1 The Transformer 1

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

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

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

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

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

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

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

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

 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

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

19 Find maximum

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

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

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

N1N1 N2N2 core Primary Secondary Single phase IpIp IsIs

Maximum power transfer to the load

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

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

Center tapped

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

Interconnection

n turns ratio

35 Topic of ELEC 3508: Power Electronics

36

37 LabVolt Module used in ELEC 3508

 Initially used from

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 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 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 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 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 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 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 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 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 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 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 Lecture 25 Transformer Voltage START Current Impedance Power

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 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 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 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 Lecture 25 Transformer Voltage Current Impedance START Power

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 Input admittance Equivalent circuit for the primary

58 Lecture 25 Transformer Voltage Current START Impedance Power

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

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

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

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

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}