TIDUEY0A November 2020 – December 2025
The value of the output capacitor and the ESR of the capacitor determine the output voltage ripple and load transient performance. This capacitor is designed to achieve the best possible output voltage ripple and load transient performance.
Equation 8 calculates the duty cycle of the buck current.
Inductor ripple current is typically defined
between 0.2 to 0.3 times the output current. Since the control loop needs to be
fast, the inductor can be picked at smaller value. So 0.2 is selected as the ripple
current coefficient. Full scale of 12 A is used in the calculation for some margin,
so the inductor ripple current is estimated to be
2.4
A.
The worst-case scenario for Continuous Conduction Mode (CCM) in a buck converter occurs when Toff is at the maximum as represented in Equation 2.
Solving the equation, the inductor value needs to be higher than 4.45 μH. An inductor with smaller inductance has a larger saturation current. In this design, a 4.7 μH inductor is selected. Equation 3 calculates the actual inductor current.
Equation 4 calculates the output capacitance without considering the ESR.
The output voltage ripple is targeted to be 0.1% of the maximum output voltage, which is 6 mV. Plug in Equation 4, and the output capacitor is calculated at 192 μF. In this design, four 47 μF and two 1-μF ceramic capacitors are placed in parallel to match the capacitance.
The ESR requirement of output capacitors is also critical to determine the total output voltage ripple. Higher ESR can lead to an overall higher output voltage ripple. To calculate the ESR requirements, use the following equation to calculate the voltage ripple due to the capacitor alone and due to ESR alone:
Solve for Resr ≅ 0.5 mΩ. This value can be achieved with paralleling ceramic capacitors; each capacitor has around 1.5 mΩ of ESR at 250 kHz switching frequency. Paralleling the connection of the capacitor can reduce the overall ESR to less than 1 mΩ.