13.9 Applications

← Back to Fundamentals of Electric Circuits Overview

13.9 Applications

Transformers are the largest, the heaviest, and often the costliest of cir cuit components. Nevertheless, they are indispensable passive devices in electric circuits. They are among the most efficient machines, 95 percent efficiency being common and 99 percent being achie vable. They have numerous applications. For example, transformers are used:

• To step up or step down voltage and current, making them useful for power transmission and distribution.
• To isolate one portion of a circuit from another (i.e., to transfer power without any electrical connection).
• As an impedance-matching device for maximum power transfer.
• In frequenc y-selective circuits whose operation depends on the response of inductances.

Practice Problem 13.14

For more information on the many kinds of transformers, a good text is W. M. Flanagan, Handbook of Transformer Design and Applications, 2nd ed. (New York: McGraw-Hill, 1993).

Figure 13.59

A transformer used to isolate an ac supply from a rectifier.

Because of these di verse uses, there are man y special designs for transformers (only some of which are discussed in this chapter): v oltage transformers, current transformers, power transformers, distribution transformers, impedance-matching transformers, audio transformers, single-phase transformers, three-phase transformers, rectifier transformers, inverter transformers, and more. In this section, we consider three important applications: transformer as an isolation de vice, transformer as a matching device, and power distribution system.

13.9.1 Transformer as an Isolation Device

Electrical isolation is said to exist between two devices when there is no physical connection between them. In a transformer, energy is transferred by magnetic coupling, without electrical connection between the primary circuit and secondary circuit. We no w consider three simple practical examples of how we take advantage of this property.

First, consider the circuit in Fig. 13.59. A rectifier is an electronic circuit that converts an ac supply to a dc supply. A transformer is often used to couple the ac supply to the rectifier. The transformer serves two purposes. First, it steps up or steps do wn the voltage. Second, it pro vides electrical isolation between the ac po wer supply and the rectifier, thereby reducing the risk of shock hazard in handling the electronic device.

As a second e xample, a transformer is often used to couple tw o stages of an amplifier, to prevent any dc voltage in one stage from affecting the dc bias of the next stage. Biasing is the application of a dc voltage to a transistor amplifier or any other electronic device in order to produce a desired mode of operation. Each amplifier stage is biased separately to operate in a particular mode; the desired mode of operation will be compromised without a transformer providing dc isolation. As shown in Fig. 13.60, only the ac signal is coupled through the transformer from one stage to the next. We recall that magnetic coupling does not exist with a dc voltage source. Transformers are used in radio and TV receivers to couple stages of high-frequency amplifiers. When the sole purpose of a transformer is to pro vide isolation, its turns ratio n is made unity . Thus, an isolation transformer has n = 1.

As a third example, consider measuring the voltage across 13.2-kV lines. It is obviously not safe to connect a voltmeter directly to such highvoltage lines. A transformer can be used both to electrically isolate the line power from the v oltmeter and to step do wn the v oltage to a safe level, as shown in Fig. 13.61. Once the voltmeter is used to measure the

Figure 13.60 A transformer providing dc isolation between two amplifier stages.

Figure 13.61

A transformer providing isolation between the power lines and the voltmeter.

secondary voltage, the turns ratio is used to determine the line v oltage on the primary side.

Determine the voltage across the load in Fig. 13.62.

Solution:

We can apply the superposition principle to find the load voltage. Let vL = vL¹ + vL2, where vL1 is due to the dc source and vL2 is due to the ac source. We consider the dc and ac sources separately, as shown in Fig. 13.63. The load voltage due to the dc source is zero, because a timevarying voltage is necessary in the primary circuit to induce a voltage in the secondary circuit. Thus, vL¹ = 0. For the ac source and a value of Rs so small it can be neglected,

$$\frac{\mathbf{V}_2}{\mathbf{V}_1} = \frac{\mathbf{V}_2}{120} = \frac{1}{3} \quad \text{or} \quad \mathbf{V}_2 = \frac{120}{3} = 40 \text{ V}$$

Hence, VL² = 40 V ac or vL² = 40 cos ωt; that is, only the ac voltage is passed to the load by the transformer. This example shows how the transformer provides dc isolation.

For Example 13.15: (a) dc source, (b) ac source.

Refer to Fig. 13.61. Calculate the turns ratio required to step down the 14.4-kV line voltage to a safe level of 120 V.

Practice Problem 13.15

Answer: 120.

13.9.2 Transformer as a Matching Device

We recall that for maximum po wer transfer, the load resistance RL must be matched with the source resistance Rs. In most cases, the two resistances are not matched; both are fixed and cannot be altered. However, an iron-core transformer can be used to match the load resistance to the source resistance. This is called impedance matching. For example, to connect a loudspeaker to an audio power amplifier requires a transformer, because the speak er’s resistance is only a few ohms while the internal resistance of the amplifier is several thousand ohms.

Consider the circuit shown in Fig. 13.64. We recall from Eq. (13.60) that the ideal transformer reflects its load back to the primary with a

Figure 13.64

Transformer used as a matching device.

Figure 13.62 For Example 13.15.

12 V dc

• ‒

scaling factor of n² . To match this reflected load RL∕n² with the source resistance Rs, we set them equal,

$$R_s = \frac{R_L}{n^2} \tag{13.73}$$

Equation (13.73) can be satisfied by proper selection of the turns ratio n. From Eq. (13.73), we notice that a step-down transformer (n < 1) is needed as the matching de vice when Rs > RL, and a step-up ( n > 1) is required when Rs < RL.

The ideal transformer in Fig. 13.65 is used to match the amplifier circuit to the loudspeak er to achie ve maximum po wer transfer. The Thevenin (or output) impedance of the amplifier is 192 Ω, and the internal impedance of the speaker is 12 Ω. Determine the required turns ratio.

Solution:

Speaker

We replace the amplifier circuit with the Thevenin equivalent and reflect the impedance ZL = 12 Ω of the speaker to the primary side of the ideal transformer. Figure 13.66 shows the result. For maximum power transfer,

$$Z_{\text{Th}} = \frac{Z_L}{n^2}$$

or $n^2 = \frac{Z_L}{Z_{\text{Th}}} = \frac{12}{192} = \frac{1}{16}$

Thus, the turns ratio is n = 1∕4 = 0.25.

Using P = I 2 R, we can sho w that indeed the po wer delivered to the speaker is much lar ger than without the ideal transformer. Without the ideal transformer, the amplifier is directly connected to the speaker. The power delivered to the speaker is

$$P_L = \left(\frac{\mathbf{V}_{\text{Th}}}{\mathbf{Z}_{\text{Th}} + \mathbf{Z}_L}\right)^2 \mathbf{Z}_L = 288 \text{ V}_{\text{Th}}^2 \mu \text{W}$$

With the transformer in place, the primary and secondary currents are

$$I_p = \frac{\mathbf{V}_{\text{Th}}}{\mathbf{Z}_{\text{Th}} + \mathbf{Z}_L/n^2}, \qquad I_s = \frac{I_p}{n}$$

Hence,

$$P_L = I_s^2 \mathbf{Z}_L = \left(\frac{\mathbf{V}_{\text{Th}}/n}{\mathbf{Z}_{\text{Th}} + \mathbf{Z}_L/n^2}\right)^2 \mathbf{Z}_L$$ $$= \left(\frac{n\mathbf{V}_{\text{Th}}}{n^2 \mathbf{Z}_{\text{Th}} + \mathbf{Z}_L}\right)^2 \mathbf{Z}_L = 1,302 \mathbf{V}_{\text{Th}}^2 \mu \mathbf{W}$$

confirming what was said earlier.

Calculate the turns ratio of an ideal transformer required to match a 8-Ω load to a source with internal impedance of 800 Ω. Find the load voltage when the source voltage is 300 V.

Answer: 0.1, 15 V.

Figure 13.66 Equivalent circuit of the circuit in Fig. 13.65; for Example 13.16.

Using an ideal transformer to match the speaker to the amplifier; for

Figure 13.65

Example 13.16.

Practice Problem 13.16

13.9.3 Power Distribution

A po wer system basically consists of three components: generation, transmission, and distrib ution. The local electric compan y operates a plant that generates se veral hundreds of me gavolt-amperes (MV A), typically at about 18 kV . As Fig. 13.67 illustrates, three-phase step-up transformers are used to feed the generated po wer to the transmission line. Why do we need the transformer? Suppose we need to transmit 100,000 VA over a distance of 50 km. Since S = VI, using a line voltage of 1,000 V implies that the transmission line must carry 100 A and this requires a transmission line of a large diameter. If, on the other hand, we use a line voltage of 10,000 V, the current is only 10 A. The smaller current reduces the required conductor size, producing considerable savings as well as minimizing transmission line I 2 R losses. To minimize losses requires a step-up transformer. Without the transformer, the majority of the power generated would be lost on the transmission line. The ability of the transformer to step up or step down voltage and distribute power economically is one of the major reasons for generating ac rather than dc. Thus, for a gi ven power, the lar ger the v oltage, the better . Today, 1 MV is the lar gest voltage in use; the le vel may increase as a result of research and experiments.

Figure 13.67

A typical power distribution system.

Source: A. Marcus and C. M. Thomson, Electricity for Technicians, 2nd edition, © 1975, p. 337. Pearson Education, Inc., Upper Saddle River, NJ.

Beyond the generation plant, the power is transmitted for hundreds of miles through an electric netw ork called the power grid. The threephase power in the po wer grid is con veyed by transmission lines hung overhead from steel towers which come in a v ariety of sizes and shapes. The (aluminum-conductor, steel-reinforced) lines typically ha ve overall diameters up to about 40 mm and can carry current of up to 1,380 A.

At the substations, distrib ution transformers are used to step do wn the voltage. The step-do wn process is usually carried out in stages. Power may be distributed throughout a locality by means of either overhead or under ground cables. The substations distrib ute the po wer to residential, commercial, and industrial customers. At the receiving end, a residential customer is e ventually supplied with 120 ∕240 V, while industrial or commercial customers are fed with higher voltages such as One may ask, How would increasing the voltage not increase the current, thereby increasing I 2 ^R losses? Keep in mind that ^I = ^Vℓ∕R, where Vℓ is the potential difference between the sending and receiving ends of the line. The voltage that is stepped up is the sending end voltage V, not Vℓ. If the receiving end is V^R, then V^ℓ = V − V^R. Since V and VR are close to each other, ^Vℓ is small even when V is stepped up. 460∕208 V. Residential customers are usually supplied by distrib ution transformers often mounted on the poles of the electric utility company. When direct current is needed, the alternating current is converted to dc electronically.

Example 13.17

A distribution transformer is used to supply a household as in Fig. 13.68. The load consists of eight 100-W bulbs, a 350-W TV, and a 15-kW kitchen range. If the secondary side of the transformer has 72 turns, calculate: (a) the number of turns of the primary winding, and (b) the current Ip in the primary winding.

Figure 13.68 For Example 13.17.

Solution:

(a) The dot locations on the winding are not important, since we are only interested in the magnitudes of the variables involved. Since

$$\frac{N_p}{N_s} = \frac{V_p}{V_s}$$

we get

$$N_p = N_s \frac{V_p}{V_s} = 72 \frac{2,400}{240} = 720 \text{ turns}$$

(b) The total power absorbed by the load is

$$S = 8 \times 100 + 350 + 15{,}000 = 16.15$$

kW

But S = VpIp = VsIs, so that

$$I_p = \frac{S}{V_p} = \frac{16,150}{2,400} = 6.729 \text{ A}$$

Practice Problem 13.17

In Example 13.17, if the eight 100-W bulbs are replaced by twelve 60-W bulbs and the kitchen range is replaced by a 4.5-kW air- conditioner, find: (a) the total power supplied, (b) the current Ip in the primary winding.

Answer: (a) 5.57 kW, (b) 2.321 A.

13.10 Summary

1. Two coils are said to be mutually coupled if the magnetic flux ϕ emanating from one passes through the other . The mutual induc tance between the two coils is given by

$$M = k\sqrt{L_1 L_2}$$

where k is the coupling coefficient, 0 < k < 1.

2. If v1 and i1 are the voltage and current in coil 1, while v2 and i2 are the voltage and current in coil 2, then

$$v_1 = L_1 \frac{di_1}{dt} + M \frac{di_2}{dt}$$

and $v_2 = L_2 \frac{di_2}{dt} + M \frac{di_1}{dt}$

Thus, the voltage induced in a coupled coil consists of self-induced voltage and mutual voltage.

• 3. The polarity of the mutually-induced v oltage is e xpressed in the schematic by the dot convention.
• 4. The energy stored in two coupled coils is

$$\frac{1}{2}L_1i_1^2 + \frac{1}{2}L_2i_2^2 \pm Mi_1i_2$$

• 5. A transformer is a four -terminal de vice containing tw o or more magnetically coupled coils. It is used in changing the current, v oltage, or impedance level in a circuit.
• 6. A linear (or loosely coupled) transformer has its coils wound on a magnetically linear material. It can be replaced by an equi valent T or Π network for the purposes of analysis.
• 7. An ideal (or iron-core) transformer is a lossless (R1 = R2 = 0) transformer with unity coupling coef ficient (k = 1) and infinite inductances (L1, L2, M → ∞).
• 8. For an ideal transformer,

$$\mathbf{V}_2 = n\mathbf{V}_1, \qquad \mathbf{I}_2 = \frac{\mathbf{I}_1}{n}, \qquad \mathbf{S}_1 = \mathbf{S}_2, \qquad \mathbf{Z}_R = \frac{\mathbf{Z}_L}{n^2}$$

where n = N2∕N1 is the turns ratio. N1 is the number of turns of the primary winding and N2 is the number of turns of the second ary winding. The transformer steps up the primary voltage when n > 1, steps it do wn when n < 1, or serv es as a matching de vice when n = 1.

• 9. An autotransformer is a transformer with a single winding common to both the primary and the secondary circuits.
• 10. PSpice is a useful tool for analyzing magnetically coupled circuits.
• 11. Transformers are necessary in all stages of po wer distribution systems. Three-phase voltages may be stepped up or do wn by threephase transformers.
• 12. Important uses of transformers in electronics applications are as electrical isolation devices and impedance-matching devices.