electrical drives: an application of power electronics
TRANSCRIPT
ELECTRICAL DRIVES: ELECTRICAL DRIVES: An Application of Power ElectronicsAn Application of Power Electronics
T1 conducts va = Vdc
Q1Q2
Va
Ia
T1
T2
D1
+
Va
-
D2
ia
+
Vdc
DC DRIVES
AC-DC-DCAC-DC-DC DC-DC: Two-quadrant Converter
Power Electronic Converters in ED Systems
Jika vdc=110 Volt dan duti cycle=0.75.
Tentukan:
(a). Va (avg, dan V(rms)
(b). Cara Kerja rangkaian untuk operasi 2 kuadran
leg A leg B
+ Va Q1
Q4
Q3
Q2
D1 D3
D2D4
+
Vdc
va = Vdc when Q1 and Q2 are ON
Positive current
Power Electronic Converters in ED SystemsDC DRIVES
AC-DC-DCAC-DC-DC DC-DC: Four-quadrant Converter
Jika vdc=110 Volt dan duti cycle=0.75.
Tentukan:
(a). Va (avg, dan V(rms)
(b). Cara Kerja rangkaian untuk operasi 4 kuadran
T1 conducts va = Vdc
Q1Q2
Va
Ia
T1
T2
D1
+
Va
-
D2
ia
+
Vdc
DC DRIVES
AC-DC-DCAC-DC-DC DC-DC: Two-quadrant Converter
Power Electronic Converters in ED Systems
Jika vdc=110 Volt dan duti cycle=0.75.
Tentukan:
(a). Va (avg, dan V(rms)
(b). Cara Kerja rangkaian untuk operasi 2 kuadran
Q1Q2
Va
Ia
T1
T2
D1
+
Va
-
D2
ia
+
Vdc
D2 conducts va = 0
Va Eb
T1 conducts va = Vdc
Quadrant 1 The average voltage is made larger than the back emf
DC DRIVES
AC-DC-DCAC-DC-DC DC-DC: Two-quadrant Converter
Power Electronic Converters in ED Systems
Q1Q2
Va
Ia
T1
T2
D1
+
Va
-
D2
ia
+
Vdc
D1 conducts va = Vdc
DC DRIVES
AC-DC-DCAC-DC-DC DC-DC: Two-quadrant Converter
Power Electronic Converters in ED Systems
Q1Q2
Va
Ia
T1
T2
D1
+
Va
-
D2
ia
+
Vdc
T2 conducts va = 0
VaEb
D1 conducts va = Vdc
Quadrant 2 The average voltage is made smallerr than the back emf, thus forcing the current to flow in the reverse direction
DC DRIVES
AC-DC-DCAC-DC-DC DC-DC: Two-quadrant Converter
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DCAC-DC-DC DC-DC: Two-quadrant Converter
+vc
2vtri
vc
+vA
-
Vdc
0
Power Electronic Converters in ED Systems
leg A leg B
+ Va Q1
Q4
Q3
Q2
D1 D3
D2D4
+
Vdc
va = Vdc when Q1 and Q2 are ON
Positive current
Power Electronic Converters in ED SystemsDC DRIVES
AC-DC-DCAC-DC-DC DC-DC: Four-quadrant Converter
leg A leg B
+ Va Q1
Q4
Q3
Q2
D1 D3
D2D4
+
Vdc
va = -Vdc when D3 and D4 are ON
va = Vdc when Q1 and Q2 are ON
va = 0 when current freewheels through Q and D
Positive current
Power Electronic Converters in ED SystemsDC DRIVES
AC-DC-DCAC-DC-DC DC-DC: Four-quadrant Converter
va = -Vdc when D3 and D4 are ON
va = Vdc when Q1 and Q2 are ON
va = 0 when current freewheels through Q and D
Positive current
va = Vdc when D1 and D2 are ON
Negative current
leg A leg B
+ Va Q1
Q4
Q3
Q2
D1 D3
D2D4
+
Vdc
Power Electronic Converters in ED SystemsDC DRIVES
AC-DC-DCAC-DC-DC DC-DC: Four-quadrant Converter
va = -Vdc when D3 and D4 are ON
va = Vdc when Q1 and Q2 are ON
va = 0 when current freewheels through Q and D
Positive current
va = -Vdc when Q3 and Q4 are ON
va = Vdc when D1 and D2 are ON
va = 0 when current freewheels through Q and D
Negative current
leg A leg B
+ Va Q1
Q4
Q3
Q2
D1 D3
D2D4
+
Vdc
Power Electronic Converters in ED SystemsDC DRIVES
AC-DC-DCAC-DC-DC DC-DC: Four-quadrant Converter
Power Electronic Converters in ED SystemsDC DRIVES
AC-DC-DCAC-DC-DC
vAB
Vdc
-Vdc
Vdc
0vB
vAVdc
0
2vtri
vc
vc
+
_
Vdc+vA
-
+vB
-
Bipolar switching scheme – output swings between VDC and -VDC
Power Electronic Converters in ED SystemsDC DRIVES
AC-DC-DCAC-DC-DCUnipolar switching scheme – output swings between Vdc and -Vdc
Vtri
vc
-vc
vc
+
_
Vdc+vA
-
+vB
-
-vc
vA
Vdc
0
vB
Vdc
0
vAB
Vdc
0
Power Electronic Converters in ED SystemsDC DRIVES
AC-DC-DCAC-DC-DC
Bipolar switching scheme
0.04 0.0405 0.041 0.0415 0.042 0.0425 0.043 0.0435 0.044 0.0445 0.045
-200
-150
-100
-50
0
50
100
150
200
Unipolar switching scheme
0.04 0.0405 0.041 0.0415 0.042 0.0425 0.043 0.0435 0.044 0.0445 0.045
-200
-150
-100
-50
0
50
100
150
200
• Current ripple in unipolar is smaller
• Output frequency in unipolar is effectively doubled
Vdc
Vdc
Vdc
DC-DC: Four-quadrant Converter
Armature current
Armature current
vc
+
Va
−
vtri
Vdc
q
Switching signals obtained by comparing control signal with triangular wave
Va(s)vc(s)DC motor
We want to establish a relation between vc and Va
?
AVERAGE voltage
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
dtqT1
dtriTt
ttri
tri
on
Tt
Vdc
0
Ttri
ton
0
1
01
qVc > Vtri
Vc < Vtrivc
dc
dT
0 dctri
a dVdtVT1
Vtri
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
-Vtri
Vtri
-Vtri
vc
d
vc
0.5
For vc = -Vtri d = 0
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
0.5
Vtri
Vtri
vc
d
vc
-Vtri-Vtri
For vc = -Vtri d = 0
For vc = 0 d = 0.5
For vc = Vtri d = 1
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
0.5
vc
d
-Vtri-Vtri
ctri
vV21
5.0d
Vtri
Vtri
vc
For vc = -Vtri d = 0
For vc = 0 d = 0.5
For vc = Vtri d = 1
Thus relation between vc and Va is obtained as:
ctri
dcdca v
V2V
V5.0V
Introducing perturbation in vc and Va and separating DC and AC components:
ctri
dcdca v
V2V
V5.0V
ctri
dca v~
V2V
v~
DC:
AC:
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
Taking Laplace Transform on the AC, the transfer function is obtained as:
tri
dc
c
a
V2V
)s(v)s(v
va(s)vc(s)DC motor
tri
dc
V2V
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
2vtri
vc
vc
vtri+
Vdc
−
q-Vdc
q
Vdc
+ VAB
vAB
Vdc
-Vdc
ctri
dcABBA v
VV
VVV
tri
cAB V2
v5.0d1d
ctri
dcdcB v
V2V
V5.0V
vB
Vdc
0
tri
cA V2
v5.0d
ctri
dcdcA v
V2V
V5.0V
vA
Vdc
0
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
Bipolar switching scheme
tri
dc
c
a
VV
)s(v)s(v
va(s)vc(s)DC motor
tri
dc
VV
Bipolar switching scheme
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
+
Vdc
−vc
vtri
qa
Vdc
-vc
vtri
qb
Leg a
Leg b
The same average value we’ve seen for bipolar !
Vtri
vc
-vc
tri
cA V2
v5.0d
ctri
dcdcA v
V2V
V5.0V
vA
tri
cB V2
v5.0d
ctri
dcdcB v
V2V
V5.0V
vB
ctri
dcABBA v
VV
VVV
vAB
Unipolar switching scheme
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
tri
dc
c
a
VV
)s(v)s(v
va(s)vc(s)DC motor
tri
dc
VV
Unipolar switching scheme
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
DC motor – separately excited or permanent magnet
Extract the dc and ac components by introducing small perturbations in Vt, ia, ea, Te, TL and m
aa
aaat edtdi
LRiv
Te = kt ia ee = kt
dtd
JTT mle
aa
aaat e~dti~
dLRi
~v~
)i~(kT
~aEe
)~(ke~ Ee
dt)~(d
J~BT~
T~
Le
ac components
aaat ERIV
aEe IkT
Ee kE
)(BTT Le
dc components
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
Perform Laplace Transformation on ac components
aa
aaat e~dti~
dLRi
~v~
)i~(kT
~aEe
)~(ke~ Ee
dt)~(d
J~BT~
T~
Le
Vt(s) = Ia(s)Ra + LasIa + Ea(s)
Te(s) = kEIa(s)
Ea(s) = kE(s)
Te(s) = TL(s) + B(s) + sJ(s)
DC motor – separately excited or permanent magnet
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
Tkaa sLR
1
)s(Tl
)s(Te
sJB1
Ek
)s(Ia )s()s(Va
+-
-
+
DC motor – separately excited or permanent magnet
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
Tc
vtri
+
Vdc
−
q
q
+
–
kt
Torque controller
Tkaa sLR
1
)s(Tl
)s(Te
sJB1
Ek
)s(Ia )s()s(Va
+-
-
+
Torquecontroller
Converter
peak,tri
dc
VV)s(Te
-+
DC motor
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
Design procedure in cascade control structure
• Inner loop (current or torque loop) the fastest – largest bandwidth
• The outer most loop (position loop) the slowest – smallest bandwidth
• Design starts from torque loop proceed towards outer loops
Closed-loop speed control – an example
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
OBJECTIVES:
• Fast response – large bandwidth
• Minimum overshoot good phase margin (>65o)
• Zero steady state error – very large DC gain
BODE PLOTS
• Obtain linear small signal model
METHOD
• Design controllers based on linear small signal model
• Perform large signal simulation for controllers verification
Closed-loop speed control – an example
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
Ra = 2 La = 5.2 mH
J = 152 x 10–6 kg.m2B = 1 x10–4 kg.m2/sec
kt = 0.1 Nm/Ake = 0.1 V/(rad/s)
Vd = 60 V Vtri = 5 V
fs = 33 kHz
Closed-loop speed control – an example
• PI controllers • Switching signals from comparison of vc and triangular waveform
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
Bode Diagram
Frequency (rad/sec)
-50
0
50
100
150From: Input Point To: Output Point
Mag
nitu
de (
dB)
10-2
10-1
100
101
102
103
104
105
-90
-45
0
45
90
Pha
se (
deg)
compensated
compensated
kpT= 90
kiT= 18000
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
Torque controller design Open-loop gain
Speed controller design
1Speedcontroller sJB
1
* T* T
–
+
Torque loop
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
Bode Diagram
Frequency (Hz)
-50
0
50
100
150From: Input Point To: Output Point
Mag
nitu
de (
dB)
10-2
10-1
100
101
102
103
104
-180
-135
-90
-45
0
Pha
se (
deg)
Open-loop gain
compensated
kps= 0.2
kis= 0.14
compensated
Speed controller design
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
Large Signal Simulation results
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45-40
-20
0
20
40
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45-2
-1
0
1
2
Speed
Torque
Modeling and Control of Electrical DrivesModeling of the Power Converters: DC drives with SM Converters
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