Buck Converter Inductor Saturation at High Input
A 24 V-to-5 V buck converter delivers 2.5 A continuous at 500 kHz using a 4.7 µH inductor, two 22 µF ceramic output capacitors, and a current-mode controller. The inductor datasheet gives 4.0 A saturation at 25 °C, derated to 3.2 A at 100 °C, and 25 mΩ DCR. The MOSFETs are rated for 5 A. At 24 V input and 2.5 A load the switch-node current ramps up to about 3.3 A and then shows a sharp spike; the output drops out of regulation, and the inductor produces an audible buzz. The same board works at 12 V input and 2.5 A, although the current waveform is close to the limit. A technician added a third 22 µF output capacitor, which reduced output ripple slightly but did not stop the dropout. The controller's current-limit threshold is 4.5 A, and the current-sense blanking time is 80 ns. The ambient inside the enclosure reaches 85 °C.
1. Calculate the inductor peak current at 24 V and at 12 V. Determine whether saturation explains the 24 V failure and the 12 V behavior.
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For a buck converter, ΔI_L = (V_in − V_out) × D/(L × f_sw), with D = V_out/V_in. At 24 V, D = 5/24 = 0.208, so ΔI_L = (24 − 5) × 0.208/(4.7 µH × 500 kHz) = 3.95/2.35 ≈ 1.68 A. The peak is 2.5 A + 0.84 A = 3.34 A. At 100 °C the saturation current is 3.2 A, so the inductor saturates. At 12 V, D = 0.417, ΔI_L = (12 − 5) × 0.417/2.35 ≈ 1.24 A, and the peak is 2.5 A + 0.62 A = 3.12 A, just below 3.2 A. That explains why 12 V appears to work but has almost no margin. The sharp current spike, buzz, and dropout are consistent with saturation: inductance collapses, the current slope increases, and the controller cannot maintain regulation. The 4.5 A current limit is above the saturation point, so it does not protect the inductor.
2. Explain why adding output capacitance did not fix the problem. Specify an inductor selection criterion and propose a replacement.
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Output capacitance controls output ripple and transient voltage, but it does not limit the current through the inductor. Once the core saturates, the inductance falls sharply, the current rises faster, and the controller may hit current limit or lose regulation; more capacitance only hides a little ripple. Select the inductor so its saturation current at the maximum operating temperature exceeds the worst-case peak with margin. A reasonable criterion is I_sat(100 °C) ≥ 1.3 × I_peak. At 24 V, I_peak is 3.34 A, so the target is about 4.3 A. A 6.8 µH inductor with I_sat ≥ 5 A at 100 °C gives ΔI_L = 3.95/(6.8 µH × 500 kHz) ≈ 1.16 A and I_peak ≈ 3.08 A, with ample margin. Check DCR and thermal rise, and verify that the higher inductance does not make the current-mode loop too slow for the load transient. The 12 V case must also be checked, not assumed safe.
3. At 36 V input the same converter must still deliver 2.5 A. Recalculate the peak current and state the additional risks.
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At 36 V, D = 5/36 = 0.139, so ΔI_L = (36 − 5) × 0.139/(4.7 µH × 500 kHz) = 4.31/2.35 ≈ 1.83 A. The peak is 2.5 A + 0.92 A = 3.42 A, which is higher than at 24 V and well above the 3.2 A hot saturation rating. The failure therefore becomes worse at high line, not better. The 6.8 µH replacement gives ΔI_L ≈ 1.27 A and I_peak ≈ 3.14 A, still below a 5 A saturation rating. Additional risks at 36 V include higher MOSFET switching loss, a shorter minimum on-time, and more high-frequency noise around the current-sense blanking window. Verify the controller's minimum controllable on-time, the switch-node ringing, and the input capacitor voltage rating. Do not assume that a converter validated at 24 V is safe at 36 V.