Figure 1.a shows a step-up or a PWM boost converter. It consists of dc input voltage source $V_{S},$ boost inductor $L,$ controlled switch $S,$ diode $D,$ filter capacitor $C,$ and load resistance $R .$

The converter waveforms in the CCM are presented in Fig.1.b. When the switch S is in the on state, the current in the boost inductor increases linearly and the diode D is off at that time. When the switch S is turned off, the energy stored in the inductor is released through the diode to the output $R C$ circuit.

Using Faraday's law for the boost inductor

$$V_{S} D T=\left(V_{O}-V_{S}\right)(1-D) T-----(1)$$

from which the dc voltage transfer function turns out to be

$$M_{V} \equiv \frac{V_{O}}{V_{S}}=\frac{1}{1-D}-----(2)$$

As the name of the converter suggests, the output voltage is always greater than the input voltage.

The boost converter operates in the CCM for $L \gt L_{b}$ where,

$$L_{b}=\frac{(1-D)^{2} D R}{2 f}-----(2)$$

For $D=0.5, R=10 \Omega,$ and $f=100 \mathrm{kHz}$ , the boundary value of the inductance is $L_{b}=6.25 \mu \mathrm{H}$ .

As shown in Fig. 1.b , the current supplied to the output $R C$ circuit is discontinuous. Thus, a larger filter capacitor is required in comparison to that in the buck-derived converters to limit the output voltage ripple. The filter capacitor must provide the output dc current to the load when the diode $D$ is off. The minimum value of the filter capacitance that results in the voltage ripple $V_{r}$ is given by $$C_{\min }=\frac{D V_{O}}{V_{r} R f}-----(3)$$

At $D=0.5, V_{r} / V_{O}=1 \%, R=10 \Omega,$ and $f=100 \mathrm{kHz}$ , the minimum capacitance for the boost converter is $C_{\min }=50 \mu \mathrm{F}$ .