Key Takeaways
- ΔI_L = (VIN−VOUT) × D / (f×L)
- Larger L → smaller ripple; higher f → smaller ripple
- Ripple magnitude directly determines the CCM/DCM boundary
Lesson 2
Calculating Inductor Current Ripple
Ripple magnitude is determined by L, f, VIN, VOUT. Formula: ΔI = (VIN−VOUT)×D / (f×L).
During switch-on, voltage across the inductor is VL = VIN − VOUT. From the fundamental inductor relation VL = L × dI/dt, the current rising slope is dI/dt = (VIN−VOUT)/L.
On-time is D×T = D/f, so current rises from minimum to maximum by: ΔI = (VIN−VOUT) × D / (f×L).
With our standard parameters: ΔI = (12−6) × 0.5 / (10000 × 0.001) = 0.3A. If we reduce inductance to 0.3mH: ΔI = 6 × 0.5 / (10000 × 0.0003) = 1.0A. Ripple increased over 3×!
This tells us: increasing L or raising f both reduce ripple. But larger L means bigger, costlier inductors; higher f means more switching losses. This is the core design tradeoff in practice.
Smaller Inductance = Larger Ripple
L=0.3mH, R=10Ω. ΔI≈1.0A, ripple is 3× larger than the previous demo.
After reading this section, run the simulation and observe the waveforms. To explore further, open the example in a standalone page.
Watch Items
- Compare i_l1 peak-to-peak: ~1.0A with L=0.3mH (much larger than 0.3A with L=1mH)
- Although ripple is larger, current minimum is still > 0 → still CCM