Digital Control of Electric Drives. Induction Motor Vector Control. Czech Technical University in Prague Faculty of Electrical Engineering

Transcription

1 Digital Control of Electric Drives Induction Motor Vector Control Czech Technical University in Prague Faculty of Electrical Engineering BE1M14DEP O. Zoubek, J. Zdenek 1

2 Induction Motor Control Methods Scalar control Vector control BE1M14DEP 2 2

3 Three-phase Inverter BE1M14DEP 3 3

4 Scalar Control Properties (U/f Method) Disadvantages of scalar (U/f) control: Very low dynamics Changes in the asynchronous machine take place at the speed given by the rotor time constant (ie, up to seconds on large machines) The input is the frequency - it is not possible to directly control of IM torque Not suitable for traction The motor torque is dependent on the slip frequency - it is still the same (the percentage with the lower output frequency of the drive increases) Advantages of scalar (U / f:) control: Simplicity of management (especially development) BE1M14DEP 4 4

5 Scalar Control - U/f Method) Scalar control: Only one variable, mostly frequency, is controlled. Voltage is tied to frequency, so the name "U / f" Voltage Rated voltage Rated speed Speed [1/min] I 1 IM B1M14DEP BE1M14DEP 5 5

6 PWM Generation hw Support Carrier Wave Code (Kód nosné vlny) Žádost o obsluhu Int Clock (Takt) Prescaler (Předdělička) Up-Down Counter (Obousměrný čítač) + Comparator (Komparátor ) Dead Time & Output Logic Modulation Wave Value (Kód modulačního průběhu ) CR Compare Register (Porovnávaná hodnota ) Program Generování mrtvých dob a výstupní logika B1M14DEP BE1M14DEP 6 6

7 PWM Generation (One possible option) +1 Modulation Wave (Modulační průběh) Carrier Wave (Nosná vlna) 0-1 Interrupt Request (Žádost o obsluhu) + PWM Output (Výstup PWM) PWM Output (Výstup PWM) Time B1M14DEP BE1M14DEP 7 7

8 Vector Control Advantages of vector control versus scalar: Unmatched higher dynamics Working from zero revolutions (for some types of vector control including standing rotor) The drive input is a torque requirement (gas lever) Fit for traction The induction motor current is fully controlled Possibility of short-term overload and work with a higher torque than the maximum torque point To switch from scalar to vector control, in most cases, only software change is sufficient B1M14DEP BE1M14DEP 8 8

9 Torque is related to currents Control Dymamics Torque of any electric machine current Dynamics of electric machine fast current changes Rapid changes in currents require a voltage (the winding of the machine is inductive) In order to increase the dynamics, it is necessary to directly control the currents (for induction motor with the short-circuit armature only stator currents are possible to control), which can not be done by the U / f control B1M14DEP BE1M14DEP 9 9

10 Vector Control Vector control: Two quantities of the induction machine are controlled separately Mostly, this is a stator current divided into two components: Current component affecting the magnetic flux Current component affecting the torque IM machine DC machine I 1 IM I M I Ψ IM = B1M14DEP BE1M14DEP 10 10

11 Space Vector Idea The space vector I 1 can express the effect of all currents flowing through all phases of the electric machine stator Symmetrical winding is assumed for a three-phase, two-pole machine (the angle of rotation is 120 ) The expression can be modified provided the center point is not connected (i.e. i a +i b +i c = 0) and after modifiying e j120 a e j240 i =R I = 3 2 K i a I=K i a i b e j 120 i c e j 240 I=K 3 2 i a j 3 2 i a 2i b i =I I = 3 2 K i bi c = 3 2 K i a 2 i b BE1M14DEP 11 11

12 Space Vector Idea The space vector I 1 can express the effect of all currents flowing through all phases of the electric machine stator Symmetrical winding is assumed for a three-phase, two-pole machine (the angle of rotation is 120 ) I=K i a i b e j 120 i c e j 240 The expression can be modified provided the center point is not connected (i.e. i a +i b +i c = 0) and after modifiying e j120 a e j240 i =R I = 3 2 K i a Při vhodné volbě K I=K 3 2 i a i α j = i a 3 2 i a 2i b i =I I = 3 2 K i bi c = 3 2 K i a 2 i b BE1M14DEP 12 12

13 Clark Transformation Three phases of the machine Two axis i a i α i β i c i b Clark's transformation converts the three phases (a, b, c) into two phases (α, β) For K = 2/3 Clark's transformation is expressed as: i α =i a i β = 3 3 (i a+2i b ) BE1M14DEP 13 13

14 Rotation of Coordinates I 1 I k Space vector I in the coordinate system tied to the stator Space vector I in coordinate system k I k = I 1 e j k Where θ k is the angle of rotation of the coordinate system k If the coordinate system (k, l) is rotated by the angle θ k against (α, β): i k =i α cosθ k +i β sinθ k i l =i α sinθ k +i β cosθ k B1M14DEP BE1M14DEP 14 14

15 Park Transfomation α θ d β ω 1 q i d =i cos i sin i q =i sin i cos BE1M14DEP 15 15

16 Used Coordinate Systems Coordinates Designation Rotation speed Coordinates tied to the stator 0 Coordinates tied to the rotor k, l ω Coordinates tied to the magnetic flux of the rotor d, q ω 1 BE1M14DEP

17 Induction Motor Torque Pro K= 2 3 P= 3 2 U 1 I 1 cos = 3 2 R U 1 I 1 BE1M14DEP 17 17

18 Induction Motor Torque For K= 2 3 P= 3 2 U 1 I 1 cos = 3 2 R U 1 I 1 Voltage equation of induction motor stator winding written by space vectors: 1 U 1 =R 1 I 1 d dt =R 1 I 1 j 1 1 BE1M14DEP 18 18

19 Induction Motor Torque Pro K= 2 3 P= 3 2 U 1 I 1 cos = 3 2 R U 1 I 1 P= 3 2 [ R 1 i 2 2 i i 1 1 i 1 ] BE1M14DEP 19 19

20 Induction Motor Torque Pro K= 2 3 P= 3 2 U 1 I 1 cos = 3 2 R U 1 I 1 P= 3 2 [ R 1 i 2 2 i i 1 1 i 1 ] Power dissipation in resistance: P=R I 2 =R i a 2 i b 2 BE1M14DEP 20 20

21 Induction Motor Torque Pro K= 2 3 P= 3 2 U 1 I 1 cos = 3 2 R U 1 I 1 P= 3 2 [ R 1 i 2 2 i i 1 1 i 1 ] Losses in the stator P=R I 2 =R i a 2 i b 2 BE1M14DEP 21 21

22 Induction Motor Torque Pro K= 2 3 P= 3 2 U 1 I 1 cos = 3 2 R U 1 I 1 P= 3 2 [ R 1 i 2 2 i i 1 1 i 1 ] Losses in the stator Power transmitted by air gap from the stator to the rotor BE1M14DEP 22 22

23 Induction Motor Torque Pro K= 2 3 P= 3 2 U 1 I 1 cos = 3 2 R U 1 I 1 BE1M14DEP 23 23

24 Induction Motor Torque Pro K= 2 3 P= 3 2 U 1 I 1 cos = 3 2 R U 1 I 1 1s = 1 p p BE1M14DEP 24 24

25 Induction Motor Torque Pro K= 2 3 P= 3 2 U 1 I 1 cos = 3 2 R U 1 I 1 BE1M14DEP 25 25

26 Induction Motor Torque M i = 3 2 p p c A B sin sin γ is the angle between the vectors A a B A B c I 1 I 1 I 1 I 1 I 2 I 2 I 2 Ψ 1 I 2 Ψ 1 Ψ µ Ψ 2 Ψ 1 Ψ µ Ψ 2 Ψ 2 L h L h / L 2 L h / L 1 L h / σl 1 L 2 M i = 3 2 p p 1 i 1 1 i 1 BE1M14DEP

27 Induction Motor Torque M i = 3 2 p p c A B sin sin γ is the angle between the vectors A a B A B c I 1 I 1 I 1 I 1 I 2 I 2 I 2 Ψ 1 I 2 Ψ 1 Ψ µ Ψ 2 Ψ 1 Ψ µ Ψ 2 Ψ 2 L h L h / L 2 L h / L 1 L h / σl 1 L 2 M i = 3 2 p p 2 d i 1 q 2 q i 1 d L h L 2 M i = 3 2 p p 1 i 1 1 i 1 BE1M14DEP

28 Stator Current Components M i = 3 2 p p 2 d i 1 q 2 q i 1 d L h L 2 BE1M14DEP 28 28

29 Stator Current Components M i = 3 2 p p 2 d i 1 q 2 q i 1 d L h L 2 Ψ 2q is equal to zero because the (d, q) axis direction is given just by the direction of Ψ 2 M i = 3 2 p p 2 i 1 q L h L 2 BE1M14DEP 29 29

30 Stator Current Components M i = 3 2 p p 2 d i 1 q 2 q i 1 d L h L 2 Ψ 2q is equal to zero because the (d, q) axis direction is given just by the direction of Ψ 2 M i = 3 2 p p 2 i 1 q L h L 2 M i 2 Ψ 2 is excitation stator current component which affect the torque BE1M14DEP 30 30

31 Stator Current Components M i = 3 2 p p 2 d i 1 q 2 q i 1 d L h L 2 Ψ 2q is equal to zero because the (d, q) axis direction is given just by the direction of Ψ 2 M i = 3 2 p p 2 i 1 q L h M i 2 L 2 Ψ 2 is excitation stator current component which affects the torque d 2d dt = R 2 L 2 L h 2d => 2 L h stator current component which affects magnetic flux BE1M14DEP

32 IM versus DC machine The stator current of the three-phase induction short-circuit machine has two stages of freedom (the third one is deleted because center of windings is drawn outside) and can be divided into two components that correspond to the armature current and the excitation of the DC motor. It is therefore possible to control independetly the torque and the excitation of induction machine. IM machine DC machine AM i a i b i a i b BE1M14DEP 32 32

33 Voltage Vector Control of an Induction Machine * * PI PI u 1d u 1q d,q θ u 1α u 1β a,b,c u 1a,b,c Inverter i a,i b,(i c ) IM IM Model d,q i α a,b,c ω 2 θ r θ i β θ IRC n IRC Decoder BE1M14DEP 33 33

34 Voltage Vector Control of an Induction Machine * * Clark Transformation u i 1d =i a i = 3 PI PI u 1q θ u 1α d,q 3 i a 2i b a,b,c u 1β u 1a,b,c Inverter i a,i b,(i c ) IM IM Model ω 2 θ r d,q θ i α i β a,b,c θ IRC n IRC Decoder BE1M14DEP

35 Voltage Vector Control of an Induction Machine * * Park Transformation PI PI i d =i cos i sin u 1d u 1q d,q θ u 1α u 1β a,b,c u 1a,b,c i q =i sin i cos střídač i a,i b,(i c ) AM IM Model ω 2 θ r d,q θ i α i β a,b,c θ IRC n IRC Decoder BE1M14DEP

36 Voltage Vector Control of an Induction Machine Induction machine I 1 -n model * * d Ψ 2 dt u 1d PI d,q střídač AM PI (L a,b,c Lu h Ψ u 1q 1β 2 ) ω 2 = L h R 2 2 L θ 2 i a,i b,(i c ) = R 2 u 1α u 1a,b,c Ψ 2 IM Model ω 2 θ r d,q θ i α i β a,b,c θ IRC n IRC Decoder B1M14DEP 36 36

37 Voltage Vector Control of an Induction Machine Induction machine I 1 -n model * * d 2 dt u 1d PI d,q střídač AM PI L a,b,c Lu h u 1q 1β 2 2 = L h R 2 2 L θ 2 i a,i b,(i c ) = R 2 u 1α u 1a,b,c 2 IM Model ω 2 θ r d,q θ i α i β a,b,c θ IRC n IRC Decoder BE1M14DEP 37 37

38 Voltage Vector Control of an Induction Machine * Induction machine I 1 -n model after suitable substitution is used i 2 = 2 * d i 2 dt PI u 1d L d,q h střídač u u a,b,c 1q i L 1d i 1β 2 2 = R 2 2 PI = R 2 θ u 1α u 1a,b,c AM L 2 i 2 i a,i b,(i c ) IM Model ω 2 θ r d,q θ i α i β a,b,c θ IRC n IRC Decoder BE1M14DEP 38 38

39 Voltage Vector Control of an Induction Machine * PI Slip frequency integrator PI * IM Model ω 2 θ r u 1d u 1q d,q d,q θ θ u 1α u 1β a,b,c i α i β u 1a,b,c střídač r = 2 dt a,b,c AM i a,i b,(i c ) θ IRC n IRC Decoder BE1M14DEP

40 Voltage Vector Control of an Induction Machine * PI d,q The IRC sensor decoder PI u u a,b,c number 1q of pulses 1β θ IRC = θ pulses per rotation p p 2 * IM Model ω 2 θ r u 1d d,q θ u 1α i α i β u 1a,b,c střídač a,b,c IM i a,i b,(i c ) θ IRC n IRC Decoder BE1M14DEP

41 Voltage Vector Control of an Induction Machine * * IM Model ω 2 PI PI θ r u 1d u 1q d,q θ θ r 2 u 1α 1 střídač a,b,c u 1β i β u 1a,b,c IRC Slip frequency 2 = 1 r IRC = i d,q α a,b,c Feeding frequency of machine AM El. angle speed of shaft i a,i b,(i c ) θ IRC n IRC Decoder BE1M14DEP

42 Voltage Vector Control of an Induction Machine * * PI PI u 1d u 1q d,q θ u 1α u 1β a,b,c u 1a,b,c Inverter i a,i b,(i c ) i IM d,q α a,b,c PI Model controllers regulate i each component i of the current separately 1q θ β Both components are DC (does not change with ω, ω 1 or ω 2 ) ω 2 The input to the θ controllers are the error values of the currents. The outputs are a r voltage requirements. u t =k p e t k i e t dt IM θ IRC n dekodér IRC BE1M14DEP

43 Voltage Vector Control of an Induction Machine * * model AM PI PI θ r u 1d u 1q d,q u 1α u 1β a,b,c u 1a,b,c Inverse Park and Clark transformations i u d,q α =u d cos u q sin u =u d sin u q cos ω 2 Inverse: sins have inverse signs θ i β Inverter i a,i b,(i c ) IM u a =u a,b,c u b = 1 2 u 3 2 u θ IRC u c = 1 2 u 3 n2 u dekodér IRC B1M14DEP 43 43

44 Voltage Vector Control of an Induction Machine BE1M14DEP 44 44

45 Three-phase Inverter BE1M14DEP 45 45

46 Space Vector Modulation Each of the three branches of the voltage converter always has only one of two transistor switched on. For the three branches it is 2 3 = 8 different states altogether. a b c Ua-b Ub-c Ua-c výstup nic U DC 0 -U DC U DC -U DC U DC +U DC U DC 0 +U DC U DC +U DC U DC -U DC nic BE1M14DEP 46 46

47 Space Vector Modulation BE1M14DEP 47 47

48 Space Vector Modulation BE1M14DEP 48 48

49 Six available active states at the output of the inverter Space Vector Modulation b U180 U120 U240 U60 U0 U300 a c BE1M14DEP 49 49

50 Space Vector Modulation To achieve a certain output should be combined over time several output vectors, including zero vector. There are many ways, how to draw a vector which differ in accuracy, computational difficulty and switching losses. However, the outputs always have character PWM modulation. b U0 U60 a The simplest method is: PWM x = u x 2U max 0.5 x {a,b, c} c BE1M14DEP 50 50

51 Space Vector Modulation BE1M14DEP 51 51

52 Space Vector Modulation 52 BE1M14DEP 52

53 Digital Control of Electric Drives Induction Motor Vector Control End BE1M14DEP 53 53