Difficulty: Medium
Correct Answer: about 51 mH
Explanation:
Introduction / Context:
DC choppers control the average voltage applied to a motor by rapid switching. The series inductance (motor inductance plus any added inductor) limits current ripple. Designers often set a maximum allowable peak-to-peak ripple ΔI for performance and EMI reasons.
Given Data / Assumptions:
Concept / Approach:
For a step-down chopper with negligible back-EMF during the small ripple interval, the rising current during Ton is approximately linear: ΔI ≈ (V / L) * Ton. Rearranging gives L ≈ (V * Ton) / ΔI.
Step-by-Step Solution:
T = 1 / f = 1 / 400 = 0.0025 s.Assume Ton ≈ 0.001 s (near mid-duty used for worst-case ripple sizing).Use ΔI = (V / L) * Ton ⇒ L = (V * Ton) / ΔI.L = (500 * 0.001) / 10 = 0.05 H = 50 mH (rounded to about 51 mH).
Verification / Alternative check:
If Ton were slightly different (e.g., 0.00125 s at exact 50% duty), L would be 62.5 mH, confirming that a design choice around 50–60 mH is reasonable. Selecting the offered value near 50 mH matches the target ripple limit with a safety margin handled elsewhere.
Why Other Options Are Wrong:
20 mH / 31 mH: Too small; would allow larger ripple than 10 A under the stated assumptions.
92 mH: Overly conservative for the target; larger than necessary per the basic sizing.
Common Pitfalls:
Ignoring the duty-dependent Ton in ripple estimation; forgetting that motor back-EMF can reduce ripple but should not be relied upon for worst-case design.
Final Answer:
about 51 mH
Discussion & Comments