Regulate 0-6V output
Is it possible to get that range?

No, the lowest output voltage you can get is 2.5 V because that's the internal reference voltage which the chip uses to set the output voltage.

How do I know that?

Let's look at the Functional Block Diagram:



R1 and R2 form the feedback divide, the feedback signal goes into the chip via the FB pin and then goes into two comparators. The overvoltage comparator we can ignore, the regulation comparator is what is important. The reference input for this regulation comparator comes from the left, from pin SS where there's a capacitor to ground for the soft start. Then there's a NMOS switch to ground, a current source and a 2.5 V. That 2.5 V is the reference voltage.

When the chip operates, the NMOS switch to ground will be open, doing nothing. So the current source will charge the soft start capacitor to 2.5 V. So after startup there will be 2.5 V at the SS pin so there will be 2.5 V as well at the input of the regulation comparator.

The feedback loop then controls the system such that there will be 2.5 V at FB (on the right) as well. This is common to almost all (switched) regulators.

Also note that in the formula to calculate R1 and R2 from Vout (see section 8.2.2.1.1 on page 14) there's a 2.5 V in the formula.

If you want Vout = 2.5 V then R1/R2 = 0 so R1 = short and R2 = not placed.

If you want Vout &lt; 2.5 V then you would need a resistor with a negative value! Obviously that is not a working solution.

Solutions do exist, a common one is to create a negative voltage (for example -2.5 V) and then use that as the negative supply rail for the chip.

Another solution is to use add an external circuit with an opamp which has inputs which can work down to GND voltage.

Both solutions aren't trivial, have issues of their own and are best left to experienced circuit designers.

If you want to learn more I suggest that you look at "lab supply" circuits as most lab supplies can go down to 0 V.