Buck converter - CCM and DCM

Let's define $$D = duty \ cycle$$ $$D' = 1-D$$

Finding output-to-input relation in case of a CCM buck is straightforward $$\begin{cases} \Delta I_L = \frac{V_I-V_O}{L}T_{on} = \frac{V_I-V_O}{Lf_{sw}}D\\ \Delta I_L = \frac{-V_O}{L}T_{off} = \frac{-V_O}{Lf_{sw}}D' \end{cases}$$ $$\frac{V_I-V_O}{Lf_{sw}}D = \frac{V_O}{Lf_{sw}}D' \qquad \rightarrow \qquad D = \frac{V_O}{V_I}$$

Output-to-input relation in a buck converter operating in DCM is quite more complicated than CCM; here the calculations. Again, D is defined as turn-on time over period and D' is turn-off time over period without the idle time. $$\begin{cases} \Delta I_L = \frac{V_I-V_O}{L}T_{on} = \frac{V_I-V_O}{Lf_{sw}}D\\ \Delta I_L = \frac{-V_O}{L}T_{off} = \frac{-V_O}{Lf_{sw}}D' \end{cases} \qquad \rightarrow \qquad \begin{cases} T_{off} = \frac{V_I-V_O}{V_O}T_{on}\\ I_{L,avg} = \frac{T_{on}+T_{off}}{2T}I_{L,pk} = \frac{V_O}{R}\\ I_{L,pk} = \frac{V_I-V_O}{L}T_{on} \end{cases}$$ $$\begin{cases} T_{off} = \frac{V_I-V_O}{V_O}T_{on}\\ I_{L,avg} = \frac{T_{on}+\frac{V_I-V_O}{V_O}T_{on}}{2T}\frac{V_I-V_O}{L}T_{on} = \frac{V_O}{R}\\ I_{L,pk} = \frac{V_I-V_O}{L}T_{on} \end{cases} \qquad \rightarrow \qquad \begin{cases} T_{off} = \frac{V_I-V_O}{V_O}T_{on}\\ I_{L,avg} = \frac{V_I T_{on}}{2T L I_O} (V_I-V_O)T_{on} = V_O\\ I_{L,pk} = \frac{V_I-V_O}{L}T_{on} \end{cases}$$ $$\begin{cases} T_{off} = \frac{V_I-V_O}{V_O}T_{on}\\ I_{L,avg} = \frac{V_I D^2}{2 f_{sw} L I_O} V_I = V_O(1+\frac{V_I D^2}{2 f_{sw} L I_O})\\ I_{L,pk} = \frac{V_I-V_O}{L}T_{on} \end{cases} \qquad \rightarrow \qquad V_O = \frac{V_I}{1+\frac{2 f_{sw} L I_O}{V_I D^2}} $$

Buck converter can operate in both CCM and DCM, according to input and output voltage, inductor value, switching frequency and load conditions. Remember that all buck converters under light or no load condition will operate in DCM, so it is important to understand when such condition is met and if it is good for our design.

Since requirements are typically fixed (input, output voltage and loads operating points) we can operate on switching frequency and inductor value, so $$\begin{cases} \Delta I_L = \frac{V_I-V_O}{L}T_{on} = \frac{V_I-V_O}{Lf_{sw}}D\\ \Delta I_L = \frac{-V_O}{L}T_{off} = \frac{-V_O}{Lf_{sw}}D' \end{cases}$$ Now we can find the critical inductance whose value set the buck to work in BCM (Boundary mode) $$\overline{i_L} - \frac{\Delta I_L}{2} = 0$$ $$\frac{V_O}{R} - \frac{V_I-V_O}{2Lf_{sw}}D = 0$$ Rearranging terms we can obtain $$L = \frac{V_I-V_O}{2V_O f_{sw}}DR = \frac{V_I-V_O}{2I_O f_{sw}}D = \frac{(1-D)R}{2f_{sw}}$$ To ensure CCM operation, we must choose an inductor larger than the one obtained from the formula, also considering inductor tolerance and saturation.