eee1 slides week15

Fundamentals of Electrical and Electronics

EngineeringPaul Leonard Atchong C. Hilario, Ph.D.

EEEI 526 [email protected]

mailto:[email protected]

Transformers

• A device based on magnetic coupling

• Used in power systems and communications

• Two coils wrapped around a single core

Transformers9.11 The Ideal Transformer

from which, for k = 1,

II h M

u VUL2

L2 (9.85)

Equation 9.85 is equivalent to

I , ^ i - l2N2. (9.86) -4 Current relationship for an ideal transformer

Figure 9.42 shows the graphic symbol for an ideal transformer. The vertical lines in the symbol represent the layers of magnetic material from which ferromagnetic cores are often made. Thus, the symbol reminds us that coils wound on a ferromagnetic core behave very much like an ideal transformer.

There are several reasons for this. The ferromagnetic material creates a space with high permeance. Thus most of the magnetic flux is trapped inside the core material, establishing tight magnetic coupling between coils that share the same core. High permeance also means high self-inductance, because L = N2V. Finally, ferromagnetically coupled coils efficiently transfer power from one coil to the other. Efficiencies in excess of 95% are common, so neglecting losses is not a crippling approximation for many applications.

Determining the Polarity of the Voltage and Current Ratios We now turn to the removal of the magnitude signs from Eqs. 9.76 and 9.77. Note that magnitude signs did not show up in the derivations of Eqs. 9.83 and 9.86. We did not need them there because we had established reference polarities for voltages and reference directions for currents. In addition, we knew the magnetic polarity dots of the two coupled coils.

The rules for assigning the proper algebraic sign to Eqs. 9.76 and 9.77 are as follows:

If the coil voltages V! and V2 are both positive or negative at the dot-marked terminal, use a plus sign in Eq. 9.76. Otherwise, use a nega-tive sign.

If the coil currents I] and I2 are both directed into or out of the dot-marked terminal, use a minus sign in Eq. 9.77. Otherwise, use a plus sign.

The four circuits shown in Fig. 9.43 illustrate these rules.

A", • N,

Ideal

Figure 9.42 • The graphic symbol for an ideal transformer.

A Dot convention for ideal transformers

+ • JV, N2\ • + \

V, I >

Ideal vi = v_2 /V, /V2"

Ni\i = ~N2l2

(a)

I2 V2

+ • yv, N2\ + f V, I II, V,

Ideal I •

W,I, = N2l2

(b)

+ • V, I,)

J

|N, N2\ • +

Ideal V, V2

/V, N2

A^I, = M,I2

(c)

l/V, N2\

Vt I, >'

I, V2

Ideal V] = _V_2

Ni N2

A7,I, = -N2l2

Figure 9.43 • Circuits that show the proper algebraic signs for relating the terminal voltages and currents of an ideal transformer.

https://upload.wikimedia.org/wikipedia/commons/thumb/6/64/Transformer3d_col3.svg/1280px-Transformer3d_col3.svg.png

https://upload.wikimedia.org/wikipedia/commons/thumb/6/64/Transformer3d_col3.svg/1280px-Transformer3d_col3.svg.png

Mutual inductance• Since the two coils are magnetically coupled,

changes in magnetic flux in one coil induces a potential difference in another coil

• The induced voltage is given by VM1 = M di2/dt, where M is the mutual inductance

• The sign will depend on how the coils were wound relative to each other

Dot convention• Choose a direction of the reference currents

• If the reference current enters the coil at the dotted terminal, the induced voltage at the other coil is positive at the dotted terminal

• If the reference current leaves the coil at the dotted terminal, the induced voltage at the other coil is negative at the dotted terminal

Example• i1 enters inductor 1 at the

dotted terminal

• voltage across inductor 2 is positive at the dotted terminal

• positive terminal is at the non-dotted terminal

• voltage measured across inductor 2 is negative

Figure 6.22 • The self- and mutually induced voltages appearing across the coils shown in Fig. 6.21.

6.4 Mutual Inductance 191

Now let's look at the sum of the voltages around each closed loop. In Eqs. 6.31 and 6.32, voltage rises in the reference direction of a current are negative:

di\ di2

dii du ioRo + L2-f - M - 1

dt dt 0.

(6.31)

(6.32)

The Procedure for Determining Dot Markings We shift now to two methods of determining dot markings. The first assumes that we know the physical arrangement of the two coils and the mode of each winding in a magnetically coupled circuit. The following six steps, applied here to Fig. 6.23, determine a set of dot markings:

a) Arbitrarily select one terminal—say, the D terminal—of one coil and mark it with a dot.

b) Assign a current into the dotted terminal and label it /D. c) Use the right-hand rule1 to determine the direction of the magnetic

field established by /D inside the coupled coils and label this field <j6D. d) Arbitrarily pick one terminal of the second coil—say, terminal A—and

assign a current into this terminal, showing the current as /A. e) Use the right-hand rule to determine the direction of the flux estab-

lished by /A inside the coupled coils and label this flux <£A. f) Compare the directions of the two fluxes <£D and <£A. If the fluxes

have the same reference direction, place a dot on the terminal of the second coil where the test current (/A) enters. (In Fig. 6.23, the fluxes <£D and c/>A have the same reference direction, and therefore a dot goes on terminal A.) If the fluxes have different reference direc-tions, place a dot on the terminal of the second coil where the test current leaves.

The relative polarities of magnetically coupled coils can also be deter-mined experimentally.This capability is important because in some situations, determining how the coils are wound on the core is impossible. One experi-mental method is to connect a dc voltage source, a resistor, a switch, and a dc voltmeter to the pair of coils, as shown in Fig. 6.24. The shaded box covering the coils implies that physical inspection of the coils is not possible. The resis-tor R limits the magnitude of the current supplied by the dc voltage source.

The coil terminal connected to the positive terminal of the dc source via the switch and limiting resistor receives a polarity mark, as shown in Fig. 6.24. When the switch is closed, the voltmeter deflection is observed. If the momentary deflection is upscale, the coil terminal connected to the positive terminal of the voltmeter receives the polarity mark. If the

eP2)

__ Arbitrarily dotted

D terminal (Stepl)

Figure 6.23 • A set of coils showing a method for determining a set of dot markings.

R

+ Switch

dc voltmeter

Figure 6.24 • An experimental setup for determining polarity marks.

See discussion of Faraday's law on page 193.

Example• i2 leaves inductor 2 at the

dotted terminal

• voltage across inductor 1 is negative at the dotted terminal

• positive terminal is at the dotted terminal

• voltage measured across inductor 1 is negative

Figure 6.22 • The self- and mutually induced voltages appearing across the coils shown in Fig. 6.21.

6.4 Mutual Inductance 191

Now let's look at the sum of the voltages around each closed loop. In Eqs. 6.31 and 6.32, voltage rises in the reference direction of a current are negative:

di\ di2

dii du ioRo + L2-f - M - 1

dt dt 0.

(6.31)

(6.32)

The Procedure for Determining Dot Markings We shift now to two methods of determining dot markings. The first assumes that we know the physical arrangement of the two coils and the mode of each winding in a magnetically coupled circuit. The following six steps, applied here to Fig. 6.23, determine a set of dot markings:

a) Arbitrarily select one terminal—say, the D terminal—of one coil and mark it with a dot.

b) Assign a current into the dotted terminal and label it /D. c) Use the right-hand rule1 to determine the direction of the magnetic

field established by /D inside the coupled coils and label this field <j6D. d) Arbitrarily pick one terminal of the second coil—say, terminal A—and

assign a current into this terminal, showing the current as /A. e) Use the right-hand rule to determine the direction of the flux estab-

lished by /A inside the coupled coils and label this flux <£A. f) Compare the directions of the two fluxes <£D and <£A. If the fluxes

have the same reference direction, place a dot on the terminal of the second coil where the test current (/A) enters. (In Fig. 6.23, the fluxes <£D and c/>A have the same reference direction, and therefore a dot goes on terminal A.) If the fluxes have different reference direc-tions, place a dot on the terminal of the second coil where the test current leaves.

The relative polarities of magnetically coupled coils can also be deter-mined experimentally.This capability is important because in some situations, determining how the coils are wound on the core is impossible. One experi-mental method is to connect a dc voltage source, a resistor, a switch, and a dc voltmeter to the pair of coils, as shown in Fig. 6.24. The shaded box covering the coils implies that physical inspection of the coils is not possible. The resis-tor R limits the magnitude of the current supplied by the dc voltage source.

The coil terminal connected to the positive terminal of the dc source via the switch and limiting resistor receives a polarity mark, as shown in Fig. 6.24. When the switch is closed, the voltmeter deflection is observed. If the momentary deflection is upscale, the coil terminal connected to the positive terminal of the voltmeter receives the polarity mark. If the

eP2)

__ Arbitrarily dotted

D terminal (Stepl)

Figure 6.23 • A set of coils showing a method for determining a set of dot markings.

R

+ Switch

dc voltmeter

Figure 6.24 • An experimental setup for determining polarity marks.

See discussion of Faraday's law on page 193.

Model of a transformer

• Mesh analysis on the a-b loop

• Mesh analysis on the c-d loop

9.10 The Transformer 335

behavior of the transformer is required in the analysis of both communication and power systems. In this section, we will discuss the sinusoidal steady-state behavior of the linear transformer, which is found primarily in communica-tion circuits. In Section 9.11, we will deal with the ideal transformer, which is used to model the ferromagnetic transformer found in power systems.

Before starting we make a useful observation. When analyzing circuits containing mutual inductance use the meshor loop-current method for writ-ing circuit equations. The node-voltage method is cumbersome to use when mutual inductance in involved. This is because the currents in the various coils cannot be written by inspection as functions of the node voltages.

The Analysis of a Linear Transformer Circuit A simple transformer is formed when two coils are wound on a single core to ensure magnetic coupling. Figure 9.38 shows the frequency-domain cir-cuit model of a system that uses a transformer to connect a load to a source. In discussing this circuit, we refer to the transformer winding con-nected to the source as the primary winding and the winding connected to the load as the secondary winding. Based on this terminology, the trans-former circuit parameters are

jR] = the resistance of the primary winding,

R2 = the resistance of the secondary winding,

Lj = the self-inductance of the primary winding,

L2 - the self-inductance of the secondary winding.

M = the mutual inductance.

The internal voltage of the sinusoidal source is Vv, and the internal impedance of the source is Zs. The impedance Z L represents the load con-nected to the secondary winding of the transformer. The phasor currents Ij and I2 represent the primary and secondary currents of the transformer, respectively.

Analysis of the circuit in Fig. 9.38 consists of finding I, and I2 as func-tions of the circuit parameters Vv, Zs, Rh Lt , L2, R2, M, ZL, and OJ. We are also interested in finding the impedance seen looking into the transformer from the terminals a,b.To find l\ and I2, we first write the two mesh-cur-rent equations that describe the circuit:

Source Transformer Load

Figure 9.38 A The frequency domain circuit model for a transformer used to connect a load to a source.

Vv = (Zv + Ri + jo)Ll)ll - ja>Ml2, (9.57)

0 = -;©Afl| + (R2 + joiL2 4- ZL)I2. (9.58)

To facilitate the algebraic manipulation of Eqs. 9.57 and 9.58, we let

Z,, = Z, + Rx + jtoLu (9.59)

Z22 = R2 + jcoL2 + ZJL, (9.60)

where Z(1 is the total self-impedance of the mesh containing the primary winding of the transformer, and Z22 is the total self-impedance of the mesh containing the secondary winding. Based on the notation introduced in Eqs. 9.59 and 9.60, the solutions for Ij and I2 from Eqs. 9.57 and 9.58 are

I. = Z]]Z22 4- o)~M

V ~> * 1 7 x'

(9.61)

JOjM

ZUZ22 + w'M io)M

-V = - 1, (9.62) '22















0 = -;©Afl| + (R2 + joiL2 4- ZL)I2. (9.58)


Z,, = Z, + Rx + jtoLu (9.59)

Z22 = R2 + jcoL2 + ZJL, (9.60)


I. = Z]]Z22 4- o)~M

V ~> * 1 7 x'

(9.61)

JOjM

ZUZ22 + w'M io)M

-V = - 1, (9.62) '22















0 = -;©Afl| + (R2 + joiL2 4- ZL)I2. (9.58)


Z,, = Z, + Rx + jtoLu (9.59)

Z22 = R2 + jcoL2 + ZJL, (9.60)


I. = Z]]Z22 4- o)~M

V ~> * 1 7 x'

(9.61)

JOjM

ZUZ22 + w'M io)M

-V = - 1, (9.62) '22

Model of a transformer• Let,

• The currents will be















0 = -;©Afl| + (R2 + joiL2 4- ZL)I2. (9.58)


Z,, = Z, + Rx + jtoLu (9.59)

Z22 = R2 + jcoL2 + ZJL, (9.60)


I. = Z]]Z22 4- o)~M

V ~> * 1 7 x'

(9.61)

JOjM

ZUZ22 + w'M io)M

-V = - 1, (9.62) '22















0 = -;©Afl| + (R2 + joiL2 4- ZL)I2. (9.58)


Z,, = Z, + Rx + jtoLu (9.59)

Z22 = R2 + jcoL2 + ZJL, (9.60)


I. = Z]]Z22 4- o)~M

V ~> * 1 7 x'

(9.61)

JOjM

ZUZ22 + w'M io)M

-V = - 1, (9.62) '22















0 = -;©Afl| + (R2 + joiL2 4- ZL)I2. (9.58)


Z,, = Z, + Rx + jtoLu (9.59)

Z22 = R2 + jcoL2 + ZJL, (9.60)


I. = Z]]Z22 4- o)~M

V ~> * 1 7 x'

(9.61)

JOjM

ZUZ22 + w'M io)M

-V = - 1, (9.62) '22

Model of a transformer• The impedance according to

the source is

• The impedance at points a-b is given by















0 = -;©Afl| + (R2 + joiL2 4- ZL)I2. (9.58)


Z,, = Z, + Rx + jtoLu (9.59)

Z22 = R2 + jcoL2 + ZJL, (9.60)


I. = Z]]Z22 4- o)~M

V ~> * 1 7 x'

(9.61)

JOjM

ZUZ22 + w'M io)M

-V = - 1, (9.62) '22

To the internal source voltage Vs, the impedance appears as V^/Ii, or

Vv ZnZ22 + a>2M2 a>2M2

J " = Am = 7 = Zu + - ^ - - (9.63) ll ^-22 ^ 2 2

The impedance at the terminals of the source is Zint — Zs, so

co2M2 „ . r a>2M2

2«b = Zu + -Z Zs = R{ + ja>Lx + — ——. (9.64) Z 2 2 (R2 + J<oL2 + Z L )

Note that the impedance Zab is independent of the magnetic polarity of the transformer. The reason is that the mutual inductance appears in Eq. 9.64 as a squared quantity. This impedance is of particular interest because it shows how the transformer affects the impedance of the load as seen from the source. Without the transformer, the load would be con-nected directly to the source, and the source would see a load impedance of ZL; with the transformer, the load is connected to the source through the transformer, and the source sees a load impedance that is a modified version of ZL, as seen in the third term of Eq. 9.64.

Reflected Impedance The third term in Eq. 9.64 is called the reflected impedance (Zr), because it is the equivalent impedance of the secondary coil and load impedance transmitted, or reflected, to the primary side of the transformer. Note that the reflected impedance is due solely to the existence of mutual induc-tance; that is, if the two coils are decoupled, M becomes zero, Zr becomes zero, and Zab reduces to the self-impedance of the primary coil.

To consider reflected impedance in more detail, we first express the load impedance in rectangular form:

Z L = RL + jXL, (9.65)

where the load reactance XL carries its own algebraic sign. In other words, XL is a positive number if the load is inductive and a negative number if the load is capacitive. We now use Eq. 9.65 to write the reflected imped-ance in rectangular form:

Zr = a?M2

R2 + RL + j{coL2 + XL)

_ OJ2M2[(R2 + RL) - ;(o>L2 + XL)] (R2 + RL)2 + (o>L2 + XLf

2 A,/2

1̂ 221 The derivation of Eq . 9.66 takes advantage of the fact that, when Z L is written in rectangular form, the self-impedance of the mesh containing the secondary winding is

-22 R2 + Rh + j(coL2 + XL). (9.67)

Now observe from Eq. 9.66 that the self-impedance of the secondary circuit is reflected into the primary circuit by a scaling factor of (wM/|Z22|)2, and that the sign of the reactive component (wL2 -I- XL) is reversed. Thus the linear transformer reflects the conjugate of the self-impedance of the secondary circuit (Zj2) into the primary winding by a scalar multiplier. Example 9.13 illustrates mesh current analysis for a cir-cuit containing a linear transformer.

To the internal source voltage Vs, the impedance appears as V^/Ii, or

Vv ZnZ22 + a>2M2 a>2M2

J " = Am = 7 = Zu + - ^ - - (9.63) ll ^-22 ^ 2 2

The impedance at the terminals of the source is Zint — Zs, so

co2M2 „ . r a>2M2

2«b = Zu + -Z Zs = R{ + ja>Lx + — ——. (9.64) Z 2 2 (R2 + J<oL2 + Z L )

Note that the impedance Zab is independent of the magnetic polarity of the transformer. The reason is that the mutual inductance appears in Eq. 9.64 as a squared quantity. This impedance is of particular interest because it shows how the transformer affects the impedance of the load as seen from the source. Without the transformer, the load would be con-nected directly to the source, and the source would see a load impedance of ZL; with the transformer, the load is connected to the source through the transformer, and the source sees a load impedance that is a modified version of ZL, as seen in the third term of Eq. 9.64.

Reflected Impedance The third term in Eq. 9.64 is called the reflected impedance (Zr), because it is the equivalent impedance of the secondary coil and load impedance transmitted, or reflected, to the primary side of the transformer. Note that the reflected impedance is due solely to the existence of mutual induc-tance; that is, if the two coils are decoupled, M becomes zero, Zr becomes zero, and Zab reduces to the self-impedance of the primary coil.

To consider reflected impedance in more detail, we first express the load impedance in rectangular form:

Z L = RL + jXL, (9.65)

where the load reactance XL carries its own algebraic sign. In other words, XL is a positive number if the load is inductive and a negative number if the load is capacitive. We now use Eq. 9.65 to write the reflected imped-ance in rectangular form:

Zr = a?M2

R2 + RL + j{coL2 + XL)

_ OJ2M2[(R2 + RL) - ;(o>L2 + XL)] (R2 + RL)2 + (o>L2 + XLf

2 A,/2

1̂ 221 The derivation of Eq . 9.66 takes advantage of the fact that, when Z L is written in rectangular form, the self-impedance of the mesh containing the secondary winding is

-22 R2 + Rh + j(coL2 + XL). (9.67)

Now observe from Eq. 9.66 that the self-impedance of the secondary circuit is reflected into the primary circuit by a scaling factor of (wM/|Z22|)2, and that the sign of the reactive component (wL2 -I- XL) is reversed. Thus the linear transformer reflects the conjugate of the self-impedance of the secondary circuit (Zj2) into the primary winding by a scalar multiplier. Example 9.13 illustrates mesh current analysis for a cir-cuit containing a linear transformer.

Reflected impedance

The ideal transformer

• Internal resistance is very low

• self-inductance is very high

• very strong coupling between the two coils

340 Sinusoidal Steady-State Analysis

jcoM

j(oLA j(oL2

IN, v,

(a)

jcoM

jcoL2

\N->

(b) Figure 9.41 A The circuits used to verify the volts-per-turn and ampere-turn relationships for an ideal transformer.

factor equal to the turns ratio (M/A^) squared. Hence we may describe the terminal behavior of the ideal transformer in terms of two characteristics. First, the magnitude of the volts per turn is the same for each coil, or

ft N2 (9.76)

Second, the magnitude of the ampere-turns is the same for each coil, or

i, MI = My L J i V l (9.77)

We are forced to use magnitude signs in Eqs. 9.76 and 9.77, because we have not yet established reference polarities for the currents and voltages; we discuss the removal of the magnitude signs shortly.

Figure 9.41 shows two lossless (R^ = R2 = 0) magnetically coupled coils. We use Fig. 9.41 to validate Eqs. 9.76 and 9.77. In Fig. 9.41(a), coil 2 is open; in Fig. 9.41(b), coil 2 is shorted. Although we carry out the following analysis in terms of sinusoidal steady-state operation, the results also apply to instantaneous values of v and L

Determining the Voltage and Current Ratios Note in Fig. 9.41(a) that the voltage at the terminals of the open-circuit coil is entirely the result of the current in coil 1; therefore

The current in coil 1 is

From Eqs. 9.78 and 9.79,

V2 = ja>Ml\.

jcoL]

(9.78)

(9.79)

(9.80)

For unity coupling, the mutual inductance equals VL]L2, so Eq. 9.80 becomes

V, = (9.81)

For unity coupling, the flux linking coil 1 is the same as the flux linking coil 2, so we need only one permeance to describe the self-inductance of each coil. Thus Eq. 9.81 becomes

V, = /V?SP

A^ V] (9.82)

or

Voltage relationship for an ideal transformer •

N2 (9.83)

Summing the voltages around the shorted coil of Fig. 9.41(b) yields

0 = -j(x)M\{ + jcoL2l2, (9.84)

The ideal transformer• Voltage at the load,

• Current at the input

• Voltage at the output (strong coupling between coils)


jcoM

j(oLA j(oL2

IN, v,

(a)

jcoM

jcoL2

\N->



ft N2 (9.76)


i, MI = My L J i V l (9.77)






V2 = ja>Ml\.

jcoL]

(9.78)

(9.79)

(9.80)


V, = (9.81)


V, = /V?SP

A^ V] (9.82)

or


N2 (9.83)


0 = -j(x)M\{ + jcoL2l2, (9.84)


jcoM

j(oLA j(oL2

IN, v,

(a)

jcoM

jcoL2

\N->



ft N2 (9.76)


i, MI = My L J i V l (9.77)






V2 = ja>Ml\.

jcoL]

(9.78)

(9.79)

(9.80)


V, = (9.81)


V, = /V?SP

A^ V] (9.82)

or


N2 (9.83)


0 = -j(x)M\{ + jcoL2l2, (9.84)


jcoM

j(oLA j(oL2

IN, v,

(a)

jcoM

jcoL2

\N->



ft N2 (9.76)


i, MI = My L J i V l (9.77)






V2 = ja>Ml\.

jcoL]

(9.78)

(9.79)

(9.80)


V, = (9.81)


V, = /V?SP

A^ V] (9.82)

or


N2 (9.83)


0 = -j(x)M\{ + jcoL2l2, (9.84)


jcoM

j(oLA j(oL2

IN, v,

(a)

jcoM

jcoL2

\N->



ft N2 (9.76)


i, MI = My L J i V l (9.77)






V2 = ja>Ml\.

jcoL]

(9.78)

(9.79)

(9.80)


V, = (9.81)


V, = /V?SP

A^ V] (9.82)

or


N2 (9.83)


0 = -j(x)M\{ + jcoL2l2, (9.84)


jcoM

j(oLA j(oL2

IN, v,

(a)

jcoM

jcoL2

\N->



ft N2 (9.76)


i, MI = My L J i V l (9.77)






V2 = ja>Ml\.

jcoL]

(9.78)

(9.79)

(9.80)


V, = (9.81)


V, = /V?SP

A^ V] (9.82)

or


N2 (9.83)


0 = -j(x)M\{ + jcoL2l2, (9.84)

9.11 The Ideal Transformer

from which, for k = 1,

II h M

u VUL2

L2 (9.85)

Equation 9.85 is equivalent to

I , ^ i - l2N2. (9.86) -4 Current relationship for an ideal transformer

Figure 9.42 shows the graphic symbol for an ideal transformer. The vertical lines in the symbol represent the layers of magnetic material from which ferromagnetic cores are often made. Thus, the symbol reminds us that coils wound on a ferromagnetic core behave very much like an ideal transformer.

There are several reasons for this. The ferromagnetic material creates a space with high permeance. Thus most of the magnetic flux is trapped inside the core material, establishing tight magnetic coupling between coils that share the same core. High permeance also means high self-inductance, because L = N2V. Finally, ferromagnetically coupled coils efficiently transfer power from one coil to the other. Efficiencies in excess of 95% are common, so neglecting losses is not a crippling approximation for many applications.

Determining the Polarity of the Voltage and Current Ratios We now turn to the removal of the magnitude signs from Eqs. 9.76 and 9.77. Note that magnitude signs did not show up in the derivations of Eqs. 9.83 and 9.86. We did not need them there because we had established reference polarities for voltages and reference directions for currents. In addition, we knew the magnetic polarity dots of the two coupled coils.

The rules for assigning the proper algebraic sign to Eqs. 9.76 and 9.77 are as follows:

If the coil voltages V! and V2 are both positive or negative at the dot-marked terminal, use a plus sign in Eq. 9.76. Otherwise, use a nega-tive sign.

If the coil currents I] and I2 are both directed into or out of the dot-marked terminal, use a minus sign in Eq. 9.77. Otherwise, use a plus sign.

The four circuits shown in Fig. 9.43 illustrate these rules.

A", • N,

Ideal

Figure 9.42 • The graphic symbol for an ideal transformer.

A Dot convention for ideal transformers

+ • JV, N2\ • + \

V, I >

Ideal vi = v_2 /V, /V2"

Ni\i = ~N2l2

(a)

I2 V2

+ • yv, N2\ + f V, I II, V,

Ideal I •

W,I, = N2l2

(b)

+ • V, I,)

J

|N, N2\ • +

Ideal V, V2

/V, N2

A^I, = M,I2

(c)

l/V, N2\

Vt I, >'

I, V2

Ideal V] = _V_2

Ni N2

A7,I, = -N2l2

Figure 9.43 • Circuits that show the proper algebraic signs for relating the terminal voltages and currents of an ideal transformer.

Announcement

• No class on November 10 (Tuesday)

• Second Exam is on November 24 (Tuesday)

Quiz

• What is the output of the circuit shown?

Fundamentals of Electrical and Electronics

EngineeringPaul Leonard Atchong C. Hilario, Ph.D.

EEEI 526 [email protected]

mailto:[email protected]

Three phase systems

• Three-phase voltage source and a three-phase load

• Voltage source consist of sinusoids with equal amplitude and frequency

• Each sinusoid is phase shifted from the next by 120 degrees

• Sum of all three voltages is 0

400 Balanced Three-Phase Circuits

Three-phase

Three-phase voltage • source

line

/ \ \

Three-phase load

Figure 11.1 • A basic three-phase circuit.

11.1 Balanced Three-Phase Voltages A set of balanced three-phase voltages consists of three sinusoidal volt-ages that have identical amplitudes and frequencies but are out of phase with each other by exactly 120°. Standard practice is to refer to the three phases as a, b, and c, and to use the a-phase as the reference phase. The three voltages are referred to as the a-phase voltage, the b-phase voltage, and the c-phase voltage.

Only two possible phase relationships can exist between the a-phase voltage and the b- and c-phase voltages. One possibility is for the b-phase voltage to lag the a-phase voltage by 120°, in which case the c-phase volt-age must lead the a-phase voltage by 120°. This phase relationship is known as the abc (or positive) phase sequence. The only other possibility is for the b-phase voltage to lead the a-phase voltage by 120°, in which case the c-phase voltage must lag the a-phase voltage by 120°. This phase relationship is known as the acb (or negative) phase sequence. In phasor notation, the two possible sets of balanced phase voltages are

Vh = V I M / - 1 2 0 '

vc = ym/+120- (11.1)

and

vm/o\ Vh = 1/,,,/ + 120°,

Vc = ^ , , , / - 1 2 0 ° . (11.2)

Equations 11.1 are for the abc, or positive, sequence. Equations 11.2 are for the acb, or negative, sequence. Figure 11.2 shows the phasor dia-grams of the voltage sets in Eqs. 11.1 and 11.2. The phase sequence is the clockwise order of the subscripts around the diagram from Va. The fact that a three-phase circuit can have one of two phase sequences must be taken into account whenever two such circuits operate in par-allel. The circuits can operate in parallel only if they have the same phase sequence.

Another important characteristic of a set of balanced three-phase voltages is that the sum of the voltages is zero. Thus, from either Eqs. 11.1 or Eqs. 11.2,

Figure 11.2 A Phasor diagrams of a balanced set of three-phase voltages, (a) The abc (positive) sequence, (b) The acb (negative) sequence.

Vfl + V„ + Vr = 0. (11.3)

Because the sum of the phasor voltages is zero, the sum of the instanta-neous voltages also is zero; that is,

v& + vh + vc = 0. (11.4)

Now that we know the nature of a balanced set of three-phase volt-ages, we can state the first of the analytical shortcuts alluded to in the introduction to this chapter: If we know the phase sequence and

one voltage in the set, we know the entire set. Thus for a balanced three-phase system, we can focus on determining the voltage (or current) in one phase, because once we know one phase quantity, we know the others.

NOTE: Assess your understanding of three-phase voltages by trying Chapter Problems 11.2 and 11.3.

11.2 Three-Phase Voltage Sources 401

11.2 Three-Phase Voltage Sources A three-phase voltage source is a generator with three separate wind-ings distributed around the periphery of the stator. Each winding com-prises one phase of the generator. The rotor of the generator is an electromagnet driven at synchronous speed by a prime mover, such as a steam or gas turbine. Rotation of the electromagnet induces a sinusoidal voltage in each winding. The phase windings are designed so that the sinusoidal voltages induced in them are equal in amplitude and out of phase with each other by 120°. The phase windings are stationary with respect to the rotating electromagnet, so the frequency of the voltage induced in each winding is the same. Figure 11.3 shows a sketch of a two-pole three-phase source.

There are two ways of interconnecting the separate phase windings to form a three-phase source: in either a wye (Y) or a delta (A) configura-tion. Figure 11.4 shows both, with ideal voltage sources used to model the phase windings of the three-phase generator. The common terminal in the Y-connected source, labeled n in Fig. 11.4(a), is called the neutral terminal of the source. The neutral terminal may or may not be available for exter-nal connections.

Sometimes, the impedance of each phase winding is so small (com-pared with other impedances in the circuit) that we need not account for it in modeling the generator; the model consists solely of ideal voltage sources, as in Fig. 11.4. However, if the impedance of each phase winding is not negligible, we place the winding impedance in series with an ideal sinusoidal voltage source. All windings on the machine are of the same construction, so we assume the winding impedances to be identical. The winding impedance of a three-phase generator is inductive. Figure 11.5 shows a model of such a machine, in which is the winding resistance, and Xw is the inductive reactance of the winding.

Because three-phase sources and loads can be either Y-connected or A-connected, the basic circuit in Fig. 11.1 represents four different configurations:

Source

Y

Y A

A

Load

Y

A

Y A

We begin by analyzing the Y-Y circuit. The remaining three arrangements can be reduced to a Y-Y equivalent circuit, so analysis of the Y-Y circuit is the key to solving all balanced three-phase arrangements. We then illus-trate the reduction of the Y-A arrangement and leave the analysis of the A-Y and A-A arrangements to you in the Problems.

Axis of a-phase winding

Axis of c-phasc windine

\ Axis of b-phase winding

Stator

Figure 11.3 A A sketch of a three-phase voltage source.

Figure 11.4 A The two basic connections of an ideal three-phase source, (a) A Y-connected source, (b) A A-connected source.

Voltage and current

• Each line can be connected to an identical load

• The current will be

400 Balanced Three-Phase Circuits

Three-phase

Three-phase voltage • source

line

/ \ \

Three-phase load

Figure 11.1 • A basic three-phase circuit.

11.1 Balanced Three-Phase Voltages A set of balanced three-phase voltages consists of three sinusoidal volt-ages that have identical amplitudes and frequencies but are out of phase with each other by exactly 120°. Standard practice is to refer to the three phases as a, b, and c, and to use the a-phase as the reference phase. The three voltages are referred to as the a-phase voltage, the b-phase voltage, and the c-phase voltage.

Only two possible phase relationships can exist between the a-phase voltage and the b- and c-phase voltages. One possibility is for the b-phase voltage to lag the a-phase voltage by 120°, in which case the c-phase volt-age must lead the a-phase voltage by 120°. This phase relationship is known as the abc (or positive) phase sequence. The only other possibility is for the b-phase voltage to lead the a-phase voltage by 120°, in which case the c-phase voltage must lag the a-phase voltage by 120°. This phase relationship is known as the acb (or negative) phase sequence. In phasor notation, the two possible sets of balanced phase voltages are

Vh = V I M / - 1 2 0 '

vc = ym/+120- (11.1)

and

vm/o\ Vh = 1/,,,/ + 120°,

Vc = ^ , , , / - 1 2 0 ° . (11.2)

Equations 11.1 are for the abc, or positive, sequence. Equations 11.2 are for the acb, or negative, sequence. Figure 11.2 shows the phasor dia-grams of the voltage sets in Eqs. 11.1 and 11.2. The phase sequence is the clockwise order of the subscripts around the diagram from Va. The fact that a three-phase circuit can have one of two phase sequences must be taken into account whenever two such circuits operate in par-allel. The circuits can operate in parallel only if they have the same phase sequence.

Another important characteristic of a set of balanced three-phase voltages is that the sum of the voltages is zero. Thus, from either Eqs. 11.1 or Eqs. 11.2,

Figure 11.2 A Phasor diagrams of a balanced set of three-phase voltages, (a) The abc (positive) sequence, (b) The acb (negative) sequence.

Vfl + V„ + Vr = 0. (11.3)

Because the sum of the phasor voltages is zero, the sum of the instanta-neous voltages also is zero; that is,

v& + vh + vc = 0. (11.4)

Now that we know the nature of a balanced set of three-phase volt-ages, we can state the first of the analytical shortcuts alluded to in the introduction to this chapter: If we know the phase sequence and

one voltage in the set, we know the entire set. Thus for a balanced three-phase system, we can focus on determining the voltage (or current) in one phase, because once we know one phase quantity, we know the others.

NOTE: Assess your understanding of three-phase voltages by trying Chapter Problems 11.2 and 11.3.

11.2 Three-Phase Voltage Sources 401

11.2 Three-Phase Voltage Sources A three-phase voltage source is a generator with three separate wind-ings distributed around the periphery of the stator. Each winding com-prises one phase of the generator. The rotor of the generator is an electromagnet driven at synchronous speed by a prime mover, such as a steam or gas turbine. Rotation of the electromagnet induces a sinusoidal voltage in each winding. The phase windings are designed so that the sinusoidal voltages induced in them are equal in amplitude and out of phase with each other by 120°. The phase windings are stationary with respect to the rotating electromagnet, so the frequency of the voltage induced in each winding is the same. Figure 11.3 shows a sketch of a two-pole three-phase source.

There are two ways of interconnecting the separate phase windings to form a three-phase source: in either a wye (Y) or a delta (A) configura-tion. Figure 11.4 shows both, with ideal voltage sources used to model the phase windings of the three-phase generator. The common terminal in the Y-connected source, labeled n in Fig. 11.4(a), is called the neutral terminal of the source. The neutral terminal may or may not be available for exter-nal connections.

Sometimes, the impedance of each phase winding is so small (com-pared with other impedances in the circuit) that we need not account for it in modeling the generator; the model consists solely of ideal voltage sources, as in Fig. 11.4. However, if the impedance of each phase winding is not negligible, we place the winding impedance in series with an ideal sinusoidal voltage source. All windings on the machine are of the same construction, so we assume the winding impedances to be identical. The winding impedance of a three-phase generator is inductive. Figure 11.5 shows a model of such a machine, in which is the winding resistance, and Xw is the inductive reactance of the winding.

Because three-phase sources and loads can be either Y-connected or A-connected, the basic circuit in Fig. 11.1 represents four different configurations:

Source

Y

Y A

A

Load

Y

A

Y A

We begin by analyzing the Y-Y circuit. The remaining three arrangements can be reduced to a Y-Y equivalent circuit, so analysis of the Y-Y circuit is the key to solving all balanced three-phase arrangements. We then illus-trate the reduction of the Y-A arrangement and leave the analysis of the A-Y and A-A arrangements to you in the Problems.

Axis of a-phase winding

Axis of c-phasc windine

\ Axis of b-phase winding

Stator

Figure 11.3 A A sketch of a three-phase voltage source.

Figure 11.4 A The two basic connections of an ideal three-phase source, (a) A Y-connected source, (b) A A-connected source.

Voltage and current

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