Im trying to play around with some buck converters and have found one that I might want to make a circuit with. The IC I want to use is the the TPS53125 in QFN package. It's a dual output buck converter IC which takes between 4.5 V and 24 V.

Data sheet

When I use equation 3 from the datasheet to find the suggested inductor value for regulating 1.8 V with a max input of 24V and a output correct of 4 A I end up with the value of 4 uH for the inductor.


enter image description here

enter image description here

4uH is somewhere in the range that I would expect and sounds correct.

My problem is that in the example circuit above they use a 1.5 uH inductor although they try to regulate 1.8 V at a max of 4 A which should use a 4 uH inductor if you follow equation 3, but instead they use 1.5 uH:

enter image description here

So now I'm very unsure who is wrong? Is the diagram using an inductor value which doesn't conform to the equation or is my calculation faulty?

  • \$\begingroup\$ Have you double-checked their value for ripple current and their operating frequency? \$\endgroup\$
    – TimWescott
    Nov 3, 2018 at 15:34
  • \$\begingroup\$ @TimWescott at the top of the Datasheet it states: 350-kHz Switching Frequency im not sure it that's the same as operating frequency but I would guess so. And for the second part of your question,- right above Equation 3 it states that: The inductance value is selected to provide approximately 30% peak to peak ripple current at maximum load. but that ripple current shouldn't affect my math since I used the far right equation in they supplied wich includes it in the ripple current by multiplying the maximum output current by 0.3 as such: "0.3xIo1" \$\endgroup\$
    – Mercury
    Nov 3, 2018 at 15:42
  • \$\begingroup\$ Then I'm not sure what is going on. It looks like they chose the inductor values for close to 100% ripple. The equation certainly looks right. \$\endgroup\$
    – TimWescott
    Nov 3, 2018 at 17:13

1 Answer 1


TI provides the sample design targeting 12V Vin. From the datasheet:

enter image description here

This explains the discrepancy between formulas (that assume 24V Vin) and the value of inductor in the example design. They should however mention this fact in their schematics, which boldly says "4.5V to 24 V", and you can try to advise them to correct the schematics.

However, I think the conversion from 24V to 1.8V is unwise, it will suffer in terms of efficiency.

CORRECTION: It is true that 12V V in doesn't change much the result. However, the whole selection of inductor uses just 30% ripple current at max load, and the datasheet has another note:

Note:The calculation above shall serve as a general reference. To further improve transient response, the output inductance could be reduced further. This needs to be considered along with the selection of the output capacitor.

So if you (I mean quality of your output capacitor) can afford 75% current swing in the inductor, then all numbers will align.

I would say this: don't confuse yourself with formulas too much, all simplified formulas are derived under many assumptions, which may or may not be exactly true. Always use the manufacturer's reference design as the main guidance. They build it, thoroughly characterized it, and sell it. More, if you want to avoid surprises, follow their recommended (tested!!!) layouts, and get their particular inductor from their BOM - not all inductors are made equal.

  • \$\begingroup\$ The way I understand your answer is that the Vin max for the circuit in the Datasheet should actually be 12V rather than 24V. But this does not solve my problem, because if you put 12V rather then 24V in the equation it gives 3.6uH which is still not what is shown in the circuit. Maybe I misunderstood your reply, in that case could you please elaborate what you mean. \$\endgroup\$
    – Mercury
    Nov 3, 2018 at 19:12
  • \$\begingroup\$ So for a device which needs 1.8V with high current what do you suggest? Using an intermediate rail? \$\endgroup\$
    – jaskij
    Nov 3, 2018 at 19:56
  • \$\begingroup\$ @JanDorniak, I am sure the system does have 5V somewhere, so yes, use intermediate rail. Or change the DC-DC converter topology to some flyback transformer type. \$\endgroup\$ Nov 3, 2018 at 20:15
  • \$\begingroup\$ @Ale..chenski you actually got me thinking about ATX here. A modern CPU needs anywhere over 100W on a power rail of less then 1.5V. I'm pretty sure they use multiphase bucks. And the input voltage is 12V \$\endgroup\$
    – jaskij
    Nov 3, 2018 at 20:22

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