7 Frequency-controlled induction motor drives
• The relationship between the synchronous
speed and the frequency
ns =
120fs
P
• 7.2 Static frequency changes
• There are two types
direct: cycloconverters
indirect: a rectification and an inverter
• Cycloconverter drive
• PWM inverter
• Direct (cycloconverter): The output
frequency has a range of from 0 to o.5fs.
0.33fs for better waveform control.
• Indirect: broadly classified depending on
the feeding source, voltage or current source
• Variable-voltage, variable-frequency drive
• Regenerative voltage-source inverter-fed ac
drive
• Current-regulated voltage-source-driven drive
• Current-source inverter-driven drive
• Classification of frequency changer
7.3 Voltage-source inverter
• A half-bridge modified McMurray inverter.
• Snubbers: across transistors or SCR to limit
dv/dt and its effects.
• (a) Half-bridge inverter charging
• (b) Half-bridge inverter communication
• The frequency of the load voltage is
determined by the rate at which T1 and T2
are enabled.
• A full-bridge inverter
• Voltage-source inverter with transistors
7.4 Voltage-source inverter
• A generic self-communicating inverter
• The gating signals and the line voltage
• The line voltage in term of the phase voltage
Vab = Vas - Vbs
Vbc = Vbs - Vcs
Vca = Vcs - Vas
• We can obtain
Vab - Vca = 2Vas - (Vbs + Vcs)
• Because a three-phase system is balance,
Vas + Vbs + Vas = 0.
• Then, we have
Vab - Vca = 3Vas
• That is,
Vas = (Vab - Vca)/3
• Similarly,
Vbs = (Vbc - Vab)/3
Vcs = (Vca - Vbc)/3
• These periodic voltage waveforms (in
Fourier components) have the following
2 3
1
1
v ab ( t ) =
v dc (sin ωs t − sin 5ωs t + sin 7ωs t − ...)
π
5
7
1
2 3
1
v bc ( t ) =
v dc {sin(ωs − 120o ) t − sin(5ωs − 120o ) t + sin(7ωs t − 120o ) − ...
π
7
5
1
2 3
1
o
o
v ca ( t ) =
v dc {sin(ωs + 120 ) t − sin(5ωs + 120 ) t + sin(7ωs t + 120o ) − ...
π
7
5
• The fundamental rms phase voltage for the
six-stepped waveform is
Vas 2 Vdc
Vph =
= ⋅
= 0.45Vdc
2 π 2
• 7.4.2 Real power
Pi = VdcIdc = 3VphIphcosφ1
⇒ Idc = 1.35Iphcosφ1
• 7.4.3 Reactive power
Qi = 3VphIphsinφ1
• 7.4.4 Speed control
• The air gap induced emf
E1 = 4.44kω1φmfsT1
• Neglecting the stator impedance, Rs+jX1s
Vph ≅ E1
• The flux is then written as
φm ≅
Vph
K bf s
where Kb = 4.44kω1T1
• If Kb is constant
φm ∝
Vph
fs
∝ K vf
• A number of control strategies about the
voltage-to-frequency ratio:
(i) Constant volts/Hz control
(ii) Constant slip-speed control
(iii) Constant air gap flux control
(iv) Vector control
• 7.4.5 Constant volts/Hz control
• Relationship between voltage and frequency
Vas1 = E1 + Is1(Rs + jX1s)
• Equivalent circuit of the induction motor
• The phase voltage in p.u.
Vasn = E1n + Is1n(Rsn + jX1sn)
• The p.u. fundamental input voltage
Vasn = IsnRsn + jωsn(λmn+L1snIsn) (p.u.)
• The normalized input-phase stator voltage
2
Vasn = ( Isn R sn )2 + ωsn
(λ mn + L1sn Isn )2 (p.u.)
• The relationship between the applied phase
voltage and frequency (for law performance)
Vas = Vo +Kvkfs
where Vo = Is1Rs
• Because of
Vas = 0.24Vdcn, Von = Vo/Vb, and
E1n = E1/Vb = Kvffs/Kvffb = fsn,
we have
Vdcn = 2.22{Von + fsn}
• Implementation of volts/Hz strategy
• Closed-loop induction motor drive constant
volts/Hz control strategy
• 7.4.6 Constant slip-speed control
• Places the drive operation on the static
torque-speed characteristic
• The speed of the induction motor
ωs = ωr + ωs1
• The slip is obtained as
ωs1
ωs1
s=
=
ωs ωr + ωs1
• Constant-slip-speed strategy( one-quadrant)
• Simplified equivalent circuit considered for
the steady state analysis
• The rotor current
Ir =
E1
R
( r + jX1r )
s
=
E1 / ωs
R
( r + jL1r )
ωs1
• The electromagnetic torque
P Pa
P I 2r R r
P I 2r R r
Te = ⋅
= 3⋅ ⋅
= 3⋅ ⋅
2 ωs
2 sωs
2 ωs1
Rr
(
)
2
E12
V
P E1
ωs1
Te = 3 ⋅ ⋅ 2 ⋅
= K tv ( 2 ) ≅ K tv ( s )2
2 ωs ( R r )2 + ( L )2
ωs
ωs
1r
ωs1
P R
3 ( r)
2 ωs1
K tv =
R
( r )2 + ( L1r )2
ωs1
• Torque vs. applied voltage for various slip
speed
• 7.4.7 Constant air gap flux control
• Equivalent separately-excited dc motor in
terms of its speed but not in terms of
decoupling of flux and torque channel.
• Constant air gar flux linkages
E
λ m = L mi m = 1
ωs
• The electromagnetic torque
Te = 3 ⋅
R
( r)
ωs1
R
( r)
ωs1
P 2
⋅ λm ⋅
= K tm
R
R
2
( r )2 + ( L1r )2
( r )2 + ( L1r )2
ωs1
ωs1
• Drive strategy for constant-air gap-fluxcontrolled induction motor drive
• 9.4.7 Control of
harmonics
• Sinusoidal pulsewidth modulation
• The switching logic for one phase
1
Vdc , v c < v*a
2
1
= − Vdc , v c > v*a
2
va0 =
• The fundamental of this midpoint voltage
*
Vdc v ap
v a 01 =
⋅
2 v cp
• 7.4.10 Steady-state evaluation with PWM
voltage
• PWM voltage generation
t (i) =
1 ± m sin[a (i)]
, + for even n; - for odd n
2f c
a (i) =
2 πi
, rad;
n
where n is the ratio between the carrier and
modulation frequencies, t(i) is ith pulse width,
m is the modulation ration, fc is the carrier
frequency.
• The pulse widths in electrical radians
pw(i) =t(i)fs, rad
where fs is the modulation frequency.
• (see the table on pp. 370)
• The d and q axes voltages
2
v qs = [ v as − 0.5( v bs + v cs )]
3
1
v ds =
[( v cs − v bs )]
3
• The model in the stator reference frames
V = (R+Lp)i + Gωri
V = [vas vds 0 0]t
i = [iqs ids iqr idr]t
⎡Rs
⎢0
R=⎢
⎢0
⎢
⎣0
⎡ 0
⎢ 0
G=⎢
⎢ 0
⎢
⎣− Lm
0
Rs
0
0
0
0
Lm
0
0
0
Rr
0
0
0
0
Lr
0⎤
0⎥
⎥
0⎥
⎥
Rr ⎦
0 ⎤
0 ⎥
⎥
− Lr ⎥
⎥
0 ⎦
⎡ Ls
⎢ 0
L=⎢
⎢ Lm
⎢
⎣ 0
0
Ls
0
Lm
Lm
0
Rr
0
0 ⎤
Lm ⎥
⎥
0 ⎥
⎥
Rr ⎦
• State-space form
& = AX + Bu
X
where A = -L-1[R+ωrG]
B = L-1
X=i
u=V
• Direct evaluation of steady-state current
• The solution of current vector
X( t ) = e At X (0) + ∫0t e A( t − τ) Bu( τ)dτ
• Discretization
X (Ts ) = e ATs X (0) + (e ATs − I) A −1Bu(0)
• The kth sampling interval
X(k+1) = ΦX(k) + Fu(k)
where Φ = eATs and F = (Φ - I)A-1B
• X(k+1) = X(0) (if k+1 = 360 electrical
degrees)
• The steady-state initial vector
X (0) = [ I − Φ ( k +1) ]−1{∑ j=0...k Φ jFu( k − j)}
• Steady-state performance
Te ( k ) =
3P
L m {iqs ( k )ids ( k ) − ids ( k )iqr ( k )}
22
ias(k) = iqs(k)
ibs(k) = -0.5iqs(k) - 0.866ids(k)
7.5 Current-source ...
• Torque is directly related to the current
rather than voltage.
• 7.5.2 ACSI (Autosequentially commutated
Current-source Inverter)
• The inductor is provided to maintain the dc
link current at a steady value.
• Current-source induction motor drive
• Commutation
• Stator current in a star connection
• Forward motoring
• Regeneration
• 7.5.3 Steady-state performance
• Equivalent circuit approach
• The rotor and magnetizing currents
Ir =
jL m
Rr
+ jL r
sωs
⋅ Is
Rr
+ jL1r
sωs
Im =
⋅ Is
Rr
+ jL r
sωs
where Is 2π 3 Idc = 0.779Idc
• The electromagnetic torque
P
L2m
R
Te = 3 ⋅ ⋅
⋅ r ⋅ Is2
2 ( R r )2 + L2 sωs
r
sωs
• The maximum torque occurs at
s = Rr/ωsLr
• Thus
2
3 P L2m
2 3 P Lm 2
Te(max) = ⋅ ⋅
⋅ Is = ⋅ ⋅
⋅ Is
2 2 ( Rr )
2 2 Lr
sωs
• 7.5.4 Direct steady-state evaluation of sixstep current-source inverter-fed induction
motor (CSIM) drive system
• Stator currents:
(i) interval I: 0<θs<60°
ias = Idc; ibs = -Idc; ics = 0.
The quadrature axis stator current
2π
2π
2
ieqs = [ias cos θs + i bs cos(θs − ) + ics cos(θs + )]
3
3
3
π
ieqsI = aI dc cos(θs + ) where
6
similarly,
π
iedsI = aI dc sin( θs + )
6
a=
2
3
(ii) interval II: 60°<θs<120°
The quadrature and driect asix stator currents
3 e
π
π π
1
idsI
ieqsII = aI dc cos( θs − ) = aI dc cos( θs + − ) = ieqsI +
2
2
6
6 3
1
π
π π
3 e
iedsII = aI dc sin( θs − ) = aI dc sin( θs + − ) = −
iqsI + iedsI
6
6 3
2
2
• The transformation matrix
π ⎤ ⎡ 1
⎡e
⎢iqs ( θs + 3 )⎥ ⎢ 2
=
⎢e
π ⎥ ⎢⎢
⎢ids ( θs + )⎥ − 3
3 ⎦ ⎢⎣ 2
⎣
3⎤
⎥ ⎡ieqs ( θs )⎤
2 ⎥⎢
⎥
1 ⎥ ⎢⎣ieds ( θs )⎥⎦
2 ⎥⎦
• Closed-loop system