Electronic – Testing high power inductors

That looks like a 220 µH inductor. These size inductors (like a lot of electronic parts) are labeled with something a lot like a floating point format in a computer. The first N-1 digits are the mantissa and the last digit is the exponent of 10 to apply to the integer mantissa value. In this case, the mantissa is 22 and the exponent 1. The value is therefore 22 x 10^1 = 220. The fact that the units are µH is obvious from the general size and windings. Some very small inductors are labeled like this in nH, but this one is obviously a lot bigger than 220 nH.

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.