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

TransformersConstruction

• 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)

In a step-down transformer, the secondary voltage is less

than the primary voltage and Nsec < Npri

Step Up Transformer

In a step-up transformer, the secondary voltage is greater

than the primary voltage and Nsec > Npri

In an isolation transformer, the secondary voltage is equal to the

primary voltage and Nsec = Npri

https://www.youtube.com/watch?v=GMePE7NZcxw

Transformer Types

Principle of Transformer Action

1 1 2 2

d de N e N

dt dt

Faraday’s Law

• 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.

In DC, the induced emfs

are transients, in steady

state:

1 20 0d

e edt

AC

Assuming:

• Core permeability constant

• No leakage flux

max max4.44 4.44p p s sE N f E N f

p p

s s

E N

E N

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

Transformer

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

𝑁1= 𝐸2

𝐸1→ 2250

90 = 𝐸2120 →

25 = 𝐸2120 → 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

𝑁1= 𝐼1

𝐼2→ 2250

90 = 𝐼1 2 →

25 = 𝐼1 2 → 𝐼1 = 25 × 2 = 50 𝐴

primary

secondary

Single Phase Transformer information:

HV 115 kV

LV 7200 V

Capacity 16,670 KVA

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

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 isƞ =

𝑃𝑜𝑢𝑡𝑃𝑖𝑛

× 100%

𝑃𝑜𝑢𝑡 = 𝑉𝑆2

𝑅 =152

100= 2.25 𝑃𝑖𝑛 = 𝑉𝑝𝐼 = 120 × 20 × 10−3 = 2.4

ƞ =𝑃𝑜𝑢𝑡𝑃𝑖𝑛

× 100% = 2.252.4 × 100% = 93.75% ≅ 94%

• 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 half-

cycle.

• 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.

In-Rush current

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 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 Systems

Three Phase Transformers

• A three-phase transformer can be built by constructing a three-

phase transformer on a common magnetic structure or can be

built by suitably connecting a bank of similar three single-

phase 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

• 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

Balanced Three-Phase System

Three Phase connections of three Single Phase Transformers

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

Connections

Connection Phase Voltage Line Voltage Phase Current Line Current

Star VP = VL ÷ √3 VL = √3 × VP IP = IL IL = IP

Delta VP = VL VL = VP IP = IL ÷ √3 IL = √3 × IP

Three-phase Voltage and Current

𝑃 = 3𝐼𝐿𝑉𝐿 or 𝑃 = 3𝐼𝑃𝐻𝑉𝑃𝐻

𝑡𝑢𝑟𝑛 𝑟𝑎𝑡𝑖𝑜 = 𝑉𝑃𝐻𝑝𝑟𝑖𝑚𝑎𝑟𝑦/𝑉𝑃𝐻𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦

N

H1 H2 H3

X1 X2 X3

CBA

a b c

W1W2 W3

W4 W5W6

HS Line 230 kV

LS Line 23 kV

Transformer 3 Ø 230/13280/23000 Gnd. Wye

25 MVA230 kV

23 kV

HS Line 115 kV

LS Line 12 kV

Transformer:

3 Ø 115/12000/6928 Gnd. Wye 40MVA

N

H1 H2 H3

X1 X2 X3

CBA

a b c

W2 W3

W5 W6

16.6/1

Voltage on HS 115 kV.

Only one voltage because it

is delta

Voltage on LS 12,000 & 6,928

because it is wye connected

Transformer

Ratio 16.6/1

W4

W1

Transformer Solutions

• Information: Δ-Υ 230/23kV 25 MVA transformer

• What is the voltage on High voltage line?

• What is the voltage on low voltage line?

• What is the voltage on low voltage phase?

• What is the voltage across W1, W2, W3?

• What is the voltage across W4, W5, W6?

• 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?

230 kV (Given)

23 kV (Given)

13,280 kV ( V 𝐿 − 𝐿 ÷ √3)

230 kV (Same as the line for Δ winding)

13,280 kV ( L-N voltage)

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?

• On the delta side of the transformer

Same as winding W4, W5, W6= 627.556 amps

The line current ÷ √3 = Winding Current ;

62.76 ÷ √3 = 36.23

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

N

H1 H2 H3

X1 X2 X3

CBA

a b c

W

1

W2 W3

W4 W5W6

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

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

primary

secondary

primary

secondary

HH H H

H H HH

H H

H H

primary

secondary

How Are They

Connected?

N

H1 H2 H3

X1 X2 X3

CBA

a b c

W1W2 W3

W4 W5W6

N

H1 H2 H3

X1 X2 X3

CBA

a b c

W

1W

2

W

3

W

4

W

5W

6

N

H1 H2 H3

X1 X2 X3

CBA

a b c

W

1W

2

W

3

W

4

W

5W

6

Both transformers are Δ-Υ but

notice the Δ windings are not

closed the same

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 MVADelta and Wye 3-phase circuits : Electronics Worksheet