It is not clear what happens if resistor Rfb1 is too high or even is an open circuit.
Q1 : What then becomes the value of Vout and can this (possibly too high) value damage the LM3492HC ?
Q2 : Is the overvoltage protection for protection of the LED string or for protection of the driver ?
For a design with a LED string voltage of 50V the Rfb1 was choosen to be 390K. With this value and 16K2 for Rfb2,
the FB input voltage becomes 1.99V. This value is according to the recommended value of 1.05V en 2.0V
However with Rfb1 = 390K the Voutmax values becomes 2.8(1 + Rfb1/Rfb2) = 70.21V.
Q3 : Can this value damage the LM3492HC ?
If we choose the LM3492, then the formula for Voutmax is different and Voutmax becomes 2.5(1 + Rfb1/Rfb2) = 64.54V.
This value is lower than the absolute value of 67V according to the datasheet.
Q4 : Is it correct to state that the LM3492HC is not suited for this design and that using the LM3492 will be a better option ?