1. An Application of Power Electronics
ELECTRICAL DRIVES:
An Application of Power Electronics

2. Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Two-quadrant Converter Va T1 +
Vdc  D1 ia Q2 Q1 Ia + Va - D2 T2
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

3. Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Four-quadrant Converter
leg A
leg B +
Vdc  D1 D3 Q1 Q3
+ Va  D4 D2 Q4 Q2
Jika vdc=110 Volt dan duti cycle=0.75.
Tentukan:
(a). Va (avg, dan V(rms)
(b). Cara Kerja rangkaian untuk operasi 4 kuadran
va = Vdc when Q1 and Q2 are ON
Positive current

4. Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Two-quadrant Converter Va T1 +
Vdc  D1 ia Q2 Q1 Ia + Va - D2 T2
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

5. Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Two-quadrant Converter Va T1 +
Vdc  D1 ia Q2 Q1 Ia + Va - D2 T2
D2 conducts  va = 0
T1 conducts  va = Vdc Va Eb
Quadrant 1 The average voltage is made larger than the back emf

6. Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Two-quadrant Converter Va T1 +
Vdc  D1 ia Q2 Q1 Ia + Va - D2 T2
D1 conducts  va = Vdc

7. Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Two-quadrant Converter Va T1 +
Vdc  D1 ia Q2 Q1 Ia + Va - D2 T2
D1 conducts  va = Vdc
T2 conducts  va = 0 Eb Va
Quadrant 2 The average voltage is made smallerr than the back emf, thus forcing the current to flow in the reverse direction

8. Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Two-quadrant Converter
2vtri vc + vA -
Vdc + vc

9. Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Four-quadrant Converter
leg A
leg B +
Vdc  D1 D3 Q1 Q3
+ Va  D4 D2 Q4 Q2
Positive current
va = Vdc when Q1 and Q2 are ON

10. Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Four-quadrant Converter
leg A
leg B +
Vdc  D1 D3 Q1 Q3
+ Va  D4 D2 Q4 Q2
Positive current
va = Vdc when Q1 and Q2 are ON
va = -Vdc when D3 and D4 are ON
va = when current freewheels through Q and D

11. Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Four-quadrant Converter
leg A
leg B +
Vdc  D1 D3 Q1 Q3
+ Va  D4 D2 Q4 Q2
Positive current
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 = when current freewheels through Q and D

12. Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Four-quadrant Converter
leg A
leg B +
Vdc  D1 D3 Q1 Q3
+ Va  D4 D2 Q4 Q2
Positive current
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 = when current freewheels through Q and D
va = when current freewheels through Q and D

22. Power Electronic Converters in ED Systems
DC DRIVES
Bipolar switching scheme – output swings between VDC and -VDC
AC-DC-DC vc + _
Vdc vA - vB
2vtri vc vA
Vdc
Vdc vB
vAB
Vdc
-Vdc

23. Power Electronic Converters in ED Systems
DC DRIVES
Unipolar switching scheme – output swings between Vdc and -Vdc
AC-DC-DC
Vtri vc
-vc
Vdc + vA - + vB - vA
Vdc vB
Vdc vc +
vAB
Vdc _
-vc

24. Power Electronic Converters in ED Systems
DC DRIVES
AC-DC-DC
DC-DC: Four-quadrant Converter
Armature
current
Vdc
Vdc
Armature
current
Vdc
Bipolar switching scheme
Unipolar switching scheme
Current ripple in unipolar is smaller
Output frequency in unipolar is effectively doubled

25. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Switching signals obtained by comparing control signal with triangular wave
Vdc + Va −
vtri q vc
We want to establish a relation between vc and Va
AVERAGE voltage
Va(s)
vc(s)
DC motor ?

26. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Ttri
ton 1
Vc > Vtri
Vc < Vtri vc
Vdc

27. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters d
0.5 vc
-Vtri
Vtri
-Vtri vc
For vc = -Vtri  d = 0

28. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters d
-Vtri
0.5 vc
-Vtri
Vtri vc
For vc = -Vtri  d = 0
For vc = 0  d = 0.5
For vc = Vtri  d = 1

29. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters d
-Vtri
0.5 vc
-Vtri
Vtri vc
For vc = -Vtri  d = 0
For vc = 0  d = 0.5
For vc = Vtri  d = 1

30. 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:
Introducing perturbation in vc and Va and separating DC and AC components:
DC:
AC:

31. 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:
va(s)
vc(s)
DC motor

32. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Bipolar switching scheme vc
vtri +
Vdc − q
-Vdc
+ VAB 
2vtri vc vA
Vdc vB
Vdc
vAB
Vdc
-Vdc

33. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Bipolar switching scheme
va(s)
vc(s)
DC motor

34. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters +
Vdc − vc
vtri qa
-vc qb
Leg a
Leg b
Unipolar switching scheme
Vtri vc
-vc vA vB
vAB
The same average value we’ve seen for bipolar !

35. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Unipolar switching scheme
vc(s)
va(s)
DC motor

36. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
DC motor – separately excited or permanent magnet
Te = kt ia
ee = kt 
Extract the dc and ac components by introducing small perturbations in Vt, ia, ea, Te, TL and m
ac components
dc components

37. 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
Vt(s) = Ia(s)Ra + LasIa + Ea(s)
Te(s) = kEIa(s)
Ea(s) = kE(s)
Te(s) = TL(s) + B(s) + sJ(s)

38. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
DC motor – separately excited or permanent magnet + -

39. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters Tc
vtri +
Vdc − q – kt
Torque controller + -
Torque
controller
Converter
DC motor

40. 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

41. 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)
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

42. 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
J = 152 x 10–6 kg.m2
B = 1 x10–4 kg.m2/sec
kt = 0.1 Nm/A
ke = 0.1 V/(rad/s)
Vd = 60 V
Vtri = 5 V
fs = 33 kHz
PI controllers
Switching signals from comparison of vc and triangular waveform

43. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Torque controller design
Open-loop gain
kpT= 90
kiT= 18000
compensated
compensated

44. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Speed controller design
Torque loop 1
Speed
controller * T* T  – +

45. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Speed controller design
Open-loop gain
kps= 0.2
kis= 0.14
compensated
compensated

46. Modeling and Control of Electrical Drives
Modeling of the Power Converters: DC drives with SM Converters
Large Signal Simulation results
Speed
Torque

47. THANK YOU