Circuitos Trifásicos Aula 8 Transformador

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Circuitos Trifásicos

Aula 8

Transformador

Engenharia Elétrica Universidade Federal de Juiz de Fora

tinyurl.com/profvariz

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Aplicações dos transformadores

Sistemas de energia

Utilizado paraelevar a tensãopara que apotência seja transmitida em alta tensão e baixas correntes=⇒redução das perdas

Naproximidade das áreas de consumotransformadoresabaixadoressão empregados visando reduzir a tensão paraníveis desejados.

Figura 1:Transformador.

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Sistemas de energia

Propicia a medição segura de altos valores de tensão e corrente. Transformador de potencial (TP)

Transformador de corrente (TC)

Figura 2:Transformador de corrente (TC).

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Sistemas eletrônicos

Propicia aisolação galvânica=⇒reduz o risco de choque elétrico Transformador comrelação de espiras unitária=⇒utilizado paraacoplar dois estágiosde um amplificador de modo aimpedir que qualquer tensão contínuado estágio anterior afete apolarizaçãodo próximo

Casamento de impedâncias=⇒máxima transferência de potência.

IMPEDANCE MATCHING, ISOLATION, AND DISPLACEMENT  945

Solutions:

a. The source current:

Is 230.8 mA

The power to the speaker:

P I2R (230.8 mA)2⋅8  426.15 mW 0.43 W or less than half a watt.

b. Zp a2ZL

a 8

and Zp (8)28  512

which matches that of the source. Maximum power transfer condi-tions have been established, and the source current is now deter-mined by

Is 117.19 mA

The power to the primary (which equals that to the secondary for the ideal transformer) is

P I2R (117.19 mA)2512  7.032 W

The result is not in milliwatts, as obtained above, and exceeds 7 W, which is a significant improvement.

c. Comparing levels, 7.032 W/426.15 mW 16.5, or more than 16 times the power delivered to the speaker using the impedance match-ing transformer.

Another important application of the impedance matching capabili-ties of a transformer is the matching of the 300- twin line transmis-sion line from a televitransmis-sion antenna to the 75- input impedance of today’s televisions (ready-made for the 75- coaxial cable), as shown in Fig. 21.9. A match must be made to ensure the strongest signal to the television receiver.

Using the equation Zp a2ZLwe find

300  a275 120 V 1024 120 V 512 512 E RT 8 1 Np Ns 120 V 520 120 V 512 8 E RT (a) (b) 8 : 1 + V_g – 8 120 V R_s 512 Zp + V_g – 8 120 V R_s 512 FIG. 21.8 Example 21.5. 75 TV input 300 :75 FIG. 21.9

Television impedance matching transformer.

Figura 3:Casamento de impedância.

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Transformador linear

Utilizado como dispositivo deacoplamento entre a fonte e a carga. O transformador e ditolinear se for enrolado em uma material magneticamente linear: Ar, plástico e madeira.

540 PART 2 AC Circuits

material—a material for which the magnetic permeability is constant. Such materials include air, plastic, Bakelite, and wood. In fact, most ma-terials are magnetically linear. Linear transformers are sometimes called air-core transformers, although not all of them are necessarily air-core. They are used in radio and TV sets. Figure 13.20 portrays different types of transformers.

A linear transformer may also be regarded as one whoseﬂux is proportional to thecurrents in its windings.

V +− I1 L1 L2 I2 ZL

M

R₁ R₂

Primary coil Secondary coil

Figure13.19 A linear transformer.

(b) (a)

Figure13.20 Different types of transformers: (a) copper wound dry power transformer, (b) audio transformers.

(Courtesy of: (a) Electric Service Co., (b) Jensen Transformers.)

We would like to obtain the input impedance Zin as seen from the source, because Zingoverns the behavior of the primary circuit. Applying KVL to the two meshes in Fig. 13.19 gives

V= (R1+ jωL1)I1− jωMI2 (13.40a) 0 = −jωMI1+ (R2+ jωL2+ ZL)I2 (13.40b)

Figura 4:Transformador linear.

As resistênciasR1e R2são incluídas para modelar asperdas no primário e no secundário, respectivamente.

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Tipos de núcleos

Núcleo de ar Núcleo de ferro

Núcleo ajustável de material magnético

TYPES OF TRANSFORMERS  959

drop, and N_p and E_p remain the same, then _mmust increase in magni-tude, as determined by Eq. (21.12):

m

The result is that B will increase, as shown in Fig. 21.35, causing H to increase also. The resulting DI could cause a very high current in the primary, resulting in possible damage to the transformer.

21.11 TYPES OF TRANSFORMERS

Transformers are available in many different shapes and sizes. Some of the more common types include the power transformer, audio trans-former, IF (intermediate-frequency) transtrans-former, and RF (radio-frequency) transformer. Each is designed to fulfill a particular require-ment in a specific area of application. The symbols for some of the basic types of transformers are shown in Fig. 21.36.

Ep 4.44 fp↓Np Knee of curve B = m A_core B = m A_core H = N1I1 I_core H = N1I1 I_core 0 FIG. 21.35

Demonstrating why the frequency of application is important for transformers.

Air-core Iron-core Variable-core FIG. 21.36

Transformer symbols.

The method of construction varies from one transformer to another. Two of the many different ways in which the primary and secondary coils can be wound around an iron core are shown in Fig. 21.37. In either case, the core is made of laminated sheets of ferromagnetic mate-rial separated by an insulator to reduce the eddy current losses. The sheets themselves will also contain a small percentage of silicon to increase the electrical resistivity of the material and further reduce the eddy current losses.

A variation of the core-type transformer appears in Fig. 21.38. This transformer is designed for low-profile (the 2.5-VA size has a maximum

Figura 5:Tipos de núcleos.

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Transformador com núcleo de ferro

Apermeabilidade no ferroµéassumida sendo constante em uma faixa de operação de tensão e corrente.

EQUIVALENT CIRCUIT (IRON-CORE TRANSFORMER)  949

mary and secondary coils of a transformer, there is a small amount of flux that links each coil but does not pass through the core, as shown in Fig. 21.16 for the primary winding. This leakage flux, representing a definite loss in the system, is represented by an inductance L_pin the pri-mary circuit and an inductance L_s in the secondary.

The resistance R_c represents the hysteresis and eddy current losses (core losses) within the core due to an ac flux through the core. The induc-tance L_m (magnetizing inductance) is the inductance associated with the magnetization of the core, that is, the establishing of the flux _m in the core. The capacitances C_pand C_sare the lumped capacitances of the pri-mary and secondary circuits, respectively, and C_w represents the equiva-lent lumped capacitances between the windings of the transformer.

Since i′_p is normally considerably larger than if_m (the magnetizing

current), we will ignore if_m for the moment (set it equal to zero),

result-ing in the absence of R_c and L_m in the reduced equivalent circuit of Fig. 21.17. The capacitances C_p, C_w, and C_s do not appear in the equivalent circuit of Fig. 21.17 since their reactance at typical operating frequencies will not appreciably affect the transfer characteristics of the transformer.

– + E_p R_p L_p R_C C_p L_m N_p N_s C_s E_s R_L R_s L_s – + i_p ifm i'p C_w Ideal transformer FIG. 21.15

Equivalent circuit for the practical iron-core transformer.

Φleakage

Φm

Φleakage

FIG. 21.16

Identifying the leakage flux of the primary.

– + E_p R_p L_p N_p N_s R_L R_s L_s – + E_s Ideal transformer a = Np N_s FIG. 21.17

Reduced equivalent circuit for the nonideal iron-core transformer.

If we now reflect the secondary circuit through the ideal transformer using Eq. (21.19), as shown in Fig. 21.18(a), we will have the load and generator voltage in the same continuous circuit. The total resistance and inductive reactance of the primary circuit are determined by

(21.22) R_equivalent R_e R_p  a2R_s

Figura 6:Fluxo de dispersão do primário.

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A correnteI1produz o fluxoφ1= φ11+ φ12. A corrente naturalI2 o fluxoφ2= φ22+ φ21.

Osfluxos de dispersãopodem ser escritosem função do coeficiente de acoplamento; φ11 = (1 − k)φ1 (1) φ22 = (1 − k)φ2 (2) 12  22  1 v 1 i i2   11  21  1 R R2 L Z 2 N 1 N

Figura 7:Transformador de dois enrolamentos.

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Asindutâncias de dispersãoestãorelacionadascom asindutâncias própriaspelas seguintes relações:

L11 ≡ (1 − k)L1 (3) L22 ≡ (1 − k)L2 (4) 12  22  1 v 1 i i2   11  21  1 R R2 L Z 2 N 1 N

Figura 8:Transformador de dois enrolamentos.

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Transformador Ideal

Considerando umtransformador não ideal

2 Z 1 I I₂ 1 V    2 V j M 1 j L j L ₂

Figura 9:Transformador não ideal.

   V1 = jωL1I1− jωMI2 jωL2I2+ Z2I2− jωMI1= 0 (5)

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Transformador Ideal

Com base em (5) pode-se estabelecer as seguintes relações

V2 V1 = jωMZ2 −ω2_(L 1L2− M2) + jωL1Z2 (6) I2 I1 = jωM jωL2+ Z2 (7) V1 I1 = Z1= jωL1+ ω2_M2 jωL2+ Z2 (8)

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Transformador Ideal

Definição

OTransformador ideal é um transformador sem perdas com coeficiente de acoplamento unitário no qual as bobinas primária e secundária possuem autoindutâncias infinitas.

As bobinas possuem reatâncias muito grandes (L1, L2, M → ∞); O coeficiente de acoplamento é unitário (k = 1);

As bobinas primária e secundária não tem perdas (R1 = R2= 0).

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Transformador Ideal

Acoplamento magnético unitárioentre as bobinas (k = 1)

M=√L1L2 (9) Portanto V2 V1 =jω √ L1L2Z2 jωL1Z2 =r L2 L1 (10)

Uma vez que

L= µN 2 iAi l (11) Logo V2 V1 = N2 N1 = a (12)

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Transformador Ideal

A equação (8) pode ser reescrita como

Z1= jωL1Z2 jωL2+ Z2 ou Z1= Z2 a2 1 + Z2 jωL2 (13) De maneira similar I2 I1 = r L1 L2 1 + Z2 jωL2 (14)

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Transformador Ideal

Considerando que notransformador idealasindutâncias própriase a mútua tendem ao infinito, embora arelaçãoentre tais indutâncias permanecefinita.

As equações (13) e (14) podem ser simplificadas resultando em

Z2 Z1 = N2 N1 2 = a2 ₍₁₅₎ I2 I1 =N1 N2 = 1 a (16)

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Transformador Ideal

Conclusões

Asrelações entre as tensõesno primário e secundário ou entre as correntesé função da relação entre onúmero de espiras das bobinas; Notransformador de isolaçãoa= 1;

Notransformador elevador, a > 1, a tensão no secundário é maior que a tensão no primário;

Notransformador abaixador, a < 1, a tensão no secundário é menor que a tensão no primário;

Amaior tensãoestá sobre a bobina com omaior número de espiras; Amaior correntecircula na bobina com omenor número de espiras; Apotência na entradado transformador ideal éigualapotência de saída do mesmo =⇒sem perdas.

a= N2 N1

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Transformador Ideal

Regras para análise de circuitos com transformadores ideais

1. Se tanto V1 quanto V2forempositivas ou ambas negativas nos terminais

pontuados, use+ana relação de tensão. Caso contrário, use−a;

2. Se tanto I1 quanto I2entrarem ou ambas deixarem os terminais

pontuados, use−ana relação de corrente. Caso contrário, use+a;

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Transformador Ideal

Relações para o transformador ideal

CHAPTER 13 Magnetically Coupled Circuits

547

For the reason of power conservation, the energy supplied to the primary must equal the energy absorbed by the secondary, since there are no losses in an ideal transformer. This implies that

v1i1 = v2i2 (13.53)

In phasor form, Eq. (13.53) in conjunction with Eq. (13.52) becomes

I1

I2

= V2

V1

= n (13.54)

showing that the primary and secondary currents are related to the turns ratio in the inverse manner as the voltages. Thus,

I2 I1 = N1 N2 = 1 n (13.55)

When n = 1, we generally call the transformer an isolation transformer. The reason will become obvious in Section 13.9.1. If n > 1, we have a step-up transformer, as the voltage is increased from primary to sec-ondary (V2 > V1). On the other hand, if n < 1, the transformer is a step-down transformer, since the voltage is decreased from primary to

secondary (V2 < V1).

A

step-down transformer

is onewhosesecondary voltage

is less than its primary voltage.

A

step-up transformer

is onewhosesecondary voltage

is greater than its primary voltage.

The ratings of transformers are usually specified as V1/V2. A transformer

with rating 2400/120 V should have 2400 V on the primary and 120 in the secondary (i.e., a step-down transformer). Keep in mind that the voltage ratings are in rms.

Power companies often generate at some convenient voltage and use a step-up transformer to increase the voltage so that the power can be transmitted at very high voltage and low current over transmission lines, resulting in significant cost savings. Near residential consumer premises, step-down transformers are used to bring the voltage down to 120 V. Section 13.9.3 will elaborate on this.

It is important that we know how to get the proper polarity of the voltages and the direction of the currents for the transformer in Fig. 13.31. If the polarity of V1or V2or the direction of I1 or I2 is changed, n in Eqs.

(13.51) to (13.55) may need to be replaced by−n. The two simple rules to follow are:

1. If V1 and V2 are both positive or both negative at the dotted

terminals, use +n in Eq. (13.52). Otherwise, use −n.

2. If I1 and I2 both enter into or both leave the dotted terminals,

use−n in Eq. (13.55). Otherwise, use +n.

The rules are demonstrated with the four circuits in Fig. 13.32.

V₁ V₂ N₁:N₂ + − + − I₁ I₂ V₂ V₁ N₂ N₁ = I2 I₁ N₁ N₂ = (a) V₁ V₂ N₁:N₂ + − + − I₁ I₂ V₂ V₁ N₂ N₁ = I2 I₁ N₁ N₂ = − (b) V₁ V₂ N₁:N₂ + − + − I₁ I₂ V₂ V₁ N₂ N₁ = − V₂ V₁ N₂ N₁ = − I₂ I₁ N₁ N₂ = (c) V₁ V₂ N₁:N₂ + − + − I₁ I₂ I₂ I₁ N₁ N₂ = − (d)

Figure13.32

Typical

circuits illustrating proper voltage polarities and current directions in an ideal transformer.

CHAPTER 13 Magnetically Coupled Circuits

547

v1i1 = v2i2 (13.53)

I1

I2

= V2

V1

= n (13.54)

I2 I1 = N1 N2 = 1 n (13.55)

A

step-down transformer

is onewhosesecondary voltage

is less than its primary voltage.

A

step-up transformer

is onewhosesecondary voltage

is greater than its primary voltage.

It is important that we know how to get the proper polarity of the voltages and the direction of the currents for the transformer in Fig. 13.31. If the polarity of V1or V2 or the direction of I1 or I2is changed, n in Eqs.

(13.51) to (13.55) may need to be replaced by−n. The two simple rules to follow are:

Figure13.32

Typical

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Transformador Ideal

Relações para o transformador ideal

CHAPTER 13 Magnetically Coupled Circuits

547

v1i1 = v2i2 (13.53)

I1

I2

= V2

V1

= n (13.54)

I2 I1 = N1 N2 = 1 n (13.55)

A

step-down transformer

is onewhosesecondary voltage

is less than its primary voltage.

A

step-up transformer

is onewhosesecondary voltage

is greater than its primary voltage.

It is important that we know how to get the proper polarity of the voltages and the direction of the currents for the transformer in Fig. 13.31. If the polarity of V1 or V2 or the direction of I1or I2 is changed, n in Eqs.

(13.51) to (13.55) may need to be replaced by −n. The two simple rules to follow are:

use −n in Eq. (13.55). Otherwise, use +n.

Figure13.32

Typical

CHAPTER 13 Magnetically Coupled Circuits

547

v1i1 = v2i2 (13.53)

I1

I2

= V2

V1

= n (13.54)

I2 I1 = N1 N2 = 1 n (13.55)

A

step-down transformer

is onewhosesecondary voltage

is less than its primary voltage.

A

step-up transformer

is onewhosesecondary voltage

is greater than its primary voltage.

It is important that we know how to get the proper polarity of the voltages and the direction of the currents for the transformer in Fig. 13.31. If the polarity of V1 or V2or the direction of I1 or I2 is changed, n in Eqs.

(13.51) to (13.55) may need to be replaced by −n. The two simple rules to follow are:

terminals, use+n in Eq. (13.52). Otherwise, use −n.

Figure13.32

Typical

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Transformador Ideal

Modelos equivalentes para o transformador ideal com fontes controladas

1 V I1aI2 V2aV1 V2 + _ _ _ + + 2 I 1 V a  1 2 I I 1 a V2 V V2 + _ _ _ + + 1 I

Figura 10:Modelos equivalentes para o transformador ideal

a= N2 N1

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Exemplo 1

Calcule V3na figura abaixo

1 R R₂ 1 I I2 V     2 V 2 j L   1 V   1: 2 N N 1 j L 3 V R₃ Figura 11:Exemplo 1.

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Exemplo 1

As equações que descrevem o circuito são

(R1+ jωL1)I1+ V1= V

(R2+ jωL2+ R3)I2= V2

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Pode-se ainda escrever as equações para o transformador ideal

V1 V2 =N1 N2 I1 I2 =N2 N1 (18)

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Exemplo 1 Assim (R1+ jωL1)I1+ V2 N1 N2 = V (19) Substituindo-se V2da malha I2 (R1+ jωL1)I1+ (R2+ jωL2+ R3)I2 N1 N2 = V (20) Ou (R1+ jωL1)I1+ (R2+ jωL2+ R3)I1 N1 N2 2 = V (21)

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Exemplo 1 Logo I1= V (R1+ jωL1) + (R2+ jωL2+ R3) N1 N2 2 (22)

O circuito equivalente abaixo representa a equação (22)

1 R 1 I V     2 1 2 2 N j L N  _ _   1 j L 3 ' V 2 1 2 2 N R N       2 1 3 2 N R N      

Figura 12:Circuito equivalente com os elementos do secundário refletidos para o primário.

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Exemplo 1

A tensão sobre R3é dada por

V3= R3I2 (23) De modo alternativo V30= R3 N1 N2 2 I1 (24) V3= V30 N2 N1 = R3I2 (25)

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Refletir elementos pelo Trafo

Conclusões (do ex. 1)

Otransformador idealpode serremovidodo circuito se todos os elementos de um ladosão transferidos, oureferidos,para o outro lado. Para referir os elementos do secundário do trafo para o lado primário:

Todas as tensões do secundário são multiplicadas por N = N1/N2;

Todas as corrente do secundário são multiplicadas por 1/N = N2/N1;

Todas as impedâncias do secundário são multiplicadas por N2_{= (N} 1/N2)2; Este procedimento é normalmente utilizado quandonão há conexões externas entre as bobinasdo transformador ideal;

Havendo tais conexões, a solução é via análise de laço ou nodal.

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Exemplo 2

Calcule a potência fornecida ao resistor de 10 Ω no circuito abaixo com transformador ideal.

Resposta: P = 5,3 W

Thus,

(b) Since both and leave the dotted terminals,

(c) The complex power supplied is

S VsI*1 (120l0)(11.09l33.69) 1,330.8l33.69 VA Vo 20I2 110.9l213.69 V I2 1 nI1 5.545l33.69 A I2 I1 I1 120l₀ Zin 120 l₀ 10.82l33.69 11.09 l_33.69 A Zin 4 j6 ZR 9 j6 10.82l33.69 13.5 Ideal Transformers 579

In the ideal transformer circuit of Fig. 13.38, find and the complex power supplied by the source.

Vo

Practice Problem 13.8

2 Ω + − 16 Ω V₁ V₂ V_o 1 : 4 + − + − I₁ I₂ V rms 240 0° + − −j24 Ω Figure 13.38

For Practice Prob. 13.8.

Answer: _429.4l_116.57 V, 17.174l26.57 kVA.

Calculate the power supplied to the 10- resistor in the ideal trans-former circuit of Fig. 13.39.

Example 13.9

20 Ω 10 Ω V₁ V₂ 2 : 1 + − + − V rms 120 0° 30 Ω + − I₁ I₂ Figure 13.39 For Example 13.9. Solution:

Reflection to the secondary or primary side cannot be done with this circuit: There is direct connection between the primary and

ale80571_ch13_555-612.qxd 11/30/11 2:07 PM Page 579

Figura 13:Exemplo 2.