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From our parts list we will add a capacitor to our switched inductor converter and get the buck converter as shown in Figure 3-5. 
Figure 3-5: Buck Converter
The buck converter goes by many names, voltage step-down converter, current step-up converter, chopper, direct converter, et .al. No matter the name, converters derived from this topology account for a substantial percentage of all converters sold. Understanding its operation is basic to switching-mode power supply design. Our first problem is to select initial values of L and C. Later the design can be optimized. In our switched inductor example we used the rule of thumb of designing L to get a peak-to-peak ripple current of 10% of the full load output current and got a value of 75 uH. We will use overshoot to get a value for C and then check the output ripple.
The problem of overshoot, Output Filter Design - Overshoot, is discussed in the problem section of the SMPS Technology Knowledge Base and it is an important consideration in output filter design -- enough so that it is prudent to consider overshoot first, rather than output ripple as is normally done, so that overshoot is not forgotten. We will use a simple rule of thumb of making the characteristic impedance of the filter, Zo, equal the load resistor, which gives an overshoot of SQRT(2)= 1.41 and a slightly over-damped filter at full load. For our example, if our full load 20 A is removed, the 5 Vdc voltage goes to (5 V)*1.41 = 7.07 V. Solving Zo=SQRT(L/C) for C = L/(Zo*Zo) = 75E-6/(0.25*0.25) = 1,200 uF In checking for ripple we will use another formula that is used enough in switching-mode power supply design that it is worth memorizing, V = (1/C)*INTEGRAL( i(t)*dt), which combined with V=L*(di/dt) allows us to calculate output ripple due to the capacitor capacitive impedance (which is often less than the contribution of capacitor equivalent series resistance (ESR) to output ripple). Example:
Solving for current as a function of time in the output capacitor
The current in the capacitor is the triangular current in the inductor which has the slope K
K = di/dt = V/L, from V=L*(di/dt)
K = (12 V - 5 V)/(75 uH) = 0.093E+6 A/s
Solving for capacitor ripple voltage
V = (1/C)*INTEGRAL ((K*t)dt)
= (K/(2*C)*t^2
where t = Ton
Ton = D*T = 0.417*(50E-6 s) = 20.85E-6 s
V = ((0.093E+6 A/s)/(2*1200E-6 F))*(20.85E-6 s)*(20.85E-6 s)
= 0.0168 V With less than 17 mV ripple due to the capacitance of C and assuming a 50 mV ripple specification, this would leave 33 mV for ripple due to the equivalent series resistance of C or for design margin. If we needed less overshoot or less ripple, we could experiment with other values of LC, however, this seems good enough for our tutorial purposes. Two useful properties for any filter are the characteristic impedance, Zo=SQRT(L/C), and the cutoff frequency, Fo = 1/(2*PI*SQRT(L*C). These can be calculated or read off of impedance graph paper. Impedance graph paper is so useful that it should always be handy. How to use it for many purposes will be discussed in this tutorial. Using impedance paper we look for the intersection of L and C and read Zo on the Ordinate and Fo on the Abscissa. Or we can calculate it. Example:
Zo = SQRT(L/C) = SQRT(75E-6/1200E-6) = 0.25 ohms
Fo = 1/(2*PI*SQRT(LC)) = 1/(6.28*SQRT(75E-6*1200E-6) = 532 Hz We will start our understanding of the buck converter by turning the switch on and off and examining what happens, first as a function of time and then in the state-plane. Do not use this information for design without independent verification of the information. Editor: Jerrold Foutz |
