0
\$\begingroup\$

For a part of my next project I'm required to make a variable power supply module. I have never designed a power supply and I would like to make my life as easy as possible. Thats why I would use LM25118 buck boost controller. The datasheet is a bit confuseing for me.

The input voltage varies from 3 to 6 volts (can deliver 4A). The output voltage needed is between 2 and 12 volts. The output voltage would be regulated with a potentiometer. I would like to look at the worst possible scenario when I have 3 volts at the input.

  1. What voltage do I have to set to still get 1A at the output, while there is a 3 V input?
  2. Is there a better solution? Like a controller from another manufacturer something else. I would need at least 1A at 5V out, 1A at 9V out would be great! (with 3V in)
\$\endgroup\$
2
\$\begingroup\$

There's nothing technically stopping you from drawing 1A from this supply if your source is able to source that much power, and you take efficiency into account, and of course if you designed the inductor, switch and smoothing capacitors accordingly (which you can do by using TI's own supply designer tool on their website, so I assume you did that and will choose proper components by simply following its design).

So, the max input power is 3V · 4A = 12W. Your max output power is 12 V · 1A = 12 W. That can't work with any supply, doesn't matter which because there's no such thing as a 100% efficient power supply.

So, get a beefier input, or reduce your output requirements.

Note that it's not easy to operate a buck-boost supply at \$V_\text{in}\approx V_\text{out}\$; if that is among your requirements, more involved architectures like Cuk or SEPIC might be better suited.

\$\endgroup\$
  • \$\begingroup\$ Thank you for your answer! It helped me a lot. I did try TI's WEBENCH but when I putt in desired 1A at any voltage range i get these two comments: "Cant find suitable FET for the design. Try reducing load current" and "Circuit calculator can't create design" \$\endgroup\$ – Lazyboy Mar 22 '18 at 18:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.