Transformers Guide: Types, Working Principle & Applications

TRANSFORMERS
Dr.R.Subasri
Professor, Kongu Engineering College,
Perundurai, Erode, Tamilnadu, INDIA
Dr.R.Subasri, KEC, Erode, India

Transformer
• A static device
• Helps in the transformation of electric power
from one circuit to other at same frequency.
• The voltage can be raised or lowered in a
circuit, but with a proportional increase or
decrease in the current ratings.
P = V I
• P is same in two circuits. Hence if V  then I

Working Principle
• Faraday’s Law of Induction
•Statically induced EMF:
•When ever there is a change in flux, an emf is
induced

Working Principle
• Mutual inductance between two circuits
which is linked by a common magnetic flux.
di
E M
dt
=Mutually Induced EMF

Working Principle
• A basic transformer consists of two coils that
are electrically separate, but are magnetically
linked through a path of reluctance
Requirements:
1. Windings :
2. Core
Primary & Secondary

Leakage Flux And Fringing
Leakage flux is defined as the magnetic flux which does not
follow the particular intended path in a magnetic circuit.
•Flux which is not used for any work is
called Leakage Flux and is denoted by φl .
•Flux utilized in the magnetic circuit is
known as Useful flux φu.
•Total flux Φ produced :
leakage coefficient or leakage factor
The useful flux when sets up in the air gap, it tends to bulge
outward (b and b’), because of this bulging the effective area of
the air gap increases and the flux density of the air gap
decreases. This effect is known as Fringing

Core Construction
❖To provide the magnetic path to channel the flux
❖ Consists of thin strips of high-grade steel,
called laminations, which are electrically separated by a
thin coating of insulating material.
❖Thickness ranges from 0.25 mm to 0.5 mm

Core Construction
• Provide a low reluctance path for the
magnetic field
• Core is designed to prevent eddy currents
-Eddy Currents: circulating electric currents within
the iron core itself.
cause heating and energy losses within the
core
decreasing the transformers efficiency.
• Laminations are electrically insulated from each other
by a very thin coating of insulating varnish or by the use
of an oxide layer on the surface.Dr.R.Subasri, KEC, Erode, India

Winding Construction
• Main current-carrying conductors wound around
the laminated sections of the core
• Copper or aluminium
• Aluminium wire is lighter and generally less
expensive than copper wire
• A larger cross sectional area of conductor must
be used to carry the same amount of current as
with copper
• Aluminium : mainly used in larger power
transformer applications.
• Copper : Small power ratings

Types of Construction
• Dependant upon how the primary and secondary
windings are wound around the central laminated
steel core.
Core Type
primary and secondary
windings are wound outside
and surround the core ring.
Shell Type
windings pass inside the steel
magnetic circuit (core) which
forms a shell around the windings
Leakage of flux is more Leakage of flux is lessDr.R.Subasri, KEC, Erode, India

Types of cooling
• Oil Filled Self-Cooled Type :
The assembled windings and core of such
transformers are immersed in a tank filled with oil for
dissipating the heat by radiation
• Oil Filled Water Cooled Type:
……..cold water keeps circulating to carry the heat
from the device.
• Air Blast Type :
The transformer is housed in a thin sheet metal box
open at both ends through which air is blown from
the bottom to the top.

Types based on voltage ratio
to “increase” the voltage on its
secondary winding with respect
to the primary
Step-up transformer
Step-down transformer
to “decrease” the voltage on its
secondary winding with respect
to the primary
(voltage ratio)

EMF Equation
• Faraday’s Law
dφ
E N
dt
= mφ φ sinωt=
mE Nωφ cosωt=
max mE Nωφ=
rms m
1
E Nωφ
2
= ω 2πf=
rms m mE 2 Nπ f φ 4.44 f N φ volts= =
Transformers DO NOT Operate on DC Voltages, ONLY AC.

Example
• A single phase transformer has 480 turns on the primary
winding and 90 turns on the secondary winding. The
maximum value of the magnetic flux density is 1.1T when
2200 volts, 50Hz is applied to the transformer primary
winding. Calculate
• 1. Maximum flux in the core 2. The cross-sectional area of the
core 3. The secondary induced emf.
rms m mE 2 Nπ f φ 4.44 f N φ volts= =
0.0206 wb
m mB φ / A= 0.0187 m2
p p
s s
E N 480
E N 90
= = Es = 412.5 V

Losses in transformer
• Iron Losses :
– Hysteresis Loss: Because of the friction of the
molecules against the flow of the magnetic lines of
force required to magnetise the core, which are
constantly changing.
Molecular friction causes heat to be developed which
represents an energy loss to the transformer.
-Eddy current Loss : flow of circulating currents induced
into the steel caused by the flow of the magnetic flux
around the core and create power loss within the core.
•Copper Loss : I2R Loss (heat loss)

Dot Convention
• Method of identifying the orientation or direction of a
transformers windings
• Either coil could be wound around the core clockwise or
anticlockwise so to keep track of their relative
orientations “dots” are used

Transformer –No Load Condition
•IO will flow through the primary coil winding due to the
presence of the primary supply voltage.
•A back EMF (ep) along with the primary winding
resistance acts to limit the flow of this primary current
•No-load primary current ( Io ) must be sufficient to
maintain enough magnetic field to produce the required
back emf
The ammeter above will
indicate a small current
IO flowing through the
primary winding even though
the secondary circuit is open
circuited

Transformer- No Load Condition
No-load primary current (Io) is made up of two
components:
1). An in-phase current, IE which supplies the core losses
(eddy current and hysteresis).
2). A current, IM at 90o to the voltage which sets up the
magnetic flux.
Io is very small compared to the transformers normal
full-load current. Dr.R.Subasri, KEC, Erode, India

Transformer – Load Condition
=k

( ) ( )
2 2
p o s o s o sI I I 2I I cos(φ φ )  = + + −
 
Use parallelogram method

A single phase transformer with a ratio of 440 / 110 V takes no-
load current of 5A at 0.2 pf lagging.If the secondary supplies a
current of 120A at a pf of 0.8 lagging, find the current taken by
the primary.
Solution
Io = 5 A ; Is = 120 A; cos o =0.2 ; cos s = 0.8
Is referred to primary Is’ = k Is p s s
s p s
V I I
V I I
= =

s
s s
p
V 110
I I 120 30A
V 440
 = = =
Use parallelogram method
( ) ( )
2 2
p o s o s o sI I I 2I I cos(φ φ )  = + + −
 
33.39 A

Transformer – On Load
with winding resistance and
Leakage flux

Shifting impedances in transformer
Impedance Ratio VV psZ ; Zs pI Is p
= =
I IZ V Vp ps s s
Z I V I Vp s p s p
=  = 
Z 2s k
Zp
=
2
p p
2
p p
R k R
X k X
 =
 =
p
s 2
p
s 2
R
R
k
X
X
k
 =
 =
Equivalent impedance referred
to secondary
Equivalent impedance referred to primary

• Note:
• When the primary voltage V1 is shifted to secondary,
then it is K V1.
• When the secondary voltage V2 is shifted to primary,
then it is V2/K.
• When the primary current I1 is shifted to secondary,
then it is I1/K.
• When the secondary current I2 is shifted to primary,
then it is I2 K.
• In general, K = V2/ V1
• When primary impedance Z1 is shifted to secondary
,then it is Z1K2
• When secondary impedance Z2 is shifted primary,then
it is Z2 /K2

• A 10KVA , 2000/400V single phase transformer has R1
= 5;X1=12 ;R2 =0.2  and X2=0.48 .Determine the
equivalent impedance of the transformer referred to
(i)primary (ii)secondary.
• Solution:
• Referred to primary:
K=400/2000 = 1/5
• =10 ; ;
• Referred to secondary:
2
01 1 2
R
R R
K
= + 2
01 1 2
X
X X
K
= + ( )2 2
01 01 01Z R X 26Ω= + =
2
02 2 1R R K R 0.4Ω= + =
2
02 2 1X X K X 0.96Ω= + =
( )2 2
02 02 02Z R X 1.04Ω= + =

Transformer -Equivalent circuit
c E W o o
M o o
I I I I cosφ
I I sin φ
= = =
=
1
c
E
1
m
M
V
G
I
V
B
I
=
=
s
2 2 2
p
V
I kI I
V
 = =

Transformer –Equivalent circuit
Equivalent circuit referred to primary
Load

Voltage regulation
• The change in secondary terminal voltage when the
transformer loading is at its maximum
• Determines the voltage drop (or increase) that
occurs inside the transformer because of the
transformers loading
• It is a measure of performance and efficiency.
• % regulation = 2o 2
2o
V V
100
V
−

No load secondary voltage Secondary voltage on load
2o 2
2 02 2 2 02 2
V V voltagedrop
=I R cosφ I X sin φ
− =

+ Lag PF
- Lead PF

Transformer efficiency
output power in Watts
%Efficiency
output power in Watts+Losses in Watts
=
By direct loading
2
1
output power W Watts
%Efficiency
Input power W Watts
=
W2
W1
2o 2
2o
V V
%regulation 100
V
−
= 
V2
No load secondary voltage

Predetermination of transformer efficiency
output power in Watts
%Efficiency
output power in Watts+Losses in Watts
=
Given:
No load iron loss = Pi
Full load Cu loss =Pc
( )
( ) 2
i c
Full load VA P.F
%Efficiency
Full load VA P.F+P P
 
=
  +
x
x x
i c
Full load VA P.F
%Full load Efficiency

=
 +
At any fractional load condition , for example 25 % of full
load,75 % full load, etc
Where x indicates the fraction
Note:
Pi remains
same at all load
conditions

Open circuit & Short circuit test
• To predetermine
(i) equivalent circuit of transformer in general
(ii) voltage regulation of transformer at a specific load
(iii) efficiency of transformer at a specific load .
Open Circuit Test
o
p
e
n
Io
Wo
Vo Io cos o
Under open circuit ,
voltage is high, Hence
Wo -wattmeter reading
can be taken as equal to
core losses
Neglect Copper Losses
Vo

• Short Circuit test
S
h
o
r
t
Under short circuit ,
voltage is less, Hence
Neglect core Losses
Ws -wattmeter reading
can be taken as equal to
copper losses
Ws
Is Vs Is cos s
Vs
s
o1
s
s
o1 2
s
V
Z
I
W
R
I
=
=
( )2 2
01 01 01X z R= −

Equivalent Circuit
% Voltage Regulation at a load PF (cos 2)
2 02 2 2 02 2
2
[I R cosφ I X sin φ ]
100
V

= 
x 2
2
Power rating in KVA
I
Secondary voltage rating V
=
% Efficiency at a load PF (cos 2)
( )
( ) 2
i c
Full load VA P.F
%Efficiency
 
=
  +
x
x x
x indicates the fractional load
i o
2
2
c s 2
s
P W
I
P W
I
=
=
2
1
2
02 01
2
02 01
V
k
V
R k R
X k X
=
=
=

• A 4 kVA , 2000/ 400 V, 50Hz single phase
transformer under the O.C & S.C tests are:
• No-load ( O.C) test : Low Voltage data : 200V , 0.7 A, 60 W
• Short circuit test :High voltage data : 9V , 6A , 21.6 W
• Solution :
• Equivalent circuit:
o
0
o o
W
cosφ
V I
=
o
0
o 0
o
0
o 0
V
R
I cosφ
V
X
I sin φ
=
=
s
o1
s
V
Z
I
= ( )2 2
01 01 01X z R= −s
o1 2
s
W
R
I
=
0.43
0.63 A
0.3 A
317.6
666.6
1.5  0.6 1.37
2
1
V
K
V
=
2
02 01
2
02 01
R K R
X K X
=
=

• % Voltage Regulation at UPF full load (cos 2=1)
2 02 2 2 02 2
2
I R cosφ I X sin φ
100
V

= 
4 kVA
400 V
2
2
Power rating in KVA
I
Secondary voltage rating V
=
10 A1.52%
• % Efficiency at UPF full load (cos 2=1)
( )
( ) 2
i c
Full load VA P.F
%Efficiency
 
=
  +
x
x x
1
60 W 2
2
c s 2
s
I
P W
I
=
60 W
4000
%Efficiency 100 97.1%
4120
=  =

All Day Efficiency-Distribution
transformer
• The step down transformers used for electric
power distribution purpose are referred as
distribution transformer.
• A distribution transformer cannot be run with
constant load throughout 24 hours.
• At day peak time it’s loading is high, whereas in
night lean time its loading may be negligible.
• So selecting a transformer depending upon its
conventional efficiency is not practical and
economical, too.

• The concept of all day efficiency is
recommended for distribution transformer
• we use the ratio of total energy delivered by
the transformer to the total energy fed to the
transformer, during a 24 hrs span of time
instead of ratio of power output and input of
the transformer.
hours)24(
kWhinInput
kWhinoutput
in wattsinput
in wattsputout
efficiencycommercialordinary
day forall =
=


• To determine the all day efficiency , it is
necessary to know how the load varies from
hour to hour during a day
• Problem:
• A 40KVA distribution transformer has iron loss
of 500W and a full load copper loss of 500W.
The transformer is supplying a lighting load
(UPF).The load cycle is as under:
• Full load for 4 hours
• Half of the full load for 8 hours
• No load for 8 hours. Calculate all day efficiency

• Output in KWhr = Output in KW x working hours
= (full load x 4 hrs) + (½ of full load x 8 hrs)
= (40 x 4)+ (½ x 40 x 8) = 320 KWh
Total losses in working hours:
• Iron loss for 24 hours (always even at no load it
appears) = 24 x 500 = 12 Kwhr
• Copper loss :
at full load : 500 x 4 hrs = 2Kwhr
at ½ of full load = ¼ x 500 x 8 = 1Kwhr
Input Kwhr = output Kwhr +losses in Kwhr
= 320 + 12 + 3 = 335 Kwhr
All day efficiency = 320 / 335 = 95.52 %

Auto transformer
• A kind of electrical transformer where primary and
secondary shares same common single winding.
Winding AB of total turns N1 - primary .
Tapped at ′C′ and the portion BC is secondary (N2)
Advantages: Higher efficiency, lower cost and small sizeDr.R.Subasri, KEC, Erode, India

Conventional TransformerAutoTransformer
Weight of copper in conventional transformer  (N1I1+N2I2)
Weight of copper in auto transformer:
Weight of copper in section LS and MS
 (N1-N2) I1 + N2(I2-I1)
1 2 1 2 2
1 1 2 2
(N 2N )I N Iwt of copper in auto.tr
1 K
wt of copper in conv.tr N I N I
− +
= = −
+
 Both sides by N2 I1
Copper saving in auto.tr = K x copper used in conv.trDr.R.Subasri, KEC, Erode, India

•An autotransformer can not safely be used for stepping
down higher voltages to much lower voltages suitable for
smaller loads.
•If the secondary circuit suffers a short-circuit condition,
the resulting primary current would be much larger than
an equivalent double wound transformer due to the
increased flux linkage damaging the autotransformer.
•If the secondary side winding becomes open-circuited,
current stops flowing through the primary winding
stopping the transformer action resulting in the full
primary voltage being applied to the secondary terminals.
•Because of lack of isolation and decreased economy,
auto-transformers are rarely used in ration greater than
4:1 except for low-power devices on low-voltage systems.
Disadvantages of Auto transformer

•To obtain partial line voltages for starting induction and
synchronous motors with squirrel-cage windings.
•To give a small boost to a distribution cable to correct for
the voltage drop.
•As furnace transformers for getting a convenient supply
to suit the furnace winding from a 230 V supply.
•As regulation transformers.
•In electrical testing laboratory.
Applications:

A three phase transformer or 3φ transformer can be
constructed either by connecting together three single-
phase transformers, thereby forming a so-called three
phase transformer bank, or by using one pre-assembled
and balanced three phase transformer which consists of
three pairs of single phase windings mounted onto one
single laminated core.
•One 'three phase transformer' occupies less space than a
gang of three 'single phase transformers'.
•Single 'three phase' unit is more economical.
• The overall bus-bar structure, switchgear and
installation of 'three phase transformer' is simpler
Advantages of three phase transformers
Three phase transformers

Cost of repair is more for three phase
transformer.
To restore the service, spare unit cost is
more compared with one single
transformer spare unit.
When these are self cooled, the
capacity of the transformer is reduced.
Disadvantgaes

Three phase transformers-Construction

The primary and secondary of the transformer can be
independently connected either in star or delta. There
are six possible connections for a 3-phase transformer
bank.
1. Δ – Δ (Delta – Delta) Connection
2. Υ – Υ (Star – Star) Connection
3. Δ – Υ (Delta – Star) Connection
4. Υ – Δ (Star – Delta ) Connection
5. Open delta or V-V connection
6. Scott or T-T connection
Three phase transformers-Connections

Service is satisfactory only if the three-phase load is
balanced ;
when the load is unbalanced, the electrical neutral will
shift from its exact centre to a point that will make the
line to neutral voltages unequal.
Star –Star Connections

Generally used in systems in which the voltages are not very high
and especially when continuity of service must be maintained even
though one of the transformers should fail.
Delta –Delta Connections

Used where the voltage is to stepped down, as for
example, at the end of a transmission line.
It is also employed in moderately low voltage distribution
circuits for stepping down from transmission voltages of
4000 or 8000 V to 230 (and 115 V).
Star –Delta Connections

This type of connections is employed where it is necessary
to step up the voltage, as for example, at the beginning of
a high-tension transmission system.
When one of the transformers in a Δ-Δ bank fails, so that
the service may be continued until the faulty transformer
is repaired or a good one is substituted.
Delta –Star Connections

When one of the transformers
in a Δ-Δ bank fails, so that the
service may be continued
until the faulty transformer is
repaired or a good one is
substituted.
Open Delta (or) V-V Connections

This connection is used to convert the three phase power into two
phase power using two single phase transformers.
One transformer called as main transformer having 50% tap and is
connected between the two lines of the three phase wires. The other
transformer called as teaser transformer having 86.6 tap and is
connected between the third phase wire and 50 percent tap of the
main transformer.
The secondary winding of each transformer provides the phases of
two phase systems.
This connection is mainly used to supply the power to the two phase
motor.
Scott (or) T-T Connections

Thank you

Transformers Guide: Types, Working Principle & Applications

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Similar to Transformers Guide: Types, Working Principle & Applications

Similar to Transformers Guide: Types, Working Principle & Applications (20)

More from Ramaiahsubasri

More from Ramaiahsubasri (9)

Recently uploaded

Recently uploaded (20)

Transformers Guide: Types, Working Principle & Applications