In the datasheet for the switching regulator I'm using, it says you can supply an optional diode to put on the BST pin of the regulator, and in Figure 2 it says IN if Vin less than 5.5V, or OUT if Vout less than 5.5V. In the picture, I am unsure which way they have it hooked up, I am assuming 'IN' since the diode is pointing towards the BST pin? The regulator is a fixed 5V output model.
So I need help on trying to understand which way to hook it up, as the datasheet doesn't say much about this diode:
An external bootstrap diode (D2 in Figure 2) is recommended if the input voltage is less than 5.5V or if there is a 5V system rail available. This diode helps strengthen gate drive at lower input voltages, result- ing in lower on-resistance and higher efficiency.
The regulator's input is on the same power source as a 12V starter motor, which is controlled by a MOSFET, there will be a voltage drop on the circuit when the FET is activated, so would hooking this diode to 'OUT' help the situation since in Figure 2 it specifies 'OUT' if Vout less than 5.5V?
Would hooking up the diode's cathode to GND and anode to the BST pin be what they consider 'OUT'? On second thought maybe I need to hook it up as 'IN' since the voltage drop happens on the input if the starter motor approaches stall current, the battery voltage can dip to 4.4V temporarily.
Sorry for the multitude of questions.
Thank you for your help.