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Power Electronics

Gate Drivers and Bootstrap

Thinking "I'll just drive the MOSFET from a microcontroller pin" is the classic power-electronics trap. The gate is a capacitor (Qg, in the nC range): switching it fast means injecting amps of peak current within nanoseconds — a GPIO's 20 mA won't do; the MOSFET turns on slowly, lingers in its linear region and cooks. The fix is a gate driver IC (e.g. IR2101): it takes a logic input and slams ±2 A into the gate. A small series gate resistor (10–100 Ω) tames dv/dt and ringing.

The real puzzle is the high-side switch: the upper NMOS in a bridge leg has its source tied not to ground but to the switching node — it jumps between 0 and Vdc. Keeping the gate on requires a voltage 10–12 V above the source, i.e. Vdc+12 V with respect to ground! The bootstrap circuit solves this with one diode and one capacitor: while the low-side switch conducts, the switching node drops to 0 and the supply charges the bootstrap capacitor (~0.1–1 µF) through the bootstrap diode; when the high-side switch turns on, that capacitor "floats up" with the source and supplies a floating 12 V to the gate.

Bootstrap has a price: the high-side switch cannot stay on forever — the capacitor leaks down; the low-side must conduct regularly (D < 100%). Bridges also generate dead time inside the driver: ~0.5–2 µs of gap at every edge so the two switches never overlap. The lab setup (STM32 + IR2101 + IRF540 + MBR560) builds exactly this chain: the MCU generates complementary PWM with dead time and the IR2101 floats the high-side gate via bootstrap. In our simulator the idealized "PWM Driver (dead-time)" block does the same job — you can see it at work in the inverter circuits.

Formulas

Ig,peak ≈ Qg / tsw
Cboot ≥ 10·Qg / ΔVboot
Pg = Qg·Vgs·fs (gate-drive loss)
Dead time: both switches briefly OFF together

⚡ Open this lesson in the simulator — free

Test Yourself

Why does the high-side NMOS of a bridge leg need a special drive circuit?
Answer: Because its source is tied to the switching node, so the gate voltage must "float" up to Vdc+12 V w.r.t. ground — The high-side source jumps 0↔Vdc; the gate must stay at source+12 V → a bootstrap or an isolated driver is required.
When does the bootstrap capacitor charge?
Answer: While the low-side switch conducts — the switching node drops to 0 and the capacitor charges through the diode — With the switching node at 0 the path Vcc → diode → Cboot is established; that is why the high-side cannot be held on at 100% duty forever.