Prerequisites: Capacitors, Diodes, MOSFETs, Half-Bridges
What is a bootstrap circuit? What is it commonly used for?
A bootstrap circuit is a circuit that provides bias (extra voltage to the gate) of a high-side MOSFET so that it can more easily turn on. A high-side MOSFET is the MOSFET in a half-bridge configuration that is connected to the power supply. The bootstrap circuit works by charging a capacitor in the off-state of the cycle (or when the low-side MOSFET is on), and in the on-state, the capacitor discharges and provides extra voltage to the gate of the high-side MOSFET.
Describe a bootstrap circuit. Explain how the bootstrap capacitor and bootstrap diode play a role in generating bootstrapped voltage rails
A bootstrap circuit consists of a bootstrap capacitor and a bootstrap diode. The bootstrap diode is connected between VDD (or VCC) which is the positive supply voltage to an IC, and the bootstrap node (HB). This diode is used to charge the bootstrap capacitor during the circuit’s off-state, and direct current when the capacitor discharges during the on-state.
The bootstrap capacitor (CBOOT in the diagram) is connected between the bootstrap node (HB) and the switching node (commonly denoted as SW in buck converters, but shown as HS in the gate driver example diagram). Figure 1 shows the circuit’s off-state when the low-side MOSFET is on, and the bootstrap capacitor is being charged through the bootstrap diode.
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Figure 2 shows the circuit’s on-state when the high-side MOSFET is on, and the bootstrap capacitor is discharging. Because of the bootstrap diode, the charge from the bootstrap capacitor is forced to go through the HB pin, and out through the HO pin of the IC, to provide voltage to the gate of the high-side MOSFET.
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The minimum capacitance of the bootstrap capacitor is calculated by dividing the MOSFET gate charge by the tolerable droop on the bootstrap capacitor, which is typically less than 5% of VCC.
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A resistor is optional to add in the circuit that can be used to limit peak currents at the bootstrap diode during start-up. Simply using Ohm’s Law, a highest bootstrap resistance means a lower current through the bootstrap diode, meaning a slower start-up time for the on-state.
Why do high-side N-channel MOSFETs require higher voltage rails to switch?
An N-channel MOSFET requires the gate-to-source voltage to be higher than the threshold voltage to close the switch (for enhancement) or open the switch (for depletion). Moreover, the voltage at the gate must be sufficiently higher than the voltage at the source to switch. When working with high-side MOSFETs, the source voltage is higher since it is connected to the power supply. Therefore in order to get a positive voltage difference high enough to exceed the threshold indicated by the MOSFET, larger voltage rails are required so the MOSFET is able to switch.
How does a bootstrap circuit play a role in dictating the gate-to-source voltage used to switch the FETs of a half-bridge?
Depending on how much extra voltage you are able to supply to the gate, this affects the gate to source voltage used to switch the MOSFETs on a half-bridge. The gate-to-source voltage cannot be too high as even with the bootstrap circuit the gate will not be able to overcome the VGS threshold. Therefore, the bootstrap circuit and VGS are taken into account together when designing a circuit so that the capacitor is able to discharge enough to help the high-side MOSFET overcome its VGS(TH).
What issues arise as a half-bridge begins to switch at a higher duty cycle?
What are some similarities and differences between bootstrap circuits and charge pumps? When would I consider using one over the other?