# Buck Converter Inductor selection Maximum Switch Current Limit vs Output current

#### hoyyoth

Joined Mar 21, 2020
273
I need to select an inductor for this buck converter. Its output current is 3 A and maximum switch current limit is 6 A. As per my knowledge I need to select an inductor whose current ratings (both DC and saturation) greater than 3 A.

My question is:

When selecting an inductor, do I need to select an inductor whose current rating (DC) greater than 6 A(maximum switch current limit) and current - saturation (Isat) greater than 3 A?

#### Ian0

Joined Aug 7, 2020
8,368
Actually it's 3A plus half the ripple current for saturation, but it's not a good idea to get too close to the saturation current as the inductance starts to reduce well below that current. There should be a graph of inductance v. current in the inductor's datasheet.
The ripple current is equal to Vt/L where V is the output voltage, t is the off time and L is the inductance.

And yes, the DC output current rating should be more than the DC output current. It's a thermal rating, so it isn't going to fail immediately if you exceed it, but it will be getting rather warm!

#### ronsimpson

Joined Oct 7, 2019
2,747
There are several different things to think about.
DC current, causes heat because of copper loss. You need an inductor that can handle and average of 3A all day long. (core saturation has nothing to do with this) Read the inductor data sheet. It might be that at Max DC current the wire is at 150C. (hot)

Because the IC can do 6A you need a core that will not saturate at 6A. This might only happen at power up or if the load has a large step. At power up the current will be 5 or 6A for mS.

#### LowQCab

Joined Nov 6, 2012
3,448
Over-Rating both the Inductor, and the Capacitor,
is likely to cause much fewer performance issues
than trying to squeak-by on the bare minimum specifications.
.
.
.

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#### Ian0

Joined Aug 7, 2020
8,368
Don't be tempted to choose a lower value of inductance because it gets you a higher current rating for the same core.
Lower inductance = more ripple. Core loss is proportional to (ripple current)^(5/2).