currents and high mechanical stresses that it is used only with small motors. 

9.9 Modes of operation

The four –quadrant diagram helps clarify the various modes of operation of a  dc  motor. The first quadrant corresponds to standard motor operation in one direction, while the third quadrant corresponds to motor operation in the reverse direction. The second and fourth quadrants correspond to generator operation.

Taking the 1st Q as reference , it is seen that motor operation in the reverse direction , 3rd Q , requires reversal of either the applied voltage or the field current. If both are reversed at the same time , motor operation will continue in the same direction, 1st Q.

If initial operation is in the 1st Q , then the 2nd Q corresponds to dynamic or regenerative braking. Plugging also shifts operation to the 2nd  Q, but attempts to continue to the 3rd; the supply is disconnected when the operating point passes through zero speed. If initial operation is in the 3rd Q, i.e. in the reverse direction, then dynamic braking, regenerative braking, and plugging occur in the 4th Q.

For example, if the motor is used in lifts or cranes, we have:

Quadrant  I  : motor raises load;

Quadrant  III: motor lowers load;

Quadrant  II : motor brakes upward(inertia)motion of load;

Quadrant  IV: load moves down by its own weight while motor applies a braking torque to keep speed constant.

9.10 Applications

DC  motors are less common than  ac  motors because  ac  motors are  cheaper, more robust, and require less maintenance, and because standard mains are  ac. However, there are two main types of application for which  dc  motors are more suitable than  ac motors: battery-operated equipment(think about motors used in vehicles) , and applications requiring accurate or flexible control of speed or torque.

Battery-operated equipment includes small portable apparatus (such as DVD players, cassette recorders, etc), cordless tools, and toys, as well as electric drives in road vehicles. These are usually permanent magnet or shunt (fixed –excitation) motors, but occasionally series motors. The high starting torque of the series motor makes it suitable for self-starter duty in cars. Electric vehicles employ permanent magnet or series motors, with speed control by means of choppers(armature voltage control) or armature series resistance.

The accurate control of   dc  motors makes them suitable for servomotor duty in automatic control systems. The motors in such applications generally have small power rating(less than 1 KW) , and are required to drive a load in accordance with a control signal applied to the armature (armature voltage control). They are usually constant field motors(PM or separately-excited) designed to have a low moment of inertia for quick response, and linear mechanical characteristics for accurate control.

The flexibility of control of  dc  motors makes them suitable for certain heavy power applications such as lifts, cranes, hoists , and electric traction(electric trains), as well as certain drives in heavy industry. These applications can involve frequent changes in speed, stops and starts, and possibly reversals. The hard c/s’s of shunt motors with armature voltage control are ideal for adjustable speed drives, while the softer c/s’s of  compound and series motors are sometimes exploited in traction(locomotives) to do without different gear ratios.