T RANSFORMERS
Transformers are key elements in power systems. In order to effectively transmit power over long distances without prohibitive line losses, the voltage from the generator (a maximum of output voltage of approximately 25-30 kV) must be increased to a significantly higher level
(from approximately 150 kV up to 750 kV). Transformers must also be utilized on the distribution end of the line to step the voltage down (in stages) to the voltage levels required by the consumer.
Transformers also have a very wide range of applications outside the power area. Transformers are essential components in the design of DC power supplies. They can provide DC isolation between two parts of a circuit. Transformers can be used for impedance matching between sources and loads or sources and transmission lines. They can also be used to physically insulate one circuit from another for safety.
Fundamentally, the transformer consists of two or more windings that are magnetically coupled using a ferromagnetic core. For a two-winding transformer, the winding connected to the AC supply is typically referred to as the primary while the winding connected to the load is referred to as the secondary . A time-varying current passing through the primary coil produces a time-varying magnetic flux density within the core. According to Faraday’s law, the time-changing flux passing through the secondary induces a voltage in the secondary terminals.

I DEAL T RANSFORMER
The basic two-winding (single-phase) transformer is shown below.
To simplify the initial analysis, the transformer will be assumed to be ideal.
The following assumptions are made in the analysis of an ideal transformer:
(1) The transformer windings are perfect conductors (zero winding resistance).
(2) The core permeability is infinite (the reluctance of the core is zero).
(3) All magnetic flux is confined to the transformer core (no leakage flux).
(4) Core losses are assumed to be zero.
The figure above shows the common convention for the primary and secondary voltage polarities and current directions. The voltage polarities and current directions shown above yield positive input power on the primary and positive output power on the secondary. The actual polarity relationship between the primary voltage and the secondary voltage is dictated by the orientation of the primary and the secondary coils. The transformer voltage relationship is obtained by applying Faraday’s law.

Applying Faraday’s law to both the primary and the secondary (noting the possibility of sources applied to either winding), yields where the line integrals of the electric fields are along the primary and secondary windings from the “ !
” terminal to the “+” terminal, and the corresponding surface integrals of the magnetic flux densities are over the cross-sections of the primary and secondary coils. The directions of the differential lengths and differential surfaces are related by the right-hand rule. The total magnetic flux passing through the primary coil also passes through the secondary coil, assuming an ideal transformer (all of the flux is confined to the transformer core).
Note that the orientations of the given coils yield differential surface vectors that point in the same direction around the magnetic circuit, such that

Dividing the equations for v
1
and v
2
gives where the ratio of the primary turns to the secondary turns (defined as the turns ratio a ) is equal to the ratio of the primary and secondary voltages.
According to the turns ratio equation, a transformer with more secondary turns than primary turns yields a secondary voltage that is larger than the primary voltage ( step-up transformer ) while a transformer with fewer secondary turns than primary turns yields a secondary voltage that is smaller than the primary voltage ( step-down transformer ).
If the orientation of one of the transformer coils is reversed, then the differential surface vectors for the primary and secondary would be in opposite directions yielding
If we apply Ampere’s law around the transformer core (clockwise, on the centerline of the core), we find
Note that with L clockwise, the normal to the surface S is inward, so that currents inward are positive and currents outward are negative. The total enclosed current is then

Assuming an ideal transformer core ( : r
= 4 ), the magnetic field inside the core is zero (similar to the fact that the electric field is zero inside a perfect conductor with F = 4 , but carries a current on the surface of the conductor).
Thus,
The conservation of power relationship for the ideal transformer (power in equals power out, given no losses) can be stated by multiplying the voltage ratio by the current ratio:
Assuming sinusoidal excitation, the ideal transformer can be analyzed using phasor techniques. The ratios of the phasor voltages and phasor currents satisfy the same turns ratio relationships as the time-domain values.

Note that the complex power relationship is also valid for the ideal transformer.

Polarity of Transformer Windings
The operation of the transformer depends on the relative orientation of the primary and secondary coils. We mark one of the terminals on the primary and secondary coils with a dot to denote that currents entering these two terminals produce magnetic flux in the same direction within the transformer core.
The equivalent circuit diagram and phasor equations for this ideal transformer are
If either coil orientation is reversed, the dot positions are reversed and the current and voltage equations must include a minus sign.

Input Impedance (Ideal Transformer)
Consider an arbitrary load ( Z
2
) connected to the secondary terminals of the ideal transformer as shown below.
impedance seen looking into the primary winding is given by
The input
Thus, the input impedance seen looking into the primary of the ideal transformer is the load impedance times a 2 . Using this property, the secondary impedance of the ideal transformer can be reflected to the primary.

In a similar fashion, a load on the primary side of the ideal transformer can be reflected to the secondary.

Example (Ideal transformer)
Determine the primary and secondary currents for the ideal transformer below if Z s
= (18 !
j4) S and Z
2
= (2+j1) S .
The load impedance reflected to the primary of the transformer is
The primary current is then
The primary voltage is
The secondary voltage is
The secondary current is

T RANSFORMER R ATING
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.
N ON -I DEAL T RANSFORMER E QUIVALENT C IRCUITS
The non-ideal transformer equivalent circuit below accounts for all of the loss terms that are neglected in the ideal transformer model. The individual loss terms in the equivalent circuit are:
R w 1
, R w 2
- primary and secondary winding resistances
(losses in the windings due to the resistance of the wires)
X l 1
, X l 2
- primary and secondary leakage reactances
(losses due to flux leakage out of the transformer core)
R c 1
- core resistance
(core losses due to hysteresis loss and eddy current loss)
X m 1
- magnetizing reactance
(magnetizing current necessary to establish magnetic flux in the transformer core)

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 below. The primed quantities represent those values that equal the original secondary quantity multiplied by a (voltages), divided by a (currents) or multiplied by a 2
(impedance components).
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 the this shunt impedance and allows for the winding resistances and leakage reactances to be combined.

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 ( V
1
and V N
2
).
This circuit can easily be modified so that it is referred to the secondary side of the transformer ( V N
1
and V
2
).

D ETERMINATION OF E QUIVALENT C IRCUIT P ARAMETERS
In order to utilize the complete transformer equivalent circuit, the values of R w 1
, R w 2
, X l 1
, X l 2
, R c 1
, X m 1
and a must be known. These values can be computed given the complete design data for the transformer including dimensions and material properties. The equivalent circuit parameters can also be determined by performing two simple test measurements. These measurements are the no-load (or open-circuit) test and the short-circuit test .
No-Load Test - The rated voltage at rated frequency is applied to the high-voltage (HV) or low-voltage (LV) winding with the opposite winding open-circuited.
Measurements of current, voltage and real power are made on the input winding (most often the LV winding, for convenience).
Short-Circuit Test - Either the LV or HV winding is short-circuited and a voltage at rated frequency is applied to the opposite winding such that the rated current results.
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).

Example (Equivalent circuit parameters / no-load / short-circuit tests)
The approximate equivalent circuit parameters for a single-phase
10kVA, 2200/220, 60 Hz transformer are required.
The rated currents and voltages for the transformer windings are:
V
V
H,rated
= 2200 V
L,rated
= 220 V I
I
H,rated
L,rated
= 10000/2200 = 4.55 A
= 10000/220 = 45.5 A
No-load and short-circuit tests are performed on the transformer with the following results:
No-load test (HV winding open, V
L
I
L
= 2.5 A, P
L
= 100 W
= V
L,rated
= 220 V)
Short-circuit test (LV winding shorted, I
H
V
H
= 150 V, P
H
= 215 W
= I
H,rated
= 4.55 A)
(a.) Determine the approximate equivalent circuit parameters from the test data (use the approximate equivalent circuit that includes core losses).
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.
Referred to LV side Referred to HV side

Equivalent circuit for no-load test (determine R cL
and X mL
)

Equivalent circuit for short-circuit test (determine R eqH
and X eqH
)
The values measured on the HV winding (primary) in the short-circuit test need to be referred to the LV side. Note that our turns ratio is given by

(a.)
Referred to LV side
(b.) To obtain the same equivalent circuit referred to the HV side
(primary), we simply multiply all impedances by a 2 (100). The resulting equivalent circuit is
Referred to HV side
(c.) From the no-load test results, the total excitation current is 2.5 A while the rated current in the LV winding is 45.5 A. Thus the excitation current is (2.5/45.5) or 5.5% of the rated current.

(d.) The power factor is defined as
For the no-load test, the power factor is
For the short-circuit test, the power factor is

T RANSFORMER V OLTAGE R EGULATION
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 ( Z primary ( Z N
2
2
) and the resulting load current ( I
, I N
2
).
2
) have been reflected to the
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.
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 refelcted secondary values divided by the turns ratio.

Thus, the percentage voltage regulation may be written in terms of the reflected secondary voltages.
According to the approximate transformer equivalent circuit, the reflected secondary voltage under no-load conditions is equal to the primary voltage, so that
The secondary voltage for the loaded condition is taken as the rated voltage.
Inserting the previous two equations into the percentage voltage regulation equation gives
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.
To determine the percentage voltage regulation, we may use the reflected secondary voltage as the voltage reference, and determine the corresponding value of * V
1 equivalent circuit.
* from the approximate

The voltages V
1
and V N
2
in the approximate equivalent circuit are related by where
The reflected secondary current can be written as
The expression for V
1
becomes
We can draw the phasor diagram relating the voltages V
1
and V N
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 * V
1 if * V
1
* < V
1, rated
* > V
1, rated
. Note that with the limits on the angles of
and negative the worst case scenario for the percentage voltage regulation occurs when or when the load has a lagging power factor with the power factor angle equal to the transformer impedance angle of Z eq 1
.

Example (Transformer voltage regulation)
Using the transformer for which the approximate equivalent circuits were found based on no-load and short-circuit test results, determine the percentage voltage regulation for (a.) a load drawing 75% of rated current at a power factor of 0.6 lagging (b.) a load drawing 75% of rated current at a power factor of 0.6 leading.
The approximate equivalent circuit for the transformer (10kVA,
2200/220, 60 Hz) referred to the high voltage winding was found to be (neglecting the excitation branch of the model)
Assume:
Primary = HV winding
Secondary = LV winding a = N
V
H,rated
I
H,rated
V
L,rated
I
L,rated
1
/ N
2
= 10
= 2200 V
= 10000/2200 = 4.55 A
= 220 V
= 10000/220 = 45.5 A
(a.) The reflected voltage V
L
N on the HV side is given by so that the circuit to be analyzed becomes

Based on the information provided in the problem statement, the magnitude of the load current (and thus the current I that of the rated value.
L
N ) is 0.75 times
The phase angle of I
L
N is given by the load power factor.
so that the phasor current I
L
N is
The voltage V
H
and V
L
N are related by
The percentage voltage regulation is thus
(b.) For the leading power of 0.6, 2 i
I
L
N is
= +53.13
o so that the phasor current and V
H
is given by

The voltage regulation is
The percentage voltage regulation results for these two cases shows that if this transformer is providing 75% of rated current (3.41 A-rms) to a load with a power factor of 0.6 lagging, and the load is suddenly removed, the load voltage magnitude rises from 220 V to 230.70V. For the load with a power factor of 0.6 leading, the load voltage magnitude drops from 220 V to 213.79 V.

T RANSFORMER E FFICIENCY
The efficiency ( 0 ) of a transformer is defined as the ratio of the output power ( P out
) to the input power ( P in
). The output power is equal to the input power minus the losses ( P loss
) in the transformer. The transformer loss power has two components: core loss ( P core
) and so-called copper loss ( P cu associated with the winding resistances. The transformer efficiency in
) percent is given by
Assuming a relatively constant voltage source on the primary of the transformer, the core loss can be assumed to be constant and equal to power dissipated in the core loss resistance ( R c 1
) of the equivalent circuit for the no-load test. The copper loss in a transformer may be written in terms of both the primary and secondary currents, or in terms of only one of these currents based on the relationship I
2
= aI
1
.
The output power of the transformer can be written in terms of the secondary voltage and current (real part of output complex power).

The transformer efficiency, written in terms of secondary values, is
It can be shown that the maximum transformer efficiency occurs when the core losses equal the copper losses and the power factor is unity.
Example (Transformer efficiency)
Using the transformer for which the approximate equivalent circuits were found based on no-load and short-circuit test results, determine (a.) the transformer efficiency at 75% of rated output power with a power factor of 0.6 lagging (b.) the output power at maximum efficiency, the value of maximum efficiency, and at the percentage of full load power where maximum efficiency occurs.
(a.) The rated power for this transformer is 10 kW at a power factor of unity. If P out
is 75% of the rated value at a power factor of 0.6, then
From the no-load test, the core losses were P core losses for this transformer are
= 100 W. The copper
The transformer efficiency is

(b.) At maximum efficiency Y P
Cu
= I
L
2 R eqL
= P core
= 100 W, PF = 1
The maximum efficiency is
The maximum efficiency occurs at