One of the most common and necessary subcircuits that appears in a wide variety of electronic systems—for example, stereos, TVs, radios, and computers—is a quality dc voltage source or power supply. The standard wall socket supplies an alternating current (ac) voltage waveform shown in Fig.7.37a,

Question

One of the most common and necessary subcircuits that appears in a wide variety of electronic systems—for example, stereos, TVs, radios, and computers—is a quality dc voltage source or power supply. The standard wall socket supplies an alternating current (ac) voltage waveform shown in Fig. 7.37a, and the conversion of this voltage to a desired dc level is done as illustrated in Fig. 7.37b. The ac waveform is converted to a quasi-dc voltage by an inexpensive ac–dc converter whose output contains remnants of the ac input and is unregulated. A higher quality dc output is created by a switching dc–dc converter. Of the several versions of dc–dc converters, we will focus on a topology called the boost converter, shown in Fig. 7.38. Let us develop an equation relating the output voltage to the switching characteristics.

Accepted Answer

Consider the boost converter in Fig. 7.38a, where switch 1 (S1) is closed and S2 is open for a time interval [latex]t_{on}[/latex]. This isolates the inductor from the capacitor, creating two subcircuits that can be analyzed independently. Note that during [latex]t_{on}[/latex] the inductor current and stored energy are increasing while at the output node, the capacitor voltage discharges exponentially into the load. If the capacitor’s time constant ([latex] au[/latex] = RC) is large, then the output voltage will decrease slowly. Thus, during [latex]t_{on}[/latex] energy is stored in the inductor and the capacitor provides energy to the load.
Next, we change both switch positions so that S1 is open and S2 is closed for a time interval [latex]t_{off}[/latex], as seen in Fig. 7.38b. Since the inductor current cannot change instantaneously, current flows into the capacitor and the load, recharging the capacitor. During [latex]t_{off}[/latex] the energy that was added to the inductor during [latex]t_{on}[/latex] is used to recharge the capacitor and drive the load. When [latex]t_{off}[/latex] has elapsed, the cycle is repeated.
Note that the energy added to the inductor during [latex]t_{on}[/latex] must go to the capacitor and load during [latex]t_{off}[/latex]; otherwise, the inductor energy would increase to the point that the inductor would fail. This requires that the energy stored in the inductor must be the same at the end of each switching cycle. Recalling that the inductor energy is related to the current by

w(t) = [latex]\frac{1}{2} Li^{2}[/latex](t)

we can state that the inductor current must also be the same at the end of each switching cycle, as shown in Fig. 7.39. The inductor current during [latex]t_{on}[/latex] and [latex]t_{off}[/latex] can be written as

[latex]i_{L}(t) = \frac{1}{L} \int_{0}^{t_{on}} v_{L}(t) dt = \frac{1}{L} \int_{0}^{t_{on}} V_{in} dt = \Big[\frac{V_{in}}{ L}\Big] t_{on} + I_{0} [/latex] 0 < t < [latex]t_{on}[/latex]

[latex]i_{L}(t) = \frac{1}{L} \int_{t_{on}}^{t_{on}+ t_{off}} v_{L}(t) dt=\frac{1}{L}\int_{t_{on}}^{t_{on}+ t_{off}} (V_{in} - V_{o}) dt[/latex] =

[latex]\Big[\frac{V_{in} - V_{o}}{L}\Big] t_{off} + I_{0}[/latex] [latex]t_{on}[/latex] < t < [latex]t_{off}[/latex] 7.33

where [latex]I_{0}[/latex] is the initial current at the beginning of each switching cycle. If the inductor current is the same at the beginning and end of each switching cycle, then the integrals in Eq. (7.33) must sum to zero. Or

[latex]V_{in} t_{on} = (V_{o} - V_{in})t_{off} = (V_{o} - V_{in})(T - t_{on})[/latex]

where T is the period (T = [latex]t_{on} + t_{off})[/latex]. Solving for [latex]V_{o}[/latex] yields

[latex]V_{o} = V_{in}\Big[\frac{T}{T - t_{on}}\Big] = V_{in}\Big[ \frac{1}{ (T - t_{on})/T}\Big] = V_{in}\Big[ \frac{1}{(1 - t_{on}/T)}\Big] = V_{in} \Big[\frac{1}{1 - D}\Big] [/latex]

where D is the duty cycle (D = [latex]t_{on}[/latex]/T). Thus, by controlling the duty cycle, we control the output voltage. Since D is always a positive fraction, [latex]V_{o}[/latex] is always bigger than [latex]V_{in}[/latex]—thus the name boost converter. A plot of [latex]V_{o}/V_{in}[/latex] versus duty cycle is shown in Fig. 7.40.

Question 7.16: One of the most common and necessary subcircuits that appear...

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