Transcript Microelectromechanical Devices

ECE 8830 - Electric Drives
Topic 14: Synchronous Motors
Spring 2004
Introduction
The stator of a synchronous motor is
identical to that of an induction motor.
However, unlike an induction motor, a
magnetic field is created by the rotor
either through the use of permanent
magnets or through a rotor winding with
slip rings and brushes. The presence of
the magnetic field on the rotor allows the
rotor to move at synchronous speed with
the stator field.
Introduction (cont’d)
The rotor shape of a synchronous motor
may be salient or non-salient, i.e. the
airgap may be non-uniform or uniform,
respectively.
Introduction (cont’d)
Synchronous motors are more expensive
than induction motors but offer the
advantage of higher efficiency, an
important advantage at high power.
Thus, synchronous motors are used for
power generation and large motor drives.
A comparison of a 6MW induction motor
and a wound field synchronous motor is
shown on the next slide.
Introduction (cont’d)
Introduction (cont’d)
Salient pole synchronous generators are
used in low speed applications, such as in
hydroelectric power stations where they
are used to match the low operating
speed of the hydraulic turbines.
Non-salient pole synchronous generators
are used in high-speed applications such
as steam-power stations to match the
high-speed steam turbines.
Introduction (cont’d)
In addition to the field winding, the rotor
of a wound field synchronous motor
usually also contains a second winding.
This armortisseur, or damper winding, is
like the short-circuited squirrel cage
bars in an induction motor.
Also, additional damper windings in the
rotor can be used to represent the
damping effects of eddy currents in the
solid iron of the rotor poles.
Introduction (cont’d)
The de-axis is aligned with the North pole
of the rotor and the qe-axis is aligned
orthogonally to the de-axis.
Note: In Ong’s book
the qe-axis leads the
de-axis whereas in
Bose’s book the qeaxis lags the de-axis.
Equivalent Circuit of Non-Salient
Pole Wound Rotor Motor
A simple per-phase equivalent circuit for
a round rotor synchronous motor can be
developed in a manner similar to the perphase equivalent circuit of the induction
motor.
The figure on the next slide shows a
transformer-coupled circuit linking the
stator and the moving rotor winding.
Equivalent Circuit of Non-Salient
Pole Wound Rotor Motor (cont’d)
The rotor is supplied with a current If
produced by a voltage Vf.
Equivalent Circuit of Non-Salient
Pole Wound Rotor Motor (cont’d)
The rotor circuit can be replaced by an ac
current source whose amplitude is If’(=nIf )
and frequency is e. That results in the
equivalent circuit below:
Equivalent Circuit of Non-Salient
Pole Wound Rotor Motor (cont’d)
Neglecting Rm and replacing the current
source and parallel inductance with a
Thevenin equivalent, we get the
equivalent circuit shown below:
Mathematical Model of the Wound
Rotor Motor (cont’d)
The winding inductances for deriving a
mathematical model of a synchronous
machine are shown below:
Mathematical Model of the Wound
Rotor Motor (cont’d)
Before writing down the equations for this
particular circuit, let us examine the
variation of inductances with rotor position.
While the rotor mmf’s will be aligned along
the d- and q-axes, this is not necessarily
true of the stator mmf’s. We therefore
resolve the stator mmf’s in these two
directions.
Mathematical Model of the Wound
Rotor Motor (cont’d)
The resolved components of the a-phase
mmf, Fa, results in flux components along
the d- and q-axes given by:
d  Pd Fa sin r
q  Pq Fa cosr
where Pd and Pq are the permeances along
the d- and q-axes.
Mathematical Model of the Wound
Rotor Motor (cont’d)
The flux linkage of these flux components
with the a-phase winding is:
aa  Ns (d sinr  q cosr )
 Ns Fa (Pd sin2 r  Pq cos2 r )
 A  B cos 2 r
where A 
Pd  Pq
2
and B 
Pd  Pq
2
Mathematical Model of the Wound
Rotor Motor (cont’d)
Similarly, the linkage of the flux
components by the b-phase windings
that is 2/3 are given by:

2

ba  Ns Fa  Pd sin  r sin  r 
3



 A

     B cos 2  r  
3
2

2


  Pq cos r cos  r 
3





Mathematical Model of the Wound
Rotor Motor (cont’d)
Based on these relationships, we can
write expressions for the self-inductance
for the a-phase (excluding leakage
inductance):
Laa  L0  Lms cos 2 r
Similar expressions can be written for the
self-inductances for the b- and c-phases
except that r is replaced by (r-2/3) and
(r-4/3), respectively.
Mathematical Model of the Wound
Rotor Motor (cont’d)
Similarly, the mutual inductances between
the a- and b- phases are given by:
L0


Lab  Lba    Lms cos 2  r  
2
3

Similarly, Lbc and Lca expressions are
obtained by replacing r by (r-2/3) and
(r-4/3), respectively.
Mathematical Model of the Wound
Rotor Motor (cont’d)
The voltage equations for the seven stator
and rotor windings can now be written as:
 v s  rs
v    0
 r 
0  i s  d   s 
  



rr  i r  dt   r 
where:
vs=[va vb vc]T, vr=[vf vkd vg vkq]T,
is=[ia ib ic]T, ir=[if ikd ig ikq]T, rs=diag[ra rb rc],
rr=diag[rf rkd rg rkq], s=[a b c]T, and
r=[f kd g kq]T.
Mathematical Model of the Wound
Rotor Motor (cont’d)
We can now write down the equations for
the flux linkages of the stator and rotor
windings as:
s  Lss i s  Lsr i r
 r   L sr  i s  L r i r
T
where
Lss=

1

1
 


 L0  Lms cos 2  r    L0  Lms cos 2  r   
 Lls  L0  Lms cos 2 r
2
3
2
3




 1


1


L

L
cos
2


L

L

L
cos
2


L

L
cos
2







0
ms
ls
0
ms
r
0
ms
r
 r

3
2

 2

 1


1


L

L
cos
2



L

L
cos
2



L

L

L
cos
2

 r 


0
ms
0
ms
ls
0
ms
r
 r

3
2

 2

Mathematical Model of the Wound
Rotor Motor (cont’d)
Lrr=
Lsr=
 Llf  Lmf
 L
kdf

 0

 0


Lsf sin  r


2

L
sin


 sf
 r
3



2

 Lsf sin  r 
3


L fkd
0
Llkd  Lmkd
0
0
Llg  Lmg
0
Lkqg
Lskd sin  r
2


L
sin


skd

 r
3


2


L
sin


skd

 r
3




0

Lgkq 

Llkq  Lmkq 
0
Lsg cos  r
2


L
cos


sg

 r
3


2


L
cos


sg

 r
3


Lskg cos  r
2


L
cos


skq

 r
3


2


L
cos


skq

 r
3











Mathematical Model of the Wound
Rotor Motor (cont’d)
We notice from these equations that the
elements of Lss and Lsr are functions of
the rotor angle r which is varying with
rotation of the rotor. This is the same
problem that we encountered with the
induction motor and the solution was to
transform to the rotor reference frame.
We do the same thing here - Park’s
transformation to the rescue once more!
Mathematical Model of the Wound
Rotor Motor (cont’d)
Recall the Park’s transform matrix is:

 cos  r

2
Tqd 0 ( r )     sin  r
3
 1

 2
2 
2  


cos  r 
cos



 r

3 
3 



2 
2  


 sin  r 

sin



 r

3 
3 



1
1

2
2

and the inverse transform matrix is:


cos  r

1
 
2
Tqd 0 ( r )    cos  r 
3
 
 
2
cos  r 
3
 
 sin  r
2



sin



 r
3


2



sin


r


3









1


1


1

Mathematical Model of the Wound
Rotor Motor (cont’d)
Applying the Park’s transform to the stator
voltage equations, we get:
1
qd 0 qd 0
v qd 0  Tqd 0rs T i
 Tqd 0
d 1
Tqd 0  qd 0
dt
If ra=rb=rc=rs, the first term simplifies to:
1
qd 0 qd 0
Tqd 0rs T i
 rs iqd 0
Mathematical Model of the Wound
Rotor Motor (cont’d)
Now,
Tqd 0
d 1
 d 1 

1 d
Tqd 0 qd 0  Tqd 0  Tqd 0  qd 0  Tqd 0 qd 0 
dt
dt

 dt

It can be shown that,
d 1
Tqd 0  qd 0
dt


 sin  r


2

  r   sin  r 
3



2

  sin   r 
3


d r
where  r 
.
dt
cos  r
2


 cos   r 
3


2


 cos   r 
3









0


0   qd 0


0

Mathematical Model of the Wound
Rotor Motor (cont’d)
and that
 d 1 
Tqd 0  Tqd 0   qd 0
 dt

Since,
1
qd 0
Tqd 0 T
 0 1 1
  r  1 0 0   qd 0
 0 0 0 
d
d
 qd 0   qd 0
dt
dt
the stator voltage equations in the rotor
qd reference frame become simply:
v qd 0
 0 1 0
d
 rs i qd 0   r  1 0 0   qd 0   qd 0
dt
 0 0 0 
Mathematical Model of the Wound
Rotor Motor (cont’d)
The flux linkages can be obtained in a
similar manner by only transforming the
stator quantities :
1
qd 0 qd 0
qd 0  Tqd 0LssT i
 Tqd 0Lsr i r
The resulting equations are:
3


q   Lls  ( L0  Lms )  iq  Lsg ig  Lskqikq
2


3


d   Lls  ( L0  Lms )  id  Lsfd i f  Lskd ikd
2


0  Llsi0
Mathematical Model of the Wound
Rotor Motor (cont’d)
The rotor winding flux linkages do not
need to be transformed. The expressions
for the rotor flux linkages are:
3
 f  Lsf id  L ff i f  L fkd ikd
2
3
kd  Lskd id  L fkd i f  Lkdkdkd ikd
2
3
g  Lsg iq  Lgg ig  Lgkq ikq
2
3
kq  Lskq iq  Lgkqig  Lkqkqikq
2
Mathematical Model of the Wound
Rotor Motor (cont’d)
The 3/2 terms in the above rotor
equations will result in non-symmetric
inductance coefficient matrices. Therefore
we multiply the actual rotor current terms
by 2/3 to produce equivalent rotor
currents which result in symmetric
inductance coefficient matrices, i.e.
2
2
i f  i f ; i kd  ikd ;
3
3
2
i g  ig
3
2
; i kq  ikq
3
Mathematical Model of the Wound
Rotor Motor (cont’d)
The rotor quantities can be referred to
the stator as described in the Ong text
pp. 267-269. The resulting equations
in the rotor’s qd0 reference frame
(with rotor quantities referred to the
stator) can be summarized as shown
on the next slide.
Mathematical Model of the Wound
Rotor Motor (cont’d)
Mathematical Model of the Wound
Rotor Motor (cont’d)
The qd0 equivalent circuits from these
equations are shown below:
Mathematical Model of the Wound
Rotor Motor (cont’d)
The power into the machine is given by:
Pin  vaia  vbib  vcic  v f i f  vgig
Using the transformations of the stator
quantities to the rotor qd0 reference
frame, this equation becomes:
3
Pin  vq iq  vd id  3v0i0  v f i f  vg ig
2


Mathematical Model of the Wound
Rotor Motor (cont’d)
With further algebraic manipulation and
removing the ohmic loss and rate of
change of magnetic energy terms, the
electromechanical power developed by
the motor can be expressed as:
3
Pem   r (d iq  q id )
2
For a P-pole motor with rotor speed rm
mechanical radians/sec. we can write:
3P
Pem 
 rm (d iq  qid )
22
Mathematical Model of the Wound
Rotor Motor (cont’d)
Thus the electromechanical torque
provided by the motor is given by:
Tem 
Pem
 rm
3P

(d iq  q id )
22
Steady State Operation
Assuming balanced, steady state conditions
with the rotor rotating at synchronous
speed, e, and the field excitation held
constant, we can write the stator phase
voltages as:
va  Vm cos et
2 

vb  Vm cos  et 

3 

4 

vc  Vm cos  et 

3 

Steady State Operation (cont’d)
The steady state stator currents flowing
into the motor are given by:
ia  I m cos et   
2 

ib  I m cos  et   

3 

4 

ic  I m cos  et   

3


Steady State Operation (cont’d)
At this stage, we do not know the relative
orientation of the rotor’s qr axis and the
synchronously rotating qe axis but since
the qr axis rotates at synchronous speed,
the relative orientation will be constant in
time.
Let us define an angle  (known as the
power or torque angle) which represents
the phase difference between VA and Ef
(=r(t)- e(t)).
Steady State Operation (cont’d)
In steady state, the qd voltage equations
of the stator windings in the rotor qd
reference frame may be written as:
vq  rsiq  e Ld id  E f
vd  rsid  e Lqiq
where Ef is the steady state field excitation
voltage on the stator side given by:
Ef
 v 'f
 e Lmd  '
r
 f



Steady State Operation (cont’d)
These equations can be used to derive an
expression for  (see Ong pp. 274-276).
The resulting expression is:
tan   
rs I m sin   e Lq I m cos 
Vm  rs I m cos   e Lq I m sin 
For a non-salient pole motor, the torque can
be shown to be given by (see Bose pp. 7980):
P
Te  3   s I s cos 
2
Steady State Operation (cont’d)
For a salient pole motor, the torque can be
shown to be given by (see Bose pg. 81):

L

L


ds
qs
 P   s  f
Te  3   
sin   s2
sin 2 
2Lds Lqs
 2   Lds

The torque- curves for the two types of
motor are shown below:
Phasor Diagrams
The phasor diagrams for motoring mode
are shown below. See Ong pg. 278 for
details.
Leading PF
Lagging PF
Simulation Model of Three-Phase
Synchronous Motor
The winding equations derived earlier
can be used for a simulation model.
The model inputs include:
1) the stator abc phase voltages,
2) the excitation to the rotor field
windings, and
3) the applied mechanical torque to
the rotor.
Simulation Model of Three-Phase
Synchronous Motor (cont’d)
The simulation model outputs are:
1) the three stator abc phase currents,
2) the field current in the rotor,
3) the electromagnetic torque generated
by the motor,
4) the speed of the rotor,
and 5) the torque angle of the motor.
Simulation Model of Three-Phase
Synchronous Motor (cont’d)
The first step in developing the simulation
model is to transform the abc phase
voltages to the qd reference frame
attached to the rotor. This can be
performed in two steps by first
transforming to the stationary reference
frame and then to the rotating reference
frame of the rotor.
Simulation Model of Three-Phase
Synchronous Motor (cont’d)
The transformation from the abc phase
voltages to the stationary reference frame
is given by:
2
1
1
v  va  vb  vc
3
3
3
1
s
vd 
(vc  vb )
3
s
q
1
v0  (va  vb  vc )
3
Simulation Model of Three-Phase
Synchronous Motor (cont’d)
The transformation from the stationary
reference frame to the rotor reference
frame is given by:
vq  v cosr (t )  v sinr (t )
s
q
s
d
vd  v sinr (t )  v cosr (t )
s
q
s
d
t
where  r (t )   r (t )dt   r (0)
0
Simulation Model of Three-Phase
Synchronous Motor (cont’d)
For the case of a wound rotor synchronous
motor with one field winding in the d-axis
and a pair of damper windings in the dand q-axes, we can write integral equations
for the winding flux linkages as:


r

 q  b  vq  r  d  s ( mq  q ) dt
b
xls


 kq' 


rs
r
 d  b  vd   q  ( md  d ) dt
b
xls




rs
 0  b   v0   0 dt
xls 

 kd' 
 
'
f
b rkq'
'
xlkq
b rkd'
'
xlkd
b rf' 
xmd
 
'


mq
kq dt
 
md
 kd' dt
xmd
'
E




  f xlf'  md f


dt

Simulation Model of Three-Phase
Synchronous Motor (cont’d)
In the previous expressions,
 mq  b Lmq (iq  ikq' )
 'f  xlf' i'f  md
 md  b Lmd (id  i  i )
'
kd
E f  xmd
v 'f
r
'
f
 q  xlsiq  mq
'
f
  x i  md
'
kd
'
'
lkd kd
  x i  mq
'
kq
 d  xlsid  md
'
'
lkq kq
 0  xlsi0
Note: All of the above equations assume
a motoring current convention.
Simulation Model of Three-Phase
Synchronous Motor (cont’d)
From the flux linkages, we can get the dand q-axis stator, and d- and q-axis and
field rotor (referred back to the stator)
winding currents as follows:
'

 q  mq
kq  mq
'
ikq 
iq 
'
xlkq
xls
id 
 d  md
xls
ikd' 
i 'f 
 kd'  md
'
xlkd
 'f   md
xlf'
Simulation Model of Three-Phase
Synchronous Motor (cont’d)
We can now get the abc phase winding
currents by a two-step transformation.
First we transform from the rotor qd0
reference frame to the stationary qd0
reference frame. This is accomplished
through the below transformation:
iqs  iq cosr (t )  id sinr (t )
i  iq sinr (t )  id cosr (t )
s
d
Simulation Model of Three-Phase
Synchronous Motor (cont’d)
The second step is to transform from the
stationary qd0 reference frame back to
abc using the following transformation:
ia  iqs  i0
1 s 1 s
ib   iq 
id  i0
2
3
1 s 1 s
ic   iq 
id  i0
2
3
Simulation Model of Three-Phase
Synchronous Motor (cont’d)
For a motor,
net acceleration torque =Tem+Tmech-Tdamp
From Newton’s 2nd law of motion applied to
a rotating body, we have:
Tem  Tmech  Tdamp

P
r (t )  e 
2J
d rm (t ) 2 J d r (t )
J

dt
P dt
t
 T
em
0
 Tmech  Tdamp  dt
Simulation Model of Three-Phase
Synchronous Motor (cont’d)
In this equation, the electromagnetic
torque produced by the motor can be
calculated as shown earlier by:
Tem 
Pem
 rm
3P

(d iq  q id )
22
Also, the torque angle  can be calculated
in simulation from the equation:
 (t )   r (t )   e (t )   r (t )  e dt   r (0)   e (0)
t
0
Simulation Model of Three-Phase
Synchronous Motor (cont’d)
The flow of
variables for
simulation of
the synchronous
motor is shown
for the q-axis
circuit here and
for the d-axis
circuit on the
next slide.
Simulation Model of Three-Phase
Synchronous Motor (cont’d)
Per-Unit Expressions for
Synchronous Motor


Normalization of machine parameters allows
for standardized comparison of different
machines. The per-unit (p.u.) expressions for
synchronous motors are as follows:
Base Power, Sb, is rated kVA of machine
Base Voltage, Vb, is peak voltage,
2Vlinetoline
i.e. Vb 
3



Base Current, Ib, is peak current =2Sb/3Vb
Base Impedance, Zb=Vb/Ib
Base Torque, Tb = SbP/2b ,where b is base
electrical angular frequency.
Synchronous Motor Parameters
Synchronous motor parameters from
manufacturers are usually in the form of
reactances, time constants, and resistances
which are derived from stator measurements.
Two time constants appear in the transient
behavior of the rotor - a subtransient period
(during the first few cycles of a short circuit
of the windings) when the current decay is
very rapid. This is followed by the transient
period in which the current decay is slower.
These two time constants are denoted as Tdo’
and Td0”.
Synchronous Motor Parameters
(cont’d)
The reactances, resistances and time
constants can be used to calculate the
synchronous motor parameters as
described on pp. 302-304 Ong.
A sample calculation of synchronous
motor parameters is also provided in
this section.
Higher-Order Models
Significant error is found in some cases
between simulations using the model
described earlier and actual synchronous
motor performance, especially in the rotor
winding components of motors with solid
iron rotors. The main source of error is
the shielding effects of the damper
winding currents and the eddy currents in
a solid rotor.
Higher-Order Models (cont’d)
A third circuit to account for the eddy
currents induced in the pole surface can
be incorporated into the model. The three
current components in the rotor slot are
shown below:
Higher-Order Models (cont’d)
The field current path is deep in the
slot and links the slot leakage flux
components shown.
The damper current path is higher up
in the slot and links r1c partially and all
of r2c directly above it.
The eddy currents flow in the surface of
the slot and partially links r2c but not
r1c.
Higher-Order Models (cont’d)
The qd0 circuits need to be modified
from the earlier circuits to incorporate
additional inductances Lr1c and Lr2c to
account for the fluxes r1c and r2c in the
rotor slot and L’1c, L’2c, and L’3c to
account for the self-leakages of the field,
damper and eddy currents.
The modified qd0 circuits are shown on
the next slide.
Higher-Order Models (cont’d)
Higher-Order Models (cont’d)
The simulation equations for the
higher order model represented by
these modified circuits are developed
in sec. 7.8.1 (pp. 307-309) in Ong’s
book.