6.

0 TRANSFORMERS
A transformer is a device that transfers electrical energy from one circuit to
another through inductively coupled conductors

1
6.1 UNDERSTAND MUTUAL INDUACTANCE

2
 UNDERSTAND
MUTUAL
INDUACTANCE

 When two coils are placed close to each other, a changing electromagnetic field
produced by the current in one coil will cause an induced voltage in the second coil
because of the mutual inductance.
 Mutual inductance is established by the inductance of each coil and by the amount of
coupling between the two coils.

3
 UNDERSTAND
MUTUAL
INDUACTANCE How Induction Works in a Transformer

Ferromagnetic core with primary coil (AC driven) and secondary coil.

 If this secondary coil experiences the same magnetic flux change as the primary and
has the same number of turns around the core, a voltage of equal magnitude and phase
to the applied voltage will be induced along its length.
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 UNDERSTAND
MUTUAL
INDUACTANCE

 Like normal (self-) inductance, it is measured in the unit of Henrys, but

unlike normal inductance it is symbolized by the capital letter “M” rather
than the letter “L”:

• The mutual inductance between two coils (L 1 and L2) may be expressed
mathematically as:

5
 UNDERSTAND
MUTUAL
INDUACTANCE

No current will exist in the secondary coil, since it is open-circuited. However, if

we connect a load resistor to it, an alternating current will go through the coil, in-
phase with the induced voltage

Resistive load on secondary has voltage and current in-phase.

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6.1.1 DISCUSS MAGNETIC COUPLING:

 Two conductors are referred to as mutual-inductively coupled or

magnetically coupled when they are configured such that change in current
flow through one wire induces a voltage across the ends of the other wire
through electromagnetic induction.

 The amount of inductive coupling between two conductors is measured by

their mutual inductance.

 The coupling between two wires can be increased by winding them into coils
and placing them close together on a common axis, so the magnetic field of
one coil passes through the other coil.

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6.1.2 DEFINE ELECTRICAL ISOLATION:

Transformers also provide an extremely useful feature called

isolation:

 Which is the ability to couple one circuit to another without the use
of direct wire connections.

 By being able to transfer power from one circuit to another without

the use of interconnecting conductors between the two circuits,
transformers provide the useful feature of electrical isolation.

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6.1.3 DEFINE COEFFICIENT OF COUPLING:

 The coefficient of coupling (k) between two coils is the ratio of the lines of
force (flux) produced by one coil linking the second coil (φ1-2) to the total
flux produced by the first coil (φ1)

 The COEFFICIENT OF COUPLING of a transformer is dependent on the

portion of the total flux lines that cuts both primary and secondary
windings.

 The coefficient of coupling depends on the physical closeness of the coils

and the type of core material on which they are wound

 Construction and core shape are also factors influencing coefficient of

coupling

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 DEFINE
COEFFICIENT
OF COUPLING

Eg:
 If 80 percent of the lines set up by the primary cut the secondary, there is 80
percent coupling, or a coefficient of coupling(K) is 0.8.

 Ideally, all the flux lines generated by the primary should cut the secondary, and
all the lines of the flux generated by the secondary should cut the primary. The
coefficient of coupling would then be one (unity), and maximum energy would
be transferred from the primary to the secondary.

 If none of the flux from one coil cuts the turns of the other coil, the coefficient of
coupling is zero.

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 DEFINE
COEFFICIENT
OF COUPLING
 The amount of coupling between the primary and secondary has a direct bearing
on the mutual inductance.

** where K is the coefficient of coupling

 Example:
What is the mutual inductance of a transformer whose primary has16 h of
inductance and whose secondary has 4h, if the coupling between the two coils is
75 percent?

The mutual inductance(M) of the transformer is 6 h.

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6.1.4 IDENTIFY THE FACTORS THAT AFFECT MUTUAL
INDUCTANCE:
 The amount of mutual inductance depends upon several factors:

 The relative position of the axes of the two coils;

 The permeability of the cores,
 The physical dimensions of the two coils,
 The number of turns in each coil,
 The distance between the coils.

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 IDENTIFY THE
FACTORS THAT
AFFECT MUTUAL
permeability of the cores
INDUCTANCE

distance between the coils

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6.2 UNDERSTAND THE CONSTRUCTION
AND THE OPERATION OF A TRANSFORMER

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6.2.1 SKETCH PARTS OF A BASIC TRANSFORMER:

 The principle parts of a transformer and their functions are:

1) The CORE, which provides a path for the magnetic lines of flux.
2) The PRIMARY WINDING, which receives energy from the ac source.
3) The SECONDARY WINDING, which receives energy from the primary
winding and delivers it to the load.
4) The ENCLOSURE, which protects the above components from dirt,
moisture, and mechanical damage.

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TRANSFORMER ENCLOSURE

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6.2.2 EXPLAIN THE IMPORTANCE OF THE CORE MATERIAL:

 The composition of a transformer core depends on such factors as voltage,

current, frequency, size limitations and construction costs.

 Typical core materials are: air, ferrite, and iron

 Air and ferrite cores are used at high frequencies (above 20 kHz).
 Iron cores are used for low frequency (below 20 kHz) and power applications.
 A soft-iron-core transformer is very useful where the transformer must be
physically small, yet efficient.

SYMBOL OF CORE MATERIAL

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 EXPLAIN
THE
IMPORTANC
E OF THE
CORE
MATERIAL

• The iron-core transformer provides better power transfer than does the air-core
transformer.

• A transformer whose core is constructed of laminated sheets of steel dissipates heat

readily; thus it provides for the efficient transfer of power.

• The purpose of the laminations is to reduce certain losses.

• The most efficient transformer core is one that offers the best path for the most lines of
flux with the least loss in magnetic and electrical energy. 18
6.2.3 DEFINE PRIMARY WINDING AND SECONDARY WINDING:

 The primary winding is the winding which receives the energy@

source; it is not always the high-voltage winding.

 The unpowered inductor in a transformer is called the secondary

winding.
 The secondary winding is the output winding where the load is
connected.

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6.2.4 DEFINE TURN RATIO:

 Turns ratio (n) is defined as the ratio of the number of turns in the secondary
winding (NSEC) to the number of turns in the primary winding (NPRI)

 The ratio between output voltage and input voltage is the same as the ratio of the
number of turns between the two windings.

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DEFINE
TURN
RATIO Example:

 Transformer Primary Voltage: 480 volts

 Transformer Secondary Voltage: 120 volts

(1Secondary turn)

This transformer has four primary turns for every one secondary turn.
Turns ratio is written as 4 to 1, or 4:1.

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DEFINE
TURN
RATIO Example.
 A transformer has 200 turns in the primary, 50 turns in the secondary, and
120 volts applied to the primary (Ep). What is the voltage across the
secondary (E s)?

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6.2.5 DETERMINE HOW THE DIRECTION OF WINDINGS AFFECTS
VOLTAGE POLARITIES:

 The direction of the windings determines the polarity of the voltage across the
secondary winding with respect to the voltage across the primary.

 The secondary voltage of a simple transformer may be either in phase or out of

phase with the primary voltage.

 Phase dots are used to indicate polarities

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DETERMINE
HOW THE
DIRECTION
OF In part (A) of the figure:
WINDINGS
AFFECTS
 Both the primary and secondary
VOLTAGE windings are wound from top to
POLARITIES
bottom in a clockwise direction,
 When constructed in this manner,
the top lead of the primary and the
top lead of the secondary have the
SAME polarity.
 This is indicated by the dots on the
transformer symbol.

Instantaneous polarity depends on direction of winding.

Part (B) of the figure:
• Illustrates a transformer in which the primary and secondary are wound in
opposite directions.
• The primary is wound in a clockwise direction from top to bottom, while
the secondary is wound in a counterclockwise direction.
• Notice that the top leads of the primary and secondary have OPPOSITE
polarities.
• This is indicated by the dots being placed on opposite ends of the
transformer symbol.
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6.3 UNDERSTAND HOW TRANSFORMER
INCREASES AND DECREASES VOLTAGE

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6.3.1 EXPLAIN HOW A STEP UP TRANSFORMER WORKS:

 Step-Up Transformer : The primary winding of a step-up transformer has fewer

turns than the secondary winding, with the resultant secondary voltage being
higher than the primary.

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6.3.2 IDENTIFY A STEP UP TRANSFORMER BY ITS TURNS RATIO:

 If there are fewer turns in the primary winding than in the secondary
winding, the secondary voltage will be higher than the secondary
circuit.

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6.3.3 STATE THE RELATIONSHIP BETWEEN PROMARY AND
SECONDARY VOLATGES AND THE TURN RATIO:

• The ratio of secondary voltage (VSEC) to primary voltage (VPRI) is equal to the
ratio of the number of turns in the secondary winding (NSEC) to the number of
turns in the primary winding (NPRI)

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Example:
 An autotransformer has a primary voltage;500V, 300 turns primary winding
and 150 turns secondary winding. Using suitable diagram, calculate the
value of secondary voltage.
Solution:

Vsec = 500V (150/300) = 250V

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6.3.4 EXPLAIN HOW A STEP DOWN TRANSFORMER WORKS:
 Step-Down Transformer : The primary winding of a step-down transformer
has more turns than the secondary winding, so the secondary voltage is lower
than the primary

30
6.3.5 IDENTIFY A STEP DOWN TRANSFORMER BY ITS TURNS
RATIO:

 This is a step-down transformer, as evidenced by the high turn count of the

primary winding and the low turn count of the secondary.
 The amount by which the voltage is stepped down depends on the turns ratio
 The turns ratio of a step-down transformer is always less than 1

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6.3.6 DESCRIBE DC ISOLATION

 A transformer does not pass dc, therefore a transformer can be

used to keep the dc voltage on the output of an amplifier stage
from affecting the bias of the next amplifier.
 The ac signal is coupled through the transformer between

amplifier stages

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6.4 UNDERSTAND THE EFFECT OF A
RESISTIVE LOAD ACROSS THE SECONDARY
WINDING

33
CURRENT:

 When a load resistor is connected to the secondary winding, there is a

current through the resulting secondary circuit because of the voltage
induced in the secondary coil
 This secondary current results in a primary current

Isec = (Npri/Nsec)Ipri Where;

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6.4.1 DETERMINE THE CURRENT DELIVERED BY THE
SECONDARY WHEN A STEP UP TRANSFORMER IS LOADED:

Step-Up Transformer

Where;

35
 DETERMINE
THE CURRENT
DELIVERED BY Example:
THE
SECONDARY
WHEN A STEP
UP Calculate all voltages and all currents in this circuit, given the component
TRANSFORME values and the number of turns in each of the transformer's windings:
R IS LOADED

Solution:

36
6.4.2 DETERMINE THE CURRENT DELIVERED BY THE SECONDARY
WHEN A STEP DOWN TRANSFORMER IS LOADED:

Where;

37
 DETERMINE
THE CURRENT
DELIVERED BY Example:
THE
SECONDARY Calculate all voltages and all currents in this circuit, given the component
WHEN A STEP
DOWN values and the number of turns in each of the transformer's windings:
TRANSFORME
R IS LOADED

Solution:

38
6.4.3 CALCULATE POWER IN A TRANSFORMER

 For an ideal transformer, the power delivered in the secondary

equals the power in the primary
 In a real transformer, some power is dissipated in the transformer,
so primary power is always greater than secondary power
 In an ideal transformer, power transfer is not related to the turns
ratio
Ppri = VpriIpri = VsecIsec = Psec

39
6.5 UNDERSTAND A NON-IDEAL
TRANSFORMER

• A nonideal (or actual) transformer differs from an ideal transformer in

that the former has hysteresis and eddy-current (core) losses, and has
resistive (i 2 R) losses in its primary and secondary windings.

• Real transformers have winding resistance, resistance in series with each

winding, resulting in less than idea secondary voltage

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6.5.1 LIST THE NON IDEAL CHARACTERISTICS

HYSTERESIS LOSSES@CORE LOSS:

 Hysteresis losses are core losses: due to the continuous reversal of the
magnetic field due to the changing direction of the primary current.
 Hysteresis loss is a heat loss caused by the magnetic properties of the
armature.
 When the armature core is rotating, its magnetic field keeps changing
direction. The continuous movement of the magnetic particles, as they try
to align themselves with the magnetic field, produces molecular friction.
This, in turn, produces heat. This heat is transmitted to the armature
windings. The heat causes armature resistances to increase.

41
 LIST THE NON
IDEAL
CHARACTERISTIC
S EDDY CURRENT:
 Eddy currents: result in more heat losses in the core material.
 Eddy currents are produced when voltage is induced in the core material itself

42
 LIST THE NON
IDEAL
CHARACTERISTIC
S LEAKAGE FLUX:
 Lines of flux generated by one winding which do not link with the other
winding are called LEAKAGE FLUX.
 Since leakage flux generated by the primary does not cut the secondary, it
cannot induce a voltage into the secondary.
 The voltage induced into the secondary is therefore less than it would be if the
leakage flux did exist.
 LEAKAGE INDUCTANCE is assumed to drop part of the applied voltage,
leaving less voltage across the primary

43
6.5.2 EXPLAIN POWER RATING OF A TRANSFORMER

 Transformers are typically rated in volt-amperes (VA), primary/

secondary voltage, and operating frequency.

Note:

Where:
Ppri = VpriIpri

Psec = VsecIsec

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6.5.3 DETERMINE EFFICIENCY OF A TRANSFORMER:

 Is usually expresses as a percentage.

@

45
 DETERMINE
EFFICIENCY OF
A
TRANSFORME
R
Example:
A 200kVA rated transformer has a full load copper loss of 1.5kW & an iron
loss of 1kW. Determine the transformer efficiency at full load and o.85 power
factor.
Full load output power = IV cos θ = 200 X 0.85 = 170 kW.
Total losses = copper loss + iron loss = 1.5kW + 1kW = 2.5kW
Input power = output power + losses = 170kW + 2.5kW = 172.5kW

46
 DETERMINE
EFFICIENCY OF
A
TRANSFORME
R
Example:
A  5:1  stepdown  transformer  has  a  fullload  secondary  current  of  20A.   A
short circuit test for copper loss at full load gives a wattmeter reading of 100
W. If RP = 0.3Ω, find RS and power loss in the secondary
Solution:
Copper  Loss = I2P  RP +I2S  RS =100  W
To find IP:

To find RS:

47
 DETERMINE
EFFICIENCY OF Example:
A
TRANSFORME  A primary and secondary winding of 500kVA transformer have each 0.5Ω
R
and 0.0012Ω resistor. Given a primary voltage 6.6kV and secondary voltage
0.4kV and core loss 3kW. Calculate the transformer efficiency in full load
condition.
Assume the power factor is 0.8.
Solution:

48
6.6 UNDERSTAND SEVERAL TYPES OF
TRANSFORMER

49
6.6.1 DESCRIBE CENTER TAPPED TRANSFORMER:

 Simply defined as a transformer with a “tap” in the center of the

secondary winding. 
 This “tap” or additional connection in the middle of the winding
can be used with, or instead of, other types of connections at the
ends of the windings. 
 This scenario provides a variety of winding ratios

50
DESCRIBE
CENTER
TAPPED Common applications of center-tapped transformers:
TRANSFOR
MER
 In a rectifier, a center-tapped transformer and two diodes can form a full-wave
rectifier that allows both half-cycles of the AC waveform to contribute to the
direct current.

 In an audio power amplifier center-tapped transformers are used to drive

push-pull output stages. This allows two devices operating in Class B to
combine their output to produce higher audio power with relatively low
distortion.
 In analog telecommunications systems center-tapped transformers can be used
to provide a DC path around an AC coupled amplifier for signaling purposes.

51
6.6.2 DESCRIBE MULTIPLE WINDING TRANSFORMER

 Multiple-winding transformers have more than one winding on a

common core.
 They are used to operate on, or provide, different operating
voltages

52
6.6.3 DESCRIBE AUTO-TRANSFORMERS:

 In an autotransformer, one winding serves as both the primary and the

secondary. The winding is tapped at the proper points to achieve the desired
turns ratio for stepping up or down the voltage

 An autotransformer can be smaller, lighter and cheaper than a standard dual-

winding transformer however the autotransformer does not provide electrical
isolation.

 Autotransformers are often used to step up or down between voltages in the

110-117-120 volt range and voltages in the 220-230-240 volt range,

53
 DESCRIBE
AUTO-
TRANSFORMER AUTOTRANSFORMER
S

• As in an ordinary transformer, the ratio of secondary to primary voltages is equal

to the ratio of the number of turns of the winding they connect to.

• For example, connecting the load between the middle and bottom of the
autotransformer will reduce the voltage by 50%.

54
Other types of transformer
Power transformer (power up to 20MVA)
Distribution transformer (11kV to 415V
(3 phase) or 240V (single phase)
Instrument transformer (alternative to
normal multimeter)