We previously introduced the buck converter, which is a converter that is capable of stepping down the voltage ideally with very high efficiency. The key steps were first to introduce a switch network. Here is a single-pole double-throw switch that in practice is built using power transistors and diodes that can switch back and forth very quickly. The output voltage of the switch, Vs of t. Then as a waveform that steps up and down. When the switch is in position one, the output voltage Vs of t is equal to the input voltage Vg. Likewise, when the switch is in position two the output voltage is equal to zero. We repetitively switch the devices or this switch network, with a period that is called the switching period T sub s. The switching frequency F sub s then is equal to one over T sub s. So, the switch is in position one for a fraction of time called the duty cycle D. D is a number between zero and one so that we're in position one for time D times Ts. Then, we're in position two for the remainder of the switching period. So that would be for a time one minus D times Ts, okay? It's traditional to use the notation that one minus D is equal to D prime and that's called the compliment of D. So, D prime is also a number that is between zero and one, okay? What do we accomplish then with this switch? It reduces the DC component of the voltage so that Vs of t has a lower DC component than does Vg. We previously calculated this DC component using Fourier series. So from Fourier series, the DC component is found by integrating the wave form over one period and dividing by the period, and that's effectively finding the average value of the waveform. So, to fine the integral, we would integrate the waveform or find the area under the curve. That area is in fact easy to find. It's simply the area of this rectangle. So, the area is equal to the base of the rectangle which is D times Ts, multiplied by the height of the rectangle which is the voltage, input voltage Vg. So, we plugged that area in for the integral and then divide by the period and what we get is that our DC component of VS is equal to the duty cycle times Vg. Then, DC component is also equal, is the average of the waveform. So, the switch network performs the function then of changing the DC component of the voltage and it actually reduces that DC component by a factor of the duty cycle D. Now, generally, we don't like to apply this switched waveform to our load. We would like to produce a nice steady output voltage without this switching. But the problem is that the switch waveform in addition to having the desired DC component also has these high frequency AC components in the Fourier series. These AC components are at the switching frequency and its harmonics. So, how should we remove those switching components and just apply the DC component to our load? Well, we remove the switching harmonics by introducing a low-pass filter. We can build a low-pass filter using elements that ideally are lossless with a circuit such as this one. We can insert an inductor and a capacitor that will perform as a low-pass filter with a cutoff frequency of that we call frequency F-nought or F-zero. So, the low-pass filter, what the low-pass filter does is to pass components in the Fourier series that are at frequencies less than the cutoff F-nought and reject the frequency components that are greater than F-nought. So, if we choose the values of L and C to make the cut-off frequency much less than the switching frequency Fs, then our low pass filter will reject the switching frequency and its harmonics but it will pass the DC component so that the output voltage V of t then will be essentially constant and equal to the DC component DVg. So then, we've succeeded in building a converter that has very high efficiency and uses elements that ideally are lossless and not only that, but this DC output voltage is adjustable. By changing simply the timing of the switching to increase or decrease the duty cycle, we can change the output voltage. So, we can control the output voltage or make it follow whatever wave form we like. Okay, that is called the buck converter. In the power business, the term buck denotes a reduction of voltage. So the what we call the conversion ratio of the converter M of D which is defined as the ratio of the DC output voltage to the DC input voltage at M of D is equal to D for the buck converter and so the output voltage is reduced. Now, this is a commonly used and well known converter, but there are many others. The boost converter is another well-known converter and it's built by interchanging the positions of the inductor and the switch. It turns out that the boost converter can increase the voltage. So, the output voltage is greater in magnitude than the input and it is controllable by adjustment of D. Another well-known converter is the buck-boost converter shown below here. The buck-boost converter can invert the polarity of the voltage. So, with the positive input, we get a negative output. The magnitude of the output voltage can be either increased or decreased by adjusting D. In fact, there are many converters known. We can build a converter by some circuit containing Ls and Cs in embedded switches and change any voltage into any other voltage we like. So, we found the output voltage of the buck converter using these arguments of Fourier analysis and low-pass filtering. But it may not be immediately obvious how to extend those arguments to find the output voltage of the boost or the buck-boost. It's actually possible but it may require some additional insight and cleverness. So, the object in chapter two is to develop a systematic approach to solve for the voltages and currents of a switching converter. Okay. So, we're going to introduce first the key small ripple approximation that lets us solve the wave forms of the circuit with relative ease and simple equations. This small ripple approximation is valid in well-designed converters. With that approximation then, we will develop the principles of inductor volt second balance and capacitor amp-second or charge balance that let us get a system of equations that can be solved for the voltages and currents of the converter. We'll also develop some simple methods for choosing the values of inductance and capacitance in the converter and I'll illustrate the use of all of these techniques with some examples.