1. ELEC 3105 Basic E&M and Power Engineering
The Transformer 1

2. Review: Faraday’s law of inductionX X
B field into page
𝒗 𝒕 L
𝒗 𝒕 =− 𝝏𝚽 𝝏𝒕 =− 𝝏 𝑩 ∙ 𝑳 𝟐 𝒏 𝝏𝒕 L
Case A: If B and L and orientation, 𝒏 , constant
𝒗 𝒕 =− 𝝏 𝑩 ∙ 𝑳 𝟐 𝒏 𝝏𝒕 =𝟎
No voltage on terminals

3. Review: Faraday’s law of inductionX X
B field into page
𝒗 𝒕 L
𝒗 𝒕 =− 𝝏𝚽 𝝏𝒕 =− 𝝏 𝑩 ∙ 𝑳 𝟐 𝒏 𝝏𝒕 L
Case B: If B changes as a function of time
with L and orientation, 𝒏 , constant
𝐵 =− 𝐵 𝑚𝑎𝑥 sin⁡(𝜔𝑡) 𝑛
Voltage on terminals due to a changing magnetic field
𝒗 𝒕 = 𝝎𝑳 𝟐 𝑩 𝒎𝒂𝒙 𝐜𝐨𝐬⁡(𝝎𝒕)

4. Review: Faraday’s law of inductionX
B field into page X
𝒗 𝒕 L
𝒗 𝒕 =− 𝝏𝚽 𝝏𝒕 =− 𝝏 𝑩 ∙ 𝑳 𝟐 𝒏 𝝏𝒕 L
Case C: If B and orientation, 𝒏 , constant
with L increasing as a function of time and
𝒗 𝒕 =𝐋𝐁 𝝏𝑴 𝝏𝒕 X X X
𝒗 𝒕 L X X X 𝑣
Voltage on terminals “Linear motor”
“Rail Gun” M

5. Review: Faraday’s law of inductionX
B field into page X
𝒗 𝒕 L
𝒗 𝒕 =− 𝝏𝚽 𝝏𝒕 =− 𝝏 𝑩 ∙ 𝑳 𝟐 𝒏 𝝏𝒕 L
Case D: If B and L constant with orientation, 𝒏 , changing as a function of time and 𝐵 𝜔𝑡
𝒗 𝒕 =𝝎 𝑳 𝟐 𝐁𝒔𝒊𝒏 𝝎𝒕 𝑛
Voltage on terminals “Rotational motor”
“Generator”
𝐵 ∙ 𝑛 =𝐵𝑐𝑜𝑠 𝜔𝑡

6. Transformer Loop on source side “Primary”
Loop on load side “Secondary” Zg
𝒗 𝒕
𝐵(𝑡)
𝒗′ 𝒕
Zload
I(t)
Transformer
Transformer
optimize coupling, perform transformation N1 N2
core
Primary
Secondary

7. Only a few field lines pass through secondary
Transformer N1 N2
core
Primary
Secondary
Core: Designed such that as much 𝑩 produced in the primary passes to the secondary
Iron core
Only a few field lines pass through secondary
Magnetic circuit guides field lines from primary to secondary

8. Transformer Voltage relation
ON PRIMARY SIDE
One loop
Two loops
Area A
Area 2A X X
=BA
=2BA
Three loops
N1 loops
Area 3A
Area N1A X X
=3BA
=N1BA

9. Transformer Voltage relation
1=BA
Flux for each loop on primary
N1 loops
Area N1A
𝒗 𝒕 X
=N1BA
Voltage produced for N1 loops
𝒗 𝒕 =− 𝝏𝚽 𝝏𝒕 = 𝑵 𝟏 − 𝝏𝑩𝑨 𝝏𝒕
ON PRIMARY SIDE
Voltage produced for one loop
𝒗 𝟏 𝒕 = 𝑵 𝟏 − 𝝏𝑩𝑨 𝝏𝒕

10. Transformer Voltage relation
ON SECONDARY SIDE
One loop
Two loops
Area A
Area 2A X X
’=BA
’=2BA
Three loops
N2 loops
Area 3A X X
Area N2A
’=3BA
’=N2BA

11. Transformer Voltage relation
N2 loops
1’=BA
Flux for each loop on secondary X
Area N2A
𝒗′ 𝒕
’=N2BA
Voltage produced for N2 loops
𝒗′ 𝒕 =− 𝝏𝚽′ 𝝏𝒕 = 𝑵 𝟐 − 𝝏𝑩𝑨 𝝏𝒕
ON SECONDARY SIDE
Voltage produced for one loop
𝒗 𝟐 𝒕 = 𝑵 𝟐 − 𝝏𝑩𝑨 𝝏𝒕

12. Transformer Voltage relation
Area N1A
Prefect flux coupling
𝒗 𝟏 𝒕 X
=N1BA X
Area N2A
𝒗 𝟏 𝒕 = 𝑵 𝟏 − 𝝏𝑩𝑨 𝝏𝒕
𝒗 𝟐 𝒕
’=N2BA 12
combine
𝒗 𝟐 𝒕 = 𝑵 𝟐 − 𝝏𝑩𝑨 𝝏𝒕
𝒗 𝟏 𝒕 𝑵 𝟏 = 𝒗 𝟐 𝒕 𝑵 𝟐
Voltage transformation

13. Transformer Current relation
𝒊 𝟏 (𝒕)
IDEAL TRANSFORMER
No power loss
𝒗 𝟏 𝒕 X N1 X
Voltage transformation N2
𝒗 𝟏 𝒕 𝑵 𝟏 = 𝒗 𝟐 𝒕 𝑵 𝟐
𝒗 𝟐 𝒕
𝒊 𝟐 (𝒕) 13
Power (IN)
Power (OUT)
𝑷 𝒊𝒏 = 𝒗 𝟏 𝒕 𝒊 𝟏 𝒕
combine
𝑷 𝒐𝒖𝒕 = 𝒗 𝟐 𝒕 𝒊 𝟐 𝒕
𝒊 𝟐 𝒕 𝑵 𝟏 = 𝒊 𝟏 𝒕 𝑵 𝟐
Current transformation

14. Transformer impedance relation
𝒊 𝟏 (𝒕)
IDEAL TRANSFORMER
No power loss
𝒗 𝟏 𝒕 X N1 X
𝒗 𝟏 𝒕 𝑵 𝟏 = 𝒗 𝟐 𝒕 𝑵 𝟐 N2
𝒗 𝟐 𝒕
Zload 14
combine
𝒊 𝟐 (𝒕)
Zeq looking into transformer
𝒊 𝟐 𝒕 𝑵 𝟏 = 𝒊 𝟏 𝒕 𝑵 𝟐
𝒁 𝒆𝒒 = 𝑵 𝟏 𝑵 𝟐 𝟐 𝒁 𝒍𝒐𝒂𝒅

15. Transformer impedance relation
𝒊 𝟏 (𝒕)
IDEAL TRANSFORMER
No power loss
𝒗 𝟏 𝒕 X N1 X N2
𝒗 𝟐 𝒕
Zload 15
𝒊 𝟏 (𝒕)
𝒊 𝟐 (𝒕)
𝒗 𝟏 𝒕
Zeq N1
𝒁 𝒆𝒒 = 𝑵 𝟏 𝑵 𝟐 𝟐 𝒁 𝒍𝒐𝒂𝒅
Remove transformer

16. Transformer Calculation Example
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.
SOLUTION PROVIDED IN CLASS

17. Transformer Calculation Example
N1 “Primary”
N2 “Secondary” 2Ω
𝒗 𝒕
𝒗′ 𝒕
32Ω
I(t)
Find N1/N2 ratio such that maximum power transfer to the load is observed.
𝒁 𝒆𝒒 = 𝑵 𝟏 𝑵 𝟐 𝟐 𝒁 𝒍𝒐𝒂𝒅
SOLUTION PROVIDED IN CLASS

18. Maximum Power to Load Work in phasor domain 𝒗 𝒕Zs
𝒗 𝒕
Zload
𝐼 = 𝑉 𝑍 = 𝑉 𝑍 𝑠 + 𝑍 𝑙𝑜𝑎𝑑
I(t)
In general:
𝑧 𝑠 = 𝑅 𝑠 +𝑗 𝑋 𝑠
𝑧 𝑙𝑜𝑎𝑑 = 𝑅 𝑙𝑜𝑎𝑑 +𝑗 𝑋 𝑙𝑜𝑎𝑑
𝐼 = 𝑉 𝑅 𝑠 + 𝑅 𝑙𝑜𝑎𝑑 +𝑗 𝑋 𝑠 + 𝑋 𝑙𝑜𝑎𝑑
Then
Find maximum
𝑃= 𝐼 2 𝑅 𝑙𝑜𝑎𝑑 2
Time average power to load

19. Maximum Power to Load
Find maximum
𝑃= 𝑉 𝑅 𝑠 + 𝑅 𝑙𝑜𝑎𝑑 𝑋 𝑠 + 𝑋 𝑙𝑜𝑎𝑑 𝑅 𝑙𝑜𝑎𝑑 2
Step 1: Make ( ) as small as possible with respect to the complex “reactance” part
𝑋 𝑙𝑜𝑎𝑑 =− 𝑋 𝑠
𝑃= 𝑉 𝑅 𝑠 + 𝑅 𝑙𝑜𝑎𝑑 𝑅 𝑙𝑜𝑎𝑑 2
Find maximum

20. Maximum Power to Load Rs= 50Ω 𝑉 𝑠 2 =200 Find maximum Rload= 50Ω
𝑃= 𝑉 𝑅 𝑠 + 𝑅 𝑙𝑜𝑎𝑑 𝑅 𝑙𝑜𝑎𝑑 2
Rs= 50Ω
𝑉 𝑠 2 =200
Find maximum
Rload= 50Ω
Maximum Power to Load
𝑧 𝑠 = 𝑅 𝑠 +𝑗 𝑋 𝑠
𝑧 𝑙𝑜𝑎𝑑 = 𝑅 𝑠 −𝑗 𝑋 𝑠

21. Transformer Types Autotransformer Ip Is N1 N2 Primary Secondary core
Starting motors ELEC 4602

22. Transformer Types
Single phase Ip Is N1 N2
core
Primary
Secondary

23. Transformer Types
Single phase

24. Transformer Types
Maximum power transfer to the load

25. Transformer Types Center tapped Ip Is1 Vs1 Ns1 N1 Is2 Ns2 Vs2 core
ground
core
Primary
Secondary
With ground Vs1 180o out of phase with Vs2. Two phase household

26. Transformer Types

27. Transformer Types Center tapped Ip Is1 Vs1 Ns1 N1 Is2 Ns2 Vs2 core
ground
core
Primary
Secondary
With ground Vs1 180o out of phase with Vs2. Two phase household

28. Transformer Types
Center tapped

29. Transformer Types NA1 NA’2 A A’ NB1 NB’2 B B’ NC1 NC’2 C C’
Three phase

30. Interconnection
Transformer Types

31. Transformer Types

32. Transformer Types

33. Transformer Types
n turns ratio

34. Three phase AC to DC converter
Topic of ELEC 3508: Power Electronics

35. Power distribution

36. Three phase power transformers
LabVolt Module used in ELEC 3508