1. 1 The Transformer 1

2. 2 X XX X X B field into page L L No voltage on terminals

3. 3 X XX X X B field into page L L Voltage on terminals due to a changing magnetic field

4. 4 X XX X X B field into page L L Voltage on terminals “Linear motor” “Rail Gun” X X X X M L X X

5. 5 X XX X X B field into page L L Voltage on terminals “Rotational motor” “Generator”

6. ZgZg I(t) Z load Loop on source side “Primary” Loop on load side “Secondary” Transformer N1N1 N2N2 core Primary Secondary Transformer optimize coupling, perform transformation

7. N1N1 N2N2 core Primary Secondary Only a few field lines pass through secondary Iron core Magnetic circuit guides field lines from primary to secondary

8. 8 X Area A One loop  =BA X Area 2A Two loops  =2BA X Area 3A Three loops  =3BA X Area N 1 A N 1 loops  =N 1 BA ON PRIMARY SIDE

9. 9 X Voltage produced for one loop Voltage produced for N 1 loops  1 =BA Flux for each loop on primary N 1 loops  =N 1 BA Area N 1 A

10. 10 X Area A One loop  ’=BA X Area 2A Two loops  ’=2BA X Area 3A Three loops  ’=3BA X Area N 2 A N 2 loops  ’=N 2 BA ON SECONDARY SIDE

11. 11 X Area N 2 A N 2 loops  ’=N 2 BA  1 ’=BA Flux for each loop on secondary Voltage produced for one loop Voltage produced for N 2 loops ON SECONDARY SIDE

12. 12 X  =N 1 BA Area N 1 A 12 X Area N 2 A  ’=N 2 BA combine Voltage transformation Prefect flux coupling

13. X N1N1 13 X N2N2 IDEAL TRANSFORMER No power loss Voltage transformation Power (IN) Power (OUT) combine Current transformation

14. X N1N1 14 X N2N2 IDEAL TRANSFORMER No power loss Z load Z eq looking into transformer combine

15. X N1N1 15 X N2N2 IDEAL TRANSFORMER No power loss Z load N1N1 Z eq Remove transformer

16.  A 10-kVA; 6600/220 V/V; 50 Hz transformer is rated at 2.5 V/Turn of the winding coils. Assume that the transformer is ideal. Calculate the following:  A) step up transformer ratio  B) step down transformer ratio  C) total number of turns in the high voltage and low voltage coils  D) Primary current as a step up transformer  E) Secondary current as a step down transformer. 16 SOLUTION PROVIDED IN CLASS

17.  Find N1/N2 ratio such that maximum power transfer to the load is observed. 17 SOLUTION PROVIDED IN CLASS 2Ω2Ω I(t) 32Ω N 1 “Primary” N 2 “Secondary”

18. 18 ZsZs I(t) Z load Work in phasor domain In general: Then Time average power to load Find maximum

19. 19 Find maximum

20. 20 Find maximum R s = 50Ω R load = 50Ω

21.  Find current in the primary 21 SOLUTION PROVIDED IN CLASS IpIp N 1 = 2 “Primary” N 2 = 1 “Secondary” ZsZs Z load I load

22. N1N1 N2N2 core Primary Secondary Autotransformer IpIp IsIs Starting motors ELEC 4602

23. N1N1 N2N2 core Primary Secondary Single phase IpIp IsIs

25. Maximum power transfer to the load

26. N1N1 N s1 core Primary Secondary Center tapped IpIp I s1 N s2 I s2 V s1 V s2 ground With ground V s1 180 o out of phase with V s2. Two phase household

28. N1N1 N s1 core Primary Secondary Center tapped IpIp I s1 N s2 I s2 V s1 V s2 ground With ground V s1 180 o out of phase with V s2. Two phase household

29. Center tapped

30. NA 1 NA’ 2 Three phase A A’ NB 1 NB’ 2 B B’ NC 1 NC’ 2 C C’

31. Interconnection

34. n turns ratio

35. 35 Topic of ELEC 3508: Power Electronics

36. 36

37. 37 LabVolt Module used in ELEC 3508

38.  Initially used from 2001 - 2006 38

39. 39 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Suppose a current i 1 flows in the primary. The current i 2 is zero (secondary is open) Total flux in primary Reluctance Self inductance of primary Lecture 24 slide 4

40. 40 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Typical transformer for low frequency applications Magnetic fields are confined almost entirely to the iron core. Lecture 22 Slide 14

41. 41 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Suppose a current i 1 flows in the primary. The current i 2 is zero (secondary is open) Let be the flux in each of the loops in the primary

42. 42 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Suppose a current i 1 flows in the primary. The current i 2 is zero (secondary is open) From magnetic circuit results Lecture 22 Slides 23 and 24 Reluctance

43. 43 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Suppose a current i 1 flows in the primary. The current i 2 is zero (secondary is open) Let be the flux in each of the loops in the secondary resulting from the current i 1 in the primary.

44. 44 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Suppose a current i 1 flows in the primary. The current i 2 is zero (secondary is open) Write k a flux coupling coefficient k < 1 but in good transformer k  1

45. 45 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Suppose a current i 1 flows in the primary. The current i 2 is zero (secondary is open) Total flux in secondary Mutual inductance

46. 46 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Suppose a current i 2 flows in the secondary. The current i 1 is zero (primary is open) Then after a few hidden slides Reluctance

47. 47 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Suppose a current i 2 flows in the secondary. The current i 1 is zero (primary is open)

48. 48 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Assume k = 1 for this transformer Let be the total flux in the primary Let be the total flux in the secondary Then Same B and A in all loops

49. 49 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Assume k = 1 for this transformer With Same B and A in all loops And Then

50. 50 Lecture 25 Transformer Voltage START Current Impedance Power

51. 51 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Magnetic fields are confined almost entirely to the iron core. Load resistor Suppose: Then expect for current and

52. 52 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Magnetic fields are confined almost entirely to the iron core. Load resistor Lecture 24 Slide 32

53. 53 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Magnetic fields are confined almost entirely to the iron core. Load resistor But

54. 54 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Magnetic fields are confined almost entirely to the iron core. Load resistor If: Then

55. 55 Lecture 25 Transformer Voltage Current Impedance START Power

56. 56 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Magnetic fields are confined almost entirely to the iron core. Load resistor Power in the load resistor is:

57. 57 Input admittance Equivalent circuit for the primary

58. 58 Lecture 25 Transformer Voltage Current START Impedance Power

59. 59 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Load resistor But We start from this slide Use in here

60. 60 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Load resistor

61. 61 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Load resistor Input impedance

62. 62 Primary, N 1 turns Iron core cross-sectional area A Secondary, N 2 turns Load resistor Input admittance

63. 63 Input admittance 0 for k = 1 Equivalent circuit for the primary

64. 64 Equivalent circuit for the primary If the impedance looking into the primary is just the Transformed load resistance. This result breaks down a suitably low frequencies. {Transformed load resistance}