# Power stage design

My question is related to how to select topology and component for the following problem:

The Specification Design a power supply:

output 5v @ 40mA & 12v@ 4 mA.

Input power supply 20v 15mA.

Questions

How could i achieve this design and guarentee only a 0,03 w of losses?

What's the best topology or cascade architecture do you recommand?

Thank you

• Theoretically, Buck converters could do it. But at this small power level, it'll be hard to keep quisicient currents in check, especially with 20 V input. If the load is constant, you can skip the control/regulation part and run both converters at constant frequency and duty cycle. Oct 12, 2022 at 12:43
• Hello Tobalt, Thank you for your response. I tried to do a simulation with ltspice but i saw that in startup it calls much current. I added a picture on that . take a look to the simulation. I'm worried if that will make some trouble Oct 12, 2022 at 13:14
• Is this an academic or practical question? If practical, I'm suspicious of the input current being specified so tightly, and why a 20V source can't afford more. If academic, what sources, selection of parts, and range of loads, are you required to test at? If precisely that output condition alone, a fixed duty cycle into synchronous MOSFETs will easily do it, and inductors and capacitors can be made as large and ideal as you like. Oct 12, 2022 at 13:38
• This is not an academic projet :( It's a real project. I've been scratching my head for more than two weeks and the goal is to power up a sensor with 12v and a microcontroller with other some stuff with 5v. Yes the maximum power supply is 20V and i can not have more than 15 mA. I tried many buck converter but none of them works. Have limited power source to 15mA in ltspice the buck dosen't start up Oct 12, 2022 at 14:02
• Do you need a regulated 12V or ist it possible to create this from a secondary winding of the 5 V buck converter?
– Jens
Oct 12, 2022 at 21:55

I was able to simulate a circuit that provides the required outputs, but with an input current of about 22 mA instead of the requested 15 mA. There is a current limiting circuit that actually allows 70 mA at startup, but then it limits to about 20 mA after 15 mSec. A comparator inhibits the buck switching converters as the capacitor C2 charges up. I'm sure it's possible to make a better current limiter and tweak some values to better approach the required parameters.

Actually, the current sense resistor accounts for about 86 mW extra power. But without it, the 20V supply draws about 29 mA due to lots of transients. I tried adding an inductor to the input but that made it worse.

I did a quick simulation without putting a lot of effort into optimizing it.

The value of the input capacitor seems to matter in the simulation, I reduced it to 2.2$$\\mu\$$F and it reduced the supply current, I don't know if this improvement will translate to hardware though. I also removed the 2 Meg resistor as it didn't seem to matter.

I chained the 5 V regulator off of the 12 V one. I tried running it from the 20 V, the supply current was slightly lower but had some spurious spikes.

It looks like it might be possible. You need around 83% overall efficiency, the LT3631 data sheet shows it doing better than that. As tobalt said, it's on the edge, you've got a pretty slim margin to work with.

Note that the average current is within your specs, but the peak is ~220 mA. If you're spec is based on peak you're probably up the proverbial creek.

• The peak current at start-up might be reduced to within spec by implementing a current-limited precharge of a large capacitor Oct 12, 2022 at 17:58
• @PStechPaul I took another look at it, when I reduced the capacitance I had been including the first 10ms before it was stable, looking at just the stable part a larger cap appears to reduce the peak current but the average stays about the same. Oct 12, 2022 at 20:24
• @GodJihyo I don't get your 83%. See my comment to the OP, above, for the calculation and the result I see. Seems closer to 90% to me.
– jonk
Oct 13, 2022 at 5:33
• @PStechPaul I Tried some circuit that i found but it's seems that they don't do the stuff. Could you draw a sample ? Oct 13, 2022 at 11:21
• @jonk For the efficiency formula η = Pout / Pin I used 20V x 15mA for Pin, you used Pout + 30mW. I'm not sure where the OP got that value, could be leaving a safety margin, could have just been spitballing it. So I guess it's 83% best case, 90% worst case. Oct 14, 2022 at 13:58