TRANSFORMERS

Transformers are key elements in power systems. In order toeffectively transmit power over long distances without prohibitive linelosses, the voltage from the generator (a maximum of output voltage ofapproximately 25-30 kV) must be increased to a significantly higher level(from approximately 150 kV up to 750 kV). Transformers must also beutilized on the distribution end of the line to step the voltage down (instages) to the voltage levels required by the consumer.

Transformers also have a very wide range of applications outside thepower area. Transformers are essential components in the design of DCpower supplies. They can provide DC isolation between two parts of acircuit. Transformers can be used for impedance matching between sourcesand loads or sources and transmission lines. They can also be used tophysically insulate one circuit from another for safety.

Fundamentally, the transformer consists of two or more windings thatare magnetically coupled using a ferromagnetic core. For a two-windingtransformer, the winding connected to the AC supply is typically referredto as the primary while the winding connected to the load is referred to asthe secondary. A time-varying current passing through the primary coilproduces a time-varying magnetic flux density within the core. Accordingto Faraday’s law, the time-changing flux passing through the secondaryinduces a voltage in the secondary terminals.

IDEAL TRANSFORMER

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 windingresistance).

(2) The core permeability is infinite (the reluctance of the core iszero).

(3) All magnetic flux is confined to the transformer core (noleakage flux).

(4) Core losses are assumed to be zero.

The figure above shows the common convention for the primary andsecondary voltage polarities and current directions. The voltage polaritiesand current directions shown above yield positive input power on theprimary and positive output power on the secondary. The actual polarityrelationship between the primary voltage and the secondary voltage isdictated by the orientation of the primary and the secondary coils. Thetransformer voltage relationship is obtained by applying Faraday’s law.

Applying Faraday’s law to both the primary and the secondary (noting thepossibility of sources applied to either winding), yields

where the line integrals of the electric fields are along the primary andsecondary windings from the “!” terminal to the “+” terminal, and thecorresponding surface integrals of the magnetic flux densities are over thecross-sections of the primary and secondary coils. The directions of thedifferential lengths and differential surfaces are related by the right-handrule. The total magnetic flux passing through the primary coil also passesthrough the secondary coil, assuming an ideal transformer (all of the fluxis confined to the transformer core).

Note that the orientations of the given coils yield differential surfacevectors that point in the same direction around the magnetic circuit, suchthat

Dividing the equations for v1 and v2 gives

where the ratio of the primary turns to the secondary turns (defined as theturns ratio a) is equal to the ratio of the primary and secondary voltages.According to the turns ratio equation, a transformer with more secondaryturns than primary turns yields a secondary voltage that is larger than theprimary voltage (step-up transformer) while a transformer with fewersecondary turns than primary turns yields a secondary voltage that issmaller than the primary voltage (step-down transformer).

If the orientation of one of the transformer coils is reversed, then thedifferential surface vectors for the primary and secondary would be inopposite directions yielding

If we apply Ampere’s law around the transformer core (clockwise, onthe centerline of the core), we find

Note that with L clockwise, the normal to the surface S is inward, so thatcurrents inward are positive and currents outward are negative. The totalenclosed current is then

Assuming an ideal transformer core (:r = 4), the magnetic field inside thecore is zero (similar to the fact that the electric field is zero inside a perfectconductor with F = 4, but carries a current on the surface of the conductor).Thus,

The conservation of power relationship for the ideal transformer (power inequals power out, given no losses) can be stated by multiplying the voltageratio by the current ratio:

Assuming sinusoidal excitation, the ideal transformer can beanalyzed using phasor techniques. The ratios of the phasor voltages andphasor currents satisfy the same turns ratio relationships as the time-domainvalues.

Note that the complex power relationship is also valid for the idealtransformer.

Polarity of Transformer Windings

The operation of the transformer depends on the relative orientationof the primary and secondary coils. We mark one of the terminals on theprimary and secondary coils with a dot to denote that currents enteringthese two terminals produce magnetic flux in the same direction within thetransformer core.

The equivalent circuit diagram and phasor equations for this idealtransformer are

If either coil orientation is reversed, the dot positions are reversed and thecurrent and voltage equations must include a minus sign.

Input Impedance (Ideal Transformer)

Consider an arbitrary load (Z2) connected to the secondary terminalsof the ideal transformer as shown below.

The inputimpedance seen looking into the primary winding is given by

Thus, the input impedance seen looking into the primary of the idealtransformer is the load impedance times a2. Using this property, thesecondary impedance of the ideal transformer can be reflected to theprimary.

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

Example (Ideal transformer)

Determine the primary and secondary currents for the idealtransformer below if Zs = (18!j4) S and Z2 = (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

TRANSFORMER RATING

Transformers carry ratings related to the primary and secondarywindings. The ratings refer to the power in kVA and primary/secondaryvoltages. A rating of 10 kVA, 1100/110 V means that the primary is ratedfor 1100 V while the secondary is rated for 110 V (a =10). The kVA ratinggives the power information. With a kVA rating of 10 kVA and a voltagerating of 1100 V, the rated current for the primary is 10,000/1100 = 9.09 Awhile the secondary rated current is 10,000/110 = 90.9 A.

NON-IDEAL TRANSFORMER EQUIVALENT CIRCUITS

The non-ideal transformer equivalent circuit below accounts for allof the loss terms that are neglected in the ideal transformer model. Theindividual 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 fluxin the transformer core)

Using the impedance reflection technique, all the quantities on thesecondary side of the transformer can be reflected back to the primary sideof the circuit. The resulting equivalent circuit is shown below. The primedquantities represent those values that equal the original secondary quantitymultiplied by a (voltages), divided by a (currents) or multiplied by a2

(impedance components).

Approximate Transformer Equivalent Circuits

Given that the voltage drops across the primary winding resistanceand the primary leakage reactance are typically quite small, the shuntbranch of the core loss resistance and the magnetizing reactance (excitationbranch) can be shifted to the primary input terminal. The primary voltageis then applied directly across the this shunt impedance and allows for thewinding resistances and leakage reactances to be combined.

A further approximation to the transformer equivalent circuit can bemade by eliminating the excitation branch. This approximation removesthe core losses and the magnetizing current from the transformer model.The resulting equivalent circuit is shown below.

Note that this equivalent circuitis referred to the primary side ofthe transformer (V1 and VN2).This circuit can easily bemodified so that it is referred tothe secondary side of thetransformer (VN1 and V2).

DETERMINATION OF EQUIVALENT CIRCUIT PARAMETERS

In order to utilize the complete transformer equivalent circuit, thevalues of Rw1, Rw2, Xl1, Xl2, Rc1, Xm1 and a must be known. These values canbe computed given the complete design data for the transformer includingdimensions and material properties. The equivalent circuit parameters canalso be determined by performing two simple test measurements. Thesemeasurements are the no-load (or open-circuit) test and the short-circuittest.

No-Load Test - The rated voltage at rated frequency is applied tothe high-voltage (HV) or low-voltage (LV) windingwith the opposite winding open-circuited.Measurements of current, voltage and real powerare made on the input winding (most often the LVwinding, for convenience).

Short-Circuit Test - Either the LV or HV winding is short-circuited anda voltage at rated frequency is applied to theopposite winding such that the rated current results.Measurements of current, voltage and real powerare made on the input winding (most often the HVwinding, for convenience, since a relatively lowvoltage is necessary to obtain rated current undershort-circuit conditions).

Example (Equivalent circuit parameters / no-load / short-circuit tests)

The approximate equivalent circuit parameters for a single-phase10kVA, 2200/220, 60 Hz transformer are required.

The rated currents and voltages for the transformer windings are: VH,rated = 2200 V IH,rated = 10000/2200 = 4.55 AVL,rated = 220 V IL,rated = 10000/220 = 45.5 A

No-load and short-circuit tests are performed on the transformer with thefollowing results:

No-load test (HV winding open, VL = VL,rated = 220 V)IL = 2.5 A, PL = 100 W

Short-circuit test (LV winding shorted, IH = IH,rated = 4.55 A)VH = 150 V, PH = 215 W

(a.) Determine the approximate equivalent circuit parameters from the testdata (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 HVside. (c.) From the no-load test results, express the excitation current as apercentage of the rated current in the LV winding. (d.) Determine thepower factor for the no-load and short-circuit tests.

Referred to LV side Referred to HV side

Equivalent circuit for no-load test (determine RcL and XmL)

Equivalent circuit for short-circuit test (determine ReqH and XeqH)

The values measured on the HV winding (primary) in the short-circuittest 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 a2 (100). Theresulting equivalent circuit is

Referred to HV side

(c.) From the no-load test results, the total excitation current is 2.5 Awhile the rated current in the LV winding is 45.5 A. Thus theexcitation 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

TRANSFORMER VOLTAGE REGULATION

For a given input (primary) voltage, the output (secondary) voltageof an ideal transformer is independent of the load attached to the secondary.As seen in the transformer equivalent circuit, the output voltage of arealistic transformer depends on the load current. Assuming that thecurrent through the excitation branch of the transformer equivalent circuitis small in comparison to the current that flows through the winding lossand leakage reactance components, the transformer approximate equivalentcircuit referred to the primary is shown below. Note that the load on thesecondary (Z2) and the resulting load current (I2) have been reflected to theprimary (ZN2, IN2).

The percentage voltage regulation (VR) is defined as the percentage changein the magnitude of the secondary voltage as the load current changes fromthe no-load to the loaded condition.

The transformer equivalent circuit above gives only the reflected secondaryvoltage. The actual loaded and no-load secondary voltages are equal to theloaded and no-loaded refelcted secondary values divided by the turns ratio.

Thus, the percentage voltage regulation may be written in terms of thereflected secondary voltages.

According to the approximate transformer equivalent circuit, the reflectedsecondary voltage under no-load conditions is equal to the primary voltage,so that

The secondary voltage for the loaded condition is taken as the ratedvoltage.

Inserting the previous two equations into the percentage voltage regulationequation gives

Note that this equation is defined in terms of the voltages given in thetransformer approximate equivalent circuit. Also note that the ratedsecondary voltage reflected to the primary is the rated primary voltage.

To determine the percentage voltage regulation, we may use the reflectedsecondary voltage as the voltage reference,

and determine the corresponding value of *V1* from the approximateequivalent circuit.

The voltages V1 and VN2 in the approximate equivalent circuit are related by

where

The reflected secondary current can be written as

The expression for V1 becomes

We can draw the phasor diagram relating the voltages V1 and VN2 todetermine how the phase angles of the load and the transformer impedanceaffect the percentage voltage regulation. Note that the percentage voltageregulation can be positive or negative and the sign of VR is affected by thephase angle in the expression above. Thus, the power factor of the loadwill affect the voltage regulation of the transformer.

The percentage voltage regulation is positive if *V1* > V1,rated and negativeif *V1* < V1,rated. Note that with the limits on the angles of

the worst case scenario for the percentage voltage regulation occurs when

or when the load has a lagging power factor with the power factor angleequal to the transformer impedance angle of Zeq1.

Example (Transformer voltage regulation)

Using the transformer for which the approximate equivalent circuitswere found based on no-load and short-circuit test results, determine thepercentage voltage regulation for (a.) a load drawing 75% of rated currentat a power factor of 0.6 lagging (b.) a load drawing 75% of rated current ata 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 tobe (neglecting the excitation branch of the model)

Assume:Primary = HV windingSecondary = LV windinga = N1/N2 = 10VH,rated = 2200 VIH,rated = 10000/2200 = 4.55 AVL,rated = 220 VIL,rated = 10000/220 = 45.5 A

(a.) The reflected voltage VLN on the HV side is given by

so that the circuit to be analyzed becomes

Based on the information provided in the problem statement, themagnitude of the load current (and thus the current ILN) is 0.75 timesthat of the rated value.

The phase angle of ILN is given by the load power factor.

so that the phasor current ILN is

The voltage VH and VLN are related by

The percentage voltage regulation is thus

(b.) For the leading power of 0.6, 2i = +53.13o so that the phasor currentILN is

and VH is given by

The voltage regulation is

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

TRANSFORMER EFFICIENCY

The efficiency (0) of a transformer is defined as the ratio of the outputpower (Pout) to the input power (Pin). The output power is equal to the inputpower minus the losses (Ploss) in the transformer. The transformer losspower has two components: core loss (Pcore) and so-called copper loss (Pcu)associated with the winding resistances. The transformer efficiency inpercent is given by

Assuming a relatively constant voltage source on the primary of thetransformer, the core loss can be assumed to be constant and equal to powerdissipated in the core loss resistance (Rc1) of the equivalent circuit for theno-load test. The copper loss in a transformer may be written in terms ofboth the primary and secondary currents, or in terms of only one of thesecurrents based on the relationship I2 = aI1.

The output power of the transformer can be written in terms of thesecondary 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 thecore losses equal the copper losses and the power factor is unity.

Example (Transformer efficiency)

Using the transformer for which the approximate equivalent circuitswere found based on no-load and short-circuit test results, determine (a.)the transformer efficiency at 75% of rated output power with a power factorof 0.6 lagging (b.) the output power at maximum efficiency, the value ofmaximum efficiency, and at the percentage of full load power wheremaximum efficiency occurs.

(a.) The rated power for this transformer is 10 kW at a power factor ofunity. If Pout is 75% of the rated value at a power factor of 0.6, then

From the no-load test, the core losses were Pcore = 100 W. The copperlosses for this transformer are

The transformer efficiency is

(b.) At maximum efficiency Y PCu = IL2ReqL = Pcore = 100 W, PF = 1

The maximum efficiency is

The maximum efficiency occurs at