A boost converter is shown below. It is called a boost converter because the output voltage is higher than the input When the switch is ON, inductor current increases. When the switch is OFF, inductor current decreases and charges up the capacitor. In that time, voltage at node A must be greater than V s for decreasing the inductor current. Assume voltage across the diode is zero, V o > V s Boost Converter 2
Boost Converter: Steady-State Analysis (2) 4 If the switch is open and D is zero, the output is the same as the input. As D increases, the denominator becomes smaller and the output becomes larger than the input. i.e. the boost converter can only produce an output voltage higher than or equal to the input voltage As D approaches to 1, the output voltage goes to infinity according to the equation. However, it is not the case for non-ideal components and this will be discussed later An alternative derivation using volt-second balance equation(conservation of flux in inductor): average inductor voltage is zero for periodic operation. i.e.
Boost Converter: Inductor Value(1) 5 Now the inductor current I L is not the same as the load current. In fact, I L is the source current I s. Since the input power equals to the output power, P s = V s I s = V s I L and P o = V o 2 /R, the inductor current can be computed as The maximum & minimum inductor current can be computed as
Boost Converter: Inductor Value(2) 6 Since the inductor current is always positive (CCM). To satisfy I Lmin must be greater than 0 The minimum inductance value required for CCM operation is
Boost Converter: Output Voltage Ripple 7 Similar to the buck converter, the output voltage cannot be kept perfectly constant with a finite capacitor value. The ripple voltage v r can be computed as follow When the switch is closed, the capacitor is discharged by the load current: The change in capacitor charge can be calculated from The ripple voltage is given as It should be noted that f s >> 1/RC where RC is the time constant of the output