TRANSFORMERS (I)

EPO540 ELECTRICAL MACHINES

prepared by Kanendra

1. INTRODUCTION

Static machine

Not an energy conversion device - necessary & essential in many energy

conversion

Consist of two or more windings coupled by a mutual magnetic field

Ferromagnetic cores usually used to provide tight magnetic coupling and high

flux densities – IRON CORE

IRON CORE used in high-power applications

AIR-CORE used in low power electronic circuits

prepared by Kanendra

Schematic representation of a two winding transformer

1. INTRODUCTION

Primary function is to change voltage level – step up/step down

Low power electronic circuit – isolation or impedance matching

Measure voltage and current – instrument transformers

prepared by Kanendra

Core type Shell Type

2. IDEAL TRANSFORMER

Ideal transformer has the following properties:

1. Winding resistances are negligible

2. All fluxes confined to the core and link both windings. No leakage fluxes are

present. Core loss = 0.

3. Permeability of core is infinite (μ = 0) & net mmf required to establish flux is

zero (∑ MMF = 0)

prepared by Kanendra

Ideal transformer

3. PRACTICAL TRANSFORMER

Practical transformers – winding resistances, not all windings link the same

flux, permeability of the core material is not infinite, core losses occur

prepared by Kanendra

3. PRACTICAL TRANSFORMER

Transformer equivalent circuit

prepared by Kanendra

E1 = E2’ = aE1

V2’ = aV2

I2’ = I2/a

Xl2’ = a2Xl2

R2’ = a2R2

Approximate equivalent circuit

4. DETERMINATION OF EQUIVALENT

CIRCUIT PARAMETERS

In order to predict the behavior of the transformer, circuit parameters (R1, X1, R2

’, Xl2’, Rc, Xm) has to be known

These parameters can be calculated from the dimensions and properties of the

materials used – design data of transformer

Difficulty in obtaining data and some required parameters

A simple and direct method is used to obtain these parameters – performing

two different tests: “No Load (Open Circuit) Test” & “Short Circuit Test”

prepared by Kanendra

5. No-Load Test

Open circuit HV side and apply low voltage on LV side

Low voltage power supply is readily available

prepared by Kanendra

𝑃𝑂𝐶 =𝑉𝑂𝐶

2

𝑅𝐶 => 𝑅𝐶 =

𝑉𝑂𝐶2

𝑃𝑂𝐶

𝐼𝐶 =𝑉𝑂𝐶

𝑅𝐶

𝐼𝑀 = 𝐼𝑂𝐶2 − 𝐼𝐶

2

𝑋𝑀 =𝑉𝑂𝐶

𝐼𝑀

Wiring diagram

Equivalent circuit

6. Short-Circuit Test

Short circuit LV side and apply low voltage on HV side

Low voltage power supply is readily available

prepared by Kanendra

𝑃𝑆𝐶 = 𝐼𝑆𝐶2𝑅𝑒𝑞=> 𝑅𝑒𝑞 =

𝑃𝑆𝐶

𝐼𝑆𝐶2

𝑍𝑒𝑞 =𝑉𝑆𝐶

𝐼𝑆𝐶

𝑋𝑒𝑞 = 𝑍𝑒𝑞2 − 𝑅𝑒𝑞

2 Wiring diagram

Equivalent circuit

7. VOLTAGE REGULATION

Loads connected to secondary are usually designed to operate at constant

voltage

However, current drawn from the transformer results in a voltage drop in the

internal impedance of the transformer (Zeq)

When switch is open (no load condition) => 𝑉2|𝑁𝐿 = 𝑉1

𝑎

When switch is closed (with load) => 𝑉2|𝐿 = 𝑉2|𝑁𝐿± ∆𝑉2

Voltage change depends on the nature of the load – due to IZ (internal

impedance of transformer)

Large voltage change is undesirable

prepared by Kanendra

7. VOLTAGE REGULATION

Figure of merit used to identify characteristics of voltage change in

transformer is called voltage regulation

𝑉𝑜𝑙𝑡𝑎𝑔𝑒 𝑟𝑒𝑔𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑉𝑅 = 𝑉2 𝑁𝐿 − 𝑉2 𝐿

𝑉2 𝐿

VR can be positive or negative, depending on the nature of the load

Voltage regulation should be as small as possible (< 10%)

Equivalent circuit referred to primary:

𝑉𝑅 = 𝑉2′ 𝑁𝐿 − 𝑉2′ 𝐿

𝑉2′ 𝐿

Load voltage is normally taken at rated voltage => 𝑉2′ 𝐿 = 𝑉2′ 𝑟𝑎𝑡𝑒𝑑

If load is removed (no load condition), 𝑉2′ 𝑁𝐿 = 𝑉1

𝑉𝑅 = 𝑉1 𝑁𝐿 − 𝑉2′ 𝐿

𝑉2′ 𝐿

prepared by Kanendra

7. VOLTAGE REGULATION

𝑉1 = 𝑉2

′ + 𝐼2′𝑅𝑒𝑞1 + j𝐼2′𝑋𝑒𝑞1

θ2 = angle of load impedance

θeq1 = angle of transformer equivalent impedance, Zeq1

Phasor diagram

prepared by Kanendra

8. EFFICIENCY

Losses in transformers are small – static device, no rotational losses

𝜂 =𝑃𝑜𝑢𝑡

𝑃𝑖𝑛=

𝑃𝑜𝑢𝑡

𝑃𝑜𝑢𝑡 + 𝑙𝑜𝑠𝑠𝑒𝑠

Losses – core losses (Pc) & copper loss (Pcu)

𝜂 =𝑃𝑜𝑢𝑡

𝑃𝑜𝑢𝑡 + 𝑃𝑐 + 𝑃𝑐𝑢

Core loss (Pc) depends on peak flux density in the core. Transformer connected

to constant voltage supply, therefore core loss is constant – obtained from No

Load Test

prepared by Kanendra

8. EFFICIENCY

Copper loss (Pcu) function of load current - determined through winding

currents and resistances

Pcu = I12 R1 + I2

2 R2

= I12 Req1 = I2

2 Req2

Therefore, given any load condition (with power factor)

Pout = V2 I2 cos ϴ2

η = V2

I2 cos ϴ2

V2 I2

cos ϴ2 + PC

+ I2

2 Req2

prepared by Kanendra

9. AUTOTRANSFORMER

prepared by Kanendra

Common winding mounted on a core, secondary is taken from tap on the

winding

Advantages:

1. Reduced weight and size

2. Lower losses

3. Lower exciting current

4. Lower cost – less copper used

5. Variable output voltage

Disadvantages:

1. Direct connection between primary and

secondary windings / no isolation

Autotransformer

𝑉1

𝑉2=

𝑁1

𝑁2= 𝑎

𝐼1

𝐼2=

1

𝑎