I'm new to using buck converters. In the past I've just used a linear voltage regulator but due to heat issues in my current project I've decided to give a buck converter a try but I'm getting really confused.

I'm trying to convert automotive voltage to 3.3 V. Automotive voltages in our typical applications can vary from around 8 V to 16 V. I'm trying to power a Onion Omega which gets really unhappy at even a small bit under 3.3 V so getting an operating voltage of around 3.35 or even up to 3.6 V would be ideal and must be able to deliver at least 500 mA, preferably 1 A to be more future proof.

I have looked at a few IC's and I started designing around a LTC3621HMS8E-3.3.

I started copying the reference design

Reference Design

then I realized that even though I'm using the 3.3 V version of the IC, that the reference design shows that it's for 2.5 V out. Reading on in the data sheet here: LTC3621

I've come to the conclusion that I'm really confused.

The buck converter explanations found here on SE make sense from a conceptual standpoint but I seem to have a disconnect when it comes to actually making one of these things work.

Maybe I've also simply chose an insufficient IC and something else is a better fit.

I'm not asking someone to build this whole thing for me but maybe someone can help explain these things a bit better and what I should be doing to get the correct passives hooked up to this IC.

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    \$\begingroup\$ Which things do you need to be explained better? Could you be a bit more specific? \$\endgroup\$ – Ale..chenski Jul 13 '18 at 4:34
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    \$\begingroup\$ Hint: Assume if we assume the 'FB' pin takes (and supplies) zero current and we have 2.5V out what voltage do you expect to see there? What does the data sheet say about this pin and what would you expect to see if you changed these values? \$\endgroup\$ – Warren Hill Jul 13 '18 at 4:42
  • \$\begingroup\$ I guess the confusion comes from the fact that the LTC3621 documentation doesn't show how to connects the fixed variant of it. This might be a challenge for novices, and the hint might be not enough. \$\endgroup\$ – Ale..chenski Jul 13 '18 at 5:26
  • \$\begingroup\$ Warren, I like your method of teaching.. Guided discovery. It's how I learn best. I now found the section that outlines this but also states that for fixed versions of the chip, to just connect Vout directly to FB. which is what The Photon suggested below. So that makes sense. now for the inductor selection.. I must be missing or mis understanding the sheet for that part. \$\endgroup\$ – Steven Lutz Jul 13 '18 at 5:28
  • \$\begingroup\$ Ali, this is making more sense when I read into the function on FB and the note that says to connect it directly to Vout. \$\endgroup\$ – Steven Lutz Jul 13 '18 at 5:29

Your datasheet covers 6 different part numbers. Some have fixed 3.3 V output, some have 5 V output, and some have adjustable output voltage. Some work with 1 MHz switching frequency and others work with 2.25 MHz switching. These PNs are further proliferated with different suffixes related to operating temperature range, mechanical package, etc.

The example circuit is for one of the adjustable-output variants.

To modify it for the 3.3-V fixed-output version, you'd eliminate the 191k resistor, and replace the 604k resistor with a wire.

You might also want to modify the inductor and capacitor selection to optimize control stability and load transient response. The details on how to do that should be found in the Applications section of the datasheet.

Also, you should be aware that this regulator has an absolute maximum input voltage of 17 V. Automotive environments are notorious not only for varying supply voltage, but also for occasional transients 10's of volts above the nominal voltage. You'll want to design a protection circuit to block these transients from reaching your converter chip.

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  • \$\begingroup\$ When you eliminate the 191k resistor, you're referring to also eliminating the path to ground correct? are you saying to just eliminate the voltage divider of the 2 resistors but keep the FB hooked up? By replacing the 604k resistors with a wire, aren't you also making the capacitor in parallel with the 604k resistor, useless? \$\endgroup\$ – Steven Lutz Jul 13 '18 at 5:12
  • \$\begingroup\$ Yes, removing the resistor means remove it and don't replace it with anything. Yes, you can remove the 22 pF capacitor also. \$\endgroup\$ – The Photon Jul 13 '18 at 5:25
  • \$\begingroup\$ OK thanks makes sense. In reading the section on choosing an inductor, I'm lost when referring to ripple current. Where does it come from and why does it matter? (sorry for the elementary question) \$\endgroup\$ – Steven Lutz Jul 13 '18 at 5:37

You could simply use a DC-DC step down module based on LM2596 and avoid you troubles for only 2 USD:

edit: fixed link

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  • \$\begingroup\$ The link that you provide is for a LM2577 based convertor, yet you reference a LM2596 in your answer. \$\endgroup\$ – Greenonline Apr 29 '19 at 15:55

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