Introduction three-phase diode bridge rectifier

Transcription

1 Introduction three-phase diode bridge rectifier

2 what is this all about? v A D1 D3 D5 + i OUT v OUT D2 D4 D6 v B i 1 i 2 i v 1 v 2 v 3

3 input voltages v 1 = V m cos (ω 0 t) ( v 2 = V m cos ω 0 t 2π 3 ( v 3 = V m cos ω 0 t 4π 3 ) ) ( v k = V m cos ω 0 t (k 1) 2π 3 ), k {1, 2, 3}

4 input voltages, waveforms

5 normalization of voltages m X v X V m m 1 = cos (ω 0 t) ( m 2 = cos ω 0 t 2π ) 3 ( m 3 = cos ω 0 t 4π ) 3

6 voltages? v k = V m cos ( ω 0 t (k 1) 2π 3 ), k {1, 2, 3}

7 voltages? v k = V m cos ( ω 0 t (k 1) 2π 3 ), k {1, 2, 3}

8 voltages? v k = V m cos ( ω 0 t (k 1) 2π 3 ), k {1, 2, 3}

9 voltages? v k = V m cos ( ω 0 t (k 1) 2π 3 ), k {1, 2, 3}

10 voltages? v k = V m cos ( ω 0 t (k 1) 2π 3 ), k {1, 2, 3}

11 voltages? v k = V m cos ( ω 0 t (k 1) 2π 3 ), k {1, 2, 3}

12 v 1, spectrum

13 v 2, spectrum

14 v 3, spectrum

15 voltages, quantitative characterization k V k RMS T HD(v k ) V 3.34 % V 2.77 % V 3.06 % all graphs and data PyLab processed

16 T HD And what is THD? Parseval s identity: T HD k=2 I2 k RMS I 1 RMS results in I 2 RMS = Ik 2 RMS assumed I 0 = 0 k=1 T HD I 2 RMS I2 1 RMS I 1 RMS simple, but important computational issues, finite sums...

17 normalization of currents and time j X i X I OUT unless otherwise noted ϕ ω 0 t good: physical dimensions lost, reduced number of variables, results are generalized, core of the problem focused bad: physical dimensions lost, perfect double-check tool is lost

18 how does it work? part 1: theory v A D1 D3 D5 + i OUT v OUT D2 D4 D6 v B i 1 i 2 i v 1 v 2 v 3

19 one of the three: D1, D3, D5

20 va

21 v A, analytical m A = max (m 1, m 2, m 3 )

22 v A, spectrum m A = 3 3 2π ( k=1 ) ( 1) k+1 9k 2 1 cos (3kω 0t)

23 what about v B? v A D1 D3 D5 + i OUT v OUT D2 D4 D6 v B i 1 i 2 i v 1 v 2 v 3

24 one of the three, again: D2, D4, D6

25 vb

26 v B, analytical m B = min (m 1, m 2, m 3 )

27 v B, spectrum m B = 3 3 2π ( k=1 ) 1 9k 2 1 cos (3kω 0t)

28 the output voltage, v OUT m OUT = m A m B = max (m 1, m 2, m 3 ) min (m 1, m 2, m 3 )

29 v OUT, spectrum m OUT = 3 3 π ( 1 2 k=1 ) 1 36k 2 1 cos (6kω 0t)

30 currents? i 1 (t) = (d 1 (t) d 2 (t)) I OUT i 2 (t) = (d 3 (t) d 4 (t)) I OUT i 3 (t) = (d 5 (t) d 6 (t)) I OUT

31 states of the diodes

32 the input currents

33 consider i1

34 spectra of the input currents

35 spectra of the input currents, analytical J 1C, k = 2 3 π j 1 (t) = + k=1 J 1C, k cos (kω 0 t) 1 k, k = 6n 1 1 k, k = 6n + 1 0, otherwise for n N 0, k > 0 double-check: P IN = π = 3 3 π = P OUT obtained using wxmaxima

36 numerical verification, Gibbs phenomenon

37 THD of the input currents 2 I k RMS = 3 I OUT 6 I k RMS, 1 = π I OUT Ik 2 RMS I2 k RMS, 1 T HD I k RMS, 1 π 2 T HD = % Parseval s identity based formula turned out to be useful

38 voltages and currents

39 some more parameters 1 X RMS 2π already used for the THD 2π 0 (x(ω 0 t)) 2 d(ω 0 t), x {i, v} S I RMS V RMS P 1 2π v(ω 0 t) i(ω 0 t) d(ω 0 t) 2π 0 P F P S DP F cos φ 1 and φ 1 is...

40 and if the voltages are sinusoidal... S = V RMS I RMS P = V RMS I 1, RMS cos φ 1 P F = P S = I 1, RMS I RMS cos φ 1 = I 1, RMS I RMS DP F DP F = cos ϕ 1 T HD = IRMS 2 ( I2 ) 2 1, RMS IRMS = 1 I 1, RMS I 1, RMS i.e. everything depends on the current waveform and its position

41 some more parameters, plain rectifier 2 I k RMS = 3 I OUT V k RMS = 1 V m 2 2 S = 3 3 I OUT 1 V m = 3 V m I OUT 2 P = V OUT I OUT = 3 3 π P F = 3 π 95.5% V m I OUT DP F = 1 actually, not so bad; T HD is the problem

42 back to the rectifier: how does it work? part 2: experiment v A D1 D3 D5 + i OUT v OUT D2 D4 D6 v B i 1 i 2 i v 1 v 2 v 3

43 input, at I OUT = 3 A

44 input, at I OUT = 3 A

45 input, at I OUT = 3 A

46 input, at I OUT = 3 A

47 input, at I OUT = 6 A

48 input, at I OUT = 6 A

49 input, at I OUT = 6 A

50 input, at I OUT = 6 A

51 input, at I OUT = 9 A

52 input, at I OUT = 9 A

53 input, at I OUT = 9 A

54 input, at I OUT = 9 A

55 output, at I OUT = 3 A

56 output, at I OUT = 3 A

57 output, at I OUT = 6 A

58 output, at I OUT = 6 A

59 output, at I OUT = 9 A

60 output, at I OUT = 9 A

61 in quantitative terms, input, 1st I OUT k I k RMS [A] V k RMS [V] S k [VA] P k [W] 0 A A A A

62 in quantitative terms, input, 2nd I OUT k P F k T HD(i k ) [%] T HD(v k ) [%] 0 A A A A

63 in quantitative terms, output I OUT [A] V OUT [V] P OUT [W] P IN [W] η [%]

64 overall impressions pretty good rectifier simple, robust, cheap good symmetry excellent DP F acceptable P F poor T HD (but not that poor) up to this point: diode bridge rectifier analyzed measurement tools developed is there a way to do something with the T HD?

65 fruitless effort #1: shaping the output current v A i OUT D1 D3 D5 + R OUT v OUT D2 D4 D6 v B i 1 i 2 i v 1 v 2 v 3

66 fruitless effort #1: waveforms

67 fruitless effort #1: quantitative T HD = 30.79% not a big deal of an improvement only one degree of freedom, i OUT shaping i 1, i 2, and i 3 is the goal two degrees of freedom needed, since i 1 + i 2 + i 3 = 0

68 fruitless effort #2: additional deegree of freedom v A i A + D1 D3 D5 R OUT 2 v OUT D2 D4 D6 R OUT 2 i B v B i 1 i 2 i 3 i N v 1 v 2 v 3

69 fruitless effort #2: waveforms

70 fruitless effort #2: neutral current

71 fruitless effort #2: quantitative T HD = 24.76% somewhat better all of i 1, i 2, and i 3 cannot be fixed by programming i A and i B in this circuit example: i 1 = i A, i 2 = i B, no way to fix i 3 gaps in the input currents in both of the patches the additional degree of freedom is taken by i N which is a disaster of itself we would need another degree of freedom to fix i N but this is a wrong approach, i N was not an issue before

72 conclusions three-phase diode bridge rectifier analyzed quantitative measures of rectifier performance introduced measurement tools developed theoretical predictions related to experiments gaps in the input currents identified as a problem how to fill in the gaps? an answer is current injection...