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Power Electronics

Buck-Boost Converter

The buck-boost is a hybrid of the buck and the boost: it can produce an output both lower and higher than its input. It is built with a single inductor, one switch, and one diode — but there is a price: the output is inverted relative to the input (negative). So from +12 V input you can produce −5 V or −20 V.

Operation has two phases: while the switch conducts, the inductor connects directly to the input and stores energy (no power reaches the output at that moment); when the switch turns off, the inductor dumps its stored energy into the load through the diode, but the current direction makes the output negative. The average output is Vout = −Vin·D/(1−D). At 50% duty |Vout| = Vin; below it steps down, above it steps up.

Where is it used? When you need to generate a negative supply from a positive source (e.g. the −V rail for op-amps), or when the input voltage wanders both below and above the output. If the inverted polarity is a problem, you move to non-inverting derivatives such as the SEPIC or Ćuk.

Formulas

Vout = −Vin · D / (1 − D)
|Vout| = Vin → D = 0.5
D = |Vout| / (|Vout| + Vin)

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Test Yourself

How is the output voltage of an ideal (inverting) buck-boost converter related to the input?
Answer: Inverted polarity (negative); can be either lower or higher — Vout = −Vin·D/(1−D): the sign is negative and the magnitude can go below or above the input via D.
In a 12 V inverting buck-boost at 60% duty, what is the output voltage?
Answer: −18 V — Vout = −12·0.6/(1−0.6) = −12·1.5 = −18 V.