Output voltage of Buck-Boost Regulator is ______

Question

Examoogle · Accepted Answer

$-\frac{D}{1-D} 	imes V_i$

Explanation: The Buck-Boost converter is a DC-DC converter that provides an output voltage magnitude that can be either greater than or less than the input voltage magnitude. The negative sign in the transfer function indicates that the output voltage has an inverted polarity relative to the input voltage.

Working Principle: During the ON state, the inductor stores energy from the source. During the OFF state, the energy stored in the inductor is released to the load through the diode, with the circuit configuration causing a voltage inversion. The average output voltage is determined by the duty cycle $D$, where $V_o = -V_i 	imes \frac{D}{1-D}$.

Key Formulas: $V_o = -V_i \cdot \frac{D}{1-D}$ — Output voltage transfer function; $D = \frac{t_{on}}{T}$ — Definition of Duty Cycle

Key Points: The output voltage magnitude is proportional to the duty cycle ratio.; The negative polarity is inherent to the standard inverting buck-boost topology.; The converter operates in continuous conduction mode (CCM) when the inductor current never falls to zero.

Advantages: Capability to both step-up and step-down voltage; Simple control circuitry for basic regulation

Disadvantages: Negative output polarity requires level shifting for some loads; Higher component stress due to pulsating input and output currents

Applications: Battery-powered systems; General purpose DC-DC power supplies

Additional Info: The duty cycle $D$ ranges from 0 to 1.. If $D  0.5$, the output is a step-up (boost).

Answer: B is correct — The output voltage of an inverting Buck-Boost regulator is given by the product of the input voltage and the ratio of duty cycle to its complement, with a negative sign indicating voltage inversion.

Exam Tip: Always check for the polarity in power electronics questions; inverting buck-boost converters always produce a negative voltage relative to the common ground.

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