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ELECTRICAL DRIVES:
An Application of Power Electronics
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Two-quadrant Converter
Va
+
T1
D1
ia
Vdc

Q2
Ia
+
T2
Q1
D2
Va
-
T1 conducts  va = Vdc
Jika vdc=110 Volt dan duti cycle=0.75.
Tentukan:
(a). Va (avg, dan V(rms)
(b). Cara Kerja rangkaian untuk operasi 2
kuadran
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Four-quadrant Converter
leg A
+
Q1
leg B
D3
D1
+
Va

Q3
Vdc

Q4
D4
Jika vdc=110 Volt dan duti cycle=0.75.
Tentukan:
(a). Va (avg, dan V(rms)
(b). Cara Kerja rangkaian untuk operasi 4
kuadran
D2
Q2
va = Vdc
when Q1 and Q2 are ON
Positive current
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Two-quadrant Converter
Va
+
T1
D1
ia
Vdc

Q2
Ia
+
T2
Q1
D2
Va
-
T1 conducts  va = Vdc
Jika vdc=110 Volt dan duti cycle=0.75.
Tentukan:
(a). Va (avg, dan V(rms)
(b). Cara Kerja rangkaian untuk operasi 2
kuadran
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Two-quadrant Converter
Va
+
T1
D1
ia
Vdc

Q2
Q1
Ia
+
T2
D2
Va
-
D2 conducts  va = 0
Va
T1 conducts  va = Vdc
Eb
Quadrant 1 The average voltage is made larger than the back emf
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Two-quadrant Converter
Va
+
T1
D1
ia
Vdc

Q2
Ia
+
T2
Q1
D2
Va
-
D1 conducts  va = Vdc
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Two-quadrant Converter
Va
+
T1
D1
ia
Vdc

Q2
Q1
Ia
+
T2
D2
Va
-
T2 conducts  va = 0
Va
D1 conducts  va = Vdc
Eb
Quadrant 2 The average voltage is made smallerr than the back emf, thus
forcing the current to flow in the reverse direction
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Two-quadrant Converter
2vtri
+
vA
vc
Vdc
-
0
+
vc
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Four-quadrant Converter
leg A
+
Q1
leg B
D3
D1
+
Va

Q3
Vdc

Positive current
va = Vdc
when Q1 and Q2 are ON
Q4
D4
D2
Q2
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Four-quadrant Converter
leg A
+
Q1
leg B
D3
D1
+
Va

Q3
Vdc

Q4
D4
Positive current
va = Vdc
when Q1 and Q2 are ON
va = -Vdc
when D3 and D4 are ON
va = 0
when current freewheels through Q and D
D2
Q2
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Four-quadrant Converter
leg A
+
Q1
leg B
D3
D1
+
Va

Q3
Vdc

Q4
D4
Positive current
va = Vdc
when Q1 and Q2 are ON
va = -Vdc
when D3 and D4 are ON
va = 0
when current freewheels through Q and D
D2
Q2
Negative current
va = Vdc
when D1 and D2 are ON
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Four-quadrant Converter
leg A
+
Q1
leg B
D3
D1
+
Va

Q3
Vdc

Q4
D4
Positive current
D2
Q2
Negative current
va = Vdc
when Q1 and Q2 are ON
va = Vdc
when D1 and D2 are ON
va = -Vdc
when D3 and D4 are ON
va = -Vdc
when Q3 and Q4 are ON
va = 0
when current freewheels through Q and D
va = 0
when current freewheels through Q and D
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
Bipolar switching scheme – output
swings between VDC and -VDC
2vtri
Vdc
+
vA
+
vB
-
-
vA
vc
Vdc
0
Vdc
vB
0
Vdc
vc
vAB
+
_
-Vdc
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
Unipolar switching scheme – output
swings between Vdc and -Vdc
vc
Vtri
-vc
Vdc
+
vA
+
vB
-
-
Vdc
vA
0
Vdc
vc
vB
+
_
-vc
0
Vdc
vAB
0
Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Four-quadrant Converter
Armature
current
200
150
Vdc
Vdc
200
Vdc
100
100
50
50
0
0
-50
-50
-100
-100
-150
-150
-200
-200
0.04
Armature
current
150
0.0405 0.041
0.0415 0.042
0.0425 0.043
0.0435 0.044
0.0445 0.045
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
Unipolar switching scheme
•
Current ripple in unipolar is smaller
•
Output frequency in unipolar is effectively doubled
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Vdc
Switching signals obtained by comparing
control signal with triangular wave
+
Va
−
vtri
q
vc
We want to establish a relation between vc and Va
AVERAGE voltage
vc(s)
?
Va(s)
DC motor
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Ttri
1
q
0
vc
1
0
1
d
Ttri
Vc > Vtri
Vc < Vtri

t
t  Ttri
q dt
t on

Ttri
ton
Vdc
1 dTtri
Va   Vdc dt  dVdc
Ttri 0
0
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
d
0.5
vc
-Vtri
Vtri
vc
-Vtri
For vc = -Vtri  d = 0
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
d
0.5
vc
-Vtri
-Vtri
Vtri
vc
Vtri
For vc = -Vtri  d = 0
For vc = 0 
d = 0.5
For vc = Vtri 
d=1
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
d
0.5
-Vtri
-Vtri
vc
Vtri
Vtri
vc
1
d  0.5 
vc
2Vtri
For vc = -Vtri  d = 0
For vc = 0 
d = 0.5
For vc = Vtri 
d=1
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Thus relation between vc and Va is obtained as:
Va  0.5Vdc 
Vdc
vc
2Vtri
Introducing perturbation in vc and Va and separating DC and AC components:
DC:
Vdc
Va  0.5Vdc 
vc
2Vtri
AC:
Vdc ~
~
va 
vc
2Vtri
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Taking Laplace Transform on the AC, the transfer function is obtained as:
v a ( s)
Vdc

v c ( s) 2Vtri
vc(s)
Vdc
2Vtri
va(s)
DC motor
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Bipolar switching scheme
Vdc
-Vdc
q
vtri
vc
2vtri
+
Vdc
vA
Vdc
+ VAB
0

−
vc
Vdc
vB
0
q
Vdc
v
d A  0.5  c
2Vtri
VA  0.5Vdc 
Vdc
vc
2Vtri
v
dB  1  d A  0.5  c
2Vtri
VB  0.5Vdc 
Vdc
vc
2Vtri
vAB
-Vdc
VA  VB  VAB 
Vdc
vc
Vtri
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Bipolar switching scheme
v a ( s) Vdc

v c ( s) Vtri
vc(s)
Vdc
Vtri
va(s)
DC motor
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Vdc
Unipolar switching scheme
vc
Leg b
Vtri
-vc
+
vtri
Vdc
qa
vc
−
vA
Leg a
vtri
d A  0.5 
vB
qb
-vc
vc
2Vtri
VA  0.5Vdc 
Vdc
vc
2Vtri
dB  0.5 
VB  0.5Vdc 
 vc
2Vtri
Vdc
vc
2Vtri
vAB
VA  VB  VAB 
The same average value we’ve seen for bipolar !
Vdc
vc
Vtri
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Unipolar switching scheme
v a ( s) Vdc

v c ( s) Vtri
vc(s)
Vdc
Vtri
va(s)
DC motor
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
DC motor – separately excited or permanent magnet
dia
v t  ia Ra  L a
 ea
dt
Te = kt ia
Te  Tl  J
dm
dt
e e = kt 
Extract the dc and ac components by introducing small
perturbations in Vt, ia, ea, Te, TL and m
ac components
~
d
i
~
~
v t  ia R a  L a a  ~
ea
dt
~
~
Te  k E ( ia )
dc components
Vt  Ia R a  E a
Te  k E Ia
~
~)
e e  k E (
Ee  k E
~)
d(
~
~
~
Te  TL  B  J
dt
Te  TL  B()
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
DC motor – separately excited or permanent magnet
Perform Laplace Transformation on ac components
~
d
i
~
~
v t  ia R a  L a a  ~
ea
dt
Vt(s) = Ia(s)Ra + LasIa + Ea(s)
~
~
Te  k E ( ia )
Te(s) = kEIa(s)
~
~)
e e  k E (
Ea(s) = kE(s)
~)
d(
~
~
~
Te  TL  B  J
dt
Te(s) = TL(s) + B(s) + sJ(s)
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
DC motor – separately excited or permanent magnet
Tl (s )
Va (s )
+
-
1
R a  sL a
Ia (s )
kT
-
Te (s )
1
B  sJ
+
kE
(s )
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
q
vtri
Torque
controller
Tc
+
+
Vdc
–
−
q
kt
DC motor
Tl (s )
Converter
Te (s )
Torque
controller
+
-
Vdc
Vtri ,peak
Ia (s )
1
R a  sL a
Va (s )
+
kT
-
Te (s )
+
-
kE
1
B  sJ
(s )
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Closed-loop speed control – an example
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
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Closed-loop speed control – an example
OBJECTIVES:
•
Fast response – large bandwidth
•
Minimum overshoot
good phase margin (>65o)
•
BODE PLOTS
Zero steady state error – very large DC gain
METHOD
•
Obtain linear small signal model
•
Design controllers based on linear small signal model
•
Perform large signal simulation for controllers verification
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Closed-loop speed control – an example
Ra = 2 
La = 5.2 mH
B = 1 x10–4 kg.m2/sec
J = 152 x 10–6 kg.m2
ke = 0.1 V/(rad/s)
kt = 0.1
Nm/A
Vd = 60 V
Vtri = 5 V
fs = 33
kHz
• PI controllers
• Switching signals from comparison
of vc and triangular waveform
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Torque controller design
Open-loop gain
Bode Diagram
From: Input Point To: Output Point
150
kpT= 90
Magnitude (dB)
100
compensated
kiT= 18000
50
0
-50
90
Phase (deg)
45
0
compensated
-45
-90
-2
10
-1
10
0
10
1
10
2
10
Frequency (rad/sec)
3
10
4
10
5
10
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Speed controller design
*
+
–
T*
Speed
controller
Torque loop
1
T
1
B  sJ

Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Speed controller design
Open-loop gain
Bode Diagram
From: Input Point To: Output Point
150
kps= 0.2
Magnitude (dB)
100
kis= 0.14
50
compensated
0
-50
0
Phase (deg)
-45
-90
-135
compensated
-180
-2
10
-1
10
0
10
1
10
Frequency (Hz)
2
10
3
10
4
10
Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Large Signal Simulation results
40
20
Speed
0
-20
-40
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
2
1
Torque
0
-1
-2
THANK YOU

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