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Magnetic circuits and power transformers

Easwaran Chandira Sekaran Associate Professor, Department of Electrical and Electronics Engineering, Coimbatore Institute of Technology, Coimbatore, INDIA

Abstract

In order to understand the principles and operation of electrical engineering, a reasonable knowledge in magnetic circuits and transformers is very much needed. In this chapter, some of the basic concepts of magnetic fields are discussed. It is shown that these concepts form the basis for development of magnetic circuits, which are fundamentals buildings blocks of transformers and electric machinery. The principle of operation of transformer types, testing, and the latest trends in transformers are discussed. Readers will gain a detailed overview of magnetic circuits and power transformers.

Keywords

magnetic flux

flux density

MMF

transformer

OC and SC test

third harmonics

11.1. Introduction

Magnetic circuits and magnetic components such as inductors and transformers are an important, integral, and indispensable part of most power electronics and renewable energy systems. Magnetic components fall into two categories:

1. Energy storage devices

2. Energy transfer devices

Energy storage devices store kinetic energy of a desirable quality of current flowing through it. Devices in this category are called inductors.

Energy transfer devices transfer the power from one energy port to another energy port without storing or losing energy in the process. These devices are called transformers.

11.2. Magnetic circuits

Magnetic circuits [1] have a magnetic material of standard geometrical shape called a core and a coil with conducting material having a number of turns (N) wound over the core. The coil is also called the exciting coil. When the current flow through the coil is zero there is no magnetic field of lines or lines of forces present inside the core. The incidence of the current in the coil produces magnetic lines of force and the path of these magnetic lines of force can be thought of as a magnetic circuit. The various terminologies used in the magnetic circuit are briefly discussed in the subsequent sections.

11.2.1. Magnetic field and magnetic flux

Magnetic fields are produced due to the movement of electrical charge and they are present around permanent magnets and current carrying conductors (magnetic circuit) as shown in Figure 11.1. In permanent magnets, revolving electrons produce an external field. If a current carrying conductor is wounded to form a coil of multiturns, the magnetic field is stronger than that of a single conductor. The magnetic field of the electromagnet is intensified when the coil is wound on an iron core. The strength of magnetic fields is varied in many applications through electromagnets. Magnetic fields form the basis for the operation of transformers, generators, and motors.

A magnetic field that can be visualized as the lines of force is referred to as magnetic flux, which is the “current” of the magnetic circuit. Magnetic flux is a measure of the amount of magnetic field passing through a given surface (such as a conducting coil). The unit of flux is the weber (abbreviated as Wb) and the mathematical symbol for the number of webers of flux is Φ.

11.2.2. Magnetomotive force

Magnetomotive force (MMF) is the flux-producing ability of an electric current in a magnetic circuit. MMF is analogous to the electromotive force (EMF) of the electric circuit. The MMF developed is proportional to the current and to the number of turns in the coil in which current is flowing.

The expression for the MMF is:

$F = N I$ $F = N I$

(11.1)

where N is the number of turns and I is the coil current in amperes and has the unit of ampere-turn.

11.2.3. Magnetic flux density, magnetic field strength, and flux linkage

The flux density is defined as the concentration of uniformly distributed flux per unit area of the cross-section through which it acts. Flux density is denoted as B and it is given by:

$B = \frac{Φ}{A}$ $B = \frac{Φ}{A}$

(11.2)

where Φ is the flux in webers and A is the cross-sectional area in square meters. The unit of the flux density is Wb/m². This is also called as tesla (T).

The magnetic field strength is defined as the force that creates the flow of lines of flux and is denoted as H, expressed in units of oersteds. H is a magnetic field strength gradient along a magnetic path roughly equivalent to the voltage drop around an electric circuit loop. It is measured in ampere-turns per meter:

$H = \frac{4 π N I}{l}$ $H = \frac{4 π N I}{l}$

(11.3)

where N is the number of turns in the winding, I is the instantaneous current flowing through the wire, and l is the mean magnetic length of the core. Magnetic field strength is directly proportional to the winding current and the number of turns in the winding, and is inversely proportional to the magnetic length over which the flux must travel. This also indicates that the value of H is independent of the material used within the core.

The relation between B and H is given by:

$B = μ H$ $B = μ H$

(11.4)

The flux passing through the surface bounded by a coil is said to link the coil, and for a coil of conductors, the flux passing through the coil is the product of the number of turns N and the flux passing through a single turn, Φ. This product is called the magnetic flux linkage of the coil, λ.

$Flux linkage (λ) = Φ N Weber-turns$ $Flux linkage (λ) = Φ N Weber-turns$

(11.5)

11.2.4. Reluctance and permeance

Reluctance is defined as the measure of how difficult it is to develop flux from the MMF in a given magnetic circuit. This is magnetic analogy to the electrical resistance. The symbol for reluctance is given by:

$R = \frac{F}{φ}$ $R = \frac{F}{φ}$

(11.6)

where

R

$R$

is reluctance in ampere-turns per weber (A-t/Wb),

F

$F$

is MMF in ampere-turns, and Φ is flux in webers.

Materials referred to as “magnetic” have relatively low reluctance when used in a magnetic circuit. In other words, with these materials in the magnetic circuit, a smaller coil with less current is needed to provide a given number of webers of flux than would be the case if the path were through air or some magnetic material like copper, glass, or plastic.

Permeance is the reciprocal of reluctance, which is a measure of the quantity of flux for a given number of ampere-turns in a magnetic circuit. Thus, the permeance is given by:

$P = \frac{1}{R}$ $P = \frac{1}{R}$

(11.7)

where P is permeance and

R

$R$

is reluctance. The unit for the permeance is webers per ampere-turn. From Equations (11.6) and (11.7):

$P = \frac{1}{R} = \frac{1}{(F / φ)}$ $P = \frac{1}{R} = \frac{1}{(F / φ)}$

(11.8)

$P = \frac{φ}{F}$ $P = \frac{φ}{F}$

(11.9)

The magnetic analogy of permeance is electrical conductance. Permeance, which increases as reluctance decreases, is an expression of ease with which the flux is developed in a given magnetic circuit for a given MMF. Thus, a magnetic circuit in which the path of the lines of force is almost through iron, the permeance of the circuit is relatively high compared with the permeance when air, plastic, or other nonmagnetic material is substituted for iron.

The space or material inside a coil is called the core. A coil that is wound around only a thin hollow tube of nonmagnetic material as shown in Figure 11.2 is simply known as an “air core” and such a coil generates only a relatively small amount of magnetic flux. To develop as much flux as possible it is preferred to use a core of magnetic material.

For any specified magnetic material, shape, and dimension reluctance and permeance are the important characters that indicate the magnetic properties of a magnetic circuit. These characters correspond to the resistance and conductance of an electric circuit.

11.3. Equivalent circuit of a core excited by an AC MMF

When a current passes through a wire, a magnetic field is set up around the wire. If the wire is wound on a rod, its magnetic field is greatly intensified. The magnetic circuit [2] is the space in which the flux travels around the coil. The magnitude of the flux is determined by the product of the current, I, and the number of turns, N, in the coil. The force, NI, required to create the flux is MMF. The relationship between flux density, B, and magnetizing force, H, for an air core coil is linear (B = μH and μ is unity for air core coil). If the coil is excited with an AC source, as shown in Figure 11.3, the relationship between B and H would have the characteristics shown in Figure 11.4. The linearity of the relationship between B and H represents the main advantage of air core coils. Since the relationship is linear, increasing H increases B, and therefore the flux in the coil, and, in this way, very large fields can be produced with large currents. There is obviously a practical limit to this, which depends on the maximum allowable current in the conductor and the resulting rise. To achieve an improvement over the air coil, as shown in Figure 11.4, the coil can be wound over a magnetic core. In addition to its high permeability, the advantages of the magnetic core over the air core are that the magnetic path length is well defined, and the flux is essentially confined to the core, except in the immediate vicinity of the winding. There is a limit as to how much magnetic flux can be generated in a magnetic material before the magnetic core goes into saturation, and the coil reverts back to an air core, as shown in Figure 11.4. In transformer design, it is useful to use flux density.

Figure 11.3 Air core coil driven from an AC source.

Figure 11.4 Relationship between B and H with AC excitation.

11.4. Principle of operation of a transformer

Transformers [3] are devices that transfer energy from one circuit to another by means of a common magnetic field. An ideal transformer in its simplest form is shown in Figure 11.5. When an AC voltage is applied to the primary winding, time-varying current flows in the primary winding and causes an AC magnetic flux to appear in the transformer core. This flux links with the secondary winding due to the mutual magnetic coupling, and induces a voltage in the secondary winding (Faraday’s law). Depending on the ratio of turns in the primary and secondary winding, the root mean square (RMS) secondary voltage can be greater or less than the RMS primary voltage.

This transformer has two air coils that share a common flux. The flux diverges from the ends of the primary coil in all directions. It is not concentrated or confined. The primary is connected to the source and carries the current that establishes a magnetic field. The other coil is open circuited. It is inferred that the flux lines are not common to both coils. The difference between the two is the leakage flux; that is, leakage flux is the portion of the flux that does not link both coils.

11.4.1. Air core transformer

Some small transformers for low-power applications are constructed with air between the two coils. Such transformers are inefficient because the percentage of the flux from the first coil that links the second coil is small. The voltage induced in the second coil is determined as follows:

$E = N \frac{d ϕ}{d t}$ $E = N \frac{d ϕ}{d t}$

(11.10)

where N is the number of turns in the coil,

\frac{d ϕ}{d t}

$\frac{d ϕ}{d t}$

is the time rate of change of flux linking the coil, and Φ is the flux in lines.

At a time when the applied voltage to the coil is E and the flux linking the coils is Φ, the instantaneous voltage of the supply are:

$e = \sqrt{2} E \cos ω t = N \frac{d ϕ}{d t}$ $e = \sqrt{2} E \cos ω t = N \frac{d ϕ}{d t}$

(11.11)

$ϕ = \frac{\sqrt{2} E}{2 π f N}$ $ϕ = \frac{\sqrt{2} E}{2 π f N}$

(11.12)

Since the amount of flux Φ linking the second coil is a small percentage of the flux from the first coil, the voltage induced into the second coil is small. The number of turns can be increased to increase the voltage output, but this will increase costs. The need then is to increase the amount of flux from the first coil that links the second coil.

11.4.2. Iron or steel core transformer

The ability of iron or steel to carry magnetic flux is much greater than air. This ability to carry flux is called permeability. Modern electrical steels have permeability in the order of 1500 compared with 1.0 for air. This means that the ability of a steel core to carry magnetic flux is 1500 times that of air. Then, the equation for the flux in the steel core is:

$ϕ = \frac{μ_{0} μ_{r} N A I}{d}$ $ϕ = \frac{μ_{0} μ_{r} N A I}{d}$

(11.13)

where μ_r is the relative permeability of steel ≈1500.

Since the permeability of the steel is very high compared with air, all of the flux can be considered as flowing in the steel and is essentially of equal magnitude in all parts of the core. The equation for the flux in the core can be written as follows:

$ϕ = \frac{0.225 E}{f N}$ $ϕ = \frac{0.225 E}{f N}$

(11.14)

where E is the applied alternating voltage, f is the frequency in hertz, N is the number of turns in the winding.

For analyzing an ideal transformer, the following assumptions are made:

• The resistances of the windings can be neglected.

• All the magnetic flux is linked by all the turns of the coil and there is no leakage of flux.

• The reluctance of the core is negligible.

The equations for sinusoidal voltage for the ideal transformer are as follows.

The primary winding of turns N_p is supplied by a sinusoidal voltage v_p:

$v_{p} = V_{pm} cos (ω t)$ $v_{p} = V_{pm} cos (ω t)$

(11.15)

From Faraday’s law, the voltage across the primary winding terminals can be written as:

$v_{p} = N_{p} \frac{d ϕ}{d t}$ $v_{p} = N_{p} \frac{d ϕ}{d t}$

(11.16)

Therefore:

$v_{p} = V_{p m} \cos (ω t) = N_{p} \frac{d ϕ}{d t}$ $v_{p} = V_{p m} \cos (ω t) = N_{p} \frac{d ϕ}{d t}$

(11.17)

Rearranging and integrating, the equation for common flux can be written as:

$ϕ = \frac{V_{p m}}{N_{p} ω} sin (ω t)$ $ϕ = \frac{V_{p m}}{N_{p} ω} sin (ω t)$

(11.18)

This common flux passes through both the windings.

11.5. Voltage, current, and impedance transformations

11.5.1. Voltage relationship

This common flux flows through the transformer core, links with the secondary winding, and induces voltage across the secondary winding according to Faraday’s law. The primary and secondary voltage relationships are specified by:

$v_{p} = N_{p} \frac{d ϕ}{d t} and v_{p} = N_{s} \frac{d ϕ}{d t}$ $v_{p} = N_{p} \frac{d ϕ}{d t} and v_{p} = N_{s} \frac{d ϕ}{d t}$

(11.19)

The polarities are defined by Lenz’s law. From the previous relationship:

$\frac{v_{p}}{v_{s}} = \frac{N_{p}}{N_{s}} = k (transformation ratio)$ $\frac{v_{p}}{v_{s}} = \frac{N_{p}}{N_{s}} = k (transformation ratio)$

(11.20)

The turns ratio or transformation ratio determines the amount of voltage transformed. If k = 1, it is an isolation transformer, if k > 1, it is a step-up transformer, and if k < 1, it is a step-down transformer.

11.5.2. Current relationship

If a load is connected to the secondary side, current passes through the secondary as the circuit is complete. The MMF corresponding to the current flowing in the secondary side is given by N_si_s. The input coil is forced to generate an MMF to oppose this MMF and therefore the resultant MMF is

F

$F$

= N_pi_p – N_si_s and is related to flux and reluctance. As the reluctance is zero for an ideal transformer,

F

$F$

= N_pi_p – N_si_s = 0.

Therefore:

$\frac{i_{s}}{i_{p}} = \frac{N_{p}}{N_{s}} = k (transformation ratio)$ $\frac{i_{s}}{i_{p}} = \frac{N_{p}}{N_{s}} = k (transformation ratio)$

(11.21)

11.5.3. Power in an ideal transformer

The power delivered to the load by the secondary winding is p_s = v_si_s and using the voltage and current relationship with the primary winding, power in the secondary is:

$1 / k v_{p} \times k i_{p} = v_{p} i_{p} .$ $1 / k v_{p} \times k i_{p} = v_{p} i_{p} .$

(11.22)

Therefore, power between primary and secondary is equal.

11.5.4. Impedance in an ideal transformer

Considering the load impedance Z_L connected across the secondary winding, the impedance across the secondary circuit is derived from the voltage and current flowing through the secondary circuit, Z_L = V_s/I_s.

Substituting for V_s and I_s:

$Z_{L} = \frac{(N_{s} / N_{p}) V_{p}}{(N_{p} / N_{s}) I_{p}} = {(\frac{N_{s}}{N_{p}})}^{2} \frac{v_{p}}{I_{p}} = k^{2} Z_{L}$ $Z_{L} = \frac{(N_{s} / N_{p}) V_{p}}{(N_{p} / N_{s}) I_{p}} = {(\frac{N_{s}}{N_{p}})}^{2} \frac{v_{p}}{I_{p}} = k^{2} Z_{L}$

(11.23)

11.6. Nonideal transformer and its equivalent circuits

An actual transformer differs from an ideal transformer in the following contexts:

• Copper losses are present in both the primary and secondary windings.

• Not all the flux produced by the primary winding links the secondary winding, and vice versa. This gives rise to some leakage of flux.

• The core requires a finite amount of MMF for its magnetization.

• Hysteresis and eddy current losses cause power loss in the transformer core.

The equivalent circuit of an ideal transformer can be modified to include these effects:

• Resistances R_p and R_s can be added on both the primary and secondary side to represent the actual winding resistances.

• The effect of leakage flux can be included by adding two inductances, L_p and L_s, respectively, in the primary and secondary winding circuits.

• Nonzero reluctance value is included by adding a magnetizing inductance, L_m. The corresponding reactance of the iron core is X_m (=2πfL_m).

• To account for the hysteresis and eddy currents, which cause iron losses in the core, a resistance R_c is added in the transformer equivalent circuit.

A simple two-winding transformer is shown in the schematic diagram of Figure 11.6. A primary winding of N_p turns is on one side of a ferromagnetic core loop, and a similar coil having N_s turns is on the other. Both coils are wound in the same direction with the starts of the coils at H₁ and X₁, respectively. When an alternating voltage V_p is applied from H₂ to H₁, an alternating magnetizing flux φ_m flows around the closed core loop. A secondary voltage V_s = V_p × N_s/N_p is induced in the secondary winding and appears from X₂ to X₁ and very nearly in phase with V_p. With no load connected to X₁–X₂, I_p consists of only a small current called the magnetizing current. When load is applied, current I_s flows out of terminal X₁ and results in a current I_p = I_s × N_s/N_p flowing into H₁ in addition to magnetizing current. The ampere-turns of flux due to current I_p × N_p cancel the ampere-turns of flux due to current I_s × N_s, so only the magnetizing flux exists in the core for all the time the transformer is operating normally.

Figure 11.6 Schematic of a two-winding transformer.

Figure 11.7 shows a complete equivalent circuit of the transformer. An ideal transformer is inserted to represent the current- and voltage-transformation ratios. A parallel resistance and inductance representing the magnetizing impedance are placed across the primary of the ideal transformer. Resistance and inductance of the two windings are placed in the H₁ and X₁ legs, respectively.

Figure 11.7 Complete transformer equivalent circuit.

11.7. Tests on transformers

Transformers are tested (IS:2026 Part I: 1977) before they reach consumers. The tests are classified as type tests, routine tests, and special tests. The tests are conducted to measure:

1. Resistance of windings

2. Voltage ratio

3. Voltage vector relationship

4. Short-circuit impedance

5. Load loss

6. Insulation resistance

7. Dielectric test

8. Temperature rise

9. Zero sequence impedance

10. Acoustic noise level

11. Harmonics

11.7.1. Design tests

Tests performed by manufacturers on prototypes or production samples are referred to as “design tests [4].” These tests may include sound-level tests, temperature-rise tests, and short-circuit current withstand tests. The purpose of a design test is to establish a design limit that can be applied by calculation to every transformer built. In particular, short-circuit tests are destructive and may result in some invisible damage to the sample, even if the test is passed successfully. The IEEE standard calls for a transformer to sustain six tests, four with symmetrical fault currents and two with asymmetrical currents. One of the symmetrical shots is to be of long duration, up to 2 s, depending on the impedance for lower ratings. The remaining five shots are to be 0.25 s in duration. The long-shot duration for distribution transformers 750 kVA and above is 1 s. The design passes the short-circuit test if the transformer sustains no internal or external damage (as determined by visual inspection) and minimal impedance changes. The tested transformer also has to pass production dielectric tests and experience no more than a 25% change in exciting current.

11.7.2. Production tests

Production tests are given to and passed by each transformer made. Tests to determine ratio, polarity or phase displacement, iron loss, load loss, and impedance are done to verify that the nameplate information is correct. Dielectric tests specified by industry standards are intended to prove that the transformer is capable of sustaining unusual but anticipated electrical stresses that may be encountered in service. Production dielectric tests may include applied voltage, induced voltage, and impulse tests.

11.7.2.1. Applied-voltage test

Standards require application of a voltage of (very roughly) twice the normal line-to-line voltage to each entire winding for 1 min. This checks the ability of one phase to withstand voltage it may encounter when another phase is faulted to ground and transients are reflected and doubled.

11.7.2.2. Induced-voltage test

The original applied-voltage test is now supplemented with an induced-voltage test. Voltage at higher frequency (usually 400 Hz) is applied at twice the rated value of the winding. This induces the higher voltage in each winding simultaneously without saturating the core. If a winding is permanently grounded on one end, the applied-voltage test cannot be performed. In this case, many IEEE product standards specify that the induced primary test voltage be raised to 1000 plus 3.46 times the rated winding voltage.

11.7.2.3. Impulse test

Distribution lines are routinely disturbed by voltage surges caused by lightning strikes and switching transients. A standard 1.2 × 50 μs impulse wave with a peak equal to the basic impulse insulation level of the primary system (60–150 kV) is applied to verify that each transformer will withstand these surges when in service.

11.7.3. Performance test

In order to determine the losses, and to calculate the efficiency and voltage regulation at different loads, open circuit, short circuit, load tests are conducted.

11.7.3.1. Open-circuit test

To carry out an open-circuit test, the low voltage (LV) side of the transformer, where rated voltage at rated frequency is applied, and the high voltage (HV) side are left opened as shown in Figure 11.8. The voltmeter, ammeter, and wattmeter readings are taken as V₀, I₀, and W₀, respectively. During this test, rated flux is produced in the core and the current drawn is the no-load current, which is quite small, about 2–5% of the rated current. Therefore, a low range ammeter and wattmeter current coil should be selected. Strictly speaking, the wattmeter will record the core loss as well as the LV winding copper loss. But the winding copper loss is very small compared to the core loss as the flux in the core is rated. In fact this approximation is built-in in the approximate equivalent circuit of the transformer, referred to as the primary side, which is the LV side in this case. The approximate equivalent circuit and the corresponding phasor diagrams are shown in Figures 11.9 and 11.10 under no-load condition.

Figure 11.8 Circuit diagram for an open-circuit test.

Figure 11.9 Equivalent circuit under no-load condition.

Figure 11.10 Phasor diagram under no-load condition.

The resistance of the primary winding is R₀. Therefore, the copper loss in the primary winding at no-load is

I_{0}^{2} R_{0}

$I_{0}^{2} R_{0}$

$Hence, the iron losses of the transformer = W - I_{0}^{2} R_{0}$ $Hence, the iron losses of the transformer = W - I_{0}^{2} R_{0}$

(11.24)

The power factor of the transformer at no-load is:

cos θ₀ = Resistance/impedance = (R₀I₀)/V₁

where V₁ is the supply voltage, indicated by the voltmeter. Alternatively, power factor:

$\cos θ_{0} = \frac{Wattmeter reading}{Voltmeter reading \times Ammeter reading} = \frac{W}{V_{1} I_{0}}$ $\cos θ_{0} = \frac{Wattmeter reading}{Voltmeter reading \times Ammeter reading} = \frac{W}{V_{1} I_{0}}$

(11.25)

From the values of the power factor cos θ₀, the magnetizing component (I_μ) and wattless component (I_w) of the no-load current (I₀) can be calculated as follows:

$I_{μ} = I_{0} \sin θ_{0}$ $I_{μ} = I_{0} \sin θ_{0}$

(11.26)

and

$I_{w} = I_{0} \cos θ_{0}$ $I_{w} = I_{0} \cos θ_{0}$

(11.27)

11.7.3.2. Short-circuit test

The connection diagram for a short-circuit test on a transformer is shown in Figure 11.11. A voltmeter, wattmeter, and ammeter are connected to the HV side of the transformer as shown. The voltage at rated frequency is applied to the HV side with the help of a variac of variable ratio autotransformer. Usually the LV side of the transformer is short circuited. Now with the help of the variac applied voltage is slowly increased (usually 5–10% of the normal primary voltage) until the ammeter gives a reading equal to the rated current of the HV side. After reaching the rated current of the HV side, the readings of all three instruments (voltmeter, ammeter, and wattmeter) are recorded. The ammeter reading gives the primary equivalent of a full load current, I_L.

Figure 11.11 Circuit diagram for a short-circuit test.

Since the impressed voltage (equal to only a few percent of the rated value) is merely that required to overcome the total impedance of the windings, the mutual flux produced in the core is only a small percentage of its normal value (because the flux is proportional to the voltage). Consequently the iron core losses are very small. The wattmeter reading W equals the total full load copper losses in both the primary and secondary windings of the whole transformer. If V_sc is the voltage required for circulating the rated load current in the short-circuited transformer, then equivalent impedance:

$Z_{01} (or Z_{02}) = V_{sc} / I_{1}$ $Z_{01} (or Z_{02}) = V_{sc} / I_{1}$

(11.28)

$W = I_{1}^{2} R_{01}$ $W = I_{1}^{2} R_{01}$

(11.29)

Resistance of the transformer:

$R_{01} = W / I_{1}^{2} = (R_{1} + R_{2} / k^{2})$ $R_{01} = W / I_{1}^{2} = (R_{1} + R_{2} / k^{2})$

(11.30)

And leakage reactance:

$X_{01} (or X_{02}) = X_{1} + X_{2} / k_{2} = \sqrt{Z_{01}^{2} - R_{01}^{2}}$ $X_{01} (or X_{02}) = X_{1} + X_{2} / k_{2} = \sqrt{Z_{01}^{2} - R_{01}^{2}}$

(11.31)

From the knowledge gained from Z₀₁ (or Z₀₂), the total voltage drop in the transformer referred to as primary (or secondary) can be computed and hence the voltage regulation of the transformer can be calculated.

11.7.3.3. Load test

To determine the total losses in a transformer, a load test is performed. The load test gives information about the rated load of the transformer and temperature rise. Efficiency and regulation can also be determined from the load test. Nominal voltage is applied across the primary and rated current is drawn from the secondary. Load is applied continuously observing the steady-state temperature rise. Based on the different insulation and cooling methods incorporated in the transformer, different loading levels are permitted for the same transformer.

Efficiency of a transformer:

1. Commercial efficiency of a transformer at a particular load and power factor is defined as the ratio of output power to input power. Thus, efficiency:

$\begin{matrix} η = \frac{Output power}{Input power} = \frac{Output power}{Output power + Losses} \\ = \frac{Output power}{Output power + (Iron + Copper) Losses} = \frac{Input power - Losses}{Input power} \\ = 1 - \frac{Losses}{Input power} \end{matrix}$ $\begin{matrix} η = \frac{Output power}{Input power} = \frac{Output power}{Output power + Losses} \\ = \frac{Output power}{Output power + (Iron + Copper) Losses} = \frac{Input power - Losses}{Input power} \\ = 1 - \frac{Losses}{Input power} \end{matrix}$

$Primary input = V_{1} I_{1} \cos θ_{0}$ $Primary input = V_{1} I_{1} \cos θ_{0}$

(11.32)

$Primary copper loss = I_{1}^{2} R_{1}$ $Primary copper loss = I_{1}^{2} R_{1}$

(11.33)

$Iron losses = (Hysteresis + Eddy current) Losses = W_{h} + W_{e} = W_{i}$ $Iron losses = (Hysteresis + Eddy current) Losses = W_{h} + W_{e} = W_{i}$

(11.34)

$∴ Efficiency η = \frac{V_{1} I_{1} \cos θ_{0} - I_{1}^{2} R_{1} - W_{i}}{V_{1} I_{1} \cos θ_{0}} = 1 - \frac{I_{1}^{2} R_{1} - W_{i}}{V_{1} \cos θ_{0}} - \frac{W_{i}}{V_{1} I_{1} \cos θ_{0}}$ $∴ Efficiency η = \frac{V_{1} I_{1} \cos θ_{0} - I_{1}^{2} R_{1} - W_{i}}{V_{1} I_{1} \cos θ_{0}} = 1 - \frac{I_{1}^{2} R_{1} - W_{i}}{V_{1} \cos θ_{0}} - \frac{W_{i}}{V_{1} I_{1} \cos θ_{0}}$

(11.35)

Differentiating with respect to I₁:

$\frac{d η}{d I_{1}} = 0 = \frac{R_{1}}{V_{1} \cos θ_{0}} + \frac{W_{i}}{V_{1} I_{1}^{2} \cos θ_{0}}$ $\frac{d η}{d I_{1}} = 0 = \frac{R_{1}}{V_{1} \cos θ_{0}} + \frac{W_{i}}{V_{1} I_{1}^{2} \cos θ_{0}}$

(11.36)

For maximum η,

\frac{d η}{d I_{1}} = 0

$\frac{d η}{d I_{1}} = 0$

or copper loss = iron loss.

Efficiency of a transformer depends on both load and power factor. Input and output power depends on power factor of load, so transformers are usually specified kVA rating only.

Load at maximum efficiency, let W_i and W_c be the full load iron loss and copper loss, respectively:

$∴ W_{c} \propto {(Full load kVA)}^{2}$ $∴ W_{c} \propto {(Full load kVA)}^{2}$

(11.37)

If x is the load, when the efficiency is maximum, then:

$W_{i} \propto x^{2} W_{c}$ $W_{i} \propto x^{2} W_{c}$

(11.38)

$∴ \frac{W_{c}}{W_{i}} = \frac{{(Full load kVA)}^{2}}{x^{2}}$ $∴ \frac{W_{c}}{W_{i}} = \frac{{(Full load kVA)}^{2}}{x^{2}}$

(11.39)

Load at maximum efficiency (x) is:

$x = (Full load kVA) {(\frac{Iron lossses W_{i}}{Full load copper losses W_{c}})}^{1 / 2}$ $x = (Full load kVA) {(\frac{Iron lossses W_{i}}{Full load copper losses W_{c}})}^{1 / 2}$

(11.40)

11.7.3.3.1. All day efficiency

Transformers are employed for energy distribution 24 h a day. In a day the secondary is loaded throughout 24 h contributing to iron losses. But only during peak loaded conditions are copper losses significant. The performance of a transformer is judged by its operational efficiency, called all day efficiency, based on the load cycle in 24 h.

$All day efficiency η_{all day} = \frac{Output in kWh}{Input in kWh}$ $All day efficiency η_{all day} = \frac{Output in kWh}{Input in kWh}$

(11.41)

11.7.3.3.2. Regulation

Whenever a transformer is loaded, terminal voltage across the secondary changes with the load variations, with the primary voltage supply held constant. The change in secondary terminal voltage from no load to full load, expressed as a percentage of no-load voltage, is known as voltage regulation of a transformer:

$% Regulation = \frac{Secondary voltage at no load - Secondary voltage at full load}{Secondary voltage at no load} \times 100$ $% Regulation = \frac{Secondary voltage at no load - Secondary voltage at full load}{Secondary voltage at no load} \times 100$

(11.42)

11.8. Transformer polarity

The phase relationship of single-phase transformer voltages is described as “polarity.” The polarity of a transformer can be either additive or subtractive. These terms describe the voltage that may appear on adjacent terminals if the remaining terminals are jumpered together. Although the technical definition of polarity involves the relative position of primary and secondary bushings, the position of primary bushings is always the same according to standards. Therefore, when facing the secondary bushings of an additive transformer, the X₁ bushing is located to the right (of X₃), while for a subtractive transformer, X₁ is farthest to the left. To complicate this definition, a single-phase pad-mounted transformer built to IEEE standard Type 2 will always have the X₂ mid-tap bushing on the lowest right-hand side of the low-voltage slant pattern. Polarity has nothing to do with the internal construction of the transformer windings, only with the routing of leads to the bushings. Polarity only becomes important when transformers are being paralleled or banked. Single-phase polarity is illustrated in Figure 11.12.

11.9. Transformers in parallel

For supplying a load in excess of the rating of an existing transformer, two or more transformers may be connected in parallel with the existing transformer. It is usually economical to install another transformer in parallel instead of replacing the existing transformer by a single larger unit. The cost of a spare unit in the case of two parallel transformers (of equal rating) is also lower than that of a single large transformer. In addition, it is preferable to have a parallel transformer because of reliability. With this, at least half the load can be supplied with one transformer out of service. For parallel connection of transformers, primary windings of the transformers are connected to source bus-bars and secondary windings are connected to the load bus-bars. There are various conditions that must be fulfilled for the successful parallel operation of transformers. These are as follows:

1. The line voltage ratios of the transformers must be equal (on each tap): If the transformers connected in parallel have slightly different voltage ratios, then due to the inequality of induced EMFs in the secondary windings, a circulating current will flow in the loop formed by the secondary windings under the no-load condition, which may be much greater than the normal no-load current. The current will be quite high as the leakage impedance is low. When the secondary windings are loaded, this circulating current will tend to produce unequal loading on the two transformers, and it may not be possible to take the full load from this group of two parallel transformers (one of the transformers may become overloaded).

2. The transformers should have equal per-unit leakage impedances and the same ratio of equivalent leakage reactance to the equivalent resistance (X/R): If the ratings of both the transformers are equal, their per-unit leakage impedances should be equal in order to have equal loading of both the transformers. If the ratings are unequal, their per-unit leakage impedances based on their own ratings should be equal so that the currents carried by them will be proportional to their ratings. In other words, for unequal ratings, the numerical (ohmic) values of their impedances should be in inverse proportion to their ratings to have current in them in line with their ratings. A difference in the ratio of the reactance value to resistance value of the per-unit impedance results in a different phase angle of the currents carried by the two paralleled transformers; one transformer will be working with a higher power factor and the other with a lower power factor than that of the combined output. Hence, the real power will not be proportionally shared by the transformers.

3. The transformers should have the same polarity: The transformers should be properly connected with regard to their polarity. If they are connected within correct polarities then the two EMFs, induced in the secondary windings that are in parallel, will act together in the local secondary circuit and produce a short circuit.

The previous three conditions are applicable to both single-phase as well as three-phase transformers. In addition to these three conditions, two more conditions are essential for the parallel operation of three-phase transformers:

4. The transformers should have the same phase sequence: The phase sequence of line voltages of both the transformers must be identical for parallel operation of three-phase transformers. If the phase sequence is incorrect, in every cycle each pair of phases will be short circuited.

5. The transformers should have the zero relative phase displacement between the secondary line voltages: The transformer windings can be connected in a variety of ways, which produce different magnitudes and phase displacements of the secondary voltage. All the transformer connections can be classified into distinct vector groups. Each vector group notation consists of an uppercase letter denoting HV connection, a second lowercase letter denoting LV connection, followed by a clock number representing LV winding’s phase displacement with respect to HV winding (at 12 o’clock). There are four groups into which all possible three-phase connections can be classified:

a. Group 1: Zero phase displacement (Yy0, Dd0, Dz0)

b. Group 2: 180° phase displacement (Yy6, Dd6, Dz6)

c. Group 3: −30° phase displacement (Yd1, Dy1, Yz1)

d. Group 4: +30° phase displacement (Yd11, Dy11, Yz11)

In the previously mentioned notations, letters y (or Y), d (or D), and z represent star, delta, and zigzag connections, respectively. In order to have zero relative phase displacement of secondary side line voltages, the transformers belonging to the same group can be paralleled. For example, two transformers with Yd1 and Dy1 connections can be paralleled. The transformers of groups 1 and 2 can only be paralleled with transformers of their own group. However, the transformers of groups 3 and 4 can be paralleled by reversing the phase sequence of one of them. For example, a transformer with Yd11 connection (group 4) can be paralleled with that having Dy1 connection (group 3) by reversing the phase sequence of both primary and secondary terminals of the Dy1 transformer.

11.10. Three-phase transformer connections

Transformer power levels range from low-power applications, such as consumer electronics power supplies, to very high power applications, such as power distribution systems. For higher power applications, three-phase transforms are commonly used. A three-phase transformer is constructed as a single unit with a bank of transformers, that is, three numbers of identical single-phase transformers connected in required form. In a single-phase transformer, only two coils, namely primary and secondary, are available whereas in a three-phase transformer there will be three numbers of primary coils and three numbers of secondary coils. The coils are connected in various methods as listed in Table 11.1 to obtain different voltage levels.

• An advantage of ∆–∆ connection is that if one of the transformers fails or is removed from the circuit, the remaining two can operate in the open ∆ or V connection. This way, the bank still delivers three-phase currents and voltages in their correct phase relationship. However, the capacity of the bank is reduced to 57.7% of its original value.

• In the Y–Y connection, only 57.7% of the line voltage is applied to each winding but full line current flows in each winding. The Y–Y connection is rarely used.

• The ∆–Y connection is used for stepping up voltages since the voltage is increased by the transformer ratio multiplied by 3.

Table 11.1

Various three-phase transformer connections

				Primary or secondary
S. no.	Primary configuration	Secondary configuration	Symbolic representation	Line voltage	Line current
1.	Delta (mesh)	Delta (mesh)	∆–∆	$V_{l} = n V_{l}$ $V_{l} = n V_{l}$	$I_{l} = \frac{I_{l}}{n}$ $I_{l} = \frac{I_{l}}{n}$
2.	Delta (mesh)	Star (wye)	∆–Y	$V_{l} = \sqrt{3} n V_{l}$ $V_{l} = \sqrt{3} n V_{l}$	$I_{l} = \frac{I_{l}}{\sqrt{3} n}$ $I_{l} = \frac{I_{l}}{\sqrt{3} n}$
3.	Star (wye)	Delta (mesh)	Y–∆	$V_{l} = \frac{n V_{l}}{\sqrt{3}}$ $V_{l} = \frac{n V_{l}}{\sqrt{3}}$	$I_{l} = \sqrt{3} \frac{I_{l}}{n}$ $I_{l} = \sqrt{3} \frac{I_{l}}{n}$
4.	Star (wye)	Star (wye)	Y–Y	$V_{l} = n V_{l}$ $V_{l} = n V_{l}$	$I_{l} = \frac{I_{l}}{n}$ $I_{l} = \frac{I_{l}}{n}$
5.	Interconnected star	Delta (mesh)	–∆
6.	Interconnected star	Star (wye)	–Y

11.11. Special transformer connection

An air core transformer is a special transformer, used in radio frequency circuits. As the name implies, its windings are wrapped around a nonmagnetic material in the form of a hollow tube. Though the degree of coupling (mutual inductance) is much less, ferromagnetic cores (eddy current loss, hysteresis, saturation, etc.) are completely eliminated. In high-frequency applications, the effects of iron losses are more problematic. An example of air core transformers is the tesla coil, which is a resonant, high-frequency, step-up transformer used to produce extremely high voltages.

A Scott-connected transformer is a type of circuit device used to convert a three-phase supply (3-ϕ, 120-degree phase rotation) into a two-phase (2-ϕ, 90-degree phase rotation) supply, or vice versa. The Scott connection evenly distributes a balanced load between the phases of the source.

11.12. Parallel operation of three-phase transformers

Ideal parallel operation between transformers occurs when (1) there are no circulating currents on open circuit, and (2) the load division between the transformers is proportional to their kVA ratings. These requirements necessitate that any two or more three-phase transformers, which are desired to be operated in parallel, should possess:

1. The same no-load ratio of transformation

2. The same percentage impedance

3. The same resistance to reactance ratio

4. The same polarity

5. The same phase rotation

6. The same inherent phase-angle displacement between primary and secondary terminals

The previously mentioned conditions are characteristic of all three-phase transformers whether two winding or three winding. With three-winding transformers, however, the following additional requirement must also be satisfied before the transformers can be designed suitable for parallel operation:

7. The same power ratio between the corresponding windings

Table 11.2 gives the possible combinations of transformers that can be operated in parallel.

Table 11.2

Parallel operation of three-phase transformers

S. no.	Vector group of transformers that will operate in parallel		Vector group of transformers that will NOT operate in parallel
1.	Transformer 1	Transformer 2	Transformer 1	Transformer 2
2.	∆∆	∆∆ or Yy	∆∆	∆y
3.	Yy	Yy or ∆∆	∆y	∆∆
4.	∆Y	∆y or Y∆	Y∆	Yy
5.	Y∆	Y∆ or ∆y	Yy	Y∆

The methods to check for synchronization of transformers are done using a synchroscope or synchronizing relay. The advantages of parallel operation of transformers are:

1. Maximize electrical system efficiency

2. Maximize electrical system availability

3. Maximize power system reliability

But during the parallel operation of three-phase transformers, the magnitude of short-circuit currents, risk of circulating currents, bus rating, and reduction in transformer impedance make the circuit complex in providing protective mechanisms.

11.13. Autotransformers

In an autotransformer, the primary and secondary windings are linked together both electrically and magnetically. Therefore it is economical for the same VA rating as windings are reduced, but the disadvantage is that it does not have isolation between primary and secondary windings. The winding can be designed with multiple tapping points, to provide different voltage points along its secondary winding. The winding diagram and the number of windings in primary and secondary (N_p and N_s, respectively), current, and voltage across primary and secondary are shown in Figure 11.13.

Figure 11.13 Winding diagram of an autotransformer.

11.14. Three-winding transformers

In certain high rating transformers, one winding in addition to its primary and secondary winding is used, called a tertiary winding transformer. Because of this third winding, the transformer is called a three-winding transformer. The advantage of using a tertiary winding in a transformer is to meet one or more of the following requirements:

1. It reduces the unbalancing in the primary due to unbalancing in three-phase load.

2. It redistributes the flow of fault current.

3. Sometimes it is required to supply an auxiliary load at different voltage levels in addition to its main secondary load. This secondary load can be taken from the tertiary winding of the three-winding transformer.

4. As the tertiary winding is connected in delta formation in a three-winding transformer, it assists in limiting fault current in the event of a short circuit from line to neutral.

5. A star–star transformer comprising three single units or a single unit with five limb core offers high impedance to the flow of unbalanced load between the line and neutral.

11.15. Instrument transformers

Instrument transformers are used for transforming the magnitude of current (typically 1 A or 5 A at secondary) and voltage (typically 120 V at secondary) from one level to another. Also instrument transformers can be used as an isolation transformer for safety purposes. Various types of instrument transformers, namely current transformers, inductive voltage transformers, capacitive voltage transformers, combined current/voltage transformers, and station service voltage transformers, are designed to transform high current and high voltage levels down to low current and low voltage outputs in a known and accurate proportion for a specific application.

Potential transformers consist of two separate windings on a common magnetic steel core. One winding consists of fewer turns of heavier wire on the steel core and is called the secondary winding. The other winding consists of a relatively large number of turns of fine wire, wound on top of the secondary, and is called the primary winding.

Current transformers are constructed in various ways. One method is quite similar to that of the potential transformer in that there are two separate windings on a magnetic steel core. But it differs in that the primary winding consists of a few turns of heavy wire capable of carrying the full load current while the secondary winding consists of many turns of smaller wire with a current carrying capacity of between 5 A and 20 A, dependent on the design. This is called the wound type due to its wound primary coil. Another very common type of construction is the so-called “window,” “through,” or donut type current transformer in which the core has an opening through which the conductor carrying the primary load current is passed. This primary conductor constitutes the primary winding of the current transformer (one pass through the “window” represents a one turn primary), and must be large enough in cross-section to carry the maximum current of the load.

The operation of instrument transformers differs from power transformers. The secondary winding of an instrument transformer has a very small impedance called burden, such that the instrument transformer operates under short-circuit conditions. The burden (expressed in ohms) across the secondary of an instrument transformer is also defined as the ratio of secondary voltage to secondary current. This helps to determine the volt-ampere loading of the instrument transformer.

The major applications of instrument transformers include:

1. Revenue metering for electric utilities, independent power producers, or industrial users

2. Protective relaying for use with switchgear to monitor system current and voltage levels

3. High accuracy wide current range use for independent power facilities

4. Station service power needs within substations or for power needs at remote sites

11.16. Third harmonics in transformers

In addition to the operation of transformers on the sinusoidal supplies, harmonic behavior is important as the size and rating of the transformer increases. In recent times, transformers have been designed to operate at closed levels of saturation in order to reduce the weight and cost of the core used. Because of this and hysteresis, the transformer core behaves as a nonlinear component and generates harmonic currents. If a sinusoidal voltage is applied to the primary of a transformer, the flux wave will vary as a sinusoidal function of time, but the no-load current wave will be distorted because the hysteresis loop contains a pronounced third harmonic. Figure 11.14 represents the hysteresis loop taken to the maximum flux density and the manner in which the shape of the magnetizing current can be obtained and plotted. In Figure 11.14, at any instant of the flux density wave the ampere-turns required to establish are read out and plotted by traversing the hysteresis loop.

Figure 11.14 (a) Hysteresis loop and (b) magnetizing current.

To produce a flux density of NP requires 0N ampere-turns per meter:

$N^{'} P^{'} = NP$ $N^{'} P^{'} = NP$

The ampere-turns per meter are plotted as N'A in Figure 11.14b. To produce a flux density MR requires 0M ampere-turns per meter:

$\begin{array}{l} 0 M = M^{'} B \\ MR = M^{'} R^{'} \end{array}$ $\begin{array}{l} 0 M = M^{'} B \\ MR = M^{'} R^{'} \end{array}$

In brief, the various abscissas of Figure 11.14a are plotted as ordinates to determine the shape of the current wave on Figure 11.14b. This is continued until a sufficient number of points are obtained. The use of a suitable constant changes the wave BAX from ampere-turns per meter to amperes. Such a wave represents the magnetizing component and the hysteresis component of the no-load current. It reaches the maximum at the same time as the flux wave, but the two waves do not go through zero simultaneously. The sinusoidal flux density represents the sinusoidal voltage, whereas the plot of magnetizing current rises sharply, saturates quickly, and is thereby distorted. This magnetizing current can be analyzed with Fourier series. The harmonic components are obtained from this Fourier analysis. The harmonic spectrum of the magnetizing current waveform reveals a very high percentage of third harmonics. These harmonic currents produce harmonic fields in the core and harmonic voltages in the windings. A relatively small value of harmonic fields produces substantial magnitudes of harmonic voltages. For example, a 10% magnitude of third harmonic flux produces a 30% magnitude of third harmonic voltage. These effects become even more pronounced for higher order harmonics. The no-load current can be considered to be made up of approximately two sine components (loss component and magnetizing component) and the nonsinusoidal component with dominant third harmonics. The sum of sine and nonsine waves forms a distorted waveform.

In the case of single-phase transformers connected to form a three-phase star-connected bank, the fundamental voltages supplied will produce voltages in the individual transformers that contain third harmonics. These third harmonics for all the three transformers will be in time phase. Each voltage between the neutral of the primaries and the lines will contain both a fundamental component and a third harmonic component, and as a result the secondary voltage of each transformer will contain both fundamental and third harmonic components.

The effects of the harmonic currents are:

1. Additional copper losses

2. Increased core losses

3. Increased neutral current and overheating of neutral conductor

4. Increased EMI with communication networks

On the other hand, the harmonic voltages of the transformer cause:

1. Increased dielectric stress on insulation

2. Resonance between winding reactance and feeder capacitance

11.17. Transformers in a microgrid

With technological improvements, design, materials, etc. of power transformers, autotransformers, and instrument transformers, performance has increased. In renewable energy integrated microgrids, AC and direct current (DC) supply are available and hence solid-state transformers and smart transformers are being developed and marketed to cater for the features of microgrids and smart grids.

11.17.1. Solid-state transformers

The advantages and limitations of conventional transformers when power quality is a major concern are as follows:

Advantages:

1. Relatively economical

2. Highly reliable

3. Quite efficient

Limitations:

1. Sensitive to harmonics

2. Voltage drop under load

3. No protection from system disruptions and overloads

4. Environmental concerns regarding mineral oil

5. Poor performance under DC-offset load unbalances

6. No power factor improvement

The solid-state transformers (SSTs) are designed with different topologies based on its application:

1. AC to AC buck converter: the salient points are that,

a. Transformation of the voltage level directly without any isolation transformer

b. Switches must be capable of blocking full primary voltage during OFF state and conducting full secondary current during ON state

c. Difficult to control series-connected devices

d. Lack of magnetic isolation

e. Inability to correct load power factor

2. SST without a DC link:

a. Transformer weight and size reduced

b. Provides isolation

c. No power factor improvement is possible

3. SST with a DC link:

a. Reduced size due to a high-frequency transformer

b. Power factor improvement is possible

c. Multilevel converter topologies can be applied to achieve high voltage levels (e.g., 11 kV, 22 kV)

d. High cost and low efficiency

e. It is a three-stage topology: most popular now

SSTs that can be widely used in microgrid applications are relatively advantageous with respect to power quality.

Advantages:

1. An excellent utilization of distributed renewable energy resources and distributed energy storage devices

2. Power factor control

3. Fast isolation under fault conditions due to a controlled SST

4. Control of both AC and DC loads can be done using the SST scheme

5. Improved power quality

6. DC and alternative frequency AC service options

7. Integration with system monitoring and advanced distribution

8. Reduced weight and size

9. Elimination of hazardous liquid dielectrics

Limitations:

1. Multiple power conversion stages can lower the overall efficiency

2. DC-link capacitors are required

3. The transformer lifetime can be shorter due to storage devices

11.17.2. The smart transformers

The smart transformer is a smart device for integrating with the distribution grid, solar-wind renewable energy storage, and electric vehicles. The components of a smart transformer include power conversion system and built-in STATCOM functions. The smart transformer can be fully controlled through Internet or wireless communication systems. It uses high-voltage semiconductor switches based on an AC/DC rectifier, DC/DC converter, high-voltage and high-frequency transformer, DC/AC inverter, and their switching control circuitry. This is expected to be a critical component in the development of smart grids.

11.18. Summary

In this chapter, the fundamentals of magnetic circuits and basic concepts of power transformers, types, and testing methods were discussed in detail. The development of equivalent circuit efficiency and regulation calculation of the transformer was elaborated.

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Table of Contents for 11: Magnetic circuits and power transformers

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Abstract

Keywords

11.1. Introduction

11.2. Magnetic circuits

11.2.1. Magnetic field and magnetic flux

11.2.2. Magnetomotive force

11.2.3. Magnetic flux density, magnetic field strength, and flux linkage

11.2.4. Reluctance and permeance

11.3. Equivalent circuit of a core excited by an AC MMF

11.4. Principle of operation of a transformer

11.4.1. Air core transformer

11.4.2. Iron or steel core transformer

11.5. Voltage, current, and impedance transformations

11.5.1. Voltage relationship

11.5.2. Current relationship

11.5.3. Power in an ideal transformer

11.5.4. Impedance in an ideal transformer

11.6. Nonideal transformer and its equivalent circuits

11.7. Tests on transformers

11.7.1. Design tests

11.7.2. Production tests

11.7.2.1. Applied-voltage test

11.7.2.2. Induced-voltage test

11.7.2.3. Impulse test

11.7.3. Performance test

11.7.3.1. Open-circuit test

11.7.3.2. Short-circuit test

11.7.3.3. Load test

11.7.3.3.1. All day efficiency

11.7.3.3.2. Regulation

11.8. Transformer polarity

11.9. Transformers in parallel

11.10. Three-phase transformer connections

11.11. Special transformer connection

11.12. Parallel operation of three-phase transformers

11.13. Autotransformers

11.14. Three-winding transformers

11.15. Instrument transformers

11.16. Third harmonics in transformers

11.17. Transformers in a microgrid

11.17.1. Solid-state transformers

11.17.2. The smart transformers

11.18. Summary

Table of Contents for
11: Magnetic circuits and power transformers