DC chopper (step-down, time-ratio control) – average and RMS load current A step-down chopper feeds a purely resistive load R from input voltage V using duty cycle α (0 < α < 1). Under ideal switching, determine the average load current and the RMS load current in terms of V, R, and α.

Introduction / Context:Choppers control the average power delivered to loads by switching the supply rapidly. For a step-down (buck) chopper with a resistive load, recognizing the on/off voltage pattern allows direct computation of average and RMS current values, crucial for device and thermal design.

Given Data / Assumptions:

• Input voltage V (constant, ideal source).
• Duty cycle α = T_on / T, with period T.
• Ideal switch, no ripple smoothing element (pure R load).
• Load voltage is V during T_on, and 0 during T_off.

Concept / Approach:

For a resistive load, i(t) = v(t)/R instantaneously. Average values depend on time-averaging of square wave segments, while RMS relies on the square root of the time-averaged square of the instantaneous values.

Step-by-Step Solution:

Verification / Alternative check:

Dimension check: both expressions have amperes. Limiting cases: α → 1 gives I_avg = I_rms = V/R (DC), α → 0 gives both → 0, as expected.

Why Other Options Are Wrong:

• Swapping average and RMS leads to incorrect scaling with √α.
• Statements with both equal to V/R ignore duty cycle.
• Both equal to αV/R incorrectly treat RMS as an average.

Common Pitfalls:

Confusing average with RMS; forgetting that RMS of a chopped square wave scales with √α, not α.

Final Answer:

I_avg = (αV)/R, I_rms = (V/R)*√α