Transformer Basics
A Transformer May Look Like
This
Any transformer consists of the following three basic parts in it.
• Primary coil
• Secondary coil
• Transformer (Magnetic) core
1. Primary coil
The primary coil is the coil to which the source is connected. It may
be the high voltage side or low voltage side of the transformer. An
alternating flux is produced in the primary coil.
2. Secondary coil
The output is taken from the secondary coil. The alternating flux
produced in the primary coil passes through the core and links with
there coil and hence emf is induced in this coil.
3. Magnetic core
The flux produced in the primary passes through this magnetic core.
It is made up of laminated soft iron core. It provides support to the
coil and also provides a low reluctance path for the flux.
Construction of Power and Distribution
Transformers
Construction
• Core Type
High voltage
• Shell Type
Less leakage flux
Type of Cooling
• Ventilated Dry-Type
Transformers
They are cooled by natural air
convection.
•
Gas-Filled Dry-Type
Transformers
Cooled with nitrogen or other gases
•
Liquid-Immersed Transformers
Hermetically sealed tanks with insulated
liquid (mineral oil, silicone oil)
Transformer Types
In a step-down transformer, the secondary voltage is less
than the primary voltage and Nsec < Npri
In a step-up transformer, the secondary voltage is greater
than the primary voltage and Nsec > Npri
Step Up Transformer
In an isolation transformer, the secondary voltage is equal to the
primary voltage and Nsec = Npri
https://www.youtube.com/watch?v=GMePE7NZcxw
Principle of Transformer Action
Faraday’s Law
d
d
e1  N1
e2  N 2
dt
dt
• The same flux (mutual flux) exists in both coils.
• Flux is generated by i1 (right-hand rule).
• The induced emf e1 and e2 are generated to
oppose the buildup of flux in its window.
• i2 is generated by the induced emf e2.
d
0
dt
In DC, the induced emfs
are transients, in steady
state:
e1  e2  0
AC
Assuming:
• Core permeability constant
• No leakage flux
E p  4.44 N p f  max
Ep
Es

Es  4.44 N s f  max
Np
Ns
Transformer
VP - is the Primary Voltage
VS - is the Secondary Voltage
NP - is the Number of Primary Windings
NS - is the Number of Secondary Windings
Φ (phi) - is the Flux Linkage
Example
A ideal transformer having 90 turns on the primary and 2250
turns on the secondary is connected to a 120 V, 60Hz source.
What is the effective voltage across the secondary terminals?
Solution:
𝑁2
𝐸2
→ 2250
𝐸2
𝑁1 =
𝐸1
90 =
120 →
𝐸2
25 =
120 → 𝐸2 = 25 × 120 = 3000 𝑉
Example
An ideal transformer having 90 turns on the primary and 2250
turns on the secondary is connected to a 200 V, 50 Hz source.
The load across the secondary draws a current of 2 A. What is
the effective value of the primary current?
Solution:
𝑁2
𝐼1
→ 2250
𝐼1
𝑁1 = 𝐼2
90 = 2 →
𝐼1
25 = 2 → 𝐼1 = 25 × 2 = 50 𝐴
Single Phase Transformer information:
HV 115 kV
LV 7200 V
Capacity 16,670 KVA
primary
Primary amps
16,670,000 ÷ 115,000 = 145 amps
Secondary amps
16,670,000 ÷ 7,200 = 2315 amps
Turns Ratio
115,000 ÷7200 = 15.97/1
Current ratio Check
2315÷145 = 15.97/1
secondary
Calculating single phase transformer MVA
Voltage X Current
115,000 X 145 amps = 16,675,000= 16.67MVA
7,200 X 2315 amps = 16,668,000 = 16.67 MVA
Transformer efficiency
The efficiency of a transformer is the ratio of power delivered
to the load (Pout) to the power delivered to the primary (Pin).
That is
𝑃𝑜𝑢𝑡
ƞ=
× 100%
𝑃𝑖𝑛
𝑃𝑜𝑢𝑡 =
𝑉𝑆 2
152
𝑃𝑖𝑛 = 𝑉𝑝 𝐼 = 120 × 20 × 10−3 = 2.4
=
=
2.25
𝑅 100
𝑃𝑜𝑢𝑡
ƞ=
× 100% = 2.25 2.4 × 100% = 93.75% ≅ 94%
𝑃𝑖𝑛
In-Rush current
• When connecting an AC source to a transformer circuit the
current called the transient component or in-rush current
• Although the in-rush component to a transformer decays rapidly,
dropping to the normal no-load current within 5 to 10 cycles, it
may exceed 25 times the full-load rating during the first halfcycle.
• This high in-rush must be taken into consideration when
selecting fuses and/or circuit breakers.
• The magnitude of the in-rush depends on the magnitude and
phase angle of the voltage wave at the instant the switch is
closed.
Transformer Polarity
• Transform polarity refers to the relative phase relationship of
transformer leads
• On power transformers the terminals are designated by the
symbols H1, and H2 for the high-voltage (HV) winding and by
X1, and X2 for the low-voltage (LV) winding.
• By convention, H1 and X1 have the same polarity.
• The transformer has either additive or subtractive polarity.
• A transformer is said to have additive polarity when terminal
H1 is diagonally opposite terminal X1
• A transformer has subtractive polarity when terminal H1 is
adjacent to terminal X1
Three-Phase Systems
• Three-phase power is preferred over single-phase power for
several important reasons:
– Three-phase motors, generators, and transformers are
simpler, cheaper, and more efficient
– Three-phase transmission lines can deliver more power for
a given weight and cost
– The voltage regulation of 3-phase transmission lines is
inherently better
Three Phase Transformers
• A three-phase transformer can be built by constructing a threephase transformer on a common magnetic structure or can be
built by suitably connecting a bank of similar three singlephase transformers
• Three-phase transformers use much less material than three
single-phase transformers for the same three-phase power and
voltage ratings.
• Three-phase transformers have all three phases wound on a
single magnetic core or in shell-type and core-type
construction.
• The principal disadvantage of a three-phase transformer,
compared with its three-transformer counterpart, is that failure of
one phase puts the entire transformer out of service
Balanced Three-Phase System
• Three-phase transformers are required to step up or step
down voltages in the various stages of power transmission.
• The primary and secondary windings may be connected in
either wye (Y) or delta (Δ) configurations.
• There are therefore four possible connections for a three-phase
transformer:
» Y -Δ
» Δ -Y
» Δ-Δ
» Y-Y
Three Phase connections of three Single Phase Transformers
Connections
Wye connected windings:
Have a common connection point
Have two voltages available “L-L & L-N”
Has only one current : Line current
Connections
Delta connected windings:
Have no common connection point
Have only one voltage available “L-L”
Delta can be closed “connected” in more than one way
Has two currents
Line current
Phase or winding current
Three-phase Voltage and Current
Connection
Phase Voltage
Line Voltage
Phase Current
Line Current
Star
VP = VL ÷ √3
VL = √3 × VP
IP = I L
IL = IP
Delta
VP = VL
VL = VP
IP = IL ÷ √3
IL = √3 × IP
𝑃=
3𝐼𝐿 𝑉𝐿
or
𝑃 = 3𝐼𝑃𝐻 𝑉𝑃𝐻
𝑡𝑢𝑟𝑛 𝑟𝑎𝑡𝑖𝑜 = 𝑉𝑃𝐻 𝑝𝑟𝑖𝑚𝑎𝑟𝑦/𝑉𝑃𝐻 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦
HS Line 230 kV
LS Line 23 kV
Transformer 3 Ø 230/13280/23000 Gnd. Wye
25 MVA
230 kV
B
A
C
H2
H1
W1
H3
W2
W3
W5
W4
W6
N
X1
X2
X3
23 kV
a
b
c
HS Line 115 kV
LS Line 12 kV
Transformer:
3 Ø 115/12000/6928 Gnd. Wye 40MVA
Voltage on HS 115 kV.
Only one voltage because it
is delta
B
A
H2
H1
Voltage on LS 12,000 & 6,928
because it is wye connected
C
W1
H3
W3
W2
16.6/1
Transformer
Ratio 16.6/1
W4
W6
W5
N
X1
a
X2
X3
b
c
Transformer Solutions
• Information: Δ-Υ 230/23kV 25 MVA transformer
230 kV (Given)
•
What is the voltage on High voltage line?
•
What is the voltage on low voltage line?
•
What is the voltage on low voltage phase?
13,280 kV ( V 𝐿 − 𝐿 ÷ √3)
•
What is the voltage across W1, W2, W3?
230 kV (Same as the line for Δ winding)
•
What is the voltage across W4, W5, W6?
13,280 kV ( L-N voltage)
•
What is the current in line A, B, C ?
•
What is the current in W1, W2, W3?
•
What is the current in W4, W5, W6?
23 kV (Given)
25 𝑀𝑉𝐴 ÷ 230,000 ÷ 3 = 62.75
25 𝑀𝑉𝐴 ÷ 230,000 ÷ √3 ÷ √3 = 36.23
25 MVA / 23 kV / √3 = 627.56 amps
Transformer Solutions
•
What is the current in line a, b, c?
•
The line current ÷ √3 = Winding Current ;
On the delta side of the transformer 62.76 ÷ √3 = 36.23
Same as winding W4, W5, W6= 627.556 amps
Calculating three phase transformer MVA
Voltage X Current X three windings
230,000 X 36.3 amps = 8,349,000 x 3 (windings)= 25,047,000 = 25 MVA
13,280 X 627.57 amps = 8,334,129 X 3 (windings)= 25,002, 388 = 25MVA
Transformer Ratio
3 Phase 230/23 kV 25 MVA
Transformer is 230/ 23 kV
Ratio is NOT this ratio but is the ratio of the voltages across the windings.
W1 230 kV, W4 13,280 Ratio is 230,000 / 13,280 = 17.32/1
A
B
C
H2
H1
H3
W
1
W2
W4
W5
W3
W6
N
Note:
W4 current/ W1 current = 627.56/36.3 = 17.3
627.56 amps X 13,280 volts = 8.3 MVA
36.3 amps X 230,000 volts = 8.3 MVA
8.3 MVA X 3 (windings) =25 MVA
X1
a
X2
X3
b
c
How Are They
Connected?
H
primary
H
H
primary
secondary
H
H
H
primary
secondary
H
H
H
secondary
H
H
H
A
B
H1
H2
W1
W2
H3
W3
W5
W4
C
W6
N
X1
a
X2
b
X3
c
Both transformers are Δ-Υ but
notice the Δ windings are not
closed the same
A
B
H1
H2
W
1
W
2
N
X1
A
H1
B
C
H2
H3
W
1
N
W
2
W
3
W
5
W
4
X1
a
a
W
6
X2
X3
b
c
H3
W
3
W
5
W
4
C
W
6
X2
X3
b
c
Transformer Information For
Practice Calculations
•
•
•
•
•
•
3Ø, Δ-Υ 115/12 kV 25MVA
3Ø, Δ-Υ 115/23 kV 40 MVA
3Ø, Δ- Δ 23/ 13.8 kV 5 MVA
3Ø Υ- Υ 4160/ 2400 5 MVA
3Ø, Δ- Δ 115/ 13.8 kV 10 MVA
1Ø Υ- Δ 230/22 kV 350 MVA
Delta and Wye 3-phase circuits : Electronics Worksheet