Basic capability of a step-up (boost) chopper What output–input voltage relationship does an ideal step-up (boost) DC chopper provide in steady state?

Introduction / Context:DC–DC choppers change the average DC voltage level. The step-up (boost) converter is widely used in regulated power supplies, electric vehicles, and renewable interfaces when the required output DC voltage must exceed the available input.

Given Data / Assumptions:

• Ideal components for conceptual understanding.
• Continuous conduction mode behavior.
• Neglecting device drops and losses.

Concept / Approach:In a boost converter, during the ON interval the inductor stores energy from the source. During the OFF interval, the inductor releases energy to the load through the diode, adding to the source voltage. This stacking effect produces an output average greater than the input average.

Step-by-Step Solution:Boost steady-state relation: V_out = V_in / (1 − D), where D is the duty ratio (0 < D < 1).Since 1 − D < 1, division by a number < 1 gives V_out > V_in.Therefore, the output is higher than the input for any nonzero duty ratio.

Verification / Alternative check:Example: D = 0.5 → V_out = V_in / 0.5 = 2 * V_in, clearly higher than the input.

Why Other Options Are Wrong:

• Lower or equal output corresponds to buck or unity-gain behavior, not boost.
• “Both higher and lower” or “all of the above” would fit buck-boost or SEPIC, not a pure boost topology.

Common Pitfalls:

• Mixing boost with buck-boost, which can invert and change magnitude in both directions.
• Ignoring the effect of losses; in practice V_out is slightly less than the ideal expression due to drops.

Final Answer:higher than input voltage