EE595S: Class Lecture Notes Chapter 13: Fully Controlled 3-Phase Bridge Converters. S.D. Sudhoff. Fall 2005

Transcription

1 EE595S: Class Lecture Notes Chapter 3: Fully Controlled 3-Phase Bridge Converters S.D. Sudhoff Fall 2005

2 3.2 Fully Controlled 3-Phase Bridge Converter Fall 2005 EE595S Electric Drive Systems 2

3 One Phase Leg Fall 2005 EE595S Electric Drive Systems 3

4 Phase Leg Equivalent Circuits Fall 2005 EE595S Electric Drive Systems 4

5 Phase Leg Voltages and Currents Switch On Current Direction v xg i xdc Upper Positive vdc vsw i xs Negative v dc + vd i xs Zero v dc i xs Lower Positive v d 0 Negative v sw 0 Zero 0 0 Neither Positive v d 0 Negative v dc + vd i xs Zero v dc / 2 0 Fall 2005 EE595S Electric Drive Systems 5

6 Line-to-Line Voltage and DC Current Line-to-Line Voltages v abs vbcs v cas = = = DC Current dc = iadc v v v ag bg cg v v v bg cg ag i + ibdc + icdc (3.2-) (3.2-2) (3.2-3) (3.2-4) Fall 2005 EE595S Electric Drive Systems 6

7 Line-to-Neutral Voltage From Fig ag bg = = Adding Or v = v + v ng = 3 ng as v v + v cg bs v v + v cs ng ng ng v ( vag + vbg + vcg ) ( vas + vbs + 3 = 3 + v ( vag vbg vcg ) v0s + v cs ) (3.2-5) (3.2-6) (3.2-7) (3.2-8) (3.2-9) Fall 2005 EE595S Electric Drive Systems 7

8 Line-to-Neutral Voltage IFF (IFF!) Then v ng = ( vag + vbg + v 3 cg ) (3.2-0) Fall 2005 EE595S Electric Drive Systems 8

9 Line-to-Neutral Voltage Backsubstitution of (3.2-0) into (3.2-5)- (3.2-7) yields 2 v = v v v (3.2-) as 3 ag 3 bg 3 cg vbs = 2 3 vbg vag 3 vcg 3 (3.2-2) v = cs 2 3 v cg 3 v ag 3 v bg (3.2-3) Fall 2005 EE595S Electric Drive Systems 9

10 Calculation of QD Voltages We do not need line-to-neutral voltages to find QD voltages! Fall 2005 EE595S Electric Drive Systems 0

11 o Voltage Source Inverter (VSI) Operation Comments: Six-step operation 20 o VSI used at one time Fall 2005 EE595S Electric Drive Systems

12 Line-to-Line Voltages Fall 2005 EE595S Electric Drive Systems 2

13 Line-to-Neutral Voltages (for v 0s =0) Fall 2005 EE595S Electric Drive Systems 3

14 Voltage Spectrum Line-to-line voltage 3 2 π v abs = v dc cos( θc + ) + π vdc cos((6 j )( θc π 6 j j= Line-to-neutral voltage (3.3-) π + )) + cos((6 j + )( θc 6 6 j + π + )) 6 v = as 2 vdc cosθc π 2 + vdc π j+ j ( ) ( ) cos((6 j ) θ ) + cos((6 + ) ) c j θc j= 6 j 6 j + (3.3-2) Fall 2005 EE595S Electric Drive Systems 4

15 Voltage Spectrum Fall 2005 EE595S Electric Drive Systems 5

16 An Example Load: Balanced, wye-connected, series RL R = 2 Ω L = mh DC Voltage: 00 V Fundamental Frequency: 00 Hz Fall 2005 EE595S Electric Drive Systems 6

17 ABC Variables for 80 o VSI Fall 2005 EE595S Electric Drive Systems 7

18 QD Variables Fall 2005 EE595S Electric Drive Systems 8

19 QD Inverter Analysis of 80 o VSI QD Voltages c v qs c v ds Average-Value Analysis c 2 = v π = 0 dc 2 = v π 2 v qs = vdc π v c ds 2 v π dc j= dc j= j 2( ) cos(6 jθ ) 2 c 36 j 2 j sin(6 jθ ) 36 2 c j (3.3-3) (3.3-4) (3.3-5) (3.3-6) Fall 2005 EE595S Electric Drive Systems 9

20 Calculation of DC Current We have P in = i Thus So idc = idc = Approximately dc v dc 3 P out = ( vqsiqs + vdsids ) ( vqsiqs + vdsids ) vdc 3 v + qsiqs v 2 vdc ds i ds (3.3-7) (3.3-8) (3.3-9) (3.3-0) (3.3-) Fall 2005 EE595S Electric Drive Systems 20

21 Return to Previous Example Change resistance to Ohm Find DC Current Voltages v qs = 63.7 V v ds = 0.0 V Currents Fall 2005 EE595S Electric Drive Systems 2

22 Return to Previous Example Results i qs =45.6 A i ds =28.7 A i dc =43.6 A Exact Answer i dc =43.9 A Fall 2005 EE595S Electric Drive Systems 22

23 3.4 Pulse-Width Modulation Current Capability Desire Fall 2005 EE595S Electric Drive Systems 23

24 Pulse-Width Modulation Fall 2005 EE595S Electric Drive Systems 24

25 Waveforms for Pulse-Width Modulation Fall 2005 EE595S Electric Drive Systems 25

26 Analysis Comparitor output c = d + Where p sw θ = ω Multiplying (3.4-) by (3.3-2) yields (next page) 2d sinc( kd)cos k k= sw θ sw (3.4-) (3.4-2) Fall 2005 EE595S Electric Drive Systems 26

27 Fall 2005 EE595S Electric Drive Systems 27 Ahhhh. (3.4-3) ) )cos( sinc( 2 ) cos((6 6 ) ( ) ) cos((6 6 ) ( cos = = + = c sw k dc c j c j j c dc as k kd V d j j j j dv v θ θ π θ θ θ π = + = = ) )cos( sinc( 2 ) ) (6 cos( 6 ) ( ) ) (6 cos( 6 ) ( ) sin( 2 c sw k dc c sw j c sw j k j dc k kd V d j k j j k j kd dv θ θ π θ θ θ θ π = = ) ) (6 cos( 6 ) ( ) ) (6 cos( 6 ) ( ) sin( 2 c sw j c sw j j k dc j k j j k j kd dv θ θ θ θ π

28 Points of Interest Fundamental Component vas fund QD Voltages = = d c 2 v qs dvdc c v ds π = 0 2 vdc cos π θ c (3.4-4) (3.4-5) (3.4-6) Fall 2005 EE595S Electric Drive Systems 28

29 An Example Load: Balanced, wye-connected, series RL R = 2 Ω L = mh DC Voltage: 00 V Fundamental Frequency: 00 Hz Duty Cycle: Switching Frequency:3000 Hz Fall 2005 EE595S Electric Drive Systems 29

30 ABC Variables for 80 o VSI Fall 2005 EE595S Electric Drive Systems 30

31 ABC Variables for Duty-Cycle Modulation Fall 2005 EE595S Electric Drive Systems 3

32 Pulse Width Modulation Good News Summary Bad News Applications Fall 2005 EE595S Electric Drive Systems 32

33 3.5 Sine-Triangle Modulation Features Adjustable Voltage Magnitude Adjustable Frequency Low Low-Frequency Harmonic Content Comments Originally Meant for Analog Implementation Digital Equivalents Exist Fall 2005 EE595S Electric Drive Systems 33

34 Gate Signal Controls Fall 2005 EE595S Electric Drive Systems 34

35 Operation of A-Phase Leg Fall 2005 EE595S Electric Drive Systems 35

36 Analysis Important Concept: The Fast Average xˆ ( t) = T sw t t T sw x( t) dt (3.5-) We can show that v ˆ ( + d ) v ag = 2 bg = 2 cg = a v ˆ ( + d ) v b v ˆ ( + dc) v 2 dc dc dc (3.5-2) (3.5-3) (3.5-4) Fall 2005 EE595S Electric Drive Systems 36

37 Analysis (2) Lets specify duty cycles as da = d cosθ d b = d cos( c d c = d cos( c c 2π θ ) 3 2π θ + ) 3 (3.5-5) (3.5-6) (3.5-7) Fall 2005 EE595S Electric Drive Systems 37

38 Analysis (3) Thus v qs dv 2 c ˆ = c ds vˆ = 0 dc (3.5-4) (3.5-5) Fall 2005 EE595S Electric Drive Systems 38

39 An Example Load: Balanced, wye-connected, series RL R = 2 Ω L = mh DC Voltage: 00 V Fundamental Frequency: 00 Hz Duty Cycle: 0.4 Switching Frequency: 3000 Hz Fall 2005 EE595S Electric Drive Systems 39

40 ABC Variables for Duty-Cycle Modulation (d=0.628) Fall 2005 EE595S Electric Drive Systems 40

41 Sine-Triangle Modulation Fall 2005 EE595S Electric Drive Systems 4

42 Limitations of Sine-Triangle Modulation Sine Triangle Modulation Range for duty cycle: Range for peak voltage (of fund): Duty Cycle Modulation: Range for duty cycle: Range for peak voltage (of fund.): Fall 2005 EE595S Electric Drive Systems 42

43 Overmodulation Fall 2005 EE595S Electric Drive Systems 43

44 Fall 2005 EE595S Electric Drive Systems 44 Analysis of Overmodulation Line-to-ground voltage (3.5-6) (3.5-7) (3.5-8) c dc dc fund avg ag d f v v v θ π )cos ( = + + = d d d d f 2arccos 4 2 ) ( 2 π < < < + > = 0 ) ( 2 a a dc a a dc ag d d v d d v v

45 Analysis of Overmodulation (2) Thus we have c vqs = f ( d) d = 0 d 0 v c ds 2vdc π (3.5-20) (3.5-2) Fall 2005 EE595S Electric Drive Systems 45

46 Overmodulation (d=2) Fall 2005 EE595S Electric Drive Systems 46

47 3.6 Third-Harmonic Injection Objective Fall 2005 EE595S Electric Drive Systems 47

48 The Plan Suppose da = db = dc = d cos( θc) d3 cos(3θ c) 2π d cos( θc ) d3 cos(3θ c) 3 2π d cos( θc + ) d3 cos( 3θ c) 3 (3.6-) (3.6-2) (3.6-3) Fall 2005 EE595S Electric Drive Systems 48

49 Results Working for a little bit Therefore v qs dv 2 c ˆ = dc (3.5-4) c ds vˆ = 0 (3.5-5) Fall 2005 EE595S Electric Drive Systems 49

50 So What? Consider this Thus 2 d 3 Result 5% Increase in Available Voltage (3.6-2) Fall 2005 EE595S Electric Drive Systems 50

51 Space Vector Modulation General Comments About Same Performance as Sine-Triangle (with 3 rd ) Fall 2005 EE595S Electric Drive Systems 5

52 Modulation Indices Instantaneous m s v s ˆq / vdc m d v s ˆ d / vdc q = s = (3.7-) (3.7-2) Fall 2005 EE595S Electric Drive Systems 52

53 Space-Vector Diagram Fall 2005 EE595S Electric Drive Systems 53

54 Space Vector Diagram Table 3.7- Modulation indexes versus state. State T / T4 T 2/ T5 3/ T6 T m q, x 0 0 o 2 / 3cos(0 ) 2 0 o 2 / 3cos(60 ) o 2 / 3cos(20 ) 4 0 o 2 / 3cos(80 ) o 2 / 3cos(240 ) 6 0 o 2 / 3cos(300 ) m d, x 2 / 3sin(0 o 2 / 3sin(60 o 2 / 3sin(20 2 / 3sin(80 2 / 3sin(240 2 / 3sin(300 ) o o o o ) ) ) ) ) Fall 2005 EE595S Electric Drive Systems 54

55 Space Vector Modulation Algorithm Step Compute Commanded Indices s* = q s (3.7-3) (3.7-4) Step 2 Compute Commanded Magnitude * = s* q m v / v s* dc m * s d = v * d / vdc 2 m ( m q ) + ( m s* d ) 2 (3.7-5) Fall 2005 EE595S Electric Drive Systems 55

56 Fall 2005 EE595S Electric Drive Systems 56 Space Vector Modulation Algorithm Step 3 Limit Magnitude of Command (3.7-6) (3.7-7) (3.7-8) 3 m max = > = max * * * max max * * *' m m m m m m m m m q q q > = max * * * max max * * *' m m m m m m m m m d d d

57 Space Vector Modulation Algorithm Step 4 Compute Sector angle( m *' *' q jm d )3 π Sector = ceil (3.7-9) Step 5 Compute State Sequence Fall 2005 EE595S Electric Drive Systems 57

58 Space Vector Modulation Algorithm Table State Sequence Sector Initial State (α ) 2 nd State ( β ) 3 rd State (γ ) Final State (δ ) Fall 2005 EE595S Electric Drive Systems 58

59 State Vector Modulation Algorithm Before continuing tβ tγ m mq, β + sw Tsw β tγ m d md, β + Tsw Tsw ˆ q = T t ˆ = Step 5 Compute Time in Each State D = mq tβ = T sw t = T, β md, γ mq, γ md, β ( m mq, γ md, γ m *' d, γ q mq, γ m *',, β *' d ) / D γ sw ( md β mq + mq md ) / D *' (3.7-0) (3.7-) (3.7-4) (3.7-2) (3.7-3) Fall 2005 EE595S Electric Drive Systems 59

60 State Vector Modulation Algorithm Step 6 Compute Transition Times ta = T t = ta t = t ( sw t β tγ ) / 2 B + C B + t β t γ (3.7-5) (3.7-6) (3.7-7) Fall 2005 EE595S Electric Drive Systems 60

61 Modeling Space Vector Modulation Comments: Thus s vˆ qs = s vˆ = ds ** q vdc m ** d vdc m (3.7-8) (3.7-9) Fall 2005 EE595S Electric Drive Systems 6

62 Hysteresis Modulation The Good Current Source Based Method High Bandwidth Control Readily Implements Current Limiting The Bad No Control Over Switching Frequency Fall 2005 EE595S Electric Drive Systems 62

63 State Transition Diagram Fall 2005 EE595S Electric Drive Systems 63

64 Typical Waveform Fall 2005 EE595S Electric Drive Systems 64

65 An Example Load: Balanced, wyeconnected, series RL R = 2 Ω L = mh DC Voltage: 00 V Fundamental Frequency: 00 Hz i * = 9.cos( θ 7.4 o ) as c Fall 2005 EE595S Electric Drive Systems 65

66 Typical Waveforms Fall 2005 EE595S Electric Drive Systems 66

67 Limitations of Hysteresis Modulation Voltage Constraints Thus v v < dc 3 spk (3.8-) Fall 2005 EE595S Electric Drive Systems 67

68 Delta Modulation Fall 2005 EE595S Electric Drive Systems 68

69 Given 3.0 Open Loop Voltage and Current Control Commanded q- and d-axis voltages DC Voltage Position of Synchronous Reference Frame Find d θ c Fall 2005 EE595S Electric Drive Systems 69

70 Open Loop Voltage Control for Consider Duty Cycle PWM v e qs e v ds = cosθce sinθce sinθ cosθ c ce vqs c ce v ds (3.0-) Where θ ce = θ c θ e (3.0-2) Fall 2005 EE595S Electric Drive Systems 70

71 Now Open Loop Voltage Control for Duty Cycle PWM v v e* qs e* ds = cosθ sinθ ce ce sinθ cosθ ce ce 2 dv π 0 dc (3.0-3) Manipulating Fall 2005 EE595S Electric Drive Systems 7

72 Open Loop Voltage Control for Duty Cycle PWM Thus ( ) 2 ( *) v e* v e 2 d = π qs + 2v dc ds (3.0-4) θ ce = angle( v e* e * qs jv ds ) (3.0-5) Fall 2005 EE595S Electric Drive Systems 72

73 Open Loop Voltage Control for Duty Cycle PWM Fall 2005 EE595S Electric Drive Systems 73

74 Open Loop Voltage Control for Sine-Triangle PWM (with 3 rd ) Fall 2005 EE595S Electric Drive Systems 74

75 Open Loop Voltage Control for Space Vector Modulation From the frame-to-frame transformation v s* qs = v e* qs e e* ds cosθ + v sinθ e (3.0-6) v s* ds = v e* qs e e* ds sinθ + v cosθ e (3.0-7) Fall 2005 EE595S Electric Drive Systems 75

76 Open Loop Current Control From inverse transformation i * abcs = K e s i e * qd 0 s (3.0-8) Fall 2005 EE595S Electric Drive Systems 76

77 Closed-Loop Current Control Current Source Based Current Control Voltage Source Based Current Control Fall 2005 EE595S Electric Drive Systems 77

78 Closed Loop Current Source Based Current Control Fall 2005 EE595S Electric Drive Systems 78

79 Analysis Start with Thus e* ** e f q = fqe fq, err τ rs + (3.-3) Fall 2005 EE595S Electric Drive Systems 79

80 Closed Loop Voltage Source Based Current Control Fall 2005 EE595S Electric Drive Systems 80

81 Analysis Assumed Load e qs e T e ds T e qs v = ω L i + L pi + e e ds e T e qs T e ds qt v = ω L i + L pi + e dt (3.-7) (3.-8) Thus i e qs = L K T r ( τ s r τ r s 2 + ) i + K L e* qs r T τ se s + r Kr L τ T e qt r (3.-9) Fall 2005 EE595S Electric Drive Systems 8

82 Pole Placement Note that Kr = LT ( s + s2) τr = + s s2 One choice s = s2 π f sw 5 (3.-0) (3.-) (3.2-2) Fall 2005 EE595S Electric Drive Systems 82