Sizing the Bootstrap Capacitor of a High-Side Gate Driver

The bootstrap capacitor sits between the driver's floating supply pin (VB) and the switch node (VS). When the low-side switch conducts, the switch node is pulled near ground and the capacitor charges from the driver supply VDD through the bootstrap diode. When the high-side switch turns on, the switch node rises to the bus voltage, the diode blocks, and the capacitor alone powers the high-side driver until the next low-side interval.

So the capacitor has one job: deliver all the charge the high side consumes during its longest on-time without its voltage dropping too far.

Total charge per high-side on-time

Q_total = Q_g + I_QBS · t_on,max + I_leak · t_on,max + Q_LS

Allowed droop

ΔV_max = V_DD − V_F − V_GS,min

Minimum capacitance

C_boot ≥ Q_total / ΔV_max

• Q_g is the total gate charge of the high-side switch at your gate voltage, from the datasheet.
• I_QBS is the quiescent current of the driver's floating section; it matters at low switching frequency or long on-times.
• I_leak covers the bootstrap diode reverse leakage and any gate-source resistor current; a 10 kΩ gate-source resistor at 12 V alone draws 1.2 mA.
• Q_LS is the level-shifter charge per cycle given in some driver datasheets, typically a few nanocoulombs.
• V_GS,min is whichever is higher: the gate voltage your switch needs for its rated R_DS(on), or the driver's high-side undervoltage lockout threshold plus margin.

1. 1A MOSFET with Q_g = 60 nC, a driver with I_QBS = 100 µA and Q_LS = 5 nC, diode leakage of 10 µA, and a longest on-time of 50 µs (10 kHz at 50 % duty).
2. 2Q_total = 60 nC + 100 µA · 50 µs + 10 µA · 50 µs + 5 nC ≈ 70.5 nC.
3. 3With V_DD = 12 V, a 0.6 V diode drop and a required minimum of about 10.9 V, the droop budget is 0.5 V.
4. 4C_boot ≥ 70.5 nC / 0.5 V ≈ 141 nF. Choose 470 nF, because a ceramic capacitor loses a large part of its capacitance under DC bias; a small 16 V X7R part at 12 V can easily lose half of it.

A common rule of thumb is to make C_boot at least ten times larger than the switch's gate capacitance. It usually gives a similar answer, but the charge method also captures quiescent current at low frequency, which the rule of thumb misses.

A larger capacitor is not free. At start-up it is empty, and the first low-side pulse must charge it from zero through the diode and any series resistor. A very large bootstrap capacitor makes that first charge slow and draws a large current spike from V_DD. Keep the V_DD decoupling capacitor at least about ten times larger than C_boot, so V_DD does not sag and trip the low-side undervoltage lockout while the bootstrap capacitor charges.

• Voltage: the diode blocks the full bus voltage plus switching overshoot whenever the high side is on.
• Speed: use an ultrafast or a high-voltage Schottky / SiC diode. Slow reverse recovery dumps charge back out of the capacitor and adds a current spike at every switching edge.
• Series resistor: a few ohms limits the peak charging current and the dV/dt-induced current spikes. Too large a value slows recharge; check that the low-side on-time can still replenish the droop.
• Integrated diodes: many drivers include a bootstrap diode or an active bootstrap FET. Check its current rating against your capacitor and switching frequency before relying on it.

The capacitor only recharges while the low-side switch is on (or while the low-side body diode conducts and pulls the switch node down). If the high side is held on continuously, as in a buck converter at maximum duty or a motor drive holding a phase high, the capacitor slowly drains and the driver eventually hits undervoltage lockout. Solutions are a minimum off-time in the PWM, a periodic refresh pulse, or an isolated or charge-pump supply for the high side.

At start-up, turn the low-side switch on first for long enough to charge the bootstrap capacitor, roughly three to five times R_boot · C_boot, before the first high-side pulse. In a motor drive, do this for all three phases before enabling modulation.

Place the bootstrap capacitor directly across the driver's VB and VS pins with the shortest possible loop. During hard turn-off of the high side, stray inductance can pull the switch node below ground for a few nanoseconds. That negative spike overcharges the bootstrap capacitor and, if it exceeds the driver's negative VS rating, can latch up or damage the driver. Short power loops, a small resistor in series with the bootstrap diode and, if needed, a clamp diode from VS to ground reduce the risk.

Dead time interacts with all of this: during dead time the switch node is moved by the load current, which decides whether the low-side body diode conducts and recharges the capacitor. The dead time calculator on this site helps you choose that interval.