Ideal ripple current on a DC to DC step down switching regulator

Choosing an inductor value for a buck regulator comes directly from V = \$\frac{\text{L di} }{\text{dt}}\$ . Where V is the voltage across the inductor, and i is the current through it. First, you want to design for the case where the inductor is in continuous conduction mode (CCM). This means that energy in the inductor doesn't run out during the switching cycle. So, there are two states, one where the switch is on, and another where the switch is off (and the rectifier is on). Voltage across the inductor during each state is essentially a constant (although it is a different value for each state). Anyway since the voltage is a constant, the inductor equation can be linearized (and rearranged to give L).

• L = \$\frac{V \text{$\Delta $t}}{\text{$\Delta $I}}\$ this is the basis for the equation you saw in the app-note.

• \$\text {$\Delta $I}\$ is something you define, not determine.

You will want to maintain CCM operation, so define \$\text {$\Delta $I}\$ as some small fraction of inductor current (I). A good choice is 10% of I. So, for your case \$\text {$\Delta $I}\$ would be 0.24A. This will also define the ripple current in the output capacitors, and less ripple current means less ripple voltage on the output.

Now you can choose an optimal value of L using \$V_{\text{in}}\$ and \$V_o\$ (and hence the duty cycle D = \$\frac {V_o} {V_ {\text {in}}}\$). But you can also make a quick over estimate for the inductance where you don't consider \$V_{\text{in}}\$ using L ~ \$\frac{10 V_o}{I_o F_{\text{sw}}}\$ (for more on this look here How to choose a inductor for a buck regulator circuit? ). An over estimate can be worthwhile, especially if you are early in development or uncertain exactly how much the output current will be (output current tends to end up higher than expected usually).

Since you are looking at Linear Tech you should (as Anindo Ghosh pointed out) also look at using their CAD support.

You've answered the question yourself - "with given values".

The amount of ripple current is defined by the buck inductor value and the volt-seconds applied to it. If you change operating point (different input voltage, different output voltage, different switching frequency) or change the inductance value, the ripple current will change.

Why is there so much ripple? That's how the circuit was designed. A DCM buck would have even higher ripple, which is a trade-off for simpler loop stabilization. If you want less ripple, make it CCM and design your slightly more complex loop accordingly.

The amount of current ripple applied to Cout is what defines the amount of ripple voltage you get on the output. So yes, in a manner of speaking, Cout "takes care of" the ripple current in so far as it deals with whatever's applied. If Cout is too small, your output won't stay in regulation.

You haven't specified enough about your application, but there is no reason why a 300mA buck operating at 247mA of ripple current couldn't be stable. It would be barely in DCM mode, but that's fine as long as your output capacitor ESR could handle the ripple and your application is OK with the ripple voltage.