A transformer transfers electrical power from one circuit to another through magnetic induction. It increases or decreases voltage levels while keeping frequency constant. An ideal transformer has no losses, but a real transformer has resistance and leakage reactance. The performance of a real transformer can be determined through open-circuit and short-circuit tests to find its equivalent circuit parameters.

1. CHAPTER TWO
Transformer
What is a transformer ?
 It is a static device that transfers electrical power from one
circuit to another through the magnetic field by changing the
level of voltage and current without changing the frequency
 Transformer works only with AC .
 The main active materials in transformer construction are
magnetic materials and conductors.
 Magnetic material consists of laminated iron core and carries
flux linked to windings
 The iron core provides a low reluctance path to the magnetic
flux thereby reducing magnetizing current
1

2. Why do we need transformers?
 The most important tasks performed by transformers are:-
a. Changing voltage and current levels in electrical power
systems:
They are used to step-up the generated voltage to an
appropriate level for power transmission.
Stepping down the transmission voltage at various
levels for distribution and power utilization.
b. Matching source and load impedances for maximum
power transfer
c. Electrical isolation (isolating one circuit from another)
2

3. Cont….
3

4. Transformer Classification
 In terms of number of phases
1- single phase transformer
2- poly phase transformer
4
In terms of their function
1 - Power transformer
2 - Distribution transformer
3 - Measuring transformers
A) Voltage transformer
B) Current transformer
4 - Autotransformer- Tapped autotransformer

5. Power Transformer
 It is connected to the output of
a generator to step up the
voltage to the transmission
level.
 This is important to reduce
transmission losses
5

6. Con’t
 The power transformer is used between the generator
and the distribution circuits, and these are usually
rated at 500 kVA and above.
 It is available for step-up operation, primarily used
at the generator and referred to as generator step-up
(GSU) transformers.
 Generator step-up transformers, used in power plants,
receive electrical energy at generator voltage and
increase it to a higher voltage for transmission lines.
 The power transmission at high voltages values of
132kv,230kv ,220kv ,400kv or 500kv.etc
6

7. Distribution transformer
 It is a transformer converting
the distribution voltage down
to the final level.
 This is important to increase
the level of current
7

8. Con’t
 Transformers smaller than 500 kVA are generally
called distribution transformers and serve residences
and small businesses.
 A step-down transformer receives energy at a higher
voltage and delivers it at a lower voltage for
distribution to various loads.
 The usual consumer voltage requirement is 220v or
400v.
8

9. Measuring transformers
A)Voltage Transformer
 are used where the voltage of an AC circuit exceeds 750 V
as it is not possible to provide adequate insulation on
measuring instruments for voltage more than this.
 Used to step down the voltage to the level of volt meter
9

10. B)Current Transformer
10
Current transformer is used with low range ammeters to
measure high current in HV ac circuits.
In addition to insulating the instruments from the HV line,
they stepping down the current in to known ratio or to a
suitable value which is with in the range of ammeter
As the name suggests, these
transformers are used in
conjunction with the relevant
instruments such as ammeters,
voltmeters, watt meters and
energy meters
In general, Instrument transformers are used for measuring
and control purposes.

11. Autotransformer :- tapped autotransformer
11
Transformers having only one winding are called
autotransformers,
An autotransformer has the usual magnetic core but
only one winding, which is common to both the primary
and secondary circuits.

12. According to transformer design or construction
 Based on the construction
transformer are two types
a) Shell type transformer
- The laminated steel core
surrounds the copper windings.
b) Core type transformer.
-The copper windings surround
the laminated steel core.
12
b) shell-type transformer
a) Core-type transformer

13. Cont….
The choice of core or shell-type transformer is
usually decided by cost, voltage ratings, weight,
KVA ratings, heat distribution etc because
similar characteristics can be obtained with both
types.
Shell type is preferred for high voltage transformers
and multi winding design
 Because the mean length of coil turn in shell-type is
longer than in a comparable core-type design
13

14. Essential Elements (Parts)of Transformer
a)Two coils having mutual inductance wound on a
laminated steel core which is known as primary
winding and secondary winding.
 The two coils are insulated from each other and
magnetic core.
 Special paper and wood are used for insulation and
internal structural support
b) A suitable container for the assembled core and
windings.
c) A suitable medium for insulating the core and its
windings from the container.
d) Suitable bushings for insulating and bringing out the
terminals of windings from the container
14

15. Accessories Of Transformer
Conservator
 This is an expansion tank.
 Used to keep the transformer tank full of oil whether
expansion or contraction of oil take place.
 It is mounted above the transformer and connected to
the tank through pipe.
Temperature gauge
 This indicates oil temperature in the tank.
 It is connected to an alarm.
Oil gauge
 This indicates level of the oil in the tank.
 Sometimes it is provided with the alarm contracts
when the oil level falls down below a minimum level,
contacts close and give an alarm. 15

16. Buckholtz relay
It is a gas operated relay, located in the pipe
connected to the conservator.
Breather
 To prevent entry of moisture in to the tank, a breather
with silica gel is provided in the transformer.
 Silica gel absorbs the moisture and allows only dry air to
enter the tank
16
When a fault occurs in the
transformer gas bubbles are
released and these operate
the relay to give an alarm signal.

17. Working Principle of transformer
 If the primary winding is connected to an AC source, an
alternating flux will be produced whose amplitude will
depend on the Primary voltage and primary number
of turns
 The mutual flux will link the secondary winding and
changes , thus an e.m.f will induce in it whose value will
depend on secondary number of turns and the
magnitude of the mutual flux.
 If the circuit of the secondary winding is closed a current
flows in it and so electrical energy is transferred
magnetically from the primary to the secondary winding.
17

18. Con’t
Hence magnetic flux linked with the secondary coil
changes an e.m.f is induced in the secondary winding..
18

19. E.m.f Equation Of Transformer
 The induced e.m.f in a transformer is proportional to
the product of number of turns N and the rate of
change of flux.
e= N*d
dt
19

20. Cont…
 The flux() increases from zero to its maximum value m
in one quarter of a cycle.
dt=1/4f second
 Average rate of change of flux =
m
1
4𝑓
= 4f m wb/s or volt
 Thus, rate of change of flux per turn means induced e.m.f
in volts
Average e.m.f /turn =4f m volts
 If flux  varies sinusoidally, then r.m.s value of induced
e.m.f is obtained by multiplying the average value with the
form factor.
Form factor = r.m.s value = 1.11
Average value
20

21. Cont…
 r.m.s value of e.m.f/turn = 1.11*average value
= 1.11* 4f m
= 4.44f m volt
 Now r.m.s value of the induced e.m.f in the whole of
primary winding(E1).
E1 = induced e.m.f/turn*No turns of primary winding.
= 4.44f N1 m = 4.44f N1BmA
 Similarly r.m.s value of e.m.f induced in secondary is
E2 = 4.44f N2 m = 4.44f N2BmA
21

22. Ideal Transformers
 An ideal transformer is a lossless device with an input
and an output winding.
It has the following properties:
- No iron and copper losses
- No or negligible winding resistance
- No leakage fluxes, i.e. fluxes set up by the primary
links to the secondary
 For ideal transformer E1=V1 and E2= V2
 The power in ideal transformer V1I1 =V2I2
E1/E2 =V1/V2 =N1/N2 =I2/I1
 Have equal e.m.f per turn, i.e. E1/N1 = E2/N2,
22

23. Con’t
23
Figure: 2-1 Ideal transformer
Voltages will be induced in these two coils:
Assume that a time varying
flux, Φm(t), is established in
the iron core.
Then the flux linkages in
each coil will be
λp = NpΦm(t)
λs = NsΦm(t)

24. Derivation of the Relationship
24
The relationships between Vp and Vs, Ip and Is, Np and
Ns are given by the equation (5) and it is known as Turn
ratio (a). Equations of turn’s ratio (a) is important to
describe ideal transformer

25. Power in an Ideal Transformer
25

26. Theory of Operation of Single-Phase
Real Transformers
26
LS
M
S
LP
M
P










Фp -total average primary flux
ф M - flux linking both primary and secondary
windings
ф LP - primary leakage flux
ф S - total average secondary flux
ф LS - secondary leakage flux

27. The voltage ratio across a real transformer
27
 From faraday’s law the primary coils voltage is
 The Secondary coils voltage is
 The primary and secondary voltages due to mutual fluxes are
Where:

28. Con’t
Combining the last two equations
Therefore:
That is, the ratio of the primary voltage to the
secondary voltage both caused by the mutual flux is
equal to the turn ratio of the transformer.
Therefore the following approximation normally holds
28

29. The current ratio on a transformer
If a load is connected to the secondary coil, there will be a
current flowing through it
A current flowing in to the doted end of a winding produces a
positive magneto-motive force Fm.
The net magneto-motive force in the core
Where Rm is the reluctance of the transformer core.
29

30. Con’t
For well designed transformer cores, the reluctance is
very small if the core is not saturated. Therefore
The last approximation is valid for well designed
unsaturated cores.
30

31. Equivalent Circuit of a Transformer
• The transformer shown diagrammatically in Fig. (a) Can be
resolved into an equivalent circuit in which the resistance and
leakage reactance of the transformer are imagined to be
external to the winding whose only function then is to
transform the parameters (Fig.(b)).
31

32. The Magnetization Current in a Real
Transformer
When an ac source is connected to the primary, a current
flows through the primary winding, even when the
secondary circuit is open circuited.
The transformer is said to be on no-load. If the
secondary current (IS) is zero, the primary current (Ip)
should be zero too.
However, when the transformer is on no-load,
excitation current flows in the primary because of the
core losses and the finite permeability of the core.
This current is the current required to produce flux in
the ferromagnetic core and is called excitation
current (Io).
32

33. Con’t
 Excitation current consists of two components:
Io = Ih+e + Im
1. The magnetization current Im, which is the current
required to produce the flux in the transformer core.
2. The core-loss current Ih+e, which is the current required to
make up for hysteresis and eddy current losses
33

34. Impedance Transformation through a
Transformer
34
 To make transformer calculations simpler, it is
preferable to transfer V, I and Z either to the primary or
to the secondary. This is important to eliminate the
transformer and we get an equivalent electrical circuit
 In this case, we would have to work in one winding
only which is more convenient
 This transformation is done by using constant K which is
called voltage transformation ratio and given by:

35. Con’t
35
Figure (c): Equivalent circuit referred to the primary
To transfer secondary parameter to the primary, R and x is
divided by K2 , voltage is divided by K, current is
multiplied by K.
Thus, E’2 =E2/K=E1 R’2=R2/K2,
V’2=V2/K X’2=X2/K2
I’2 =KI2 Z’2=Z2/K2

36. Con’t
36
Where, R01 equivalent resistance referred to the primary
X01 equivalent reactance referred to the primary
Z01 equivalent impedance referred to the primary
Similarly to transfer primary parameter to the secondary, R
and x is multiplied by K2 , voltage is multiplied by K, current
is divided by K.
Thus, R’1=R1K2, E’1 =E1K =E2
X’1=X1K2 V’1=V1K
Z’1=Z1K2 I’1 =I1/K

37. Con’t
37
Figure (d): Equivalent circuit referred to the secondary
Where, R02 equivalent resistance referred to the secondary
X02 equivalent reactance referred to the secondary
Z02 equivalent impedance referred to the secondary
The same relationship is used for shifting external load
impedance to the primary.
i.e. Z’L=ZL/K2

38. Con’t
• The secondary circuit is shown in Fig. (a) and its
equivalent primary values are shown in Fig.(b).
38
The total equivalent circuit of the transformer is obtained
by adding in the primary impedance as shown in Fig. (c).
This is known as the exact equivalent circuit but it
presents a somewhat harder circuit problem to solve.

39. Con’t
39
A simplification can be made by transferring the exciting
circuit across the terminals as in Fig. (d) or in Fig.(e).

40. Con’t
Further simplification may be achieved by omitting I0
altogether as shown in Fig.(f).
Where:-
From Fig.(c) shown below it is found that total
impedance between the input terminal is
40

41. Con’t
41
Z1= R1 + jX1
Z'2= R'2 + jX'2
This is so because there are two parallel circuits, one
having an impedance of Zm and the other having Z2
′ and ZL
′
in series with each other.

42. 42
The performance of a transformer can be calculated on
the basis of its equivalent circuit parameters, these are:
equivalent resistance R01 and R02,
equivalent leakage reactance X01 and X02),
core-loss conductance G0 (or resistance R0) and
magnetizing susceptance Bm (or reactance Xm).
 These parameters of a transformer can be experimentally
determine with only two tests….
- Open-circuit test
- Short-circuit test
Transformer Tests

43. 43
These tests are very economical and convenient, because
they give the required information without actually loading
the transformer.
Con’t
1.Open-Circuit Test is used to determine:
no-load input power or core loss,
no load power factor (p.f)
no-load exciting current I0 w/c is helpful in finding Xm & R0
To carry out open circuit test,
1. Secondary winding which is (HV) is left open.
2. (LV) or primary winding is connected to its supply
of normal voltage and frequency.
A wattmeter W, voltmeter V and an ammeter A are
connected in the low voltage winding.

44. 44
With normal voltage applied to the primary, normal flux
will be set up in the core, hence normal iron losses will occur
which is recorded by the wattmeter.
The primary current on no load is usually less than 5% of the
full load, Cu loss is negligibly small in primary and nil in
secondary (it being open) and there fore can be neglected.
Hence, the wattmeter reading represents practically the core
loss Pi under no-load condition.
Input voltage (Voc), input current (Ioc), and no-load input
power (Pi) to the transformer are measured.
Con’t

45. 45
Short circuit test
.
In short circuit test, the LV (secondary) winding, is short-
circuited by a thick conductor (or an ammeter which may
serve the additional purpose of indicating rated load current).
A low voltage (5 to 10% of normal Vp) is applied to the
primary and is continuously increased till full-load currents
are flowing in both primary and secondary
Input voltage, input current, and input power to the
transformer are measured.
Since, in short circuit test, the applied voltage is a
small percentage of the normal voltage, the mutual flux
Φ produced is also a small percentage of its normal
value.

46. Short Circuit Test
46
Hence, core losses are very small with the result that the
wattmeter reading represent the full-load Cu loss or I2R
loss for the whole transformer
 i.e. both primary Cu loss and secondary Cu loss.
Input voltage (Vsc), input current (Isc), and input power
Pcu to the transformer are measured.
Where, I1 and Pcu is the ammeter and watt meter reading respectively

47. Efficiency of transformer
 The efficiency of transformer depends on loss.
 The loss occurring in a transformer can be divided into
two parts
1. Copper (I2R) losses: are the resistive heating losses in
the primary and secondary windings of the transformer as
(I2
1R1 +I2
2 R2 ).
• They are proportional to the square of the current in
the windings.
2. Iron loss: in the core due to hysteresis and eddy currents
a) Eddy current losses:- are resistive heating losses in the
core of the transformer due to circulating currents set up in
the core.
47

48. Con’t
 In typical transformers under rated current and voltage
the magnetizing current Im does not exceed 1% of the
rated current.
 Under rated current, total voltage drops on the winding
resistances and leakage inductances do not exceed in
typical transformers 6% of the rated voltage.
 The efficiency of transformer depends on loss.
48

49. Rating of transformer
 As seen copper loss depends on the current
• and iron loss on voltage.
 Hence total loss of transformer depends on volt-
ampere(VA) and not on phase angle between current and
voltage i.e. it is independent on load power factor.
 That is why rating of transformer is in KVA and not in
KW
 A transformer is described by its rated apparent power.
49

50. Transformer Voltage Regulation
 Because a real transformer has series impedance within it, the
output voltage of a transformer varies with the load even if the
input voltage remains constant.
 The voltage regulation of a transformer is the change in the
magnitude of the secondary terminal voltage from no-load
to full-load.
Voltage regulation=Vs(no load)-Vs(full load)*100%
Vs(rated voltage)
%Voltage regulation=Vp (no load)-Vp (full load)*100%
Vp ( rated voltage)
Where:- Vs = Secondary voltage
Vp= Primary voltage
 The purpose of voltage regulation is to determine the
percentage of voltage drop between no load and full load
50

51. Con’t
The secondary terminal voltage falls as the load on the
transformer is increased when power factor is lagging and
it increases when the power factor is leading.
In other words, secondary terminal voltage not only
depends on the load but also on power factor.
The voltage regulation is positive when the power factor
is lagging and negative when the power factor is leading
51

52. Three phase transformer
• Majority of Electric power generation/distribution systems
in the world are 3-phase systems.
• Large scale generation of electric power is usually 3-phase at
generated voltages of 13.2 kV or somewhat higher
• Transmission is generally accomplished at higher voltages of
110, 132, 275, 400 and 750 kV
• For these purpose 3-phase transformers are necessary to step
up the generated voltage to that of the transmission level.
• The transformers for such circuits can be constructed either as
a 3-phase bank of independent identical transformers or
• As a single transformer wound on a single 3-legged core
(lighter, cheaper, more efficient).
52

53. Con’t
• Generation, transmission and distribution of electric
energy is invariably done through the use of three-phase
systems because of its several advantages over single-phase
systems.
• For 3-phase up or down transformation, three units of single-
phase transformers or one unit of 3-phase transformer may be
used
• When three identical units of 1-phase transformers are used as
shown in Figure a, the arrangement is usually called a bank of
three transformers or a 3-phase transformer bank.
• A single 3-phase transformer unit may employ 3–phase core-
type construction Figure b or three phase shell type
construction
53

54. Con’t
A single-unit 3-phase core-type transformer uses a three-limbed
core, one limb for each phase winding as shown in Fig. b
Actually, each limb has the L.V. winding placed adjacent to
the laminated steel core and then H.V. winding is placed over
the L.v. winding.
Appropriate insulation is placed in between the core and L.v.
winding and also in between the two windings.
54
P S P S P S
A B C
a b c
Input
Output
I III
II
P
S
P
S
P
S
Figure (a) 3-phase transformer both windings in star; b) three-phase core-type transformer

55. Three-Phase Transformer Connections
There are various methods available for transforming
3-phase voltages to higher or lower 3-phase voltages
i.e. for handling a considerable amount of power.
 The most common connections are
a) Y − Y (Star/Star)
b) Δ − Δ (Delta/Delta)
c) Y − Δ (Star/Delta)
d) Δ − Y (Delta/star)
55

56. Star/Star or Y/Y Connection
 This connection is used for small, high-voltage
transformers.
 Star-star connection is rarely used in practice
 This connection works satisfactorily only if the load is
balanced.
 With the unbalanced load to the neutral, the neutral point
shifts thereby making the three line-to-neutral voltages
unequal or (IR + IY+ IB 0)
 There is a phase shift of 30° between the phase voltages
and line voltages both on the primary and secondary
sides.
56

57. Star/Star or Y/Y Connection
57
Fig. Y − Y Connection

58. Star/Star or Y/Y Connection
58

59. Delta-Delta or (Δ − Δ) Connection
 This connection is used for large, low-voltage
transformers
 The secondary voltage triangle abc occupies the same
relative position as the primary voltage triangle ABC i.e.
there is no angular displacement between the two.
 Moreover, there is no internal phase shift between phase
and line voltages on either side as was the case in Y – Y
connection
59

60. Delta-Delta or (Δ − Δ) Connection
60
Figure Δ − Δ connection

61. Delta-Delta or (Δ − Δ) Connection
61

62. Advantages of (Δ − Δ) Connection
 No difficulty is experienced from unbalanced loading
as was in the case of Y − Y connection.
 The three-phase voltages remain practically constant
regardless of load imbalance.
 An added advantage of this connection is that if one
transformer becomes disabled, the system can continue
to operate in open-delta or in V − V although with
reduced available capacity.
 The reduced capacity is 58% and 66.7% of the normal
value.
62

63. Star/Delta or Y/Δ Connection
 This connection is commonly used at the end of transmission line
to step down the voltage from a high level to a medium or low
level
 The ratio between the secondary and primary line voltage is 1/3
times the transformation ratio of each transformer.
 There is a 30 degree phase shift b/n VL prim and VL sec
• The primary winding is Y-connected with grounded neutral as
shown in Fig.. Then, the ratio is
63
Figure Star- Delta connection
V
3
V I
aI
3
a
3
V
aI
I
=

64. Star/Delta or Y/Δ Connection
64

65. Delta/Star or Δ/Y Connection
 This connection is generally employed to step up the voltage
to a high level at the beginning of H.V transmission Line.
 The primary and secondary line voltages and line currents are
out of phase with each other by 30°
 The ratio of secondary to primary voltage is3 times the
transformation ratio of each transformer.
65
Figure Delta- Star Connection
3
I
V
I
a
V
a
V
3
3
aI
3
aI

66. Delta/Star or Δ/Y Connection
66

67. Three phase Voltages and Currents in Star
and Delta Connections
67

68. Inrush Currents In 3-phase Transformers
Transient inrush current
 An inrush current is a transient current with high amplitude
that may occur when a transformer is energized under no load
or lightly loaded conditions.
 Inrush currents can exceed the nominal current and may
achieve the rated value of the short-circuit current of the
power transformer and result in false operation of protective
devices.
 In actual practice, a transient phenomenon in the form of
inrush current is unavoidable since the instant of switching in
three phase transformer cannot be easily controlled
 Let us see the following figure
68

69. Con’t
69
When the transformer is switched off, the excitation
current follows the hysteresis curve to zero, whereas the
flux density value changes to a non-zero value Br.

70. Con’t
• If the transformer was not switched off, excitation current
(i) and flux density would have followed the doted curves.
• That is; these two waveforms are shifted 90o and the flux
level is at its negative peak when the voltage is at its zero
point. However, during energization the flux has to start at
zero.
70

71. Transformer design aspects
 Design is a creative physical realization of theoretical
concepts
 Engineering design is application of science, technology
and invention to produce a machine which can perform a
specific task with optimum economy and efficiency.
 The major considerations to develop a good design are
1) Cost
2) Durability
3) Compliance with performance criteria as per the
specification.
 These requirements are conflicting and usually it is difficult
to meet all of them. 71

72. Con’t
It is impossible to design a machine which is cheap
and is also durable at the same time.
A machine which is expected to have long life span
must use high quality materials which are
expensive.
The basic structural parts of a transformer which
engineers should design carefully are;
- Magnetic parts (iron core)
- Conductor parts (windings)
- Insulating parts (dielectrics)
- Ventilation and cooling parts(thermal)
- Mechanical parts
72

73. 1. Specifications
 In order to design a transformer, a designer needs to know
the following important specifications;
1. Capacity ( power rating) of the transformer, in KVA;
2. Voltage ratings of primary and secondary windings, in KV;
3. Number of phases, 1 - phase or 3 – phase;
4. Working Frequency, in Hz;
5. Type of connections in 3ϕ-transformers , STAR or DELTA;
6. Tapping, if any
7. Type of iron assembly, Core or Shell type;
8. Type of transformer, Power or Distribution;
9. Operating average ambient temperature, in oC; (generally
40oC)
73

74. Con’t
10. Type of cooling to be used;
a) Cooling medium; air, Oil, water
b) Circulation type; Natural, forced
c) Simple cooling, mixed cooling
11. Temperature rise above ambient, depending on
insulation class;
12. Voltage regulation;
a) % or P.U. at full load, at unity P.F. or 0.8 P.F. lag
b) impedance, % or P.U.
c) reactance, % or P.U.
13. No – load current;
a) Amperes
b) % 0f rated current at rated voltage and rated frequency;
14. Efficiency, in % or P.U. at full load, ½ load, ¾ load at
unity P.F. and 0.8 P.F 74

75. • The design of di-electrical, thermal and mechanical parts
is based upon knowledge of decades experience and
practice.
• However, the design of both electric and magnetic
circuits is based upon well established basic laws.
Induction, interaction and alignment laws.
• Transformer action is based on induction laws where the
coil is stationary and the flux is varying.
75
2. Basic principles
turns
of
number
N
for
volts
dt
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76. Single phase transformer components
76
Single phase transformer Assembled single phase transformer iron core

77. a. Single phase transformer
Let the basic quantities required for transformer design are designated as
follows;
S – output , KVA
Φm - main flux, wb
Bm – max. flux density, wb/m2
δ – current density, A/m2
Agi – gross core area, m2
Ai – net core area = staking factor x gross core area, m2
Ac – area of copper in the window, m2
Aw – window area, m2
D – distance between core centers, m
d – diameter of circumscribing circle, m
Kw – window space factor
f – frequency, Hz
Et – emf per turn, v
77

78. Con’t
• Np, Ns – number of turns in prim. & second. windings
respectively.
• Ip, Is – current in prim. & second. windings respectively, A
• Vp, Vs – terminal voltage of prim. &second. windings, V
• ap, as – area of conductors of prim. & second. windings, m2
• li – mean length of flux path in iron, m
• Lmi – length of mean turn of transformer windings, m
• Gi – weight of active iron, kg
• Gc – weight of copper, kg
• Pi – loss in iron per kg, w (specific iron loss)
• Pc – loss in copper per kg, w (specific copper loss)
Then;
78

79. Con’t
• Induced Voltage per turn in a single phase transformer with N
number of turns is:
𝐸𝑡=
𝑉1
𝑁1
= 4.44fϕ𝑚, Volts
 Total copper area in single phase transformer window (one primary
& one secondary)
𝐴𝑐= Copper area of prim. + Copper area of second. Winding
= Prim. turns x Area of prim. Conductor + Second. turns x
Area of Second. Conductor
𝐴𝑐= 𝑁𝑝 𝑎𝑝 + 𝑁𝑠 𝑎𝑠
 Taking the current density δ to be the same in prim. And second
winding: 𝒂𝒑=
𝑰𝒑
𝜹
, 𝒂𝒔 =
𝑰𝒔
𝜹
79

80. Con’t
80

81. a) Rating of single phase transformers in KVA
81

82. b . Three phase transformers
82

83. 83
Assembled iron core of a 3-phase transformer

84. 3. Output equation – EMF per turn of a winding
• The out put in KVA of a transformer can be related to emf
per turn, there by we will get starting point for the design of
a transformer.
Emf per turn
• We know 𝑉1 = 4.44fϕ𝑚𝑁1−−−−−−−−−−−−−− ( 1)
• So emf/turn = 𝐸𝑡=
𝑉1
𝑁1
= 4.44fϕ𝑚--------------- (2)
• Consider the out put of a single phase transformer:
• S= 𝑉1 𝐼1x 10−3 = (4.44fϕ𝑚𝑁1) 𝐼1x 10−3 KVA
= 𝐸𝑡 𝑁1 𝐼1x 10−3
KVA ------------------------- (3)
 In the design, the ratio of total magnetic loading and
electric loading may be kept constant 84

85. Cont…
• Thus, the designer has to relate the dimensions and the
material in such a way so as to obtain the desired output
and performance at the lowest cost
• Magnetic loading = Φm
• Electric loading = 𝑁1 𝐼1
So
Φm
𝑁𝐼
= constant (say r) ,and 𝑁1 𝐼1=
Φm
𝑟
put in eq. (3)
S= 𝐸𝑡
Φm
𝑟
x 10−3
KVA
Or S= 𝐸𝑡
𝐸𝑡
4.44𝑓𝑟
x 10−3
KVA using equation (2)
𝐸𝑡
2
= (4.44fr x 10−3) x S
Or 𝑬𝒕= 𝑲𝒕 𝑺 Volt/Turn
85

86. Cont…
Where 𝑲𝒕 = 4.44𝑓𝑟𝑥10−3 is a constant and values are:
• Kt = 0.6 to 0.7 for 3-phase core type power transformer
• Kt = 0.45 for 3-phase core type distribution transformer
• Kt = 1.3 for 3-phase shell type transformer
• Kt = 0.75 to 0.85 for 1-phase core type transformer
• Kt = 1.0 to 1.2 for 1-phase shell type transformer
• As
Φm
𝑁𝐼
= r is a constant and K is also a constant depending on
type, service condition and method of construction etc. It can
be taken from catalogue tables
86

87. Optimum Design
- A transformer may be designed to make one of the following
quantities as minimum;
i) Total volume ii) Total weight
iii) Total cost iV) Total losses
- In general these requirements are contradictory and normally
it is possible to satisfy only one of them.
- All these quantities vary with the ratio of Φm/NI = r
- To Design for minimum cost, the cost of iron must be equal
to the cost of conductors.
- To design for minimum loss or maximum efficiency, variable
losses (I2R loss) must be equal to constant losses ( Iron loss).
 Design work can now be started by suitably choosing values
of Bm, δ, and Kw and determining Ai and Aw.
87

88. 4. Design of core section (limb and yoke)
• For small size transformers simple rectangular core section
can be used having circular or rectangular coils.
• As the size of the transformer increases, usually circular coils
are used because of its superior mechanical stability
during short circuit situations.
• For circular coils, the shape of the core section has to be
selected so that no useful space is wasted and the amount of
copper used is minimum ( minimum permeter). This could be
achieved if the core section is a circle.
• Since transformer iron cores are assembled from large
number of laminations, the approximate circular shape of the
core is achieved by using stepped core sections. (square, two
stepped, three stepped, …. many stepped)
88

89. Estimation of core x-sectional area Ai
• We know
𝐸𝑡= 𝐾𝑡 𝑆 −−−−−−−−−−−− − 1
𝐸𝑡= 4.44fϕ𝑚
Or 𝐸𝑡= 4.44f𝐴𝑖 𝐵𝑚----------------------------(2)
So 𝐴𝑖 =
𝐸𝑡
4.44𝑓𝐵𝑚
------------------------------(3)
• Now the core may be following types
89

90. 90
Transformer core is prepared by staking together thin
sheets of laminations giving the required core section.
These laminations are insulated on both sides by varnish
and the assembled area includes the insulation as well.

91. Cont….
• Thus the gross area is related with the net area by a factor
𝐾𝑠 (0.85- 0.9) called stacking Factor.
• Yoke area Ay is generally taken 10% to 15% higher than
core section area (Ai), it is to reduce the iron loss in the
yoke section.
• But if we increase the core section area (Ai) more copper
will be needed in the windings and so more cost.
• Further length of the winding will increase winding
resistance so more cu loss
• Therefore, Ay = (1.10 to 1.15) Ai
91

92. 92
Assembled magnetic core

93. 5. Selection of design constants
• Designing work depends on proper selection of design
constants; Bm, δ, Kw and Aw.
 Choice of flux density (Bm) – depends on
1) type of iron;
 hot rolled silicon steel 1.1 – 1.5 T,
 cold rolled silicon steel 1.5-1.7 T,
2) power rating of the transformer ;
 Higher values for higher ratings,
 lower value for lower ratings.
3) service condition;
for power transformers higher values of Bm are selected
for distribution transformers, lower values are selected
93

94. Cont…
• Note: - If we choose higher values of Bm, the net
core area Ai reduces, which helps to reduce the
diameter of the core and the length of the mean
turn.
• This results in saving of cost of iron and
copper.
• However, the max value of Bm is limited by the
saturation character of the selected iron type.
• Higher values of Bm causes increased iron losses
which may require intensive cooling.
94

95. Choice of current density (δ)
Higher values of current density helps to design a
transformer with lower dimensions.
 However, it is mainly limited by
 Heat produced during nominal operation and
efficiency of the transformer.
Type of transformer ( Power or distribution) because
of different requirement of i2r and iron losses.
 Permissible values can be taken from catalogues
considering cooling system to be employed.
95

96. Choice of window space factor (Kw)
It is the ratio of copper area in the window to the total
window area. Kw =Ac/A w
The total window area includes the copper area, the
insulation and open areas for air or oil.
The amount of copper and insulation used depends on
the KVA and voltage rating respectively and there
fore, the choice of Kw
• For higher rating Kw = 0.15 to 0.20
96

97. Choice of window dimensions (Aw)
 Too a narrow window results in increased height of
window where distance between adjacent limbs/winding
is less.
 This reduces leakage reactance.
 If lower height is chosen, the window width increases
resulting increased distance between adjacent
limbs/windings and increased leakage reactance.
 Thus, height and width of the window is taken
considering desirability of leakage reactance.
 Usually, Hw/W w is taken as 2 to 4. Aw = Hw x Ww
97

98. 6. Design of windings
• The number of turns of the LV and HV windings can
be found from the voltage per turn, Et =
𝑽
𝑵
• Turns of HV side is calculated using the nearest integer
value of 𝑁𝐿𝑉
𝑵𝑯𝑽 =
𝑽𝑯𝑽
𝑽𝑳𝑽
𝑵𝑳𝑽
Tapping is provided to the HV side. Therefore 𝑁𝐻𝑉 is to
be increased according to the percentage of tapping
required
98
𝑵𝑳𝑽 =
𝑽𝑳𝑽
𝑬𝒕
(approximated to nearest integer )

99. Cont…
• We place first half of LV on one limb and rest half of
LV on other limb to reduce leakage flux.
• So, arrangement is LV insulation then half LV turns
then HV insulation and then half HV turns.
99

100. Cont…
100
• Current in LV and HV winding is given by
𝐼𝐿𝑉 =
𝐾𝑉𝐴𝑝ℎ 𝑥 103
𝑉𝐿𝑣.𝑝ℎ
, 𝐼𝐻𝑉 =
𝐾𝑉𝐴𝑝ℎ 𝑥 103
𝑉𝐻𝑣.𝑝ℎ
• The cross sectional area for LV and HV winding are:
𝑎𝑙𝑉 =
𝐼𝐿𝑉
δ
, 𝑚𝑚2, 𝑎ℎ𝑉=
𝐼ℎ𝑣
δ
𝑚𝑚2
Nearest size is to be selected from the standard table of
conductors.

101. Choice of type of winding
There are different types of windings employed in
transformers such as cylindrical, helical, cross
over, and continuous disc type.
• Choice of type of winding depends on voltage and
current ratings.
Once the type is properly selected, the winding is
designed to fit to the window dimension with
proper provision for cooling ducts, insulation and
clearances.
101

102. 7. Design of insulation
 The fundamental considerations in the design of
transformer insulation are that of voltages between
different parts. i.e. insulation
- between core and LV winding
- between LV and HV windings
- between HV windings on two consecutive limbs
- between yoke and LV as well as HV windings.
 Different types of insulators like pressboard, paper, oil
immersed paper, oil, Bakelite, etc can be used
102

103. 8. Design of cooling systems
• For large power transformers, tubes and radiators
are employed to circulate natural or forced cooling
medium. ( air, oil, water etc.)
• In addition to the above main parts, tank and other
mechanical parts are designed as required
depending on the power and type of the transformer
103

104. Auto-transformer
It is a transformer with one winding only, part of this
being common to both primary and secondary.
The primary and secondary are not electrically
isolated from each other.
But its theory and operation are similar to those of a
two-winding transformer. Because of its one winding,
it uses less copper and hence is cheaper than the 2 –
winding transformer.
As shown in Fig.(a), AB, is primary winding having
N1 turns and BC is secondary winding having N2
turns.
104
Special type of transformer

105. Cont…
105
Figure Step down and step up autotransformer
Autotransformers are used:
1. To give small boost to a distribution cable to correct the
voltage drop
2. As auto-starter transformers to give up to 50 to 60 % of
full voltage to an induction motor during starting

106. Advantages of Auto-transformer
Saving in winding material (Cu or Al), since the
secondary winding is part of the primary winding.
Lower copper loss, therefore efficiency is higher than
in the two winding transformer.
Lower leakage reactance's, lower exciting current.
Variable output voltage can be obtained.
Disadvantages of Auto-Transformer
There is a direct connection between the primary and
secondary sides.
The short-circuit current is much larger than for
normal two-winding transformer 106

107. Instrument Transformers
 These are special type of transformers constructed with
accurate ratio employed in conjunction with standard low-
range a.c. instruments to measure voltage, current, power
and energy.
 Used to reduce voltage or current to provide metering or
protection
 These instrument transformers are of two kinds:
1. Current Transformer (or Series Transformer): to measure
high ac currents
2. Potential Transformer (or Parallel Transformer): to
measure high ac voltages
107

108. Cont…
 CTs are used with low-range ammeters to
measure currents in HV ac circuits where it is not
practicable to connect ammeters directly to the
lines.
• In addition to insulating the instrument from the
HV line, they step down the current in a known
ratio.
• i.e used to measure AC current of large magnitude
with low range ammeter by stepping down a
current in a known ratio.
 PTs are used with low range voltmeter to measure
high ac voltages.
108

109. Uses of Instrument Transformer
It is used for the following three purposes
1. To insulate the HV circuit from the measuring
instrument in order to protect the measuring
instruments from burning
2. To make it possible to measure HV with low range
voltmeter and high current with low range ammeter.
3. These instrument transformers are also used in
controlling and protecting circuits, to operate relays,
circuit breakers etc.
 The working of these transformers are similar as that
of ordinary transformers.
109

110. Current transformer(CT)
• CT has a primary coil of one or more turns of thick wire
connected in series with the line whose current is to be
measured.
• The secondary consists of a large number of turns of fine
wire and is connected across the ammeter terminals.
• The primary winding carries a full load current and this
current is stepped down to a suitable value which is within
the range of ammeter.
110

111. Cont…
 The operation of CT differs slightly from the power
transformer.
 In case of CT, the secondary winding has a very small
impedance or “Burden” , so the CT operates on short
circuit conditions.
 Burden across the secondary of CT is the ratio of
secondary voltage to secondary current. ZL = Vs / Is
Safety Precautions
It should be noted that, since the ammeter resistance is
very low, the current transformer normally works short
circuited.
111

112. Cont…
 If for any reason the ammeter is taken out of the
secondary winding, then this winding should be short
circuited by short circuiting switch (S).
 This is because, in C.T. 𝑰𝟏 is determined entirely by the
load on the system and not by the load on its own
secondary
• If this is not done, then due to the absence of counter
amp-turns of the secondary, the unopposed primary
m.m.f. will set up an abnormally high flux in the core
which will produce excessive core loss with subsequent
heating and a high voltage across the secondary terminals.
• Hence, the secondary of a CT should never be left open
under any circumstances. 112

113. Cont…
• If we know
𝑰𝟏
𝑰𝟐
of C.T. and reading of ammeter, line
current is given by multiplying current transformation
ratio by ammeter reading.
𝑰𝟏
𝑰𝟐
= K
Where : 𝑰𝟏 is line current and 𝑰𝟐 is ammeter reading,
Then 𝑰𝟏 = 𝑰𝟐 x K
 Eg. If the C.T. has primary to secondary current ratio of
100: 5, it steps up the voltage 20 times where as it steps
down the current 1/20 th of its actual value.
113

114. Voltage or Potential Transformers (VT or PT)
• PTs are extremely accurate-ratio step-down transformers
and are used in conjunction with standard low-range
voltmeters (usually 150-V) to measure high AC voltage.
 The HV to be measured is fed to the primary of P.T. which
is stepped down and measured by a low range volt meter on
the secondary.
 Since their secondary windings are required to operate
instruments or relays or pilot lights, their ratings are
usually of 40 to 100 W.
114

115. Cont…
• Thus, the purpose of PT is to insulate the HV line from the
voltmeter and to step down the voltage to the level of
voltmeter
• Note: By dividing the reading of voltmeter by voltage
transformation ratio, we get the true voltage on the HV
side.
 i.e If we know
𝑽𝟐
𝑽𝟏
=K of the P.T. and reading of voltmeter.
 The line voltage on the HV side is calculated by:
𝑽𝟏 =
𝑽𝟐
𝑲
 For safety, the secondary should be completely insulated
from the HV primary and should be grounded for affording
protection to the operator.
115

116. Advantages of Instrument Transformer
1. The measuring instruments can be placed far
away from the HV side by connecting long wires
to the instrument transformer. This ensures the
safety of instruments as well as the operator.
2. This instrument transformers can be used to
extend the range of measuring instruments like
ammeters and voltmeters.
Disadvantages of Instrument Transformer
 The only main draw back is that these
instruments can not be used in DC circuits
116

117. Application CT and PT
The following are the applications
1. CT and VT are used in panel board of sub station
or grid station to measure the bus bar current or
voltage which is very high.
2. CT and VT are widely used in power measuring
circuits. The current or voltage coil of the
wattmeter is connected with CT or VT.
3. CT and VT are also used in power houses, sub
stations etc. in conjunction with the relays for
protection.
117

118. 118