The determination of the buck inductor value - its inductance - operated in continuous conduction mode or CCM is a fairly simple exercise. Once you have the main operating parameters in hand like switching frequency and input/output voltages, you can determine the inductance value by choosing a ripple current.
As you know, the current in an inductor \$L\$ will linearly ramp and down if you apply a voltage \$V\$ across its terminals for a short time duration (we assume this duration to be way smaller than the inductor time constant involving its series resistance \$r_L\$). The current ramps up with a slope \$S_{on}=\frac{V}{L}\$ and ramps down with a slope \$S_{off}=-\frac{V}{L}\$. The slope is expressed in [A]/[s]. Therefore, if you multiply a current slope by a time, you obtain a current.
In a CCM buck converter, when the switch turns on, a voltage equal to \$(V_{in}-V_{out}\$) is applied across the inductor during the on-time, \$t_{on}\$. The current starts from a valley current and reaches a peak value at which the switch turns off. Then, the voltage across the inductor reverses to \$V_{out}\$ and the current falls from its peak to the valley point during the off-time, \$t_{off}\$. This is what I show in the below illustration:
The peak-to-peak current, from valley to peak, is called the inductor ripple current and that is the one you have to pick. Years ago, people were choosing 10% of the output current for various reasons. One of the them was core losses linked to the high-frequency current swing. A low ripple current implies big inductors and a slow response - an inductance naturally opposes current changes. These days, considering good-quality magnetic materials, the ripple can be significantly increased and a 40% ripple is a commonly-seen value, leading to slim inductors and fast converters. This is what I have chosen for the calculations below:
From the illustration I gave, you saw that the average inductor current is actually the output current. So the triangular wave shape is centered around the output current. From this observation, the peak and valley currents are equal to the output current plus/minus half of the ripple current. Then to obtain the inductor value, you see that the current will ramp up during \$t_{on}\$ and ramp down during \$t_{off}\$. If you know the on- and off-slopes and the ripple current, you can solve for the inductor value which is 150 µH in this particular example.
For the capacitance, there are many ways to determine it and one of them, that I like, is to consider a low-impedance square-wave source feeding the load via a \$LC\$ filter. Once you have the inductor value, you can choose a cutoff frequency based on the accepted voltage ripple. This is what I used in my book on SMPS. In the sheet, I used the more commonly-used approach using charge balance. Please note that this is true for a ripple solely due to the capacitance value and no ESR is considered here.
To check if it works, a simple SIMPLIS simulation template will do well:
You can see how simulation matches the values that I calculated. You can also observe an ac source feeding the model and it's there to obtain the control-to-output transfer function, the corner stone of any serious compensation strategy. If you want simulate some of these structures freely with the demo version Elements, you can freely download my ready-made templates from my webpage.
