Output current effect on inductor value selection for SMPS

Question

Working on designing a power supply based on the TPS5403 SMPS. Datasheet here:

https://www.ti.com/lit/ds/symlink/tps5403.pdf

Datasheet is different than what I'm used to. Previous SMPS I've implemented usually had tables for recommended inductors, but this one has a formula:

The Io variable is the output current. The datasheet references 2A in their example. I found this unusual since the SMPS is only rated for 1.7A. For my application I expect to need less than 1.2A and it could be substantially lower at times.

I noticed if I change the Io value in the formula the recommended inductor changes a lot. A lower Io results in a much higher recommended inductance. What is the meaning of this? Should I always use Io=2A like the example they provide? 1.2A? Something else? Other power supplies I've implemented don't recommend changing the inductor based on output current.

Answer 1

This is a pretty common formula used for calculating what the output inductor value should be. There’s a few design considerations that come into choosing an inductor value.

• Size
• Cost
• Output ripple current (so you’re operating in CCM during nominal operation).
• Output ripple voltage.

The inductor controls the output ripple current of the converter. The higher the inductance value, the lower the ripple will be. If the ripple amplitude is greater than the output current, your converter will operate in discontinuous conduction mode (DCM) to try to maintain the output voltage, but this will affect the stability and efficiency of the converter. In most applications you want to avoid this.

It’s common practice to choose the ripple current to be around 10-40% of the maximum DC output current. This way you don’t usually use a bigger inductor than you need, but you’ll usually be operating in continuous conduction mode (CCM). Once you have this value it’s a good idea to rerun the calculation with \$I_O\$ as your minimum expected operational current and make sure the output ripple current ratio is less than 200%. That way you make sure you stay in CCM during operation.

Another reason to have a larger inductor is because it lowers your output ripple voltage. Your output capacitors also affect this, so there’s usually a trade off between capacitor size and inductor size to find a good balance to get the output ripple voltage within spec.

For your case I would probably choose the smallest inductor you can that would keep the converter in CCM under most operational cases and then pick capacitors to reduce the output ripple to where you want and keep the output stable. Once you have a good value you can choose an inductor that has reasonable DC resistance and make sure the max saturation current is greater than the maximum DC output current plus the output current ripple amplitude.

I would also simulate the design once I was done to double check it will behave the way I expect. You can use the free TINA-TI Spice simulator to simulate TI’s parts.

Answer 2

Io*LIR is basically the ripple current in the inductor. It is reasonable to have this 30 % of the maximum load current, but in efficiency or cost considerations may be optimized with a different value.

As with many electronics calculations, a significant safety factor or overdesign parameter gives margin for unexpected performance or other requirements.

That is why they use 2 A for a 1.7 A part. If you know your current is max. 1.2 A, you can use a different L; this will have a small effect on your DCDC efficiency as well as the quality of the output (ripple, and response to load steps).

A lower value of inductor may have slightly less efficiency (but that depends on the balance between inductor and IC losses).

Note that the next line in that datasheet has an incorrect calculation for deltaIL -- it is not divided by I0, but by L.