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Erasmus LLP Intensive Programme
Powering the Future
With Zero Emission and
Human Powered
Vehicles
Antoni Garcia Espinosa
UPC
Erasmus LLP Intensive Programme
Direct Torque Control
DTC
Erasmus LLP Intensive Programme
Introduction
 This
Thi technique
t h i
involves
i
l
the
th respective
ti control
t l off torque
t
and stator flux within two distinct bands. This means
that the torque and flux are constrained to lie
respectively within a set of upper and lower limits.
 Contrary
C t
t the
to
th PWM technique,
t h i
th switching
the
it hi
frequency is not constant, but depends upon the
instantaneous values of the torque TM developed by
the motor and the flux S of the stator.
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 Three-phase converter wherein the instantaneous position
of the switches is determined by the flux S of the stator and
the torque TM developed by the motor. The flux is allowed to
have any value between A and B. The torque is allowed to
h
have
any value
l b
between
t
TA and
d TB.
 The nominal value of S corresponds to the average value of
A and B .
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Controlling the speed
 Speed
S
d control
t l iis d
done via
i th
the ttorque TM. Thus
Th when
h th
the
speed is lower than a desired value, the control circuit
raises the levels of both TA and TB. Consequently,
Consequently the
torque developed by the motor is suddenly below TB
and the system
y
reacts to increase the torque. Therefore
the motor accelerates. When the speed has attained
the desired value TM fluctuates between the new
settings
tti
TA and
d TB. During
D i thi
this iinterval,
t
l th
the same
switches continue to keep the flux within the levels A
and B.
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Producing a magnetic field in
a 2-phase motor
 We
W representt the
th stator
t t windings
i di
by
b two
t
phases
h
X and
d
Y that are at right angles to each other. Each pole
contains 10 turns
turns, thus making a total of 20 turns
between terminals x1 - x2 and y1 - y2. The windings
respectivelyy produces X and Y. Suppose the nominal
flux per pole is 25 mWb.
 Windings are connected to a 200 V dc source by way
of a converter that comprises four switches.
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Producing a magnetic field in
a 2-phase motor
 Terminals
T
i l x1 - x2 : (+
( -)) (( +)) (+
( +)) (( -)) When
Wh the
th polarity
l it
is the same the terminals are short-circuited. So, there
are only three distinct ways to make the connections
connections.
The same remarks apply to terminals y1 - y2 . Therefore
there are 9 distinct ways
y wherebyy the X and Y windings
g
can be connected to the (+) (-) terminals of the source.
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Producing a magnetic field in
a 2-phase motor
The four switches offer nine ways of
connecting the windings X and Y to the
200 V dc source. At anyy given
g
instant at
least one of the two windings is in shortcircuit.
Schematic diagram of a 2-phase
2 phase induction
motor. The value and direction of fluxes X
and Y depends upon the volt-seconds
applied
pp
to the respective
p
winding.
g
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 Rate
R t off change
h
off flux:
fl
 X E X

t
N
 In our case EX=E
Ed= 200 V and N=20
N 20 turns:
 X
200 V
Wb
mWb

 10
 10
t
20 turns
s
ms
 When EX is zero (terminals in short-circuit), the flux
does nott change,
d
h
it remains
i att th
the value
l it h
had
d when
h
the short-circuit took place.
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Step 1: 0<t<2,5 ms; EX= 200 V; EY=0 V
 Suppose
S ppose that the initial flflux in the motor is zero;
ero so that
X = Y =0. We then close the switches so that EX= 200
V and EY=0
0V
V. Flux X will begin to increase to the right
at a rate of 10 mWb/ms. It will reach its nominal value
of 25 mWb after a interval of 2,5 ms.
 We do not want to increase the flux beyond this value,
and so we short-circuit
short circuit terminals x1 and x2 at the end of
this step.
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After a interval of 2,5
2 5 ms:
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Step 2: 2,5<t<5
2 5<t<5 ms; EX= 0 V; EY=200 V
 While terminals x1 and x2 are still short-circuited, we
close the switches so that EY=200 V. Flux Y is initially
zero, but immediately begins to increase, directly
upwards The applied voltage is kept unchanged until
upwards.
Y reaches its nominal value of 25 mWb. The time
required
q
to reach this value is ∆t=25 mWb/(10
(
mWb/ms)= 2,5 ms.
 Since flux Y must nor exceed 25 mWb
mWb, terminals y1
and y2 are short-circuited so that EY=0 V at the end of
p
this step.
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Step 3: 5<t<10 ms; EX= -200
200 V; EY=0 V
 During this step, we apply a negative voltage EX across
winding X. Consequently, flux X will tend to be directly
to the left. Increasing at a rate of 10 mWb/ms. Since
the initial value of X is 25 mWb
mWb, it will become to zero
after an interval of 2,5 ms.
 If we keep the switches in the same state, flux X will
continue to build up towards to the left. It will reach its
nominal value of 25 mWb after a further interval of 2
2,5
5
ms. At this very moment, we short-circuit terminals x1
and x2 so that the flux stops changing.
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Step 4: 10<t<15 ms; EX= 0V; EY=-200 V
 We now apply voltage EY=-200 V across coil Y.
Consequently, flux Y tends to be oriented downwards,
increasing negatively at a rate of 10 mWb/ms.
 After 5 ms, flux Y= -25
25 mWb and we short-circuited
short circuited the
terminals y1 and y2.
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Step 5: 15<t<20 ms; EX= 200 V; EY=0 V
 Flux X increases, while being directed to the right.
When it reaches +25 mWb,, terminals x1 and x2 are still
short-circuited so that the flux ceases to increase any
more.
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Step 6: 20<t<22,5 ms; EX= 0V; EY= 200 V
 Flux
Fl Y is
i di
directly
tl upwards,
d iincreasing
i positively
iti l att a
rate of 10 mWb/ms. Since its initial value is -25 mWb, it
will reach 0 mWb after an interval of 2
2,5
5 ms
ms.
 We see that the resultant flux S has made a complete
turn.
We see that the resultant flux S has made a complete turn in 20 ms
which corresponds to a speed of 50 turns per second or 3000 rev/min.
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Changes in X and Y produced by
successively applying +/-200 V to the X
and
d Y windings.
i di
This diagram shows the value and
position of the flux S at different
instants. It also enables us to visualize
the components X and Y.
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Problems
 For example at instant t=6
t 6 ms where X = +15
15 mWb
and Y = 25 mWb, the resulting flux is:
S 

2
X
 Y
2

152  252  29.1 Wb
 At the four corners the resulting flux is:
S 

2
X
 Y
2

252  252  25  2  35 Wb
 This is 40% greater than the nominal flux and this
situation must be corrected.
 That is the purpose of the tolerant band.
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Controlling the magnetic flux
 We can obtain a more uniform magnetic flux by
imposing upper and lower limits to its value.
 Suppose, for example, that we wish to restrict the
values of flux between 25 mWb and 28 mWb (1 pu and
1,12 pu) . To do so we draw two circles having radii
corresponding to 25 mWb and 28 mWb, respectively.
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Controlling the magnetic flux
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Controlling the magnetic flux
 Starting from point 1, we keep X constant and apply
EY= 200 V to coil Y. As we have already stated, this
causes flux Y to increase upwards at the rate of 10
mWb/ms As soon as the resultant flux is equal to 28
mWb/ms.
mWb (point 2 of the external circle), we reduce EY to
zero byy short-circuiting
g y1 and y2.
 Next we apply EX= -200 V, which diminishes X without
affecting Y. Having arrived at point 3 on the internal
circle where S= +25 mWb, we reduce EX to zero. Then
we
eb
briefly
e y app
apply
y EY= 200
00 V,, which
c raises
a ses the
e flux
u to
o
point 4, after we set EY= 0 V. Again applying EX= -200
V, the flux is displaced to the left. Arriving at point 5
where S= 25 mWb, we set EX= 0 V.
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Controlling the magnetic flux
 The bang-bang method of control forces the flux S to
stay
t within
ithi th
the lilimits
it off 25 mWb
Wb and
d 28 mWb.
Wb Th
The
short vertical and horizontal zig-zag lines show the path
followed by the flux during one revolution
revolution. The value
and position of the flux are indicated at each instant by
the position and amplitude of the vector S.
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Controlling the magnetic flux
 Proceeding
P
di
thi
this way between
b t
the
th limits
li it iimposed
db
by th
the
two circles, we obtain the revolving flux S which
passes through points 1
1, 2
2, 3
3, …18,
18 19
19, 20
20.
 The result is 20 commutations per revolution compared
tto 4 in
i th
the previous
i
case. IIn thi
this manner, the
th amplitude
lit d
of the flux is kept at 26,5 mWb+/- 6%.
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Controlling
g the magnetic
g
flux
The chopped voltage waveshapes
look like PWM, but the switching
periods are not constant as they
would be if the PWM technique were
used
used.
We can reduce the difference between
the upper and lower values of flux S by
reducing the tolerance band
band. For
example by choosing S between 25
and 26.5 (1 pu and 1.06 pu) we can
attain a precision of +/- 3 %
%. However
However,
this requires 44 commutations per
revolution. Since one revolution is made
in 20 ms
ms, this amounts to 44/20 ms =
2200 commutations per second.
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Controlling the speed of rotation
 It is important to note that the increase of the number of
commutations does not affect the time needed to
complete one revolution. As long as the minimum flux
level S is 25 mWb,
mWb the time to complete one revolution
will always be 20 ms. Consequently, the average speed
of rotation will always be 3000 rev/min. To generalize,
the speed of rotation nR, is given by the expression:
nR 
k Ed
S
nR: speed of rotation of the flux S [rev/min]
Ed: dc voltage of the source [V]
S: rated flux per pole [Wb]
p
upon
p the construction of the motor; number
k: Constant that depends
of turns per pole, etc..
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Controlling the speed of rotation
 To increase the speed
speed, we can rise the dc voltage above
200 V for example to 300 V, thus the speed will increase
from 3000 rev/min to 4500 rev/min. In practice, however the
supply voltage Ed is fixed.
 Another way is to reduce the flux S, this is equivalent to
reduce the diameter of the circles, while keeping the voltage
Ed at 200 V.
 A third method consists of introducing “zeros” during
which moments the voltages EX and EY are simultaneously
kept at zero (both windings in short-circuit). During this
intervals, the flux S remains frozen in the space. This
increases the time required to complete one turn
turn.
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Controlling the speed of rotation
 If we introduce 40 zeros each of 2 ms duration, the time
required to make one turn is 20 ms+40x2 ms= 100 ms
ms.
This corresponds to an average speed of 600 rev/min.
 Note that when EX and EY are simultaneously zero, the
flux S is momentarily in space. However as soon as
the short
short-circuit
circuit is removed
removed, the flux again rotates at a
speed of 3000 rev/min. Thus the flux can be made to
constantly advance and stop, its speed fluctuating
between 3000 rev/min and zero.
 The windings are short-circuited
short circuited when the motor
torque TM is greater than torque TA. This means that
the “zeros”
zeros are generated by the control process
itself.
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Programming the logic switching
procedure
 Knowing
K
i th
thatt the
th flux
fl limits
li it are 25 mWb
Wb tto 28 mWb
Wb and
d
the Torque limits: TA to TB
 How can we program this switching process to attain
these objectives?
 Suppose: the flux S has the momentary value and
p
position
indicated by
y flux vector S1 and that is rotating
g
at counterclockwise at 3000 rev/min. and the rotor is
turning in the same counterclockwise direction, but a
constant
t t speed
d off 600 rev/min.
/ i
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Programming the logic switching
procedure
 Since S1 is less than 25 mWb
mWb, windings X and Y must
be activated so as bring the flux within the desired
zone. Five options
p
are then p
possible:
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Programming the logic switching
procedure
1. Apply
pp y a voltage
g of 200 V to winding
g X. This will move
the flux to the right.
2 Apply a voltage of -200
2.
200 V to winding X
X. This will move
the flux to the left. REJECTED
3. Apply a voltage of 200 V to winding Y. This will move
the flux upwards.
4. Apply a voltage of -200 V to winding Y. This will move
the flux downwards. REJECTED
5. Apply zero voltage to both windings by short-circuiting
them. REJECTED
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Programming the logic switching
procedure
 To determine which one of the two is preferred we
must look at the position of the motor torque TM.
 If we apply 200 V to winding X, the flux S1 while
increasing to S1A it will move clockwise, which it is
opposite to the direction of the rotation of the rotor. This
choice is appropriate if torque TM is, at this moment,
greater than the maximum allowable value of TA.
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Programming the logic switching
procedure
 To determine which one of the two is preferred we must
l k att th
look
the position
iti
off the
th motor
t torque
t
TM.
 If we apply 200 V to winding Y, the flux S1 while increasing
to S1A it will move in the same direction of the rotor
rotor. Since
the flux is moving much faster than the rotor, the effect will
be to accelerate the rotor. This is the option that should be
taken provided that the torque at this moment TM is, at this
moment, less than TB.
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Programming the logic switching
procedure
 To determine which one of the two is preferred we must
look at the position of the motor torque TM.
 If the torque TM is already within the tolerance band
(between TA and TB), we will select to apply 200 V to
winding Y because it produces a torque in the same
direction as the rotation of the rotor.
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Programming the logic switching
procedure
C
Case
off fl
flux vector
t S2.
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Programming the logic switching
procedure
C
Case
off fl
flux vector
t S2.
 Since it is g
greater than 28 mWb windings
g X and Y must
be exited so that it falls inside the tolerance band.
 2.2 - Apply a voltage of -200 V to winding X.
X
 3.- Apply a voltage of 200 V to winding Y.
 5.-Apply zero voltage to both windings by shortcircuiting
g them,, rejected
j
because it should freeze the
flux in this current position.
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Programming the logic switching
procedure
C
Case
off fl
flux vector
t S2.
 If we select option
p
1 ((Apply
pp y a voltage
g of 200 V to
winding X) a braking torque will be developed.
 If we choose option 4 (Apply a voltage of -200 V to
winding Y) this will produce an accelerating torque on
the rotor.
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Programming the logic switching
procedure
C
Case
off fl
flux vector
t S3.
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Programming the logic switching
procedure
Case of flux vector S3.
 If the torque TM is less than TB , the torque must be
raised using
g option 1 ((Applyy a voltage
g of 200 V to
winding X).
 If the torque TM is greater than TA, we will choose
option 2 (Apply a voltage of -200 V to winding X). A
strong braking torque will result because the flux cuts
the rotor bars at a speed of 3000+600=3600 rev/min.
 If we select option
p
5 ((both windings
g in short-circuit)) the
flux will be stationary in the space. Since the motor
continues to turn at 600 rev/min a breaking torque will
b again
be
i applied
li d to
t the
th rotor.
t
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Instantaneous slip and production of
torque
 The flux continually advances and stops so that its
average speed
d iis ffar b
below
l
th
the iinstantaneous
t t
speed
d off
3000 rev/min. In fact, the average speed is slightly
greater than the 600 rev/min of the rotor
rotor.
a: Voltage and current induced in the rotor bars.
b: Torque TM developed by the motor
motor, and the upper and lower limits
TA and TB.
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Control of 3
3-phase
phase motors
 Switching
S it hi combinations
bi ti
ffor th
the windings
i di
off a 3
3-phase
h
motor and corresponding positions of the stator flux S.
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Control of 3
3-phase
phase motors
 Hexagonal
H
l path
th ffollowed
ll
db
by th
the flflux S in
i a6
6-step,
t
33
phase converter. The amplitude and position of the
instantaneous flux are indicated by the vector S. The
six arrows emanating from the center show the six
directions in which the flux can be moved.
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Control of 3
3-phase
phase motors
 Hexagonal path followed by the flux S in a 6-step, 3phase converter. The amplitude and position of the
instantaneous flux are indicated by the vector S.
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Control of 3
3-phase
phase motors
 Hexagonal path followed by the flux S in a 6-step, 3phase converter. The amplitude and position of the
instantaneous flux are indicated by the vector S.
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Control of 3
3-phase
phase motors
 The ratio of the maximum flux S to the minimum is
imposed by the geometry of the hexagon, namely
2/sqrt(3)=1.155
Hexagonal
g
p
path followed by
y the flux S in a 6-step,
p, 3-phase
p
converter.
The amplitude and position of the instantaneous flux are indicated by
the vector S. The six arrows emanating from the center show the six
directions in which the flux can be moved
moved.
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Commutation process
 The rotor and the flux rotates counterclockwise. The
windings
i di
are connected
t d tto a converter
t th
thatt offers
ff
6
combinations wherein the windings are excited and one
combination has all three windings in short-circuit
short circuit.
 If TM<TB then C+
When S1 is less than B, the flux can be increased by selecting one
of the following options: A (+), B (-), C (+), or C (-).
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Commutation process
 If
TM>T
TA then
th A
A+
When S1 is less than B, the flux can be increased by selecting one
of the following options: A (+), B (-), C (+), or C (-).
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Commutation process
 If TB<T
TM<T
TA then
th B
B-
When S1 is less than B, the flux can be increased by selecting one
of the following options: A (+),
(+) B (-),
( ) C (+),
(+) or C (-).
()
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Block diagram for the Direct Torque
Control strategy
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Block diagram for the Direct Torque
Control strategy
 The controlling variables of DTC are flux linkage and
torque, and the principle is to directly select voltage
vectors according to the difference between reference
and actual value of torque and flux linkage.
linkage
 The torque and flux errors are compared in two-level
(or three-level for torque) hysteresis comparators. Then
depending on the comparators outputs dl and dT and
the sector of the angle r a voltage vector is selected
from an optimal switching table.
 Because the selection referees to a hysteresis output, it
is of no matter if the errors are large or small -the
output will be the same
same.
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Block diagram for the Direct Torque
Control strategy
 The
Th stator
t t flux
fl controller
t ll imposes
i
the
th time
ti
duration
d ti off
the active voltage vectors, which move the stator flux
along the reference trajectory
trajectory, and the torque controller
determinates the time duration of the zero voltage
vectors, which keep the motor torque in the defined-byy
hysteresis tolerance band.
 At every sampling time the voltage vector selection
block chooses the inverter switching state
((Sa,Sb,Sc),
) which reduces the instantaneous flux
and torque errors.
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Erasmus LLP Intensive Programme
Hysteresis comparators
flux comparator
torque comparator
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Erasmus LLP Intensive Programme
Hysteresis comparators
 The
Th flux
fl and
d torque
t
errors, Dl and
d DT
DT, are applied
li d tto
respective bang-bang controllers. The flux controller's
output signal,
signal dl
dl, can assume the values of 0 and 1
1,
and that, dT, of the torque controller can assume the
values of -1, 0, and 1.
 Selection of the inverter state is based on values of dl,
dT and the sector of vector plane in which the stator
flux vector is currently located (see next switching
tables),
) as well as on the direction of rotation of the
motor.
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Erasmus LLP Intensive Programme
Inverter optimal switching table
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Erasmus LLP Intensive Programme
S1 switch ON, S3 switch OFF, S5 switch OFF
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Erasmus LLP Intensive Programme
Switching State
V
On-state Switch
000
0
S4, S6, S2
100
1
S1, S6, S2
110
2
S1, S3, S2
010
3
S4, S3, S2
011
4
S4, S3, S5
001
5
S4, S6, S5
101
6
S1, S6, S5
111
7
S1, S3, S5
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Erasmus LLP Intensive Programme
Switching table
 Counterclockwise
C
t l k i rotation
t ti
d
1
0
dT
1
0
-1
1
0
-1
Sector 1
2
7
6
3
0
5
Sector 2
3
0
1
4
7
6
Sector 3
4
7
2
5
0
1
Sector 4
5
0
3
6
7
2
Sector 5
6
7
4
1
0
3
Sector 6
1
0
5
2
7
4
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Erasmus LLP Intensive Programme
Switching table
 Clockwise
Cl k i rotation
t ti
d
1
0
dT
1
0
-1
1
0
-1
Sector 1
6
7
2
5
0
3
S t 2
Sector
5
0
1
4
7
2
Sector 3
4
7
6
3
0
1
Sector 4
3
0
5
2
7
6
Sector 5
2
7
4
1
0
5
Sector 6
1
0
3
6
7
4
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Erasmus LLP Intensive Programme
 EXAMPLE Th
The iinverter
t ffeeding
di a counterclockwise
t l k i
rotating motor in a DTC control is in (100). The stator
flux is too high
high, and the developed torque is too low
low,
both control errors exceeding their tolerance ranges.
With the angular
g
p
position of stator flux vector of 130°,
what will be the next state of the inverter?
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61
Erasmus LLP Intensive Programme
flux comparator
torque comparator
 The output signals of the flux and torque controllers are
d = 0 and dT = 1. The stator flux vector, ls, is in sector
3 of the vector plane.
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Erasmus LLP Intensive Programme
 The
Th output
t t signals
i
l off the
th flux
fl and
d torque
t
controllers
t ll
are
d= 0 and dT= 1. The stator flux vector, s, is in sector 3
of the vector plane.
plane Thus
Thus, according to the switching
table, the inverter will be switched to state 5.
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Erasmus LLP Intensive Programme
d
1
0
dT
1
0
-1
1
0
-1
Sector 1
2
7
6
3
0
5
Sector 2
3
0
1
4
7
6
Sector 3
4
7
2
5
0
1
Sector 4
5
0
3
6
7
2
Sector 5
6
7
4
1
0
3
Sector 6
1
0
5
2
7
4
 The torque is increased (the resultant flux has the
same direction of the motion) and the flux will remain
within the tolerance bands as depicts the figure.
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Erasmus LLP Intensive Programme
 Example
E
l 2
2. Th
The iinverter
t ffeeding
di a counterclockwise
t l k i
rotating motor in a DTC is in state 1. The stator flux is
too high,
high and the developed torque is tolerable
tolerable. With
the angular position of stator flux vector of 130°, what
will be the next state of the inverter?
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Erasmus LLP Intensive Programme
d
dT
Sector 1
Sector 2
Sector 3
Sector 4
Sector 5
Sector 6
1
1
2
3
4
5
6
1
0
7
0
7
0
7
0
0
-1
6
1
2
3
4
5
1
3
4
5
6
1
2
0
0
7
0
7
0
7
-1
5
6
1
2
3
4
 In
I this
thi second
d example,
l d = 0 and
d dT = 0.
0 Th
Thus,
according to the switching table, state 0 is imposed
which maintains the direction of the flux in the actual
position and no torque is added.
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Erasmus LLP Intensive Programme
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Erasmus LLP Intensive Programme
Principle of DTC
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Erasmus LLP Intensive Programme
Principle of DTC
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Erasmus LLP Intensive Programme
DTC Simulation
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Erasmus LLP Intensive Programme
Equations of Permanent Magnet
Synchronous Motor (PMSM)
disd
vsd  rs  isd  Ld 
  er  Lq  isq
dt
disq
vsq  rs  isq  Lq 
  er  Ld  isd   er  PM
dt
 sd  Ld  isd   PM
 sq  Lq  isq
3
Te   p   PM  isq+Ld  Lq   isd  isq 
2
 s   sd2   sq2
dr 1
  Te  D  r  Tload 
dt
J
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Erasmus LLP Intensive Programme
Torque in Permanent Magnet
Synchronous Motor (PMSM)

3
Te   p   s  i s
2
3P
  isq
22
 : rotor flux in rotor reference frame
isq : torque producing stator current component
Te 

 sd  isd 
3
3
Te   p   s  i s   p      
2
2
 sq   isq 
3
Te   p   sdd isq   sq isdd
2
3
Te   p  Ld  isd   PM  isq  Lq  isq  isd
2
3
Te   p   PM  isq+Ld  Lq   isd  isq 
2






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Erasmus LLP Intensive Programme
IN PMSM WITH SURFACE
MOUNTED MAGNETS
Ld  Lq
3
Te   p  PM  isqq
2
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Erasmus LLP Intensive Programme
M i
Maximum
Torque
T
per Ampere,
A
i*d=0:
i*d 0
Te * 
isq * 
3
 p  PM  isq *
2
2  Te * Lq
3  p  PM
s* 
2
sd

2
sq

  Ld  isd *   PM



0

 2  Te * Lq
 s *  
 3  p  PM
2

   L  i *2 
q
sq


 PM 2  Lq  isq *2
2

   PM 2

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Erasmus LLP Intensive Programme
– VDC +
s*
 2   e *  Lq

 3  p 
PM


   PM 2


m*
h=0.05 |Sn|
|S|*
2
|s| = f(( )
+
e
*
-
|S|
h=0.08 n
VSI
SWITCHING
TABLE
SPEED
+
REGULATOR
+
m
+
-

SECTOR
|S|

ia
ib
FLUX,
TORQUE and
SECTOR
ESTIMATORS
m
va
vb
ENC
S
N
PULSES
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