Why is this boost converter application circuit done this way?
Most (all that I used) of the impulse power converters have VOUT connected to the coil. This one has VIN connected to the coil.


Apparently you are used to buck converterns, then.  It might be worth a web search.  A buck converter works by switching power on and off to the coil, so the coil input sees an average voltage less than the supply.  Coils try to maintain zero volts across themselves, so the circuit tends to filter this less-than-input voltage to the output, which is also less than the supply.

A boost converter, as you saw, has the input connected to the supply, and switches the coil output between the voltage output and ground.  The coil output voltage then averages to less than the circuit output voltage.  This forces the average coil current up to the point where it can hold the output voltage higher than the input.

(This is an oddball description -- do a web search; the Wikipedia article is good.  It'll break down the switching cycle and explain how it all works).


  Also why is this converter powered (VCC) from VOUT? How is it supposed to work? Usually you power PC from VIN.


First, it's an IC, so they can always bootstrap the power supply from LX without putting that fact into their block diagram.  LX is supplied from a coil, so it's always seeing an average voltage equal to the input, even if it may be switched between ground and Vout.

Second, this converter is designed to work down to 0.9V (which is a sensible voltage for a converter running from a single cell).  But FETs like more voltage than that.  By running from Vout, they get a higher voltage on everything (most particularly, on the gates of the output transistors).  That's going to make everything work better -- especially by reducing Rds-on.