Power Electronics

Synchronous Rectification and Dead Time

In a classic buck, the freewheel diode burns roughly 0.7 V × load current every cycle. Replace the diode with a low-RDS(on) MOSFET (the synchronous switch) and the same path closes through a resistance in the mΩ range.

The gain is dramatic especially at low output voltages: at a 5 V output, a 0.7 V diode drop is a 12% loss; a synchronous MOSFET brings that down to 1-2%.

Because the two switches share one leg, dead time is mandatory; during the dead-time interval the current still flows through the MOSFET’s body diode (briefly reverting to diode-level loss). Modern controllers optimize dead time down to nanoseconds.

Formulas

P_diode = VF · Iload · (1−D)

P_sync = I²·RDS(on)·(1−D)

⚡ Open this lesson in the simulator — free

Test Yourself

In a 5 V / 2 A buck, replacing the 0.7 V diode with a 20 mΩ synchronous MOSFET reduces the loss to roughly what? (D≈0.4)

Answer: 840 mW → 48 mW — Diode: 0.7·2·0.6=0.84 W; MOSFET: 4·0.02·0.6=48 mW.

Where does the current flow during the dead time?

Answer: Through the low-side MOSFET’s body diode — Inductor current is uninterruptible; the body diode conducts briefly.

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