9.8 Applications

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9.8 Applications

In Chapters 7 and 8, we sa w certain uses of RC, RL, and RLC circuits in dc applications. These circuits also have ac applications; among them are coupling circuits, phase-shifting circuits, filters, resonant circuits, ac bridge circuits, and transformers. This list of applications is ine xhaustive. We will consider some of them later. It will suffice here to observe two simple ones: RC phase-shifting circuits, and ac bridge circuits.

9.8.1 Phase-Shifters

A phase-shifting circuit is often emplo yed to correct an undesirable phase shift already present in a circuit or to produce special desired effects. An RC circuit is suitable for this purpose because its capacitor causes the circuit current to lead the applied v oltage. Two commonly used RC circuits are shown in Fig. 9.31. (RL circuits or any reactive circuits could also serve the same purpose.)

In Fig. 9.31(a), the circuit current I leads the applied voltage Vi by some phase angle θ, where 0 < θ< 90°, depending on the values of R and C. If XC = −1∕ωC, then the total impedance is Z = R + jXC, and the phase shift is given by

$$\theta = \tan^{-1} \frac{X_C}{R}$$

(9.70)

This shows that the amount of phase shift depends on the v alues of R, C, and the operating frequenc y. Since the output v oltage Vo across the resistor is in phase with the current, Vo leads (positive phase shift) Vi as shown in Fig. 9.32(a).

In Fig. 9.31(b), the output is taken across the capacitor. The current I leads the input v oltage Vi by θ, but the output voltage vo(t) across the capacitor lags (negative phase shift) the input v oltage vi(t) as illustrated in Fig. 9.32(b).

Figure 9.32 Phase shift in RC circuits: (a) leading output, (b) lagging output.

We should keep in mind that the simple RC circuits in Fig. 9.31 also act as voltage dividers. Therefore, as the phase shift θ approaches 90°, the output vo ltage Vo approaches zero. F or this reason, these simple RC circuits are used only when small amounts of phase shift are required. If it is desired to ha ve phase shifts greater than 60°, simple RC networks are cascaded, thereby providing a total phase shift equal to the sum of the indi vidual phase shifts. In practice, the phase shifts due to the stages are not equal, because the succeeding stages load down the earlier stages unless op amps are used to sepa rate the stages.

Solution:

If we select circuit components of equal ohmic v alue, say R = ∣XC∣ = 20 Ω, at a particular frequenc y, according to Eq. (9.70), the phase shift is exactly 45°. By cascading two similar RC circuits in Fig. 9.31(a), we obtain the circuit in Fig. 9.33, providing a positive or leading phase shift of 90°, as we shall soon show. Using the series-parallel combination technique, Z in Fig. 9.33 is obtained as

$$\mathbf{Z} = 20 \| (20 - j20) = \frac{20(20 - j20)}{40 - j20} = 12 - j4 \ \Omega$$

(9.13.1)

Using voltage division,

$$\mathbf{V}_1 = \frac{\mathbf{Z}}{\mathbf{Z} - j20} \mathbf{V}_i = \frac{12 - j4}{12 - j24} \mathbf{V}_i = \frac{\sqrt{2}}{3} \angle 45^\circ \mathbf{V}_i \quad (9.13.2)$$

and

$$\mathbf{V}_o = \frac{20}{20 - j20} \mathbf{V}_1 = \frac{\sqrt{2}}{2} \underline{45^\circ} \mathbf{V}_1$$

(9.13.3)

Substituting Eq. (9.13.2) into Eq. (9.13.3) yields

$$\mathbf{V}_o = \left(\frac{\sqrt{2}}{2} \angle 45^\circ\right) \left(\frac{\sqrt{2}}{3} \angle 45^\circ \mathbf{V}_i\right) = \frac{1}{3} \angle 90^\circ \mathbf{V}_i$$

Thus, the output leads the input by 90° b ut its magnitude is only about 33 percent of the input.

Design an RC circuit to pro vide a 90° lagging phase shift of the out - Practice Problem 9.13 put voltage relative to the input v oltage. If an ac v oltage of 60 V rms is applied, what is the output voltage?

Answer: Figure 9.34 shows a typical design; 20 V rms.

Figure 9.33

An RC phase shift circuit with 90° leading phase shift; for Example 9.13.

Figure 9.35 For Example 9.14.

Example 9.14 For the RL circuit shown in Fig. 9.35(a), calculate the amount of phase shift produced at 2 kHz.

Solution:

At 2 kHz, we transform the 10- and 5-mH inductances to the corresponding impedances.

$$10 \text{ mH} \Rightarrow X_L = \omega L = 2\pi \times 2 \times 10^3 \times 10 \times 10^{-3}$$ $$= 40\pi = 125.7 \text{ }\Omega$$ $$5 \text{ mH} \Rightarrow X_L = \omega L = 2\pi \times 2 \times 10^3 \times 5 \times 10^{-3}$$ $$= 20\pi = 62.83 \text{ }\Omega$$

Consider the circuit in Fig. 9.35(b). The impedance Z is the parallel combination of j125.7 Ω and 100 + j62.83 Ω. Hence,

$$\mathbf{Z} = j125.7 \parallel (100 + j62.83)$$

\n

$$\mathbf{Z} = j125.7 \parallel (100 + j62.83)$$

\n

$$= \frac{j125.7(100 + j62.83)}{100 + j188.5} = 69.56 / 60.1^{\circ} \ \Omega$$

(9.14.1)

Using voltage division,

$$\mathbf{V}_1 = \frac{\mathbf{Z}}{\mathbf{Z} + 150} \mathbf{V}_i = \frac{69.56 / 60.1^{\circ}}{184.7 + j60.3} \mathbf{V}_i$$

= 0.3582 / 42.02° $\mathbf{V}_i$ (9.14.2)

and

$$\mathbf{V}_o = \frac{j62.832}{100 + j62.832} \mathbf{V}_1 = 0.532 \, \text{/} 57.86^{\circ} \, \mathbf{V}_1 \tag{9.14.3}$$

Combining Eqs. (9.14.2) and (9.14.3),

Vo = (0.532 ⧸ 57.86°)(0.3582 ⧸ 42.02°)Vi = 0.1906 ⧸ 100° V*ⁱ*

showing that the output is about 19 percent of the input in magnitude but leading the input by 100°. If the circuit is terminated by a load, the load will affect the phase shift.

Practice Problem 9.14 Refer to the RL circuit in Fig. 9.36. If 10 V is applied to the input, find the magnitude and the phase shift produced at 5 kHz. Specify whether the phase shift is leading or lagging.

Answer: 1.7161 V, 120.39°, lagging.

9.8.2 AC Bridges

An ac bridge circuit is used in measuring the inductance L of an inductor or the capacitance C of a capacitor. It is similar in form to the Wheatstone bridge for measuring an unknown resistance (discussed in Section 4.10) and follows the same principle. To measure L and C, however, an ac

source is needed as well as an ac meter instead of the galvanometer. The ac meter may be a sensitive ac ammeter or voltmeter.

Consider the general ac bridge circuit displayed in Fig. 9.37. The bridge is balanced when no current flows through the meter. This means that V1 = V2. Applying the voltage division principle,

$$\mathbf{V}_1 = \frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} \mathbf{V}_s = \mathbf{V}_2 = \frac{\mathbf{Z}_x}{\mathbf{Z}_3 + \mathbf{Z}_x} \mathbf{V}_s$$

(9.71)

Thus,

$$\frac{\mathbf{Z}_2}{\mathbf{Z}_1 + \mathbf{Z}_2} = \frac{\mathbf{Z}_x}{\mathbf{Z}_3 + \mathbf{Z}_x} \qquad \Rightarrow \qquad \mathbf{Z}_2 \mathbf{Z}_3 = \mathbf{Z}_1 \mathbf{Z}_x \tag{9.72}$$

or

$$Z_x = \frac{Z_3}{Z_1} Z_2 \tag{9.73}$$

This is the balanced equation for the ac bridge and is similar to Eq. (4.30) for the resistance bridge except that the R’s are replaced by Z’s.

Specific ac bridges for measuring L and C are shown in Fig. 9.38, where Lx and Cx are the unknown inductance and capacitance to be measured while Ls and Cs are a standard inductance and capacitance (the values of which are kno wn to great precision). In each case, tw o resistors, R1 and R2, are varied until the ac meter reads zero. Then the bridge is balanced. From Eq. (9.73), we obtain

$$L_x = \frac{R_2}{R_1} L_s \tag{9.74}$$

and

$$C_x = \frac{R_1}{R_2} C_s \tag{9.75}$$

Notice that the balancing of the ac bridges in Fig. 9.38 does not depend on the frequency f of the ac source, since f does not appear in the relationships in Eqs. (9.74) and (9.75).

Specific ac bridges: (a) for measuring L, (b) for measuring C.

Figure 9.37

A general ac bridge.

Example 9.15 The ac bridge circuit of Fig. 9.37 balances when Z1 is a 1-kΩ resistor, Z² is a 4.2-kΩ resistor, Z3 is a parallel combination of a 1.5-MΩ resistor and a 12-pF capacitor, and f = 2 kHz. Find: (a) the series com ponents that make up Zx, and (b) the parallel components that make up Zx.

Solution:

• 1. Define. The problem is clearly stated.
• 2. Present. We are to determine the unknown components subject to the fact that they balance the given quantities. Given that a parallel and series equivalent exists for this circuit, we need to find both.
• 3. Alternative. Although there are alternative techniques that can be used to find the unknown values, a straightforward equality works best. Once we have answers, we can check them by using hand techniques such as nodal analysis or just using PSpice.
• 4. Attempt. From Eq. (9.73),

$$Z_{x} = \frac{Z_{3}}{Z_{1}} Z_{2}$$

(9.15.1)

where Zx = Rx + jXx,

$$\mathbf{Z}_1 = 1000 \,\Omega, \qquad \mathbf{Z}_2 = 4200 \,\Omega \tag{9.15.2}$$

and

$$\mathbf{Z}_3 = R_3 \parallel \frac{1}{j\omega C_3} = \frac{\frac{R_3}{j\omega C_3}}{R_3 + 1/j\omega C_3} = \frac{R_3}{1 + j\omega R_3 C_3}$$

Since R3 = 1.5 MΩ and C3 = 12 pF,

Since

$$R_3 = 1.5 \text{ M}\Omega$$

and $C_3 = 12 \text{ pF}$ ,
\n

$$\mathbf{Z}_3 = \frac{1.5 \times 10^6}{1 + j2\pi \times 2 \times 10^3 \times 1.5 \times 10^6 \times 12 \times 10^{-12}} = \frac{1.5 \times 10^6}{1 + j0.2262}$$

or

$$Z_3 = 1.427 - j0.3228 M\Omega$$

(9.15.3)

(a) Assuming that Zx is made up of series components, we substitute Eqs. (9.15.2) and (9.15.3) in Eq. (9.15.1) and obtain

$$R_x + jX_x = \frac{4200}{1000}(1.427 - j0.3228) \times 10^6$$

= (5.993 - j1.356) MΩ (9.15.4)

Equating the real and imaginary parts yields Rx = 5.993 MΩ and a capacitive reactance

$$X_x = \frac{1}{\omega C} = 1.356 \times 10^6$$

or

$$C = \frac{1}{\omega X_x} = \frac{1}{2\pi \times 2 \times 10^3 \times 1.356 \times 10^6} = 58.69 \text{ pF}$$

(b) Zx remains the same as in Eq. (9.15.4) but Rx and Xx are in parallel. Assuming an RC parallel combination,

$$\mathbf{Z}_x = (5.993 - j1.356) \text{ M}\Omega$$ $$= R_x \parallel \frac{1}{j\omega C_x} = \frac{R_x}{1 + j\omega R_x C_x}$$

By equating the real and imaginary parts, we obtain

By equating the real and imaginary parts, we obtain
\n

$$R_x = \frac{\text{Real}(\mathbf{Z}_x)^2 + \text{Imag}(\mathbf{Z}_x)^2}{\text{Real}(\mathbf{Z}_x)} = \frac{5.993^2 + 1.356^2}{5.993} = 6.3 \text{ M}\Omega$$

\n

$$C_x = -\frac{\text{Imag}(\mathbf{Z}_x)}{\omega[\text{Real}(\mathbf{Z}_x)^2 + \text{Imag}(\mathbf{Z}_x)^2]}$$

\n

$$= -\frac{-1.356}{2\pi(2000)(5.917^2 + 1.356^2)} = 2.852 \mu\text{F}$$

We have assumed a parallel RC combination which works in this case.

5. Evaluate. Let us now use PSpice to see if we indeed have the correct equalities. Running PSpice with the equivalent circuits, an open circuit between the “bridge” portion of the circuit, and a 10-volt input voltage yields the following voltages at the ends of the “bridge” relative to a reference at the bottom of the circuit:

FREQ	VM($N_0002)	VP($N_0002)
2.000E + 03	9.993E + 00	-8.634E - 03
2.000E + 03	9.993E + 00	-8.637E - 03

Because the voltages are essentially the same, then no measurable current can flow through the “bridge” portion of the circuit for any element that connects the two points together and we have a balanced bridge, which is to be expected. This indicates we have properly determined the unknowns.

There is a very important problem with what we have done! Do you know what that is? We have what can be called an ideal, “theoretical” answer, but one that really is not very good in the real world. The difference between the magnitudes of the upper impedances and the lower impedances is much too large and would never be accepted in a real bridge circuit. For greatest accuracy, the overall magnitude of the impedances must at least be within the same relative order. To increase the accuracy of the solution of this problem, I would recommend increasing the magnitude of the top impedances to be in the range of 500 kΩ to 1.5 MΩ. One additional real-world comment: The size of these impedances also creates serious problems in making actual measurements, so the appropriate instruments must be used in order to minimize their loading (which would change the actual voltage readings) on the circuit.

6. Satisfactory? Because we solved for the unknown terms and then tested to see if they worked, we validated the results. They can now be presented as a solution to the problem.

Practice Problem 9.15 In the ac bridge circuit of Fig. 9.37, suppose that balance is achie ved when Z1 is a 4.8-k Ω resistor , Z2 is a 10- Ω resistor in series with a 0.25-μH inductor, Z3 is a 12-kΩ resistor, and f = 6 MHz. Determine the series components that make up Zx.

Answer: A 25-Ω resistor in series with a 0.625-μH inductor.

9.9 Summary

1. A sinusoid is a signal in the form of the sine or cosine function. It has the general form

$$v(t) = V_m \cos(\omega t + \phi)$$

where Vm is the amplitude, ω = 2πf is the angular frequency, (ωt + ϕ) is the argument, and ϕ is the phase.

2. A phasor is a complex quantity that represents both the magnitude and the phase of a sinusoid. Given the sinusoid v(t) = Vm cos(ωt + ϕ), its phasor V is

$$\mathbf{V}=V_m\big/\phi$$

• 3. In ac circuits, v oltage and current phasors al ways ha ve a fixed relation to one another at an y moment of time. If v(t) = Vm cos(ωt + ϕv) represents the v oltage through an element and i(t) = Im cos(ωt + ϕi) represents the current through the element, then ϕi = ϕv if the element is a resistor , ϕi leads ϕv by 90° if the element is a capacitor , and ϕi lags ϕv by 90° if the element is an inductor.
• 4. The impedance Z of a circuit is the ratio of the phasor voltage across it to the phasor current through it:

$$\mathbf{Z} = \frac{\mathbf{V}}{\mathbf{I}} = R(\omega) + jX(\omega)$$

The admittance Y is the reciprocal of impedance:

$$\mathbf{Y} = \frac{1}{\mathbf{Z}} = G(\omega) + jB(\omega)$$

Impedances are combined in series or in parallel the same w ay as resistances in series or parallel; that is, impedances in series add while admittances in parallel add.

• 5. For a resistor Z = R, for an inductor Z = jX = jωL, and for a capacitor Z = −jX = 1∕jωC.
• 6. Basic circuit la ws (Ohm’s and Kirchhof f’s) apply to ac circuits in the same manner as they do for dc circuits; that is,

$$\mathbf{V} = \mathbf{Z}\mathbf{I}$$ $$\Sigma \mathbf{I}_k = 0 \quad \text{(KCL)}$$ $$\Sigma \mathbf{V}_k = 0 \quad \text{(KVL)}$$

• 7. The techniques of voltage/current division, series/parallel combination of impedance/admittance, circuit reduction, and Y-∆ transformation all apply to ac circuit analysis.
• 8. AC circuits are applied in phase-shifters and bridges.