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Figure 8.1

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Figure 8.1

Typical examples of second-order circuits: (a) series RLC circuit, (b) parallel RLC circuit, (c) RL circuit, (d) RC circuit.

conditions of the ener gy storage elements and by step inputs. Later we e xamine other types of second-order circuits, including op amp circuits. We will consider PSpice analysis of second-order circuits. Finally, we will consider the automobile ignition system and smooth ing circuits as typical applications of the circuits treated in this chapter. Other applications such as resonant circuits and filters will be covered in Chapter 14.

8.2 Finding Initial and Final Values

Perhaps the major problem students f ace in handling second-order circuits is finding the initial and final conditions on circuit variables. Students are usually comfortable getting the initial and final values of v and i but often have difficulty finding the initial values of their derivatives: dv∕dt and di∕dt. For this reason, this section is e xplicitly devoted to the subtleties of getting v(0), i(0), dv(0)∕dt, di(0)∕dt, i(∞), and v(∞). Unless otherwise stated in this chapter , v denotes capacitor v oltage, while i is the inductor current.

There are two key points to keep in mind in determining the initial conditions.

First—as always in circuit analysis—we must carefully handle the polarity of voltage v(t) across the capacitor and the direction of the current i(t) through the inductor. Keep in mind that v and i are defined strictly according to the passive sign convention (see Figs. 6.3 and 6.23). One should carefully observe how these are defined and apply them accordingly.

Second, keep in mind that the capacitor voltage is always continuous so that

v(0+)=v(0−)(8.1a)v(0^+) = v(0^-) \tag{8.1a}

and the inductor current is always continuous so that

i(0+)=i(0−)(8.1b)i(0^+) = i(0^-) \tag{8.1b}

where t = 0− denotes the time just before a switching event and t = 0+ is the time just after the switching event, assuming that the switching event takes place at t = 0.

Thus, in finding initial conditions, we first focus on those variables that cannot change abruptly, capacitor voltage and inductor current, by applying Eq. (8.1). The following examples illustrate these ideas.

The switch in Fig. 8.2 has been closed for a long time. It is open at t = 0. Example 8.1 Find: (a) i(0+), v(0+), (b) di(0+)∕dt, dv(0+)∕dt, (c) i(∞), v(∞).

Solution:

(a) If the switch is closed a long time before t = 0, it means that the circuit has reached dc steady state at t = 0. At dc steady state, the inductor acts like a short circuit, while the capacitor acts like an open circuit, so we have the circuit in Fig. 8.3(a) at t = 0−. Thus,

i(0−)=124+2=2 A,v(0−)=2i(0−)=4 Vi(0^{-}) = \frac{12}{4+2} = 2 \text{ A}, \qquad v(0^{-}) = 2i(0^{-}) = 4 \text{ V}

As the inductor current and the capacitor voltage cannot change abruptly,

i(0+)=i(0−)=2i(0^+) = i(0^-) = 2

A, v(0+)=v(0−)=4v(0^+) = v(0^-) = 4 V

(b) At t = 0+, the switch is open; the equivalent circuit is as shown in Fig. 8.3(b). The same current flows through both the inductor and capacitor. Hence,

iC(0+)=i(0+)=2 Ai_C(0^+) = i(0^+) = 2 \text{ A}

Since C dv∕dt = iC, dv∕dt = iC∕C, and