13.4 Linear Transformers

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13.4 Linear Transformers

Here we introduce the transformer as a ne w circuit element. A transformer is a magnetic device that takes advantage of the phenomenon of mutual inductance.

A transformer is generally a four-terminal device comprising two (or more) magnetically coupled coils.

As shown in Fig. 13.19, the coil that is directly connected to the voltage source is called the primary winding. The coil connected to the load is called the secondary winding. The resistances R1 and R2 are included to account for the losses (po wer dissipation) in the coils. The transformer is said to be linear if the coils are w ound on a magnetically linear material—a material for which the magnetic permeability is constant. Such materials include air, plastic, Bakelite, and wood. In fact, most materials 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.

Figure 13.20

Different types of transformers: (a) copper wound dry power transformer, (b) audio transformers. (a) © Electric Service Co., Cincinnati Ohio, (b) © Jensen Transformers, Inc., Chatsworth, CA

A linear transformer may also be regarded as one whose flux is proportional to the currents in its windings.

It should be noted that the result in Eq. (13.41) or (13.42) is not affected by the location of the dots on the transformer, because the same result is produced when M is replaced by −M.

The little bit of experience gained in Sections 13.2 and 13.3 in analyzing magnetically coupled circuits is enough to con vince anyone that analyzing these circuits is not as easy as circuits in previous chapters. For this reason, it is sometimes convenient to replace a magnetically coupled circuit by an equi valent circuit with no magnetic coupling. We want to replace the linear transformer in Fig. 13.21 by an equi valent T or Π circuit, a circuit that would have no mutual inductance.

The v oltage-current relationships for the primary and secondary coils give the matrix equation

$$\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} j\omega L_1 & j\omega M \\ j\omega M & j\omega L_2 \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix}$$

(13.43)

By matrix inversion, this can be written as

version, this can be written as
\n

$$\begin{bmatrix}\nI_1 \\ I_2\n\end{bmatrix} = \begin{bmatrix}\n\frac{L_2}{j\omega(L_1L_2 - M^2)} & \frac{-M}{j\omega(L_1L_2 - M^2)} \\ \frac{-M}{j\omega(L_1L_2 - M^2)} & \frac{L_1}{j\omega(L_1L_2 - M^2)}\n\end{bmatrix} \begin{bmatrix}\nV_1 \\ V_2\n\end{bmatrix}$$

\n(13.44)

Our goal is to match Eqs. (13.43) and (13.44) with the corresponding equations for the T and Π networks.

For the T (or Y) network of Fig. 13.22, mesh analysis pro vides the terminal equations as

$$\begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix} = \begin{bmatrix} j\omega(L_a + L_c) & j\omega L_c \\ j\omega L_c & j\omega(L_b + L_c) \end{bmatrix} \begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix}$$

(13.45)

Figure 13.21 Determining the equivalent circuit of a linear transformer.

Figure 13.22 An equivalent T circuit.

If the circuits in Figs. 13.21 and 13.22 are equivalents, Eqs. (13.43) and (13.45) must be identical. Equating terms in the impedance matrices of Eqs. (13.43) and (13.45) leads to

$$L_a = L_1 - M, \qquad L_b = L_2 - M, \qquad L_c = M \qquad (13.46)$$

For the Π (or Δ) network in Fig. 13.23, nodal analysis gi ves the terminal equations as

$$\begin{bmatrix} \mathbf{I}_1 \\ \mathbf{I}_2 \end{bmatrix} = \begin{bmatrix} \frac{1}{j\omega L_A} + \frac{1}{j\omega L_C} & -\frac{1}{j\omega L_C} \\ -\frac{1}{j\omega L_C} & \frac{1}{j\omega L_B} + \frac{1}{j\omega L_C} \end{bmatrix} \begin{bmatrix} \mathbf{V}_1 \\ \mathbf{V}_2 \end{bmatrix}$$

(13.47)

Equating terms in admittance matrices of Eqs. (13.44) and (13.47), we obtain

$$L_{A} = \frac{L_{1}L_{2} - M^{2}}{L_{2} - M}, \qquad L_{B} = \frac{L_{1}L_{2} - M^{2}}{L_{1} - M}$$ $$L_{C} = \frac{L_{1}L_{2} - M^{2}}{M}$$

(13.48)

Note that in Figs. 13.22 and 13.23, the inductors are not magnetically coupled. Also note that changing the locations of the dots in Fig. 13.21 can cause M to become − M. As Example 13.6 illustrates, a ne gative value of M is ph ysically unrealizable b ut the equi valent model is still mathematically valid.

In the circuit of Fig. 13.24, calculate the input impedance and current I1. Take Z1 = 60 − j100 Ω, Z2 = 30 + j40 Ω, and ZL = 80 + j60 Ω.

Figure 13.24

Solution:

From Eq. (13.41),

$$Zin = Z1 + j20 + \frac{(5)2}{j40 + Z2 + ZL}$$

= 60 - j100 + j20 + $\frac{25}{110 + j140}$
= 60 - j80 + 0.14/ $-51.84^{\circ}$
= 60.09 - j80.11 = 100.14/ $-53.1^{\circ}$ Ω

Figure 13.23 An equivalent Π circuit.

Example 13.4

For Example 13.4.

Thus,

$$\mathbf{I}_1 = \frac{\mathbf{V}}{\mathbf{Z}_{in}} = \frac{50/60^{\circ}}{100.14 / -53.1^{\circ}} = 0.5 / 113.1^{\circ} \text{ A}$$

Find the input impedance of the circuit in Fig. 13.25 and the current from the voltage source. Practice Problem 13.4

Answer: 8.58⧸ 58.05° Ω, 4.662⧸− 58.05° A.

2 H

Determine the T-equivalent circuit of the linear transformer in Fig. 13.26(a).

(a) 10 H 4 H a b c d (b) a b c d 2 H I¹ I² 8 H 2 H

Solution:

Given that L1 = 10, L2 = 4, and M = 2, the T-equivalent network has the following parameters:

$$L_a = L_1 - M = 10 - 2 = 8$$

H

$$L_b = L_2 - M = 4 - 2 = 2$$

H,

$$L_c = M = 2$$

H

The T-equivalent circuit is shown in Fig. 13.26(b). We have assumed that reference directions for currents and voltage polarities in the primary and secondary windings conform to those in Fig. 13.21. Otherwise, we may need to replace M with −M, as Example 13.6 illustrates.

Example 13.5

Practice Problem 13.5

For the linear transformer in Fig. 13.26(a), find the Π equivalent network.

Answer: LA = 18 H, LB = 4.5 H, LC = 18 H.

Example 13.6
--------------	—	—	—

Solve for I1, I2, and Vo in Fig. 13.27 (the same circuit as for Practice Prob. 13.1) using the T-equivalent circuit for the linear transformer. Example 13.6

Solution:

Notice that the circuit in Fig. 13.27 is the same as that in Fig. 13.10 except that the reference direction for current I2 has been reversed, just to make the reference directions for the currents for the magnetically coupled coils conform with those in Fig. 13.21.

We need to replace the magnetically coupled coils with the T-equivalent circuit. The relevant portion of the circuit in Fig. 13.27 is shown in Fig. 13.28(a). Comparing Fig. 13.28(a) with Fig. 13.21 sho ws that there are tw o differences. First, due to the current reference direc tions and voltage polarities, we need to replace M by −M to make Fig. 13.28(a) conform with Fig. 13.21. Second, the circuit in Fig. 13.21 is in the time-domain, whereas the circuit in Fig. 13.28(a) is in the frequencydomain. The difference is the factor jω; that is, L in Fig. 13.21 has been replaced with jωL and M with jωM. Since ω is not specified, we can assume ω = 1 rad/s or any other value; it really does not matter. With these two differences in mind,

$$L_a = L_1 - (-M) = 8 + 1 = 9 \text{ H}$$ $$L_b = L_2 - (-M) = 5 + 1 = 6 \text{ H}, \qquad L_c = -M = -1 \text{ H}$$

Thus, the T-equivalent circuit for the coupled coils is as shown in Fig. 13.28(b).

Inserting the T-equivalent circuit in Fig. 13.28(b) to replace the two coils in Fig. 13.27 gives the equivalent circuit in Fig. 13.29, which can be solved using nodal or mesh analysis. Applying mesh analysis, we obtain

$$j6 = I_1(4 + j9 - j1) + I_2(-j1)$$

(13.6.1)

and

$$0 = I1(-j1) + I2(10 + j6 - j1)$$

(13.6.2)

From Eq. (13.6.2),

$$\mathbf{I}_1 = \frac{(10 + j5)}{j} \mathbf{I}_2 = (5 - j10)\mathbf{I}_2$$

(13.6.3)

Figure 13.28

For Example 13.6: (a) circuit for coupled coils of Fig. 13.27, (b) T-equivalent circuit.

For Example 13.6.

Substituting Eq. (13.6.3) into Eq. (13.6.1) gives

$$j6 = (4 + j8)(5 - j10)\mathbf{I}_2 - j\mathbf{I}_2 = (100 - j)\mathbf{I}_2 \simeq 100\mathbf{I}_2$$

Since 100 is very large compared with 1, the imaginary part of (100 − j) can be ignored so that 100 − j ≃ 100. Hence,

$$I_2 = \frac{j6}{100} = j0.06 = 0.06 / 90^{\circ} A$$

From Eq. (13.6.3),

$$I_1 = (5 - j10)j0.06 = 0.6 + j0.3$$

A

and

$$\mathbf{V}_o = -10\mathbf{I}_2 = -j0.6 = 0.6 \div 90^\circ \text{ V}$$

This agrees with the answer to Practice Prob. 13.1. Of course, the direction of I2 in Fig. 13.10 is opposite to that in Fig. 13.27. This will not affect Vo, but the value of I2 in this example is the negative of that of I² in Practice Prob. 13.1. The advantage of using the T-equivalent model for the magnetically coupled coils is that in Fig. 13.29 we do not need to bother with the dot on the coupled coils.

Solve the problem in Example 13.1 (see Fig. 13.9) using the T-equivalent model for the magnetically coupled coils.

Practice Problem 13.6

Answer: 13⧸ −49.4° A, 2.91⧸ 14.04° A.