How this instrument works
A buck converter is a switching regulator that steps a DC voltage down by chopping it, not by burning off the difference as heat the way a linear regulator does. A transistor switch connects the input to an inductor for a fraction D of every switching cycle, then disconnects it for the remaining (1 − D); a diode or second transistor picks up the current during that off time so the inductor never sees an open circuit. Averaged over one cycle, the output settles at exactly D times the input — a duty cycle of 0.4 means the switch is closed 40 percent of the time, and the output lands at 40 percent of Vin.
The multiplication falls out of volt-second balance, the rule that in steady state an inductor's average voltage over one full switching cycle must be zero — otherwise its current would ramp away forever. While the switch is closed, the inductor sees (Vin − Vout) for a time D·T; while it's open, it sees −Vout for (1 − D)·T. Setting those two volt-second products equal and solving cancels the switching period T entirely, leaving Vout = D·Vin — a result that depends only on the duty cycle, not on the inductor's size or the switching frequency.
The formula assumes an ideal switch and continuous conduction — the inductor current never falls to zero between pulses. Real converters lose a percent or two to the switch's on-resistance, the diode's forward drop, and the inductor's own resistance, so a bench measurement usually reads a shade below D·Vin. At light load, current can hit zero before the next cycle starts — discontinuous conduction mode — and the clean D·Vin relationship no longer holds; output then depends on load current, inductance, and switching frequency too.
- Enter Input voltage (Vin) — the DC rail feeding the converter's switch, such as a battery pack or an unregulated 12 V bus.
- Enter Duty cycle (D) as a fraction between 0 and 1 — the share of each switching period the high-side switch stays closed, not a percentage.
- Read Output voltage (Vout) — the steady-state DC level the converter delivers to its load once the inductor current settles.
- Switch Vin or Vout between mV and V if your working numbers are more convenient in millivolts.
Worked example — 12 V bucked to 4.8 V at 40% duty cycle
Set Input voltage to 12 V — a typical automotive or small-solar rail — and Duty cycle to 0.4, meaning the switch closes for 40 percent of every switching period. The formula gives Vout = D × Vin = 0.4 × 12 = 4.8 V, a comfortable level for charging a lithium-ion cell or feeding a 5 V logic rail after a small regulator cleans up the ripple.
Push Duty cycle to its extremes and the linearity holds exactly: at D = 1 the switch never opens, so Vout = 1 × 12 = 12 V, the input passed straight through; at D = 0 the switch never closes, so Vout = 0 × 12 = 0 V, no path for current at all. Between those bounds, Vout rises in a straight line with D — unlike a boost converter, whose Vout = Vin ⁄ (1 − D) curves sharply upward and climbs toward infinity as D approaches 1.
Questions
Where does Vout = D·Vin actually come from?
From volt-second balance on the inductor: in steady state, the average voltage it sees over one switching cycle must equal zero, or its current would ramp away without limit. Solving (Vin − Vout)·D = Vout·(1 − D) for Vout cancels the switching period entirely and leaves Vout = D·Vin — the reason the ratio depends only on duty cycle, never on inductance or frequency.
Can duty cycle be a value like 1.5 or −0.2?
No. Duty cycle is the fraction of one switching period the switch stays closed, so it's bounded between 0 and 1 by definition — 150% or −20% of a cycle has no physical meaning. This calculator flags a negative entry as invalid; values above 1 simply describe a switch that's on more often than the cycle allows, which no real PWM controller can generate.
Do I choose the duty cycle myself, or does the converter set it?
In a finished design, the converter's feedback loop sets it, not you. A PWM controller compares Vout against a reference and continuously adjusts D — many thousands of times a second — to hold the output steady as Vin and load current drift. This calculator runs the relationship the other way: given a duty cycle, what output would it produce, which is exactly the question a designer asks while checking a datasheet's expected range.
Why is Vout always a little lower than D·Vin on a real board?
Because the formula assumes an ideal, lossless switch. Real MOSFETs have on-resistance, the freewheeling diode (or synchronous FET) drops some voltage, and the inductor's own copper carries resistance — each subtracts a small amount from the ideal figure. A well-designed converter still lands within a percent or two of D·Vin at typical loads; the gap widens mainly at very light load or very high current.
How is a buck converter different from a boost converter?
A buck steps voltage down with a linear relationship, Vout = D·Vin; a boost steps it up with a reciprocal one, Vout = Vin ⁄ (1 − D). That reciprocal curve is why boost design gets touchy near D = 1 — small changes in duty cycle swing the output wildly — while a buck's straight-line relationship stays well-behaved and predictable across its whole range.
What happens if the inductor current reaches zero before the next switching cycle?
The converter enters discontinuous conduction mode, and Vout = D·Vin stops holding exactly. This typically happens at light load with a small inductor: current ramps down to zero and stays there for part of the cycle, so the volt-second balance used to derive the formula no longer applies, and output also depends on inductance, switching frequency, and load current.