1
Some Comments on the Phasor Diagram for
Round Rotor Synchronous Machines
a
x
x
Magnetic Axis of
Phase a
x
x
Magnetic Axis of
Armature Reaction Field
a'
Magnetic Axis of Field Wdg.
'
Figure 1

r
f0
Bso
Bf0
Round rotor synchronous machine conceptual view.
Let us try to derive the phasor diagram from basic principles involving the
physical device. Figure 1 depicts the basic elements of a two pole round rotor
synchronous machine. Only one phase is shown but it is assumed that three
phases exist. The angle  specifies an angular measure along the stator. The
phase a stator winding is assumed to be sinusoidally distributed with a distribution of turns given by
N se
- cos 
N    = -------2
(1)
4
N se = --- k 1 N s

(2)
where
is the effective number of turns, k 1 is the winding factor and N s is the number
of series connected turns.
The field winding produces a rotating magnetic field in the gap with value
12
B f    r  = – B f 0 sin   –  r 
(3)
To check, when  r = 0, and  = -  /2


B f  0 – --- = – B f 0 sin  – --- – 0 = B f 0




2
2
(4)
The total flux produced by one pole of the field winding is

f0
= A p  B f 0 sin   – 0  d = 2B f 0 A p
(5)
0
where A p is the area of one pole.
Assume that the number of stator and field turns are equal. If not the field
quantities can be referred to the stator by the turns ratio in the usual manner,
i.e. same as for a transformer.
The corresponding field flux linkages expressed in terms of stator turns is
N se
-  = N se B f 0 A p
 f 0 = -------2 s0
(6)
thus the flux linkage varies as a function of time as
 f   r  =  f 0 sin  r =  f 0 sin   r t 
(7)
A sinusoidal variation corresponds to – j f 0 on a phasor diagram as shown in
Figure 2.
The flux linkage  f   r  is produced by current flow in the field winding
according to
 f = L mf sin   r I f
(8)
where I f is assumed to be constant. Since the
As the field flux rotates voltage induced in the stator phase a with the value
d
E i = ---------f- =  r  f 0 cos  r =  r L mf I f cos 
dt
(9)
A cosine variation corresonds to a phasor oriented on the real axis as shown on
Figure 2, leading the corresponding flux linkage vector by 90 electrical
degrees.
Assume now that the excitation creates current flow. By convention this
current will be assumed to be directed into the machine. Figure 3 shows the cir-
Some Comments on the Phasor Diagram for Round Rotor Synchronous Machines
13
Im
Re
q-axis
Ei
~
Is
-jf0
d-axis
Figure 2
Partial phasor diagram of a salient pole synchronous
machine.
cuit encountered by the current flow. The reactance x m corresponds to the
inductance associated with the flux produced by the stator crossing air gap
while the reactance x ls corresponds to inductance associated with flux which
does not cross the air gap. Finally there is a resistance corresponding to the stator copper loss. The placement of the three elements in the series circuit is
actually arbitrary but follows convention.
+
~
Vs
_
Figure 3
+
~
Is
jxm
jxls
rs
~
V's
(s)
_
+
+
~
Vm
(m)
_
~
_
E i = r  f = r Lm I f
(f)
Round rotor synchronous machine equivalent circuit.
The voltage drop across the magnetizing reactance is plotted first. This voltage
is clearly perpendicular to the current. The sum of this voltage drop and the
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field voltage produces the key quantity, the air gap voltage. Corresponding to
the voltage drop across this inductor is the associated flux linkage. Just like the
field flux linkage and field voltage, this flux linkage must lag the voltage by 90
electrical degrees. The total flux linkage associated with the air gap is the vector sum of these two flux linkages. They are plotted in Figure 4. Note that the
flux linkage  ms is parallel to the current vector, that the net air gap flux linkage is perpendicular to the air gap voltage vector V m and that the total flux
linkage vector is perpendicular to V s' .
~
rsIs
~
~
Vs
~
jxlsIs
V's
~
Im
Vm
'


~
 f = – j f 0

Ei
m
ms
Re
q-axis
~
~
~
~
jxmIs
~
Is
s
~
ls
d-axis
Figure 4
Completed phasor diagram for a round rotor synchronous
machine.
After having completed this diagram it is important torealize that spatial
information has been obtain from the phasor diagram. Specifically, note that
the flux linkage vector ̃ f indicates where the rotor is located with respect to
the stator magnetic axis of phase a. The flux linkage vector ̃ ms shows the spatial position of flux density produced by the stator with repect to the field flux
density. The flux linkage vector ˜m indicate where the flux density is a maximum and suggests where the iron is most highly saturated.
Some Comments on the Phasor Diagram for Round Rotor Synchronous Machines
15
More information concerning the field quantities and their contribution to
torque if we begin with the relationship that describes the power flow into the
field voltage source E' i . That is
3 P Ei I s
T e = ---  --- --------- cos 
2  2  r
(10)
From Eq. 8,
E i = j r L mf I f = j r  f
(11)
3 P
3 P

T e = ---  ---   f 0 I s  cos   + --- =  – ---  ---   f 0 I s  sin 
 2  2 

2 2
2
(12)
 f 0 = 2N se B f 0 A p
(13)
3
T e = ---  P   N se B f 0 A p I s  sin  –  
2
(14)
3
T e =  P  A p  B f 0   --- N se I s sin  –  
2

(15)
in which case
From Eq. 4
so that
Rearranging
The quantity (3/2)NsIs is the total MMF acting in the gap. The MMF acting
on only one pole is this value divided by P . Therefore
3
 --- N se I s
2
2
T e = P A p B f 0  ----------------- sin  –  
 P 


(16)
The MMF per pole is equavalent to the product of field intensity H times the
gap g (or effective gap to be precise). Eq. 16 becomes
2
T e = P gA p B f 0  H s0  sin  –  
(17)
Finally the total surface area of the rotor A = D is l i is the area of one pole Ap
times the number of pole P. The expression for torque becomes finally,
16
T e = PB f 0 H s0  Ag  sin  –   = PB f 0 H s0  D is l i g  sin  –  (18)
The quantity in the parenthesis is nothing more than the volume of the active
gap V gap . The torque is also expressed as
T e = PB f 0 H s0 V gap sin  –  
(19)
or, if one uses vector notation,
T e = P  B f 0  H s0 V gap
(20)
Hence, the production of torque can be viewed as the interaction of the stator
field intensity with the rotor magnetic flux density. Torque is produced by the
product of the peak flux density times the component of stator field intensity
normal to the flux density. The torque is in the direction to pull the field winding (rotor) in the direction of the stator field intensity. With the vector definition of Eq. 20, the resultant torque becomes a vector in the direction of the axis
of rotation (as would be expected).
[1]
T.A. Lipo, “Introduction to AC Machine Design”, Wisconsin Power Electronic Research Center, University of Wisconsin, 3rd Edition,