Electronic – Switching regulator: How to determine maximum inductor current

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.

With a fixed supply voltage and a fixed inductance and a known saturation current, you not only have the maximum energy that the coil can store:

$$E = 0.5 I_{sat}^2 L$$

but also how long it takes to charge the coil with that amount of energy:

$$t_{charge} = \frac{I_{sat} L}{V_{supply}}$$

The power transferred is equal to the energy per cycle divided by the cycle period, \$t_{charge} + t_{discharge}\$.

The discharge time depends on the output voltage, with higher voltages giving shorter times, but in any case, the total period cannot be shorter than \$t_{charge}\$, so this puts an upper limit on the maximum power transfer.

Continuous conduction mode does not change this; while it allows higher switching frequencies, the energy transferred per cycle is proportionally less.