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ELECTRICAL DRIVES: An Application of Power Electronics.

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Presentation on theme: "ELECTRICAL DRIVES: An Application of Power Electronics."— Presentation transcript:

1 ELECTRICAL DRIVES: An Application of Power Electronics

2 T1 conducts  v a = V dc Q1Q2 VaVa IaIa T1 T2 D1 +Va-+Va- D2 iaia + V dc  DC DRIVES AC-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

3 leg A leg B + V a  Q1 Q4 Q3 Q2 D1 D3 D2 D4 + V dc  v a = V dc when Q1 and Q2 are ON Positive current Power Electronic Converters in ED Systems DC DRIVES AC-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

4 T1 conducts  v a = V dc Q1Q2 VaVa IaIa T1 T2 D1 +Va-+Va- D2 iaia + V dc  DC DRIVES AC-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

5 Q1Q2 VaVa IaIa T1 T2 D1 +Va-+Va- D2 iaia + V dc  D2 conducts  v a = 0 VaVa EbEb T1 conducts  v a = V dc Quadrant 1 The average voltage is made larger than the back emf DC DRIVES AC-DC-DC DC-DC: Two-quadrant Converter Power Electronic Converters in ED Systems

6 Q1Q2 VaVa IaIa T1 T2 D1 +Va-+Va- D2 iaia + V dc  D1 conducts  v a = V dc DC DRIVES AC-DC-DC DC-DC: Two-quadrant Converter Power Electronic Converters in ED Systems

7 Q1Q2 VaVa IaIa T1 T2 D1 +Va-+Va- D2 iaia + V dc  T2 conducts  v a = 0 VaVa EbEb D1 conducts  v a = V dc 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-DC DC-DC: Two-quadrant Converter Power Electronic Converters in ED Systems

8 DC DRIVES AC-DC-DC DC-DC: Two-quadrant Converter + vcvc 2v tri vcvc +vA-+vA- V dc 0 Power Electronic Converters in ED Systems

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

10 leg A leg B + V a  Q1 Q4 Q3 Q2 D1 D3 D2 D4 + V dc  v a = -V dc when D3 and D4 are ON v a = V dc when Q1 and Q2 are ON v a = 0 when current freewheels through Q and D Positive current Power Electronic Converters in ED Systems DC DRIVES AC-DC-DC DC-DC: Four-quadrant Converter

11 v a = -V dc when D3 and D4 are ON v a = V dc when Q1 and Q2 are ON v a = 0 when current freewheels through Q and D Positive current v a = V dc when D1 and D2 are ON Negative current leg A leg B + V a  Q1 Q4 Q3 Q2 D1 D3 D2 D4 + V dc  Power Electronic Converters in ED Systems DC DRIVES AC-DC-DC DC-DC: Four-quadrant Converter

12 v a = -V dc when D3 and D4 are ON v a = V dc when Q1 and Q2 are ON v a = 0 when current freewheels through Q and D Positive current v a = -V dc when Q3 and Q4 are ON v a = V dc when D1 and D2 are ON v a = 0 when current freewheels through Q and D Negative current leg A leg B + V a  Q1 Q4 Q3 Q2 D1 D3 D2 D4 + V dc  Power Electronic Converters in ED Systems DC DRIVES AC-DC-DC DC-DC: Four-quadrant Converter

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22 Power Electronic Converters in ED Systems DC DRIVES AC-DC-DC v AB V dc -V dc V dc 0 vBvB vAvA 0 2v tri vcvc vcvc + _ V dc +vA-+vA- +vB-+vB- Bipolar switching scheme – output swings between V DC and -V DC

23 Power Electronic Converters in ED Systems DC DRIVES AC-DC-DC Unipolar switching scheme – output swings between V dc and -V dc V tri vcvc -v c vcvc + _ V dc +vA-+vA- +vB-+vB- -v c vAvA V dc 0 vBvB 0 v AB V dc 0

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

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

26 V dc 0 T tri t on 0 1 V c > V tri V c < V tri vcvc Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters

27 -V tri V tri -V tri vcvc d vcvc 0.5 For v c = -V tri  d = 0 Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters

28 Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters 0.5 V tri vcvc d vcvc -V tri For v c = -V tri  d = 0 For v c = 0  d = 0.5 For v c = V tri  d = 1

29 Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters 0.5 vcvc d -V tri V tri vcvc For v c = -V tri  d = 0 For v c = 0  d = 0.5 For v c = V tri  d = 1

30 Thus relation between v c and V a is obtained as: Introducing perturbation in v c and V a and separating DC and AC components: DC: AC: Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters

31 Taking Laplace Transform on the AC, the transfer function is obtained as: v a (s)v c (s) DC motor Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters

32 2v tri vcvc vcvc v tri + V dc − q -V dc q V dc + V AB  v AB V dc -V dc vBvB V dc 0 vAvA 0 Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters Bipolar switching scheme

33 v a (s)v c (s) DC motor Bipolar switching scheme Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters

34 + V dc − vcvc v tri qaqa V dc -v c v tri qbqb Leg a Leg b The same average value we’ve seen for bipolar ! V tri vcvc -v c vAvA vBvB v AB Unipolar switching scheme Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters

35 v a (s)v c (s) DC motor Unipolar switching scheme Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters

36 DC motor – separately excited or permanent magnet Extract the dc and ac components by introducing small perturbations in V t, i a, e a, T e, T L and  m T e = k t i a e e = k t  ac components dc components Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters

37 Perform Laplace Transformation on ac components V t (s) = I a (s)R a + L a sIa + E a (s) T e (s) = k E I a (s) E a (s) = k E  (s) T e (s) = T L (s) + B  (s) + sJ  (s) DC motor – separately excited or permanent magnet Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters

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

39 TcTc v tri + V dc − q q + – ktkt Torque controller Torque controller Converter - + DC motor Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters

40 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 Drives Modeling of the Power Converters: DC drives with SM Converters

41 OBJECTIVES: Fast response – large bandwidth Minimum overshoot good phase margin (>65 o ) 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 Drives Modeling of the Power Converters: DC drives with SM Converters

42 Ra = 2  La = 5.2 mH J = 152 x 10 –6 kg.m 2 B = 1 x10 –4 kg.m 2 /sec k t = 0.1 Nm/A k e = 0.1 V/(rad/s) V d = 60 VV tri = 5 V f s = 33 kHz Closed-loop speed control – an example PI controllers Switching signals from comparison of v c and triangular waveform Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters

43 compensated k pT = 90 k iT = Modeling and Control of Electrical Drives Modeling of the Power Converters: DC drives with SM Converters Torque controller design Open-loop gain

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

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

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

47 THANK YOU


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