I am designing a buck switching circuit. The power supply component is rated for 6A, but my design will typically use 0.2 A (yes there is a good reason that I am selecting a component with a much higher rating). For selecting the inductor, the datasheet of the switching regulator says: "An inductance that gives a ripple current of 10% to 30% of the maximum output current (6 A) is a good starting point". In other words, they suggest K (ratio of Iripple to Imax) between 0.1 and 0.3. I am wondering though, are they using 6A as Imax because that is the rated current of the component, or because that is what the example application circuit is designing for? Should I use 6A for Imax or 0.2A?

Edit: I guess i was hoping the question could be answered in general but I guess the answer depends on the regulator. I am using a TI TPS7H4010-SEP. This is still a general question I have for any design though.

  • \$\begingroup\$ Most probably rated current. Ask the manufacturer if not explicitly stated in the datasheet. \$\endgroup\$
    – winny
    Jul 27, 2023 at 17:35
  • 1
    \$\begingroup\$ What regulator? What application? \$\endgroup\$ Jul 27, 2023 at 18:20
  • \$\begingroup\$ I guess i was hoping the question could be answered in general but I guess the answer depends on the regulator. I am using a TI TPS7H4010-SEP. This is still a general question I have for any design though. \$\endgroup\$
    – Jon
    Jul 28, 2023 at 19:07
  • \$\begingroup\$ Rad-hard? Yikes! Ah, well I take it that's your rationale for part reuse. Understandable. ;) \$\endgroup\$ Jul 28, 2023 at 19:31
  • \$\begingroup\$ Yikes is right! But compared to most rad-hard parts this one takes a pretty small board footprint and and relatively low cost and available. \$\endgroup\$
    – Jon
    Jul 28, 2023 at 21:37

2 Answers 2


The reason the device matters, is because many regulators available today have internal compensation and limiting. Which is the case for the TPS7H4010-SEP. Notice in the block diagram:

Block diagram
From: TPS7H4010-SEP Radiation Hardened 3.5-V to 32-V, 6-A Synchronous Step-Down Voltage Converter in Space Enhanced Plastic p.13

EA (error amplifier) goes to an internal Rc + Cc network, which is summed with a slope compensation (ramp) waveform, and compared to the high-side current sense. This is a classic peak-current-mode control, and the gain of the current sense, and transfer function of the error amp, are fixed internally.

Using values other than intended (6A nominal output, basically L and C as recommended in the application section) will most likely result in instability. This is further discussed in §7.3.10.

You can simply operate it at light load, which will give burst mode operation (unless you force CCM, but that would increase current consumption substantially). If the resulting output ripple is too much, you may consider adjusting it upward slightly then following with an LDO (assuming you can accommodate one in the design). You will still have the same high current limit, which may be a safety concern on an otherwise 200mA-design supply. Whether this suggests adding a fuse, using an LDO with more suitable current limit, or rejecting the design entirely (i.e. use a more appropriate regulator), depends on other constraints.

In contrast, a more flexible controller might be designable for any load/limit current, transistor, and compensation rate; this is both more work (the designer must select and test many more component values), and generally takes more layout area (which I'm guessing is at a premium here). Whether this is an acceptable alternative (i.e., perhaps changing both/all such regulators to a controller-based design, assuming a suitable controller can be found), again depends on overall project constraints.

  • \$\begingroup\$ Great answer! Thank you for the detailed response! I will definitely be using burst mode. An LDO is not preferred because I want to keep power consumption (and component count) minimal. I'll proceed with the design and try to limit the voltage ripple with capacitance. Thanks! \$\endgroup\$
    – Jon
    Jul 28, 2023 at 21:41
  • \$\begingroup\$ You're welcome. Don't forget to upvote and/or select as answer, any answers you find satisfactory. \$\endgroup\$ Jul 28, 2023 at 21:48

I believe they use it based on the 6A application circuit design. But it is possible that your buck is not stable with an output current that low. If your regulator is designed to output 6A, it might not behave properly that low. My best advice for you is to buy a demoboard (or design it) and try it for yourself.


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