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I'm working on the LT8607 Step-down converter to convert from 12V to 5V/500mA output. The circuit schematic is as below. enter image description here

What is the purpose of using the BST capacitor as shown in the schematic.

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    \$\begingroup\$ Pretty sure it's to generate a high enough voltage to drive the upper FET. N-channel fets used on the high side require a voltage higher than the input voltage to drive them properly. \$\endgroup\$ – Hearth Nov 23 '18 at 3:35
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    \$\begingroup\$ The capacitor is a part of a bootstrap gate driver, which allows to generate a gate voltage (referenced to ground) greater than the supply voltage. Operation of the bootstrap gate driver explained here. \$\endgroup\$ – Nick Alexeev Nov 23 '18 at 3:36
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    \$\begingroup\$ the above two are absolutely correct. To elaborate, this is done since N-Channel devices can be used to get higher system efficiencies. \$\endgroup\$ – Luke Gary Nov 23 '18 at 4:13
  • \$\begingroup\$ Both of the top two are absolutely correct, and at least one of them should make their comment an answer! I couldn't explain it better myself. \$\endgroup\$ – TimWescott Nov 23 '18 at 4:15
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Any converter using an n-channel device for high-side switching needs to have internal boost converter, often a simple charge-pump, working alongside it. This is because an n-channel FET requires that its gate be driven higher than its source to be on, and when on, the source is at approximately the drain voltage, which is the supply voltage in this particular use case.

Since having to give multiple voltage supplies to a converter is inconvenient, most such devices have the circuitry for a charge-pump voltage doubler built in.

The reason n-channel, and not p-channel, FETs are used for this is that, for reasons relating to semiconductor physics (best answered in another question), n-channel devices are more efficient and faster than p-channel ones of the same size. If one were to use a p-channel FET for this task, it would not require a bootstrap capacitor, but the tradeoff would be lower efficiency. The cost of one capacitor, and a slightly more complex and maybe slightly larger die size, is usually considered worth the efficiency saved.

For more information on bootstrap capacitors, @NickAlexeev linked this in a comment above, which gives a good explanation of what exactly is going on here.

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