Konverter thyristor (kuliah ke 5)

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Konverter thyristor (kuliah ke 5)

  1. 1. Konverter Thyristor Pekik Argo Dahono
  2. 2. Konverter Satu-Fasa Setengah-Gelombang is T io + vo vs R Tegangan keluaran rata - rata : (a) Skema 1 π vs Vo = 2π ∫α 2Vs sin (ωt )d (ωt ) is0 = 2Vs (1 + cos α ) 2π π 2π ωt vo io0 π 2π ωt α (b) Bentuk gelombang.
  3. 3. Gate Signal Generation AC line Saw - tooth vsw Gate signal Comparator Generator vsyn and logic vc vsyn 0 vsw vc Gate signal
  4. 4. Linearizing the Phase-Control Characteristicsin ωt + ∫ − FF + + 1
  5. 5. Konverter Satu-Fasa Setengah-Gelombang is io vs is + vL T vs vd vo 0 R π 2π ωt vd•Arus masukan dan keluaran mempynyai iobentuk yang sama dan diskontinyu•Harmonisa arus masukan mengandungkomponen dc dan semua orde harmonisa π 2π ωt 0•Nilai rata-rata tegangan keluaran tidak αhanya ditentukan oleh sudut penyalaantetapi juga parameter beban.
  6. 6. Konverter Satu-Fasa Setengah-Gelombang is io + vL vs T vs vd vo is FD R 0 π 2π ωt vdNilai rms arus masukan lebih kecil dibanding arus keluaran ioArus masukan mengandung semua ordeharmonisa termasuk komponen dc π 2π ωt 0Nilai rata-rata tegangan keluaran hanya ditentukan αoleh sudut penyalaan
  7. 7. Konverter Satu-Fasa Setengah Gelombang is io + vL T vsvs vd vo is FD R 0 π 2π ωt vd io π 2π ωt 0 α
  8. 8. Konverter Satu-Fasa Setengah-Gelombang is vs is + Ls T vs FD vd Io 0 π 2π ωt vd ioInduktansi sumber menyebabkannilai rata-rata tegangan keluaran π 2π ωt 0 α μberubah sebagai fungsi arus beban 2Vs Vd = (1 + cos α ) − fLs I o 2π
  9. 9. Konverter Satu-Fasa Setengah-Gelombang is io + vL T vs Vovs vd Vo is 0 π 2π ωt vdNilai rata-rata tegangan keluaran Vodipengaruhi oleh emf beban io 0 α π 2π ωt
  10. 10. Konverter Satu-Fasa Jembatan io T1 T3 is + Nilai rata - rata tegangan keluaran : vs vo 1 π Vo = ∫ 2Vs sin (ωt )d (ωt ) R π α 2Vs (1 + cos α ) T2 T4 = π vs is Arus sumber hanya mengandung harmonisa orde ganjil ωt Tegangan dan arus beban mempunyai jumlah pulsa dua dan hanya mengandung harmonisa orde genap voα io0 π 2π ωt T1 & T 4 T2 &T3
  11. 11. Konverter Thyristor Satu-Fasa io T1 T3 Ld is Tegangan keluaran rata-rata: + 2 2 Vo = V s cos α vs vo R π Arus thyristor rata-rata: IT = I o / 2 T2 T4 vs Arus rms sumber : is Is = Io ωt PF sumber : Vo I o 2 2 PF = = cos α vo Vs I s π α io Tegangan keluaran bisa diatur dari minus maksimum sampai plus maksimum0 π 2π ωt tetapi arus selalu positip. T1 & T 4 T2 &T3
  12. 12. Konverter Thyristor Satu-Fasa vs is T1 T3 ωt is +vs vo Io α vo Io T2 T4 0 π ωt 2π
  13. 13. Konverter Thyristor Satu-Fasa vs vs is is ωt ωt vo vo Io Io 0 π 2π ωt 0 π 2π ωt vT 1vT 1 ωt ωt π α= π vs 4 α= 2 is vs is ωt ωt Io 0 π 2π ωt Io π 2π vo 0 ωt vovT 1 ωt vT 1 3π α= 4 α =π
  14. 14. Analisis Daya dan HarmonisaDaya Aktif :P = Vs I s1 cos αQ = Vs I s1 sin α PAnalisis Harmonisa Arus Masukan : Q ∞is = 2 ∑ I k sin (kωt ) α k = 2 n −1 0 π π 4 π /2 I o sin (kωt )d (ωt ) 2π ∫0Ik = 2 = 4Io [1 − cos(kπ / 2)] 2kπ 2 2I1 = Io πI k = I1 / k
  15. 15. Pengaruh Induktansi Sumber T1 T3 is vs is + Lsvs vo Io ωt T2 T4 α μ vo Io 0 π 2π ωt 2 2 Vo = Vs cos α − 4 fL s I o π
  16. 16. Sudut Pemadaman vs vs is is ωt ωt α μ vo α μ Io Io vo 0 π 2π ωt 0 π 2π ωtvT 1 vT 1 γ 2 2 2 2 Vo = Vs cos α − 4 fLs I o Vo = Vs cos α − 4 fLs I o π π
  17. 17. Konduksi Tak Kontinyu vs is io Ld ωt T1 T3 is + Vo vovs vo α Vo io T2 T4 0 π 2π ωt
  18. 18. Penyearah Satu-Fasa Setengah-Terkendali io T1 T3 Ld is Tegangan keluaran rata-rata: + 2 vs vo R Vo = V s (1 + cos α ) π Arus thyristor rata-rata: T2 T4 π −α IT = I o 2π vs Arus rms sumber : is 1/ 2 ⎛π −α ⎞ Is = Io ⎜ ⎟ ωt ⎝ π ⎠ PF sumber : α vo Vo I o 2 1 + cos α PF = = Vs I s π ⎛ π − α ⎞1 / 2 ⎜ ⎟ ⎝ π ⎠0 π 2π ωt
  19. 19. Penyearah Satu-Fasa Setengah-Terkendali io io Ld T1 T3 Ld T1 D3 is is + +vs vo R vs vo R D2 D4 T2 D4
  20. 20. Konverter Thyristor Tiga-Fasa Setengah-Gelombang vun vvn vwn 0 π 2π ωt iw T3 io iu T1 u R T2 LoadAC source n vo iv T w v S
  21. 21. Persamaan TeganganNilai rata - rata tegangan : 5π +α 2Vln s sin (ωt )d (ωt ) 3Vo = 2π ∫π +α 6 6 3 2 πVo = Vlls cos α 0≤α ≤ 2π 6Vo = 2π 6 [ ( Vlls 1 + cos π + α 6 )] π 6 ≤α ≤ 5π 6
  22. 22. Konverter Thyristor Tiga-Fasa Setengah- Gelombang iw T3 io iu T1 α vun vvn vwn u R T2 LoadAC source n vo iv 0 π T 2π ωt w v S iuJumlah pulsa tegangan keluaran sama dengan tiga ωtArus sekunder trafo mengandung komponen dc. iv iw Nilai rata - rata tegangan 3 2 iR Vo = Vlls cos α 2π
  23. 23. Konverter Thyristor Tiga-Fasa Setengah-Gelombang α vun vvn vwn α vun vvn vwn 0 π 0 2π ωt π 2π ωt T1 T2 T3 T3 T1 T2vT 1 ωt vT 1 ωt
  24. 24. Konverter Thyristor Tiga-Fasa io α vwn vun vvn T1 T3 T5 iu u 0 π 2π ωt iv von v R iw vd w ωt T2 T4 T6 iuJumlah pulsa tegangan keluaran sama denganenam.Harmonisa arus masukan adalah 5,7, 11, 13,…
  25. 25. Konverter Thyristor Tiga-Fasa Beban Resistifα
  26. 26. Konverter Thyristor Tiga-Fasa Beban Resistif Tegangan rata - rata : π Vo = 3 π ∫6 2 +α π +α ( 6 ) 2Vll sin ωt − π d (ωt ) 3 2 π Vo = Vll cos α 0 ≤α ≤ π 3 π sin (ωt )d (ωt ) 3 2 Vo = Vll ∫π π 3 +α 3 2 ⎡ ⎛π ⎞⎤ π 2π = Vll ⎢1 + cos⎜ + α ⎟⎥ ≤α ≤ π ⎣ ⎝3 ⎠⎦ 3 3
  27. 27. Konverter Tiga-Fasa Jembatan Beban Induktif vun vun iu iu 2π ωt 2π ωt 0 0 π π vun vun iu iu 2π ωt 2π ωt 0 0 π π
  28. 28. AnalisisNilai rata - rata tegangan output : Faktor daya : 3 2Vo = Vllscosα 3 π PF = cos αArus masukan : π ∞iu = 2 ∑ Ik sin (kωt ) k = 2n -1 2 2 π /2I1 = π ∫ Io sin (ωt )d (ωt ) π /6 6I1 = Io πI k = I1 / kI k = 0 untuk k kelipatan 3.Iu = I o 2 / 3
  29. 29. Pengaruh Induktansi Sumber vun iu 2π ωt 0 π T1 T3 T5 iu u ivn v vuv vuw iw w Ls T6 T2 T4 3 2 Vo = Vll cos α − 6 fL s I o π
  30. 30. Half-Controlled Thyristor Converters io io T1 T3 T5 D1 L T1 T2 T3 iu L u iu u iv vo v R iv vo v R iw iw w w T2 T4 T6 D2 D1 D2 T3
  31. 31. Half-Controlled Rectifier α vun vvn vwn α vvn vwn vun 0 π 0 2π ωt π 2π ωt vd vd ωt ωt ωt ωtiu iu iwiw iRiR
  32. 32. Konverter Tiga-Fasa Jembatan Half-Controlled ωt ωt ωt ωt iu iu iw iw iR iR
  33. 33. Application Considerations• Single-phase rectifiers generate input harmonics at the order of 2p±1, where p is the pulse number.• The displacement power factor is reduced when the output voltage is reduced.• Commutation generates voltage notches across the source.• Input harmonics can be reduced by increasing the pulse number.
  34. 34. Current Controller• A thyristor converter is usually operated as a current source• A thyristor converter cannot be controlled faster than the thyristor can respond• After a thyristor is turned on, the thyristor can only be turned off by the input line voltage.• By operating as a current source, the thyristor converter is inherently overcurrent protected.• A current source can paralleled easily with other current sources
  35. 35. Current controller for Phase- Controlled Rectifiers es Ls io L vd Load Gate driver Reference + PID cos−1 current − Actual current
  36. 36. Current-Controlled Phase- Controlled Rectifiers Vo (s ) + 1 + − 1 I L (s ) *I L (s) PID 3 2 1 + sTd Vll π sL −
  37. 37. Current Control of Phase-Controlled Rectifiers
  38. 38. Dual Thyristor Converter ia *ia + Current cos−1 controller −
  39. 39. Application Considerations• At present, thyristor converters are used only for large power applications.• The AC side always need reactive power under both rectifier and inverter operations.• The AC side current has high harmonic content. The harmonic order is pk±1 where p is pulse number and k is integer. The harmonic current can be reduced by increasing the pulse number.• Thyristor converter also generates voltage nothches due to the commutation.• It is recommended to use a special feeder (or it is better if using a dedicated transformer) to supply a thyristor converter.

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