Buck-Boost Converter
Calculate the output voltage of a buck-boost DC-DC converter in continuous conduction, V_out = V_in × D ÷ (1 − D), from the input voltage V_in and the duty cycle D (0 to 1). The result, in volts, can be lower (D < 0.5) or higher (D > 0.5) than the input — the buck-boost converter steps voltage down or up depending on the duty cycle, with inverted output polarity in the classic topology. It is used when the input voltage can vary above and below the desired output (discharging batteries, universal supplies). Enter the input voltage and the duty cycle.
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Buck-boost converter
The buck-boost converter combines both capabilities: it can step the voltage down or up, depending on the duty cycle. Its relation is V_out = V_in × D ÷ (1 − D). The turning point is D = 0.5: below it (D < 0.5) the output is lower than the input (step-down mode); above it (D > 0.5) the output is higher (step-up mode); at D = 0.5 output equals input. In the classic inverting topology the output has reversed polarity with respect to the input (a +12 V source produces −V at the output), which is why it is also called an inverting converter; non-inverting variants such as the SEPIC and the Ćuk keep the polarity. The buck-boost really earns its place when the input voltage swings around the desired output voltage — the typical case of a battery: a Li-ion cell runs from ~4.2 V when charged down to ~3.0 V when depleted, and if the circuit needs a steady 3.3 V, sometimes the converter has to step down (4.2 → 3.3) and sometimes step up (3.0 → 3.3). Only a converter that does both keeps the output regulated over the whole discharge curve. That is why the buck-boost and its variants show up so often in battery-powered portable devices and in 'universal input' power supplies. Enter the input voltage and the duty cycle.
Related Tools
Buck Converter (Step-Down)
Calculate the output voltage of a buck (step-down) DC-DC converter in continuous conduction, V_out = D × V_in, from the duty cycle D (0 to 1) and the input voltage V_in. The result, in volts, is always less than or equal to the input — the buck converter lowers voltage efficiently (without dissipating the excess, unlike a linear regulator), by switching rapidly and filtering with an inductor and capacitor. Varying the duty cycle adjusts the output from 0 to V_in. It is the most common topology in switching power supplies and point-of-load regulators. Enter the duty cycle and the input voltage.
Boost Converter (Step-Up)
Calculate the output voltage of a boost (step-up) DC-DC converter in continuous conduction, V_out = V_in ÷ (1 − D), from the input voltage V_in and the duty cycle D (0 to 1). The result, in volts, is always greater than the input — the boost converter raises voltage by storing energy in an inductor and releasing it in series with the source. As D approaches 1, the output tends to infinity (limited by real losses). It is used in supplies that must step up voltage (LEDs, batteries, power factor correction) and in photovoltaic systems. Enter the input voltage and the duty cycle.
Output Voltage Ripple
Calculate the output voltage ripple of a switching converter, ΔV = I ÷ (f × C), from the output current I, the switching frequency f and the output capacitance C. The result, in volts, is the residual oscillation superimposed on the DC output voltage, caused by the filter capacitor charging and discharging each switching cycle. Higher frequency and capacitance reduce the ripple. Keeping the ripple within limits (typically <1% of the output) is essential to supply sensitive circuits. Enter the current, the switching frequency and the capacitance.
PWM Output Voltage
Calculate the average output voltage of a PWM signal, V_out = (duty ÷ 100) × V_sup, from the duty cycle (in %) and the supply voltage V_sup. The result, in volts, is the effective average voltage delivered to a load (motor, LED, heater) by rapidly switching the supply on and off. Varying the duty cycle from 0 to 100% varies the average voltage from 0 to V_sup, allowing power control without dissipating energy in resistors — the basis of motor speed and LED brightness control in microcontrollers. Enter the duty cycle and the supply voltage.
Rectifier Average Voltage
Calculate the average (DC) voltage of a full-wave rectifier, V_dc = 2 × V_p ÷ π, from the peak voltage V_p of the sinusoidal input. The result, in volts, is the mean value of the pulsating voltage at the rectifier output before filtering — about 63.7% of the peak. A full-wave rectifier (bridge or center-tap) flips the negative half-cycles, doubling the ripple frequency and raising the average voltage compared with a half-wave rectifier (V_p/π). It is the basis of designing DC power supplies from the AC mains. Capacitor filtering then raises the voltage further (close to the peak). Enter the peak voltage.
Square-Wave RMS Current
Calculate the RMS value of a pulsing square-wave current, I_rms = I_p × √D, from the peak current I_p and the duty cycle D (fraction of the period the current flows). The result, in amperes, is the RMS current that determines the actual heating (I²R losses) of a component that conducts in pulses — such as a transistor or winding in a switching converter. Unlike the average value, the RMS is what matters for sizing conductors, resistances and dissipation. The smaller the duty cycle, the lower the RMS for the same peak current. Enter the peak current and the duty cycle.
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