Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
Two Pole Three Phase Synchronous Motors
Fig.1 Two pole, three phase synchronous motor
 The stator windings of the synchronous machine are embedded in slots around the
inside circumference of the stationary member. In a 2 pole machine each phase
winding of the three phase stator winding is displaced 120 degrees with respect to
each other as shown in the figure above.
Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
 The field or fd winding is wound on the rotating member. The as, bs, cs, and/d axes
denote the positive direction of the flux produced by each of the windings. The as, bs,
and cs windings are identical in that each winding has the same resistance and the
same number of turns.
 When a machine has three identical stator windings arranged as shown in Fig. above,
it is often referred to as a machine with symmetrical stator windings.
 It is assumed that each coil spans  radians of the stator for a 2-pole machine. One
side of the coil (coil side) is represented by a
indicating that the assumed positive
direction of current is down the length of the stator (into the paper). The
indicates
that the assumed positive direction of current is out of the paper.
 Each coil contains nc conductors. Therefore, in the case of the as winding, positive
current flows in a conductor of coil a1 , then axially down the length of the stator and
'
back the length of the stator and out at coil side a1 . This is repeated for nc conductors.
'
 The last conductor of the coil a1 - a1 is then placed in the appropriate slot so as to
'
start coil a 2 - a 2 wherein the current flows down the stator via coil side a2 and then
'
'
back through a 2 and so on until a 4 .
 The bs and cs windings are arranged similarly, and the last conductors of coil
'
'
'
sides a 4 , b4 , and c 4 are connected together to form the wye-connected stator.
 Each coil consists of nc conductors, each of which makes up an individual single
conductor coil within the main coil. Thus the number of turns of each winding is
determined by the product of nc and the number of coils or the product of nc and the
number of coil sides carrying current in the same direction.
 In the case of the/d winding, each coil (f1-f1’, for example) consists of nf conductors.
 The coil sides of each phase winding are considered to be distributed uniformly over
60° of the stator circumference.
 Another practical consideration is the so-called fractional-pitch winding. The
windings shown in Fig above span  radians for the 2-pole machine. This is referred
to as full-pitch winding. In order to reduce voltage and current harmonics, the
Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
windings are often wound so that they span slightly less than  radians for a 2-pole
machine. This is referred to as a fractional-pitch winding (Short Pitch Winding).
 For the purposes of deriving an expression for the air-gap MMF, it is convenient to
employ the so-called developed diagram of the cross-sectional view of the machine
shown in Fig. 1. The developed diagram is shown in Fig. 2. The length of the air gap
between the stator and rotor is exaggerated in Figs. 1 and 2 for clarity.
Fig 2 Development diagram for 2 pole 3 phase Synchronous motor
 The angular displacement along the stator circumference is denoted  s and r ,
along the rotor circumference. The angular velocity of the rotor is  r , and  r is
the angular displacement of the rotor. For a given angular displacement relative to
the as axis we can relate  s , r and  r as
s  r   r ...........................(A)
 Due to the high permeability of the stator and rotor steel the magnetic fields
essentially exist only in the air gap and tend to be radial in direction due to the
short length of the air gap relative to the inside stator diameter ( Fig 3).
Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET


 Therefore, in the air gap the magnetic field intensity H and the flux density B have a

component only in the ar direction, and the magnitude is a function of the angle  s
whereupon


H r ,s , z  H r s ar And


 
 
 
Br s  0 H r s
Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
 With the assumed direction of the current /as, the magnetic field intensity H in the air
gap due to the as winding is directed from the rotor to the stator for —
from the stator to the rotor for


<  s < and
2
2

3
< s <
2
2
 Ampere's law may now be used to determine the form of the air-gap MMF due to the
as winding. In particular, Ampere's law states that
 
H
 .dL  i ....................................................(1)
Where i is the net current enclosed within the closed path of integration. Let us
consider the closed path of integration depicted in Fig. 3b. Applying Ampere's law
around this closed path denoted as abcda, where the path bc is at  s   .we can
4
write and evaluate (1)
   
r   g  
4 4

 
r 
4
r 0 
 
H dL  
H0 dL  0 ...................(2)
4
r 0  g ( 0 )
   
H  g    H0 g 0   0
4 4
 
 
Where r   and r 0 the radius of the rotor at points b and a and g   and g 0
4
4
are the corresponding air-gap lengths at points c and d respectively.

 The magneto motive force is defined as the line integral of H .Therefore, the terms on
the left-hand side of (2) may be written as MMFs
 
MMF  + MMF(0) = 0.................................(3)
4
Where MMF(0)  -H(0)g(0)
 Let us now consider the closed path aefda where path ef occurs at  s 
7
 here
16
Ampere's law gives
7 
MMF   + MMF(0) = -nci as .................................(4)
 16 
Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
 We can now start to plot the air-gap MMF due to the as winding. For the lack of a
better guess let us assume that MMF (0) = 0. With this assumption, (3) and (4) tell us
that the MMF is zero from s  0 to where our path of integration encircles the first
 
'
coil side a1 .If we continue to perform line integrations starting and ending at a and
each time progressing further in the  s direction, we will obtain the plot shown in
Figure 4.
Fig 4.Plot of the air-gap MMF due to the as winding with the assumption that
MMF (0) is zero.
 In any practical electric machine, the air-gap length is either
 (a) constant as in the case of a round-rotor machine or
 (b) A periodic function of displacement about the air gap as in the
case of salient-pole machine.
In particular, for a 2-pole machine
g  r   g  r   
substituting (A) in above equation w e get
g  s   r   g  s   r    .........................(5)
Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
Above equations is satisfied if the air-gap MMF has zero average value and, for a 2-pole
machine,
MMF (s )  MMF (s   ).................(6)
 Hence, the air-gap MMF wave for the as winding, which is denoted as MMFas, is
shown in figure 5. It is clear that the MMF due to the bs winding, MMFbs, is identical
to MMFas but displaced to the left by 120°. MMFcs is also identical but displaced
240° to the left.
Figure 5 Air-gap MMF due to as winding
 Actually most electric machines, especially large machines, are designed so that the
stator windings produce a relatively good approximation of a sinusoidal air-gap MMF
with respect to  s so as to minimize the voltage and current harmonics. In order to
establish a truly sinusoidal MMF waveform the winding must also be distributed
sinusoidally
 A sinusoidally distributed as winding and a sinusoidal air-gap MMFas are depicted in
Fig. 6. The distribution of the as winding may be written as
 N as  N p sin s

 N as   N p sin s
0  s   
...............(7)
  s  2 
Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
Where
Np is the maximum turn or conductor density expressed in turns per radian
Nr represents the number of turns of the equivalent sinusoidally distributed winding, then

N s   N p sin s ds
0
 N p  cos s 0  N p  cos    cos 0

 N p   1   1
 2 N p ........................................................(8)
Where Ns is the average value of the sinusoidally distributed winding.
Fig 6. MMF due to equivalent as winding.
The MMF waveform of the equivalent as winding is (from Fig 5 and Fig 6)
N
MMFas  s ias cos s .....................(9)
2
It follows that
N
2 

MMFbs  s ibs cos s 

2
3 

............(9a)
Ns
2  

MMFcs 
ics cos s 

2
3  

Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
Let us now express the total air-gap MMF produced by the stator currents. This can be
obtained by adding the individual MMFs, given by
MMFs 
Ns 
2 
2 


ias cos s  ibs cos s 
  ics cos s 
 .......(10)

2 
3 
3 


The above expressions is valid for 2 pole machine in general if P is the number of poles
of the machine then the above equations can be modified as given below



Ns
P
2



MMFbs 
ibs cos s 
........(11)
P
3 
2
N
2  
P
MMFcs  s ics cos s 

P
3 
2
MMFas 
Ns
P 
ias cos s 
P
2 
WINDING INDUCTANCES AND VOLTAGE EQUATIONS
Once the concepts of the sinusoidally distributed winding and the sinusoidal air-gap
MMF have been established, the next step is to determine the self- and mutual
inductances of the machine windings. As in the previous section, it is advantageous to use
the elementary 2-pole, 3-phase synchronous machine to develop these inductance
relationships. Here the damper windings are not considered but this section can be
extended with damper windings.
SYNCHRONOUS MACHINE
 In a magnetically linear system the self-inductance of a winding is the ratio of the
flux linked by a winding to the current flowing in the winding with all other winding
currents zero. Mutual inductance is the ratio of flux linked by one winding due to
current flowing in a second winding with all other winding currents zero including
the winding for which the flux linkages are being determined. For the purpose of
analysis the development diagram is again repeated as shown in the figure given
below
Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
 For this analysis it is assumed that the air-gap length may be approximated as given
below
g r  
1
......................................(12)
1   2 cos 2r
Where 1   2  is g min and
1   2  is g max
When r  0
g r  
1
1
1
1



 g max
1   2 cos 2r 1   2 cos 2(0) 1   2 g min
When r 
g r  

2
1

1   2 cos 2r
1
 
1   2 cos 2 

1
1

 g min
1   2 g max
2
Which is evidnent from the above figure
or
1
g s   r  
.......................(13)s  r   r 
1   2 cos 2s   r 
 MMF can be defined as
Br   0
MMF
.......................................(14)
g
Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
 It is convenient to express the stator MMF in terms of MMF  S  we can write
Br s ,  r    0
MMF s 
.......................................(15)
g s   r 
 The air-gap flux density due to current in the as winding ias  with all other currents
zero may be obtained by substituting (9) and (12) in to (15) we get
MMF s 
N
Br s , r    0
 0 s ias cos s 1   2 cos 2s   r       (16)
g s   r 
2
Similarly, the flux density with all currents zero except ibs  is
Br s , r   0
MMF s 
N
2 

 0 s ibs coss 
1   2 cos 2s   r       (17)
g s   r 
2
3 

Similarly, the flux density with all currents zero except ics  is
Br s , r   0
MMF s 
N
2 

 0 s ics coss 
1   2 cos 2s   r       (18)
g s   r 
2
3 

 In the case of salient-pole synchronous machines the field winding is generally
uniformly distributed and the poles are shaped to approximate a sinusoidal
distribution of air-gap flux due to current flowing in the field winding. In the case of
round-rotor synchronous machines the field winding is arranged to approximate more
closely a sinusoidal distribution.
 We will assume that the damper windings (additional rotor windings) may also be
approximated by sinusoidally distributed windings. Thus, the air-gap MMF due to
current i fd  flowing in the fd winding may be expressed as,
MMF fd  
Nf
2
i fd sin r .....................(19)
 Hence the air-gap flux density due to i fd with all other currents zero may be expressed
as
Br  r   0
Nf
MMF s 
  0
i fd sin r 1   2 cos 2r       (20)
g s   r 
2
Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
 Let us consider the flux linkages of a single turn of a stator winding which spans 
radians and which is located at an angle  s . In this case the flux is determined by
performing a surface integral over the open surface of the single turn. In particular
 s ,  r  
 s 
 B  , rl d ..................(21)
r
r
s
Where
 is the flux linking a single turn oriented s from the as axis
l is the axial length of the air gap of the machine
r is the radius to the mean of the air gap
 is the dummy vari able of integratio n
 Computation of the flux linkages of an entire winding involves a double integral. As
an example, let us determine the total flux linkages of the as winding due to current
flowing only in the as winding. Here
as  Llsias   N as s  s , r d
 s 
 L lsias   N as s   Br  , r rl .d .ds ......(22)
s
Where Lls is the stator leakage inductance due primarily to leakage flux at the end turns.
 Substituting (7ii) in (22) instead (with Np replaced by
Ns
) and (16) for Br  , r  in
2
the above equation we get
2
N
as  Llsias   s sin s
 2
 s 


s
0
Ns
ias cos  1   2 cos 2   r .rl .d .ds ........( 22)
2

N 


 Llsias   s  0 rl  1  2 cos 2 r ias ....................................................(23)
2


 2 
2
 We know that Lasas 
as
ias
therefore dividing (23) by ias we get,

N 


L asas  Lls   s  0 rl  1  2 cos 2 r ....................................................(24)
2


 2 
2
Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
 The mutual inductance between the as and bs windings may be determined by first
computing the flux linking the as winding due to current flowing only in the bs
winding. Thus
 s 
as   N as s   Br  , r .rl .d .ds .......(25)
s
Substituti ng (7) and (17) in the above expression we get
2
N
as    s sin s
 2
 s 


0
s
Ns
2 

ibs cos  
1   2 cos 2   r .rl .d .ds .......(26)
2
3 

 Therefore, the mutual inductance between the as and bs windings is obtained by
evaluating (26) and dividing it by ibs we get,

 
N  

Lasbs   s  0 rl 1   2 cos 2 r  ........(27)
3 

 2  2

2
 Therefore, the mutual inductance between the as and bs windings is obtained by
evaluating (20) expressed in terms of s   r in to (25) Thus
as 
2
N
 2s sin s
 s 


s
0
Nf
2
i fd sin    r 1   2 cos 2   r .rl .d .ds ..................(28)
Evaluating and dividing by i fd yeilds
 
 N  N 

Lasfd   s  f 0 rl  1  2  sin  r .....................................................................( 29)
2 

 2  2 
 The self-inductance of the field winding may be obtained by first evaluating the flux
linking the fd winding with all currents equal to zero except ifd. Thus, with the fd
winding considered as sinusoidally distributed and the air-gap flux density expressed
by (20) Thus
3 2
 fd  Llfd i fd 


2
Nf
2
cos r
 r 


r
0
Nf
2
i fd sin  1   2 cos 2 .rl .d .dr .............(30)
From which
2
 Nf 
 

 0 rl  1  2 ....................................................................(31)
L fdfd  Llfd  
2 

 2 
Where L lfd is the leakage inductane of the field winding
Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
 The remaining self- and mutual inductances may be calculated using the same
procedure. We can express these inductances compactly by defining
Let
2
N 
LA   s  0 rl1............................(32)
 2 

N 
LB   s  0 rl 2 ...........................(33)
2
 2 
 
 N  N f 

0 rl  1  2 .........(34)
Lsfd   s 
2 

 2  2 
2
Lmfd
 Nf
 
 2
2

 

 0 rl  1  2 ..............(35)
2 


 The machine inductances may now be expressed as
Lasas  Lls  LA  LB cos 2 r ............................(36)
2 

Lbsbs  Lls  LA  LB cos 2 r 
.................(37)
3 

2 

Lcsc s  Lls  LA  LB cos 2 r 
.................(38)
3 

L fdfd  Llfd  Lmfd .............................................(39)
1


Lasbs   LA  LB cos 2 r  .....................(40)
2
3

1


Lascs   LA  LB cos 2 r  .....................(41)
2
3

1
Lbscs   LA  LB cos 2 r   .....................(42)
2
Lasfd  Lsfd sin  r .............................................( 43)
2

Lbsfd  Lsfd sin  r 
3

2

Lcsfd  Lsfd sin  r 
3


.................................(44)


.................................(45)

 The voltage equations for the elementary synchronous machine is given by
Lecture Notes
M.Kaliamoorthy A.P/EEE, PSNACET
das
..................................( 46)
dt
d
Vbs  rs ias  bs ..................................( 47)
dt
d
Vcs  rs ias  cs ..................................( 48)
dt
d fd
V fd  rfd i fd 
...............................(49)
dt
Vas  rs ias 
Where rs is the resistance of the stator winding and rf is the resistance of the field
winding. The flux linkages are expressed as
as  Lasasias  Lasbsibs  Lascsics  Lasfdi fd ........(50)
bs  Lbsasias  Lbsbsibs  Lbscsics  Lbsfdi fd ........(51)
cs  Lcsasias  Lcsbsibs  Lcsc s ics  Lcsfd i fd ........(52)
ds  Ldsasias  Ldsbsibs  Ldscsics  Ldsfdi fd .......(53)