13.5 Ideal Transformers

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13.5 Ideal Transformers

An ideal transformer is one with perfect coupling ( k =1). It consists of two (or more) coils with a lar ge number of turns w ound on a common core of high permeability. Because of this high permeability of the core, the flux links all the turns of both coils, thereby resulting in a perfect coupling.

To see ho w an ideal transformer is the limiting case of tw o coupled inductors where the inductances approach infinity and the coupling is perfect, let us reexamine the circuit in Fig. 13.14. In the frequency domain,

$$V_1 = j\omega L_1 I_1 + j\omega M I_2$$

\n

$$V_2 = j\omega M I_1 + j\omega L_2 I_2$$

\n(13.49b)

_____

From Eq. (13.49a), I1 = (V1 − jωMI2)∕jωL1 (we could have also used this equation to develop the current ratios instead of using the conservation of power which we will do shortly). Substituting this in Eq. (13.49b) gives

$$\mathbf{V}_2 = j\omega L_2 \mathbf{I}_2 + \frac{M\mathbf{V}_1}{L_1} - \frac{j\omega M^2 \mathbf{I}_2}{L_1}$$

But M = √ ____ L1L2 for perfect coupling (k = 1). Hence,

$$\mathbf{V}_2 = j\omega L_2 \mathbf{I}_2 + \frac{\sqrt{L_1 L_2} \mathbf{V}_1}{L_1} - \frac{j\omega L_1 L_2 \mathbf{I}_2}{L_1} = \sqrt{\frac{L_2}{L_1}} \mathbf{V}_1 = n\mathbf{V}_1$$

where n =  √ L2∕L1 and is called the turns ratio. As L1, L2, M → ∞ such that n remains the same, the coupled coils become an ideal transformer. A transformer is said to be ideal if it has the following properties:

• 1. Coils have very large reactances (L1, L2, M → ∞).
• 2. Coupling coefficient is equal to unity (k = 1).
• 3. Primary and secondary coils are lossless (R1 = 0 = R2).

An ideal transformer is a unity-coupled, lossless transformer in which the primary and secondary coils have infinite self-inductances.

Iron-core transformers are close approximations to ideal transformers. These are used in power systems and electronics.

as shown in Fig. 13.31, the same magnetic flux ϕ goes through both windings. According to F araday’s law, the v oltage across the primary winding is

$$v_1 = N_1 \frac{d\phi}{dt} \tag{13.50a}$$

while that across the secondary winding is

$$v_2 = N_2 \frac{d\phi}{dt} \tag{13.50b}$$

Dividing Eq. (13.50b) by Eq. (13.50a), we get

$$\mathsf{w}\mathsf{c}\mathsf{g}\mathsf{c}\mathsf{t}$$ $$\frac{v_2}{v_1} = \frac{N_2}{N_1} = n \tag{13.51}$$

where n is, again, the turns ratio or transformation ratio. We can use the phasor voltages V1 and V2 rather than the instantaneous values v1 and v2. Thus, Eq. (13.51) may be written as

$$\frac{\mathbf{V}_2}{\mathbf{V}_1} = \frac{N_2}{N_1} = \mathbf{0}$$

(13.52)

Figure 13.30 (a) Ideal transformer, (b) circuit symbol for an ideal transformer.

Figure 13.31 Relating primary and secondary quantities in an ideal transformer.

For the reason of po wer conservation, the ener gy supplied to the pri mary must equal the ener gy absorbed by the secondary , since there are no losses in an ideal transformer. This implies that

$$\sum_{i=1}^{n} v_i = v_1 i_1 = v_2 i_2 \quad \text{(power conserved)} \quad (13.53)$$

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

$$\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{\mathbf{V}_2}{\mathbf{V}_1} = \mathbf{\hat{u}}$$

(13.54)

$$\frac{\mathbf{I}_2}{\mathbf{I}_1} = \frac{N_1}{N_2} = \frac{1}{n}$$

(13.55)

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

A step-down transformer is one whose secondary voltage is less than its primary voltage.

A step-up transformer is one whose secondary 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 v oltage so that the po wer 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 V1 or V2 or 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 termi nals, use +n in Eq. (13.52). Otherwise, use −n.
• 2. If I1 and I² both enter into or both lea ve the dotted terminals, use −n in Eq. (13.55). Otherwise, use +n.

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

Figure 13.32 Typical circuits illustrating proper voltage polarities and current directions in an ideal transformer.

Using Eqs. (13.52) and (13.55), we can al ways express V1 in terms of V2 and I1 in terms of I2, or vice versa:

$$V_1 = \frac{V_2}{n}$$

or $V_2 = nV_1$ (13.56)

$$\mathbf{I}_1 = n\mathbf{I}_2 \qquad \text{or} \qquad \mathbf{I}_2 = \frac{\mathbf{I}_1}{n} \tag{13.57}$$

The complex power in the primary winding is

$$\hat{\mathbf{S}}_1 = \mathbf{V}_1 \mathbf{I}_1^* = \frac{\mathbf{V}_2}{n} (n \mathbf{I}_2)^* = \mathbf{V}_2 \mathbf{I}_2^* = \mathbf{S}_2 \tag{13.58}$$

The input impedance as seen by the source in Fig. 13.31 is found from

$$Z_{\text{in}} = \frac{V_1}{I_1} = \frac{1}{n^2} \frac{V_2}{I_2}$$

(13.59)

$$V_2/I_2 = Z_L$$

, so that

$$\mathbf{Z}_{\text{in}} = \frac{\mathbf{Z}_L}{n^2}$$

(13.60)

The input impedance is also called the reflected impedance, inasmuch

as it appears as if the load impedance is reflected to the primary side. This ability of the transformer to transform a gi ven impedance into another impedance provides us a means of impedance matching to ensure maximum power transfer. The idea of impedance matching is v ery useful in practice and will be discussed more in Section 13.9.2.

In analyzing a circuit containing an ideal transformer , it is com mon practice to eliminate the transformer by reflecting impedances and sources from one side of the transformer to the other . In the circuit of Fig. 13.33, suppose we want to reflect the secondary side of the circuit to the primary side. We find the Thevenin equi valent of the circuit to the right of the terminals a-b. We obtain VTh as the open-circuit voltage at terminals a-b, as shown in Fig. 13.34(a).

Figure 13.33 Ideal transformer circuit whose equivalent circuits are to be found.

Notice that an ideal transformer reflects an impedance as the square of the turns ratio.

Figure 13.34

n2

by n.

(a) Obtaining VTh for the circuit in Fig. 13.33, (b) obtaining ZTh for the circuit in Fig. 13.33.

Because terminals a-b are open, I1 = 0 = I2 so that V2 = Vs2. Hence,

$$V_{\text{Th}} = V_1 = \frac{V_2}{n} = \frac{(V_{s2})}{n}$$

(13.61)

$$I_1 = nI_2$$

and $V_1 = V_2/n$ , so that

$$Z_{\text{Th}} = \frac{V_1}{I_1} = \frac{V_2/n}{nI_2} = \frac{Z_2}{n^2} \qquad V_2 = Z_2 I_2 \tag{13.62}$$

Figure 13.35

by reflecting the secondary circuit to the primary side.

Equivalent circuit for Fig. 13.33 obtained

The general rule for eliminating the transformer and reflecting the second-

n2Z¹ Z² nVs¹ Vs² c d V2 + ‒ ‒ ⁺ ‒

by reflecting the primary circuit to the secondary side.

Equivalent circuit for Fig. 13.33 obtained

According to Eq. (13.58), the po wer remains the same, whether calcu lated on the primary or the secondary side. But realize that this reflection approach only applies if there are no e xternal connections between the primary and secondary windings. When we have external connections between the primary and secondary windings, we simply use regular mesh and nodal analysis. Examples of circuits where there are external connections between the primary and secondary windings are in Figs. 13.39 and 13.40. Also note that if the locations of the dots in Fig. 13.33 are changed, we might have to replace n by −n in order to obey the dot rule, illustrated in Fig. 13.32.

, multiply the primary voltage by n, and divide the primary current

Example 13.7

An ideal transformer is rated at 2400 ∕120 V, 9.6 kVA, and has 50 turns on the secondary side. Calculate: (a) the turns ratio, (b) the number of turns on the primary side, and (c) the current ratings for the primary and secondary windings.

Solution:

(a) This is a step-down transformer, since V¹ = 2,400 V > V² = 120 V.

$$n = \frac{V_2}{V_1} = \frac{120}{2,400} = 0.05$$

^⇒ 0.05 =  ___ ⁵⁰

N1

(b)

or

$$N_1 = \frac{50}{0.05} = 1,000 \text{ turns}$$

(c)

$$S = V_1 I_1 = V_2 I_2 = 9.6$$

kVA. Hence,

n = ___ N2 N1

$$I_1 = \frac{9,600}{V_1} = \frac{9,600}{2,400} = 4 \text{ A}$$ $$I_2 = \frac{9,600}{V_2} = \frac{9,600}{120} = 80 \text{ A} \qquad \text{or} \qquad I_2 = \frac{I_1}{n} = \frac{4}{0.05} = 80 \text{ A}$$

Practice Problem 13.7 The primary current to an ideal transformer rated at 2200∕110 V is 25 A. Calculate: (a) the turns ratio, (b) the kVA rating, (c) the secondary current.

Answer: (a) 1∕20, (b) 55 kVA, (c) 500 A.

For the ideal transformer circuit of Fig. 13.37, find: (a) the source current I1, (b) the output voltage Vo, and (c) the complex power supplied by the source. Example 13.8

Solution:

(a) The 20-Ω impedance can be reflected to the primary side and we get

$$Z_R = \frac{20}{n^2} = \frac{20}{4} = 5 \ \Omega$$

Thus,

$$\mathbf{Z}_{in} = 4 - j6 + \mathbf{Z}_R = 9 - j6 = 10.82 \underline{/ -33.69^\circ} \,\Omega$$

\n

$$\mathbf{I}_1 = \frac{120 \underline{/ 0^\circ}}{\mathbf{Z}_{in}} = \frac{120 \underline{/ 0^\circ}}{10.82 \underline{/ -33.69^\circ}} = 11.09 \underline{/ 33.69^\circ} \,\text{A}$$

(b) Because both I1 and I2 leave the dotted terminals,

$$\mathbf{I}_2 = -\frac{1}{n}\mathbf{I}_1 = -5.545/33.69^\circ \text{ A}$$ $$\mathbf{V}_o = 20\mathbf{I}_2 = 110.9/213.69^\circ \text{ V}$$

(c) The complex power supplied is

$$S = V_s I_1^* = (120/0°)(11.09/-33.69°) = 1,330.8/-33.69° VA$$

In the ideal transformer circuit of Fig. 13.38, find Vo and the complex power supplied by the source. Practice Problem 13.8

Answer: 429.4⧸ 116.57° V, 17.174⧸ −26.57° kVA.

Calculate the power supplied to the 10-Ω resistor in the ideal transformer circuit of Fig. 13.39.

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

secondary sides due to the 30-Ω resistor. We apply mesh analysis. For mesh 1,

$$-120 + (20 + 30)I_1 - 30I_2 + V_1 = 0$$

or

$$50I_1 - 30I_2 + V_1 = 120 \tag{13.9.1}$$

For mesh 2,

$$-\mathbf{V}_2 + (10 + 30)\mathbf{I}_2 - 30\mathbf{I}_1 = 0$$

or

$$-30I_1 + 40I_2 - V_2 = 0 \tag{13.9.2}$$

At the transformer terminals,

$$V_2 = -\frac{1}{2} V_1 \tag{13.9.3}$$ $$\mathbf{I}_2 = -2\mathbf{I}_1 \tag{13.9.4}$$

(Note that n = 1∕2.) We now have four equations and four unknowns, but our goal is to get I2. So we substitute for V1 and I1 in terms of V2 and I2 in Eqs. (13.9.1) and (13.9.2). Equation (13.9.1) becomes

$$-55I_2 - 2V_2 = 120 \tag{13.9.5}$$

and Eq. (13.9.2) becomes

$$15I_2 + 40I_2 - V_2 = 0 \Rightarrow V_2 = 55I_2 \qquad (13.9.6)$$

Substituting Eq. (13.9.6) in Eq. (13.9.5),

$$-165I_2 = 120 \qquad \Rightarrow \qquad I_2 = -\frac{120}{165} = -0.7272 \text{ A}$$

The power absorbed by the 10-Ω resistor is

$$P = (-0.7272)^{2}(10) = 5.3$$

W

Find V*^o* Practice Problem 13.9 in the circuit of Fig. 13.40.

For Practice Prob. 13.9.

Answer: 96 V.

13.6 Ideal Autotransformers

Unlike the conventional two-winding transformer we have considered so far, an autotransformer has a single continuous winding with a connection point called a tap between the primary and secondary sides. The tap is often adjustable so as to provide the desired turns ratio for stepping up or stepping down the voltage. This way, a variable voltage is provided to the load connected to the autotransformer.

An autotransformer is a transformer in which both the primary and the secondary are in a single winding.

Figure 13.41 sho ws a typical autotransformer . As sho wn in Fig. 13.42, the autotransformer can operate in the step-down or step-up mode. The autotransformer is a type of po wer transformer. Its major advantage o ver the tw o-winding transformer is its ability to transfer larger apparent po wer. Example 13.10 will demonstrate this. Another advantage is that an autotransformer is smaller and lighter than an equivalent two-winding transformer. However, since both the primary and secondary windings are one winding, electrical isolation (no direct electrical connection) is lost. (We will see how the property of electri cal isolation in the conventional transformer is practically employed in Section 13.9.1.) The lack of electrical isolation between the primary and secondary windings is a major disadvantage of the autotransformer.

Some of the formulas we deri ved for ideal transformers apply to ideal autotransformers as well. F or the step-do wn autotransformer cir cuit of Fig. 13.42(a), Eq. (13.52) gives

$$\frac{\mathbf{V}_1}{\mathbf{V}_2} = \frac{N_1 + N_2}{N_2} = 1 + \frac{N_1}{N_2}$$

(13.63)

As an ideal autotransformer, there are no losses, so the comple x power remains the same in the primary and secondary windings:

$$S_1 = V_1 I_1^* = S_2 = V_2 I_2^*$$

(13.64)

Equation (13.64) can also be expressed as

$$V_1I_1=V_2I_2$$

or

$$\frac{V_2}{V_1} = \frac{I_1}{I_2}$$

(13.65)

Thus, the current relationship is

$$\frac{\mathbf{I}_1}{\mathbf{I}_2} = \frac{N_2}{N_1 + N_2}$$

(13.66)

For the step-up autotransformer circuit of Fig. 13.42(b),

$$\frac{\mathbf{V}_1}{N_1} = \frac{\mathbf{V}_2}{N_1 + N_2}$$