Key Takeaways
- ΔVout = ΔI_L / (8×f×C)
- Larger C → smaller voltage ripple; higher f → smaller ripple
- L controls current ripple, C controls voltage ripple — clear division of labor
Lesson 5
Output Voltage Ripple: The Decisive Role of the Capacitor
The inductor determines current ripple, the capacitor determines voltage ripple. Formula: ΔVout ≈ ΔI_L / (8×f×C).
In CCM, what flows into the output capacitor is the AC component of inductor current (the triangular ripple portion). The capacitor integrates this triangular current, producing the output voltage ripple.
Mathematical result: ΔVout = ΔI_L / (8×f×C). Note the factor of 8 — because integrating the triangular current gives a peak-to-peak ripple of ΔI/(8fC), not the intuitive ΔI/(fC).
Example with L=1mH, C=10μF: ΔVout = 0.3/(8×10000×0.00001) ≈ 0.375V. For a 6V output this is 6.25% ripple — quite large. With C=100μF, ripple drops to 37.5mV (0.6%).
Summary of two core Buck steady-state design formulas: inductor determines current ripple ΔI = (VIN−VOUT)×D/(f×L), capacitor determines voltage ripple ΔV = ΔI/(8×f×C). Choosing L and C gives you control over ripple levels.
Small Capacitor = Large Voltage Ripple
C=10μF, L=1mH. Output ripple≈0.375V (6.25%), clearly visible.
After reading this section, run the simulation and observe the waveforms. To explore further, open the example in a standalone page.
Watch Items
- out waveform: with C=10μF, ripple is ~0.35V, clearly visible oscillation
- Try changing C to 100μF — ripple will decrease by 10×