Hi
True shutdown is a term that means the output of a dc/dc converter is collapsed to ground (0V) when the part is in shutdown. If you have low current circuitry hanging off the output of a dc/dc converter, then put the part in shutdown, the output voltage will remain high until the output cap discharges. With a low current circuit this could be quite some time - in other words, your circuit is still operating when you think it has shut down.
True shutdown disconnects the output from the input and shorts out the output cap.
Now for a lesson in chip design: The input of most chips have a diode from the input to the Vcc pin and one from ground up to the input. Thus if the input is taken higher than the Vcc pin by more than 1 diode drop, the top diode conducts and turns on the chip. This is not good news. This is where your abs max conditions come from in most chip datasheets. Hold this thought for later...
Now for Shutdown: They sometimes build the silicon connected to the drain of a FET. The source of the FET is connected to ground. To bring the chip to life, switch on the FET, pulling the chip ground down to 0V. To put it in shutdown, switch the FET off. Thus the whole circuit floats with no reference to ground.
Back to the fate of our input diodes..... If the chip is floating (in shutdown), and you apply a voltage to the input, you cannot get any conduction path into the input diodes. Alternatively, if you shut down the chip by removing the Vcc voltage, then apply an input voltage, your diodes conduct. This is why they use such a strange way to shut down the chip.
I dont fully understand the question you are asking, but this explains what shutdown is and what true shutdown means..
Bill Naylor - chip designer to the stars