How does a bootstrap circuit work?

When the low-side switch turns ON, the switch node (VS) drops to ground. The bootstrap diode forward-biases, charging the bootstrap capacitor (CBOOT) from VCC to approximately VCC minus the diode drop. When the high-side switch turns ON, VS rises to the bus voltage, and the bootstrap capacitor provides a floating supply voltage (VBOOT = VS + VCC) to the high-side gate driver. The capacitor must hold enough charge for the entire high-side ON time.

The low-side must turn ON periodically to refresh the bootstrap capacitor. Not suitable for 100% high-side duty cycle. Maximum high-side ON time limited by capacitor droop. Bootstrap diode must handle bus voltage (reverse blocking) and have fast recovery to prevent shoot-through current. At high bus voltages (>400V), the diode reverse recovery charge becomes significant.

Power Electronics

Bootstrap

Q: Bootstrap capacitor sizing?

CBOOT must supply the gate charge (Qg) of the high-side FET plus driver quiescent current (Iq) during the ON time, with acceptable voltage droop (typically less than 0.5V). Formula: CBOOT >= (Qg + Iq * tON) / delta_V. For a GaN FET with Qg=5 nC, Iq=1 mA, tON=5 us, delta_V=0.5V: CBOOT >= (5e-9 + 1e-3 * 5e-6) / 0.5 = 20 nF. Use 100 nF minimum with low ESR ceramic.

Bootstrap is a circuit technique that generates a floating gate drive supply for high-side N-channel MOSFETs or GaN HEMTs in half-bridge topologies. A diode and capacitor charge during the low-side ON period, then provide gate drive voltage referenced to the floating switch node during high-side ON. This eliminates the need for an isolated gate drive power supply, reducing cost and size in power converters, motor drives, and Class-D amplifiers.

Category: Power Electronics

Understanding Bootstrap Circuits

In a half-bridge, the high-side FET source terminal floats at the switch node voltage, which swings from ground to the bus voltage. To turn ON the high-side N-channel FET, the gate must be driven 5-12 V above the source, requiring a floating supply. The bootstrap capacitor, pre-charged to VCC during the low-side ON period, provides this floating supply.

Modern gate driver ICs (e.g., TI UCC27211, Infineon IR2110, EPC EPC2152) integrate the bootstrap diode and level-shift circuitry, requiring only an external capacitor. For GaN HEMTs with low gate charge (1-10 nC), small 100 nF capacitors suffice.

Bootstrap Capacitor Sizing

C_BOOT ≥ (Q_g + I_q·t_ON) / ΔV
Q_g = gate charge (nC)
I_q = driver quiescent current
t_ON = max high-side ON time
ΔV = acceptable voltage droop

Example (GaN, Q_g=5 nC):
C ≥ (5 nC + 5 nC) / 0.5 V = 20 nF
Use 100 nF ceramic (margin)

High-Side Drive Comparison

Method	Cost	Max Duty	Isolation
Bootstrap	Low	<99%	No
Charge pump	Low	100%	No
Isolated supply	High	100%	Yes
P-channel FET	Medium	100%	No

Common Questions

Frequently Asked Questions

How it works?

Low-side ON: diode charges cap to VCC. High-side ON: cap provides floating gate drive. Simple, cheap, no isolation needed.

Cap sizing?

C ≥ (Qg + Iq·tON)/ΔV. GaN (5 nC): 20 nF min, use 100 nF ceramic. Low ESR critical for fast switching.

Limitations?

Must periodically turn ON low-side to refresh cap. Can't do 100% duty. Diode must handle bus voltage reverse blocking.

Related Terms

← Bootlace Lens Border Router →

Power Electronics

Request a Quote

Need gate drivers, power converter modules, or GaN half-bridge solutions? Contact our team.

Get in Touch