For a step-up/step-down chopper, if the duty cycle 0.5 then

Question

For a step-up/step-down chopper, if the duty cycle > 0.5 then

Examoogle · Accepted Answer

$V_o > V_s$

Explanation: In a step-up/step-down (Buck-Boost) chopper, the output voltage is determined by the duty cycle $\alpha$. When the duty cycle $\alpha$ is greater than 0.5, the output voltage magnitude becomes greater than the source voltage, effectively operating in the boost (step-up) mode.

Working Principle: The buck-boost converter operates by storing energy in an inductor during the ON period and releasing it to the load during the OFF period. The transfer function $V_o = V_s \cdot \frac{\alpha}{1-\alpha}$ dictates the behavior. When $\alpha = 0.5$, $V_o = V_s$; for $\alpha > 0.5$, the ratio $\frac{\alpha}{1-\alpha}$ exceeds 1, resulting in $V_o > V_s$.

Key Formulas: $V_o = V_s \cdot \frac{\alpha}{1-\alpha}$ — Relationship between output voltage, source voltage, and duty cycle $\alpha$; $\alpha = \frac{T_{on}}{T}$ — Definition of duty cycle

Key Points: If $\alpha; If $\alpha = 0.5$, the output voltage is equal to the source voltage.; The output voltage in a buck-boost converter is inverted in polarity relative to the input.

Advantages: Wide range of voltage regulation; Single switch topology simplifies control

Disadvantages: Output polarity is inverted; High peak currents in the switch and diode

Applications: DC-DC power supplies; Battery-powered portable electronics

Additional Info: The Buck-Boost converter is often referred to as an inverting regulator.. Option B is incorrect because it describes the step-down region where $\alpha

Answer: C is correct — For a buck-boost chopper, the gain is given by $\frac{\alpha}{1-\alpha}$, which exceeds unity when the duty cycle $\alpha$ is greater than 0.5.

Exam Tip: Remember the boundary condition $\alpha=0.5$ for Buck-Boost converters; it acts as the transition point between step-down and step-up modes.

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