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.

enter image description here


1 Answer 1


The anode of the diode is not shown connected anywhere. You can connect it to either the IN or OUT (obviously if you connect it nowhere you can omit the diode).

The cathode always goes to BST if the diode is present.

The purpose of this diode is to charge C23 when the output switch line (SW) is near to ground. Since the internal MOSFETs are n-channel the gate voltage must be higher than the source voltage in order to turn the MOSFETs on.

Since your input voltage can be higher than 5.5V you have two options only- leave the diode out or connect it to your 5.0V output.

  • \$\begingroup\$ This clears it up for me now, so basically the IN means the input pin, and OUT means the output after the inductor, correct? Thanks again. \$\endgroup\$
    – klcjr89
    Mar 27, 2017 at 4:27
  • \$\begingroup\$ @aviatorken89 Yes, correct. \$\endgroup\$ Mar 27, 2017 at 4:28
  • \$\begingroup\$ I know a capacitor should be used to alleviate short term voltage drops/dips, but does the D2 diode basically do the same in this case? \$\endgroup\$
    – klcjr89
    Mar 27, 2017 at 4:29
  • \$\begingroup\$ Also, my input voltage is 12V nominally, but with the motor started, it can briefly dip the battery to 4.4V, so would it be better to have D2 on the IN or the OUT? \$\endgroup\$
    – klcjr89
    Mar 27, 2017 at 4:32
  • \$\begingroup\$ C23 is the power supply for the high side driver. The diode charges it when the low side driver is on. I don't think you can use D2 on the IN if the input goes above 5.5V, but it's not clear on the limitation. BTW, if you use a diode, use one of the recommended (fast) diodes. A 1N400x diode will not work well. \$\endgroup\$ Mar 27, 2017 at 4:36

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