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.
Result
—
Buck converter (step-down)
The buck converter is the most widely used switch-mode DC-DC topology, and its job is to step down a DC voltage efficiently. In continuous conduction the relation is elegantly simple: V_out = D × V_in, where D is the duty cycle (the fraction of the period, from 0 to 1, during which the switch stays on) and V_in the input voltage. Since D ≤ 1, the output is always lower than or equal to the input. The big advantage over a linear regulator (which drops the voltage by burning the excess as heat, at poor efficiency) is that the buck does not dissipate the difference: it chops the input at high frequency (switching on and off very fast) and uses an inductor and a capacitor to filter those pulses into a smooth DC voltage — the energy is transferred, not wasted, with efficiencies of 90-98%. Varying the duty cycle by PWM adjusts the output continuously from 0 up to V_in, which enables closed-loop regulation (a controller measures the output and trims D to hold it steady even as load or input change). The buck is everywhere: point-of-load regulators on motherboards (12 V → 1.2 V for the CPU), chargers, laptop power supplies, converters inside embedded systems. This formula is only the starting point of a design; full sizing involves the inductance (current ripple), the capacitance (voltage ripple), the losses and the control loop. Enter the duty cycle and the input voltage.
Related Tools
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.
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.
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.
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.
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.
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.
The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.