I want to convert unregulated 10-14V DC to 5V DC and 12V DC regulated(constant) supply;This is a buck step down regulated voltage supply IC it outputs the voltage that I require. Is this IC all I need? Or there will be some more components needed?

Will this IC have:

  1. Under-voltage and short-circuit protection.

  2. Thermal shutdown prevents damage from overheating.

  3. Reverse-voltage protection.

I have seen buck step down regulators which have many more components, than just a IC. For example this: http://www.pololu.com/product/2110. What is the use of other components, Are all the other components used to provide the above mentioned 3 points?

  • \$\begingroup\$ What about the current needed to be supplied by the outputs to their respective loads? \$\endgroup\$
    – Andy aka
    Mar 7, 2014 at 11:08
  • \$\begingroup\$ Do the application notes for that chip, or the example circuits in its datasheet, show additional components? \$\endgroup\$
    – user16324
    Mar 7, 2014 at 11:11
  • \$\begingroup\$ @Andyaka What about it ? I will be providing power to circuit using a standard 12V 3.3A DC adapter. \$\endgroup\$
    – rajat
    Mar 7, 2014 at 11:12
  • \$\begingroup\$ Power out to the load - read what I asked again - I never mentioned power in. \$\endgroup\$
    – Andy aka
    Mar 7, 2014 at 11:25
  • \$\begingroup\$ The current needed to be supplied is around 3A. \$\endgroup\$
    – rajat
    Mar 7, 2014 at 11:35

1 Answer 1


Getting 5V from a supply that is ranging from 10V to 14V is no problem for the LM2576 and it has thermal and current limit protection.It doesn't have under-voltage-lock-out but it does have an on-off pin that can be used (with an external circuit) to provide this function.

enter image description here

It doesn't have reverse input voltage protection either - you should consider using a diode for this.

Getting 12V from a supply that ranges from 10V to 14V is more complicated and you can do this with a buck-boost type controller. Buck on its own can only produce voltages that are smaller than its input voltage - that is why the 5V is relatively easy. Note that there is a section in the LM2576's data sheet that talks about buck-boost but this is to produce a negative output voltage only and you should not get confused with the device - it cannot produce a boosted positive output easily.

You could use the LM2576 to produce 5V then have a boost regulator on the 5V output to produce 12V - boost regulators do what they say - they boost the voltage output and therefore the input supply voltage to them needs to be below the output voltage they produce.

Regarding the other device linked - without a circuit diagram it's guesswork as to what all the components do - UVLO would hardly account for anything more than two resistors on a lot of chips available (such as Linear Technology's offerings) and reverse voltage protection can be achieved with a maximum of three components usually.

The LT1074 can supply 5A at 5V and, given your input power is able to supply 3.3A, the 1074 may be more suitable: -

enter image description here

It's got enough capability to tee it's output to a boost regulator for the 12V rail providing you manage the output load currents. There's also the LT8471 that can provide two outputs - this picture shows the general idea: -

enter image description here

The negative 12V supply is easily made a positive supply - read the data sheet if you are interested.

  • \$\begingroup\$ Hmm, regarding the under-voltage lockout I am just wondering why its Vth = Vz1 + 2*Vbe. I would've said just Vz1 + Vbe. \$\endgroup\$
    – Rev
    Mar 7, 2014 at 21:28
  • \$\begingroup\$ @Rev1.0 yeah I agree - I'm confused as well - can't see why it is 2Vbe. \$\endgroup\$
    – Andy aka
    Mar 7, 2014 at 21:44
  • \$\begingroup\$ @Rev1.0 I bet the 10k and 20k resistors have something to do with this - if both were 10k then I could see it being 2Vbe! \$\endgroup\$
    – Andy aka
    Mar 8, 2014 at 11:23

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