"当chip不工作时,内置FET不工作,外面那个电感saturate,怎么可能还有输出。"
when the chip is disabled, the gate goes low, and the drain goes high
impedance. When that happens, the current goes through the inductor and the
diode to the load.
another way to look at it: notice that the Out pin supplies Vdd for the chip
and is connected to Vout? if Vout went low when the chip is disabled, how
would the chip have started?
BTW, the chip can be easily converted to a gated PFM regulator, aka LT1073
style.