TRANSFORMER RATINGS

Transformers carry ratings related to the primary and

secondary windings. The ratings refer to the power in kVA and

primary/secondary voltages.

A rating of 10 kVA, 1100/110 V means that the primary is

rated for 1100 V while the secondary is rated for 110 V (a

=10). The kVA rating gives the power information.

With a kVA rating of 10 kVA and a voltage rating of 1100 V,

the rated current for the primary is 10,000/1100 = 9.09 A while

the secondary rated current is 10,000/110 = 90.9 A.

NON-IDEAL TRANSFORMER EQUIVALENT

CIRCUITS

The non-ideal transformer equivalent circuit shown in Fig. 1

accounts for all of the loss terms that are neglected in the ideal

transformer model. The individual loss terms in the equivalent

circuit are:

Rw1, Rw2 - primary and secondary winding resistances.

(losses in the windings due to the resistance of the

wires)

Xl1, Xl2 - primary and secondary leakage reactances. (losses

due to flux leakage out of the transformer core)

Rc1 - core resistance. (core losses due to hysteresis loss and

eddy current loss)

Xm1 - magnetizing reactance. (magnetizing current

necessary to establish magnetic flux)

Fig. 1. Non-ideal transformer equivalent circuit.

Using the impedance reflection technique, all the quantities on

the secondary side of the transformer can be reflected back to

the primary side of the circuit. The resulting equivalent circuit

is shown in the figure below. The primed quantities represent

those values that equal the original secondary quantity

multiplied by a (voltages), divided by a (currents) or multiplied

by a2 (impedance components).

(1)

(2)

(3)

(4)

(5)

The primed quantities represent those values that equal the

original primary quantity divided by a (voltages), multiplied by

a (currents) or divided by a2 (impedance components).

(6)

(7)

(8)

(9)

(10)

APPROXIMATE TRANSFORMER EQUIVALENT

CIRCUITS

Given that the voltage drops across the primary winding

resistance and the primary leakage reactance are typically quite

small, the shunt branch of the core loss resistance and the

magnetizing reactance (excitation branch) can be shifted to the

primary input terminal. The primary voltage is then applied

directly across this shunt impedance and allows for the winding

resistances and leakage reactances to be combined as shown in

the figure below.

Simplifying the impedances in the series arms

We obtain the equivalent primary winding resistance and

primary leakage reactance given by

(11)

(12)

A further approximation to the transformer equivalent circuit

can be made by eliminating the excitation branch. This

approximation removes the core losses and the magnetizing

current from the transformer model. The resulting equivalent

circuit is shown below.

Note that this equivalent circuit is referred to the primary side

of the transformer (V1 and V’2). This circuit can easily be

modified so that it is referred to the secondary side of the

transformer (V’1 and V2) as shown in the figure below.

We obtain the equivalent secondary winding resistance and

secondary leakage reactance given by

(13)

(14)

TRANSFORMER VOLTAGE REGULATION

For a given input (primary) voltage, the output (secondary)

voltage of an ideal transformer is independent of the load

attached to the secondary. As seen in the transformer

equivalent circuit, the output voltage of a realistic transformer

depends on the load current. Assuming that the current through

the excitation branch of the transformer equivalent circuit is

small in comparison to the current that flows through the

winding loss and leakage reactance components, the

transformer approximate equivalent circuit referred to the

primary is shown below. Note that the load on the secondary

(Z2) and the resulting load current (I2) have been reflected to

the primary (Z’2, I’2).

The percentage voltage regulation (VR) is defined as the

percentage change in the magnitude of the secondary voltage

as the load current changes from the no-load to the loaded

condition.

(15)

The transformer equivalent circuit above gives only the

reflected secondary voltage. The actual loaded and no-load

secondary voltages are equal to the loaded and no-loaded

reflected secondary values divided by the turns ratio that are

given by

(16)

Thus, the percentage voltage regulation may be written in

terms of the reflected secondary voltages.

(17)

According to the approximate transformer equivalent circuit,

the reflected secondary voltage under no-load conditions is

equal to the primary voltage, so that

(18)

The secondary voltage for the loaded condition is taken as the

rated voltage.

(19)

Inserting the previous two equations into the percentage

voltage regulation equation gives

(20)

Note that this equation is defined in terms of the voltages given

in the transformer approximate equivalent circuit. Also note

that the rated secondary voltage reflected to the primary is the

rated primary voltage.

(21)

To determine the percentage voltage regulation, we may use

the reflected secondary voltage as the voltage reference,

(22)

and determine the corresponding value of V1 from the

approximate equivalent circuit.

The voltages V1 and V’2 in the approximate equivalent circuit

are related by

(23)

where

(24)

The reflected secondary current can be written as

(25)

The expression for V1 becomes

(26)

We can draw the phasor diagram relating the voltages V1 and

V’2 to determine how the phase angles of the load and the

transformer impedance affect the percentage voltage

regulation. Note that the percentage voltage regulation can be

positive or negative and the sign of VR is affected by the phase

angle in the expression above. Thus, the power factor of the

load will affect the voltage regulation of the transformer.

The percentage voltage regulation is positive if |V1|> V2,rated and

negative if |V1| < V2,rated. Note that with the limits on the angles

of

(27)

(28)

The worst case scenario for the percentage voltage regulation

occurs when

(29)

or when the load has a lagging power factor with the power

factor angle equal to the transformer impedance angle of Zeq1.

LOSSES IN THE TRANSFORMER

1. Copper Loss

The copper loss is due to the power wasted in the form of I2R

loss due to the resistances of the primary and secondary

windings. The copper loss depends on the magnitude of the

currents flowing through the windings.

The copper loss in a transformer may be written in terms of

both the primary and secondary currents as

(30)

or in terms of only one of these currents based on the

relationship I2 = aI1. In terms of I1, we have

(31)

and then in terms of I2

(32)

The copper losses are denoted as PCu. If the current through the

windings is full load current, we get copper losses at full load.

If the load on transformer is half then we get copper losses at

half load which are less than full load copper losses. Thus

copper losses are called variable losses. The transformer VA

rating is V1I1 or V2I2. We can say that the copper losses are

proportional to the square of the current

(33)

Copper loss or I2R loss in the windings is a variables loss as

this loss depends upon the value of load current, I. It may be

noted that sometimes the windings of transformers are made of

aluminum wires to reduce the cost of production. The copper

loss for a given fraction of the load can be defined as

Let x - the I2R loss full-load in Watts.

At half load, its load value will be x/4 Watts.

At one third of the full load, its load value will be x/9 Watts.

Therefore, copper loss is proportional to the square of the given fraction of the load.

2. Core Loss

Core loss or iron loss is called constant loss as this loss does

not depend upon the amount of electrical load connected to the

transformer. This means that the core loss remains the same as

on no-load and on full-load or on any other load.

Core loss is composed of hysteresis loss and eddy current loss.

2.1. Hysteresis Loss

The power loss is dissipated as heat from the core.

It is due to the alternate magnetization of the atoms forming

domains in the magnetic material of the core. Each domain

behaves as a very small magnet oriented in different directions

as shown in Fig. 2(a).

Due to the application of magnetizing force in Fig. 2(b), these

tiny magnets orient themselves in the direction of

magnetization. If the magnetizing force is alternating, the small

magnets will orient themselves alternately in opposite

directions.

Fig. 2. (a) Orientation of grains in different directions in an

unmagnetized magnetic material and (b) Orientation of grains an

application of magnetic force.

2.2.Eddy Current Loss

The two thick laminated sheets forming a part of the

transformer-core limb is shown in Fig. 3. However, the

laminated sheets are actually very thin and insulated from each

other.

In these sheets, EMF is induced due to the presence of an

alternating flux in the core. The EMF induced in laminated

sheet will produce circulating currents as shown. These are

called eddy current and they produces power loss in the

resistance of all iron path.

To reduce the eddy current loss in the core, the core is made up

of thin laminated sheets (instead of a solid mass) so that the

resistance to the eddy-current path is increased and hence the

value of the eddy current is reduced.

Fig. 3. Eddy currents flowing in laminated sheets of a transformer

core when current flows through the winding.

DETERMINATION OF EQUIVALENT CIRCUIT

PARAMETERS

The equivalent circuit model for the non-ideal transformer is

shown in Fig. 4.

An ideal transformer with resistors and inductors in parallel

and series replaces the non-ideal transformer. This model is

called the high side equivalent circuit model because all

parameters have been moved to the primary side of the ideal

transformer.

Series Parameters

The series resistance, Req, is the resistance of the copper

winding.

The series inductance, Xeq, accounts for the flux leakage.

That is, a small amount of flux travels through the air

outside the magnetic core path.

Shunt parameters

The shunt resistance, Rc, represents the core loss of the

magnetic core material due to hysteresis.

The shunt inductance, Xm, called the magnetizing

inductance, accounts for the finite permeability of the

magnetic core.

Fig. 4. The equivalent model for the non-ideal transformer.

The parallel parameter values are found with no load connected

to the secondary (open circuit) and the series parameter values

are found with the secondary terminals shorted (short circuit).

It is possible to make the tests on either the primary or the

secondary.

For the open (no-load) circuit test shown in Fig. 6, the series

parameters are neglected for convenience. This is reasonable

since the voltage drops are across Req and Xeq are normally

small.

Measurements of current, voltage and real power are made on

the input winding (most often the LV winding, for

convenience).

Fig. 6. Equivalent open (no-load) circuit test.

The purpose of short circuit test is to determine the series

branch parameters of the equivalent circuit as shown in Fig. 7.

The excitation current which is only 1% or less even at rated

voltage becomes negligibly small during this test and it is

neglected. The shunt branch is thus assumed to be absent.

Measurements of current, voltage and real power are made on

the input winding (most often the HV winding, for

convenience, since a relatively low voltage is necessary to

obtain rated current under short-circuit conditions).

Fig. 7. Equivalent short circuit test.

EFFICIENCY OF A TRANSFORMER

Due to the losses in a transformer, the output power of a

transformer is less than the input power supplied. Therefore

Pout = Pin – Total losses (34)

Pin = Pout + Total losses (35)

The efficiency of the device is defines as the ratio of the power

output to power input. So for a transformer the efficiency can

be expressed as

(36)

Now, we let where is the load power

factor.

The transformer supplies full load of current I2 and with

terminal voltage V2. Then the copper losses PCu on full load is

given by

(37)

Then the efficiency is given by

(38)

But V2I2 = VA rating of a transformer

(39)

This is full load percentage with I2 = full-load secondary

current

But if the transformer is subjected to fractional load then using

the appropriate values of various quantities, the efficiency can

be obtained.

Let n = fraction by which load is less than full load given by

(40)

For example, if transformer is subjected to half load then

When load changes, the load current changes by same

proportion

(41)

Similarly the output also reduces by the same fraction. The

fraction of VA rating is available at the output.

Similarly as copper losses are proportional to square of current

then,

(42)

So the copper losses get reduced by n2.

In general for fractional load, the efficiency is given by

(43)

where n is the fraction by which load is less than full load.

CONDITION FOR MAXIMUM EFFICIENCY

When a transformer works on a constant input voltage and

frequency then efficiency varies with the load. As load

increases, the efficiency increases. At a certain load current, it

achieves a maximum value. If the transformer is loaded further

efficiency starts decreasing. The graph of efficiency against

load current I2 as shown in the Fig. 8.

The load current at which the efficiency attains maximum

value is denoted as I2,ηmax and maximum efficieny is denoted as

ηmax.

Fig. 8. The graph of efficiency against load I2.

The efficiency is a function of load i.e. load current I2

assuming cos φ2 constant. The secondary terminal voltage V2 is

also assumed constant for maximum efficiency

Given the efficiency

Perform the differentiation of η with respect to I2 then equate to

0, we have

Canceling out from both the terms we get,

So to achieve maximum efficiency is that

Core Losses = Copper Losses

(44)

Solving for the load current at maximum efficiency.

For ηmax, but

but

(45)

Let (I2) FL be the full load current and divide Eq. (44) by (I2)

FL

(46)

For constant V2, the kVA supplied is function of load current

(47)

Substituting condition for ηmax in the expression of efficiency,

we can write expression for ηmax as

as (48)

ALL-DAY EFFICIENCY

For a transformer, the efficiency is defined as the ratio of

output power to input power. This is its power efficiency. But

power efficiency is not the true measure of the performance of

some special types of transformers such as distribution

transformers.

For instance, distribution transformers used for supplying

lighting loads have their primaries energized all the 24 hours in

a day but the secondaries supply little or no load during the

major portion of the day. It means that a constant loss (i.e., iron

loss) occurs during the whole day but copper loss occurs only

when the transformer is loaded and would depend upon the

magnitude of load.

Consequently, the copper loss varies considerably during the

day and the commercial efficiency of such transformers will

vary from a low value (or even zero) to a high value when the

load is high.

The performance of such transformers is judged on the basis of

energy consumption in kWh during the whole day (i.e., 24

hours). This is known as all-day or energy efficiency.

The all-day efficiency is defined as

(49)

SAMPLE PROBLEMS

1. The approximate equivalent circuit parameters for a single

phase 10 kVA, 2200/220 V, 60 Hz transformer are

required.

No-load and short circuit tests are performed on the

transformer with the following results:

For the open (no-load) circuit test:

HV winding open

The voltage is at full-load rated voltage on the LV side:

220 V

Current on the LV side: 2.5 A

Power on the LV side: 100 W (This is what you call

core loss)

For the short circuit test:

LV winding shorted

The current is at full-load rated current on the HV side:

10,000/2,200 = 4.55 A

Voltage on the HV side: 150 V

Power on the HV side: 215 W

(a) Determine the approximate equivalent circuit parameters

from the test data. Draw the equivalent circuit for this

transformer referred to the LV side.

(b) Draw the equivalent circuit for this transformer referred to

the HV side.

(c) From the no-load test results, express the excitation current

as a percentage of the rated current in the LV winding.

(d) Determine the power factor for the no-load and short-circuit

tests.

(e) Determine the percentage voltage regulation for a load

drawing 75% of rated current at a power factor of 0.6

lagging.

(f) Determine the percentage voltage regulation for a load

drawing 75% of rated current at a power factor of 0.6

leading.

(g) Determine the transformer efficiency at 75% of rated output

power with a power factor of 0.6 lagging.

(h) Determine the output power at maximum efficiency and the

value of maximum efficiency at unity power factor.

(i) Determine the percentage of full load power where

maximum efficiency occurs.

2. A 100 kVA distribution transformer supplying light and fan

loads has full-load copper loss iron loss of 1.5 and 2 kW

respectively. During 24 h in a day the transformer is loaded

as follows:

6 AM to 10 AM (4 h) Half load

10 AM to 6 PM (8 h) One-fourth load

6 PM to 10 PM (4 h) Full-load

10 PM to 6 AM (8 h) Negligible load

Calculate the all-day efficiency of the transformer.

3. A 400 kVA, distribution transformer has full load iron loss

of 2.5 kW and copper loss of 3.5 kW. During a day, its load

cycle for 24 hours is

6 hours 300 kW at 0.8 lagging pf

10 hours 200 kW at 0.7 lagging pf

4 hours 100 kW at 0.9 lagging pf

4 hours no load

Determine its all day efficiency.