Ripple in inductor current of a DC chopper feeding an R–L–E load For a step-down chopper with duty ratio a (0 < a < 1) supplying an R–L–E load, at which value of a is the steady-state inductor current ripple maximized?

Introduction / Context:In chopper (DC–DC) converters, inductor current ripple is a key design parameter influencing current stress, output voltage ripple (through R drop), and EMI. For a fixed switching period and input voltage, current ripple varies with the duty ratio a.

Given Data / Assumptions:

• Ideal step-down (buck) chopper behavior for ripple estimation.
• Switching period T_s fixed; inductor L constant.
• R–L–E load; effect of E and R on average current is not needed for locating the ripple maximum.

Concept / Approach:Over one switching period, the inductor voltage is approximately: during ON, v_L(on) ≈ V_in − V_o; during OFF, v_L(off) ≈ −V_o (neglect device drops). For a buck, V_o ≈ a * V_in in steady state. The triangular ripple amplitude ΔI is proportional to the area under v_L/L in each subinterval.

Step-by-Step Solution:Let ΔI_up = (V_in − V_o) * (a * T_s) / L = (V_in − aV_in) * aT_s / L = V_in * a(1 − a) * T_s / L.Let ΔI_down = |v_L(off)| * ((1 − a) T_s) / L = V_o * (1 − a) T_s / L = V_in * a(1 − a) * T_s / L.Net peak-to-peak ripple ΔI = ΔI_up = ΔI_down = V_in * a(1 − a) * T_s / L.The factor a(1 − a) is maximized at a = 0.5, giving maximum ripple.

Verification / Alternative check:The parabola a(1 − a) has its vertex at a = 0.5 with value 0.25, confirming the maximum ripple occurs at 50% duty.

Why Other Options Are Wrong:

• a = 1 or a → 0: ripple tends to zero as the inductor sees nearly constant voltage or zero output.
• a = 0.8 or a < 0.5: ripple is non-maximal except at exactly 0.5.

Common Pitfalls:

• Confusing current ripple with output voltage ripple; though related, their maxima can differ with nonidealities.
• Forgetting that device drops slightly skew the symmetry; however, the maximum remains near a ≈ 0.5.

Final Answer:a = 0.5