2026-05-10
When you build a half-bridge or high-side switch with an N-channel MOSFET, you hit a fundamental problem: to turn the FET fully on, the gate must sit roughly 10 V above the source. But the source is tied to the load, which swings up to V+ when the FET conducts. So the gate driver needs to produce a voltage higher than the supply rail itself. The elegant, cheap solution is the bootstrap capacitor.
The trick: when the low-side FET is on, the switch node (SW) is pulled to ground. A diode from a fixed VCC (typically 12 V) charges a capacitor (Cboot) tied between SW and a "boost" pin (BST). Cboot now holds ~12 V across it. When the high-side FET turns on and SW flies up to V+, the BST pin rides along — it now sits at V+ + 12 V. The high-side gate driver references its supply to this floating rail, and can drive the gate 12 V above the source. Voilà: full enhancement on a high-side N-channel device.
Real-world example: the IR2104 half-bridge driver with a Cboot of 100 nF, charging through a 1N4148 diode from a 12 V rail, drives an IRF540N switching a 24 V motor at 20 kHz. Every low-side conduction window refreshes Cboot.
Sizing Cboot — rule of thumb: the bootstrap cap must supply the gate charge Qg plus leakage during the high-side on-time, while not drooping more than ~5%. Use:
Cboot ≥ Qg / ΔV
For an IRF540N (Qg ≈ 70 nC) tolerating ΔV = 0.5 V droop, Cboot ≥ 140 nF. Pick 220 nF or 1 µF ceramic for margin. Multiply by 10× if the driver has high quiescent current or you operate at low duty cycles.
Critical design pitfalls:
Bootstrap circuits dominate motor drivers, class-D amps, and synchronous buck converters because they're cheap — one diode, one cap, no isolated supply.
