Steady-state current in an R–L–E load fed by a DC chopper When a DC chopper supplies an R–L–E load (resistance R, inductance L, back EMF E), how does the steady-state load current behave?

Introduction / Context:
In chopper-fed DC drives and supplies, the input switch alternates between on and off states. An R–L–E load includes a source EMF (e.g., motor back EMF), resistance, and inductance. Understanding current ripple under PWM operation is essential for sizing L, estimating ripple, and predicting torque pulsations in motors.

Given Data / Assumptions:

• Ideal chopper with duty ratio D.
• Load has finite inductance L (not infinite).
• Periodic steady state reached.

Concept / Approach:

With finite L, current cannot change instantaneously but changes gradually. During the switch ON interval, the net applied voltage across L may be (Vdc − E − iR), leading to rising current; during OFF, the freewheel path applies approximately (−E − iR) (or similar, depending on topology), causing current to fall. Hence the current oscillates between a minimum and maximum around an average value determined by D, R, L, and E.

Step-by-Step Solution:

Verification / Alternative check:

In the limit L → ∞ (theoretical), ripple tends to zero and current approaches constant; for finite L, ripple persists and is a design parameter set by switching frequency and inductance.

Why Other Options Are Wrong:

“Remains constant” requires infinite L; “constant if R is constant” is incorrect since L dictates ripple; current is not zero during off-time because the inductor freewheels current.

Common Pitfalls:

Assuming zero current during off time; forgetting that inductor enforces current continuity via freewheel path.

Final Answer:

varies between maximum and minimum values