I think I understand the buck converter operation (continuous conduction mode) except I'm a bit confused about the ripple current. Here is what I know so far
I understand everything up until the last statement in red. I can't seem to understand it. I'd appreciate some clarifications on ripple current. (continuous conduction mode for the inductor is assumed)
Assuming you know the relationship between Vin and Vout , the buck is in steady-state and continuous conduction.
During the charging phase, the current can be compute from:
$$V_{L}=V_{in}-V_{out}=L\frac{\mathrm{d}i }{\mathrm{d} t}$$
From our sequence commutation waveform analysis(drawing waveform for both on and off cycle), we can say:
$$ V_{in}-V_{out}=L \frac{I_{Lmax}-I_{Lmin}}{t_{on}} = L \frac{\Delta I_{L}}{t_{on}}$$
knowing:
$$ \alpha V_{in}= V_{out} $$
$$ \alpha=\frac{ t_{on}}{T} $$
We can manipulate the equation and isolate the current ripple
$$ V_{in}- \alpha V_{in}= L \frac{\Delta I_{L}}{\alpha T}$$
$$ \Delta I_{L} = \frac{V_{in}\alpha (1- \alpha)}{F L} $$
MathieuL's Answer covers the mathematical derivation.The ripple current in the inductor is important to know since the Inductor basically carries the average load current + ripple allowed.
Based on the ripple allowed, you can estimate your Inductor value (refer to the last equation in MathieuL's post).
Also, from this ripple current you can find the output Capacitor value.
