The gate drive voltage is 10 V, and currents other than the gate charge are ignored. When a high-side MOSFET with a gate charge of 100 nC is driven by a bootstrap circuit, which of the following is the minimum capacitance that keeps the drop in the bootstrap capacitor voltage at one turn-on to 0.5 V or less?

Answer: C. 200 nF

When the MOSFET turns on, the bootstrap capacitor gives the gate a charge Qg, and its voltage drops by ΔV = Qg/C. Keeping ΔV ≤ 0.5 V requires C ≥ Qg/ΔV = 100 nC ÷ 0.5 V = 200 nF. In practice about ten times this value (about 2 µF) is often chosen, allowing for the current drawn by the gate-drive circuit, leakage currents and component tolerances. The capacitor is charged while the lower switch is on, so the circuit is not suited to applications that keep the upper switch on for a long time.

• A. This divides the charge by the 10 V drive voltage, giving a capacitance whose voltage would fall all the way to zero.
• B. This multiplies the charge by the voltage drop, and the units do not even match.
• C. Correct. 100 nC ÷ 0.5 V = 200 nF.