The network in Fig. 7.50 models an automobile ignition system. The voltage source represents the standard 12-V battery. The inductor is the ignition coil, which is magnetically coupled to the starter (not shown). The inductor’s internal resistance is modeled by the resistor, and the switch is

Question

The network in Fig. 7.50 models an automobile ignition system. The voltage source represents the standard 12-V battery. The inductor is the ignition coil, which is magnetically coupled to the starter (not shown). The inductor’s internal resistance is modeled by the resistor, and the switch is the keyed ignition switch. Initially, the switch connects the ignition circuitry to the battery, and thus the capacitor is charged to 12 V. To start the motor, we close the switch, thereby discharging the capacitor through the inductor. Assuming that optimum starter operation requires an overdamped response for [latex]i_{L}[/latex](t) that reaches at least 1 A within 100 ms after switching and remains above 1 A for between 1 and 1.5 s, let us find a value for the capacitor that will produce such a current waveform. In addition, let us plot the response, including the time interval just prior to moving the switch, and verify our design.

Accepted Answer

Before the switch is moved at t = 0, the capacitor looks like an open circuit, and the inductor acts like a short circuit. Thus,

[latex]i_{L}(0^{-}) = i_{L}(0^{+})[/latex] = 0 A and [latex]v_{C}(0^{-}) = v_{C}(0^{+})[/latex] = 12 V

After switching, the circuit is a series RLC unforced network described by the characteristic equation

[latex]s^{2} + \frac{R}{L} s + \frac{1}{LC}[/latex] = 0

with roots at s = –[latex]s_{1}[/latex] and –[latex]s_{2}[/latex]. The characteristic equation is of the form

(s + [latex]s_{1}[/latex])(s + [latex]s_{2}[/latex]) = [latex]s^{2} + (s_{1} + s_{2})s + s_{1}s_{2}[/latex] = 0

Comparing the two expressions, we see that

[latex]\frac{R}{L} = s_{1} + s_{2}[/latex] = 20

and

[latex]\frac{1}{LC} = s_{1} s_{2}[/latex]

Since the network must be overdamped, the inductor current is of the form

[latex]i_{L}(t) = K_{1} e^{-s_{1}t} + K_{2} e^{-s_{2}t}[/latex]

Just after switching,

[latex]i_{L}(0^{+}) = K_{1} + K_{2}[/latex] = 0

or

[latex]K_{2} = -K_{1}[/latex]

Also, at t = [latex]0^{+}[/latex], the inductor voltage equals the capacitor voltage because [latex]i_{L}[/latex] = 0 and therefore [latex]i_{L}[/latex]R = 0. Thus, we can write

[latex]v_{L}(0^{+}) = L \frac{di_{L}(0^{+})}{dt} ⇒ - s_{1} K_{1} + s_{2} K_{1} = \frac{12}{L}[/latex]

or

[latex]K_{1} = \frac{60}{s_{2} - s_{1}}[/latex]

At this point, let us arbitrarily choose [latex]s_{1}[/latex] = 3 and [latex]s_{2}[/latex] = 17, which satisfies the condition [latex]s_{1}+s_{2}[/latex] = 20, and furthermore,

[latex]K_{1} = \frac{60}{s_{2} - s_{1}} = \frac{60}{14} = 4.29[/latex]

C = [latex]\frac{1}{Ls_{1} s_{2}} = \frac{1}{(0.2)(3)(17)}[/latex] = 98 mF

Hence, [latex]i_{L}[/latex](t) is

[latex]i_{L}(t) = 4.29[e^{-3t} - e^{-17t} ][/latex] A

Figure 7.51a shows a plot of [latex]i_{L}[/latex](t). At 100 ms, the current has increased to 2.39 A, which meets the initial magnitude specifications. However, 1 second later, at t = 1.1 s, [latex]i_{L}[/latex](t) has fallen to only 0.16 A—well below the magnitude-over-time requirement. Simply put, the current falls too quickly. To make an informed estimate for [latex]s_{1} [/latex] and [latex]s_{2}[/latex], let us investigate the effect the roots exhibit on the current waveform when [latex]s_{2}[/latex] > [latex]s_{1}[/latex].
Since [latex]s_{2}[/latex] > [latex]s_{1}[/latex], the exponential associated with [latex]s_{2}[/latex] will decay to zero faster than that associated with [latex]s_{1}[/latex]. This causes [latex]i_{L}[/latex](t) to rise—the larger the value of [latex]s_{2}[/latex] , the faster the rise. After 5(1/[latex]s_{2}[/latex]) seconds have elapsed, the exponential associated with [latex]s_{2}[/latex] is approximately zero and [latex]i_{L}[/latex](t) decreases exponentially with a time constant of [latex] au[/latex] = 1/[latex]s_{1}[/latex]. Thus, to slow the fall of [latex]i_{L}[/latex](t), we should reduce [latex]s_{1}[/latex]. Hence, let us choose [latex]s_{1}[/latex] = 1. Since [latex]s_{1}+s_{2}[/latex] must equal 20, [latex]s_{2}[/latex] = 19. Under these conditions

C = [latex]\frac{1}{Ls_{1} s_{2}} = \frac{1}{ (0.2)(1)(19)}[/latex] = 263 mF

and

[latex]K_{1} = \frac{60}{s_{2} - s_{1}} = \frac{60}{18}[/latex] = 3.33

Thus, the current is

[latex]i_{L}(t) = 3.33 [e^{-t} - e^{-19t}][/latex] A

which is shown in Fig. 7.51b. At 100 ms the current is 2.52 A. Also, at t = 1.1 s, the current is 1.11 A—above the 1-A requirement. Therefore, the choice of C = 263 mF meets all starter specifications.

Question 7.20: The network in Fig. 7.50 models an automobile ignition syste...

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