Synchronous Buck Bootstrap Circuit Calculator

Synchronous Buck Bootstrap Gate Drive Sizing Fundamentals

In high-efficiency synchronous buck converters and half-bridge inverters, both high-side and low-side switches are N-channel MOSFETs. Driving the high-side N-FET into full saturation requires elevating its gate voltage 10V above the floating switch node ((V_{SW})). This is achieved using a flying bootstrap capacitor ((C_{\text{boot}})) and bootstrap diode.

1. Bootstrap Capacitance Minimum Sizing Formula

During the high-side ON-time (t_{\text{on}}), the bootstrap capacitor must supply all gate charge (Q_g), driver quiescent current (I_{qbs}), and level-shifter charge (Q_{ls}) without drooping by more than (\Delta V_{\text{boot}}):

$$Q_{\text{total}} = Q_g + I_{qbs} \cdot t_{\text{on}} + Q_{ls}$$

$$C_{\text{boot(min)}} = \frac{Q_{\text{total}}}{\Delta V_{\text{boot}}}$$

Best engineering practice dictates using a ceramic capacitor ((X7R)) with an actual effective capacitance equal to at least 10 to 15 times the MOSFET equivalent input capacitance to ensure minimal voltage droop.

2. Why Series Resistor (R_{\text{boot}}) is Essential

When the low-side switch turns ON, the switch node plunges to ground, recharging (C_{\text{boot}}). Without a series resistor (R_{\text{boot}}) (typically 2.2Ω to 5.1Ω), the instantaneous inrush current can exceed 10A, causing high-frequency ringing and driving the switch node negative ((V_{SW} < -2\text{V})), which can induce parasitic latch-up in the gate driver IC.

3. Low-Side Refresh Time & 100% Duty Cycle Limits

Because (C_{\text{boot}}) recharges exclusively when the switch node is pulled low, a pure bootstrap driver cannot operate at 100% continuous duty cycle. The low-side MOSFET must remain ON for at least 3 time constants ((t_{\text{refresh}} \ge 3 \cdot R_{\text{boot}} C_{\text{boot}})) each switching cycle to replenish depleted charge.