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DC-DC Converters

Problem

A 24 V battery powers a 12 V / 2 A LED strip via a buck converter. Switching frequency 100 kHz, L = 100 µH.

Step 1 — Duty Cycle

D = \frac{V_{out}}{V_{in}} = \frac{12}{24} = 0.5

Step 2 — Inductor Current Ripple

\Delta I_L = \frac{(V_{in} - V_{out}) \cdot D}{f_{sw} \cdot L} = \frac{(24 - 12) \times 0.5}{100 \times 10^3 \times 100 \times 10^{-6}} = 0.60\,\text{A}

Step 3 — CCM Check

I_{out} = 2\,\text{A} > \frac{\Delta I_L}{2} = 0.30\,\text{A} \quad \Rightarrow \quad \text{CCM confirmed}

Step 4 — Minimum Inductance

L_{min} = \frac{(V_{in} - V_{out}) \cdot D}{2 \cdot f_{sw} \cdot I_{out}} = \frac{(24 - 12) \times 0.5}{2 \times 100 \times 10^3 \times 2} = 15\,\mu\text{H}

Figure 1. Inductor current waveform — 2 A average, 0.6 A ripple (CCM)

The 100 µH inductor is well above the 15 µH minimum — the converter runs solidly in CCM with only 0.6 A of ripple on a 2 A load.