Explanation of LED driver circuit
It's a buck-boost configuration. If it was connected in boost then the input voltage could not exceed the LED voltage or they would burn out. There's more voltage across the inductor and the switch has to withstand more voltage with buck-boost. 








simulate this circuit &ndash; Schematic created using CircuitLab


The voltage feedback is to protect the IC if the LED string should become open circuit for some reason, otherwise the voltage at Cout would quickly rise to destructive levels. The transistor effectively level shifts the overvoltage protection so that it measures the voltage above the positive rail (more than about 20V plus one Vbe above PVin will cause 50uA to flow, which will drop 1.25V across the 24.9K resistor, which in turn will cause it to shut down). 
RS2 is the sense resistor for measuring the LED current. RS1 is the sense resistor for measuring the inductor current.