Buck converter design: duty, inductor, capacitor
Calculate the duty cycle, inductor ripple current and output capacitor for a buck converter that steps 12 V down to 5 V.
Conditions
Vin = 12 V, Vout = 5 V, Iout = 2 A, fsw = 500 kHz
1. Duty cycle
D = Vout / Vin = 0.42
2. Inductor
Take the ripple current as 20 to 40% of the output current. With ΔI = 0.6 A
L = (Vin - Vout) × D / (fsw × ΔI) = 7 × 0.42 / (500k × 0.6) ≈ 9.8 µH → 10 µH
Saturation current Isat > Iout + ΔI/2 = 2.3 A or more; with margin, choose a 3 A part.
3. Output capacitor
Ripple voltage ΔV ≈ ΔI / (8 × fsw × C) + ΔI × ESR
With a 22 µF MLCC (allowing for the loss of effective capacitance), 0.6 / (8 × 500k × 22µ) ≈ 7 mV. An MLCC has a low ESR, so the ripple is small.
RoadmapSee where this note sits in Power Electronics→4. Input capacitor
Every time the switch turns on it draws a pulse of current from the input. Put an MLCC with a sufficient RMS ripple-current rating close to the IC pins.
In practice
The quick route is to start from the design equations and recommended parts in the converter IC's datasheet and the vendor's design tool (TI WEBENCH and the like), and confirm with the calculation above. A synchronous buck (a MOSFET on the low side too) has no diode loss and is more efficient.