Low-side SR carries current during (1−D). At high duty cycles Q2 conducts most of the time — minimise Rds_LS over switching FOM.
How a Synchronous Buck Converter Works
A buck (step-down) converter reduces a higher DC input voltage to a lower regulated DC output using two MOSFETs (or a MOSFET and diode), an inductor, and an output capacitor. The high-side switch (Q1) closes for a fraction D of each switching cycle, allowing current to ramp up through the inductor and store energy. When Q1 opens, the low-side switch (Q2) closes to provide a current path from ground — the inductor’s stored energy maintains current flow to the output.
The fundamental voltage relationship is V_out = D × V_in. A 50% duty cycle on a 12 V input produces 6 V output. Unlike the boost converter, the buck has no right-half-plane zero, making it substantially easier to stabilise with conventional voltage-mode or current-mode control.
Synchronous vs Asynchronous: A synchronous buck replaces the catch diode with a second MOSFET (the synchronous rectifier, SR). Because the SR FET has far lower conduction loss than a diode (I²×Rds vs Vf×I), synchronous designs achieve significantly higher efficiency — typically 92–97% vs 85–92% for asynchronous. Use this calculator’s toggle to compare both modes.
Frequently Asked Questions
What is the duty cycle formula for a buck converter?
Ideal duty cycle: D = V_out / V_in. With efficiency η: D = V_out / (V_in × η). For a 12 V to 5 V converter at 90% efficiency: D = 5 / (12 × 0.9) = 0.463 (46.3%). The ON time is t_on = D / f_sw and the OFF time is t_off = (1−D) / f_sw.
How do I size the inductor for a buck converter?
Use L_min = V_out × (1−D) / (2 × I_out × f_sw). Multiply by 1.3 for margin. Example: 5 V output, D=0.42, 3 A load, 400 kHz → L_min = 5 × 0.58 / (2 × 3 × 400,000) = 1.21 μH → use 1.8 μH. The inductor saturation current must exceed I_peak = I_out + ΔiL/2.
Why is the synchronous rectifier (low-side MOSFET) important?
In an asynchronous buck, the catch diode drops ~0.5–0.7 V, wasting P = Vf × I_out × (1−D) in heat. For a 5 V, 3 A converter at D=0.42: diode loss = 0.5 × 3 × 0.58 = 0.87 W, which is 5.8% of output power. A synchronous MOSFET with 12 mΩ Rds(on) loses only 3² × 0.012 × 0.58 = 62 mW — over 14× less loss.
What is dead time in a synchronous buck converter?
Dead time is a brief interval (typically 10–50 ns) between Q1 turning off and Q2 turning on, and vice versa. This prevents shoot-through — both MOSFETs conducting simultaneously, shorting V_in to GND. During dead time, current flows through Q2’s body diode (at Vf loss). Dead time loss = Vf × I_L × 2 × t_dead × f_sw. Minimise dead time without risking shoot-through for best efficiency.
Calculations are theoretical estimates. Actual performance depends on component parasitics, PCB layout, thermal management and control loop design.
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