How this instrument works
An inverting buck-boost converter is the single-inductor, single-switch topology that both scales a DC voltage's magnitude and reverses its sign relative to circuit ground. The transistor switch ties one end of the inductor to Input voltage for a fraction D of each switching cycle, charging it; a diode oriented with its cathode at the switch node — not its anode, as in a boost converter — then carries the inductor's current into the output capacitor when the switch opens, pulling that capacitor's terminal below ground instead of above it. The result reads out as Output voltage (inverted): a negative number even though every input to the formula is positive.
The magnitude comes from the same volt-second balance that governs every switch-mode converter, applied with a sign flip. While the switch is closed for D·T, the inductor sees +Vin across it and stores Vin·D·T volt-seconds. While it is open for (1 − D)·T, the inductor's voltage reverses to match Vout directly, discharging Vout·(1 − D)·T volt-seconds. Steady state demands the two cancel: Vin·D = −Vout·(1 − D), which rearranges to Vout = −Vin·D ⁄ (1 − D). Set D to exactly 0.5 and the magnitude of Vout equals Vin precisely — the crossover point where the converter stops behaving like a step-down and starts behaving like a step-up.
The formula assumes an ideal switch, a lossless diode, and continuous conduction, where inductor current never reaches zero between pulses. A bench measurement lands a little less negative than the calculation because the diode's forward drop and the switch's on-resistance eat into the output. The denominator also sets a hard limit: as D climbs toward 1, (1 − D) shrinks toward zero and the predicted output races toward negative infinity, so practical designs stay below roughly 0.8–0.9 duty cycle, where the diode's reverse voltage and the inductor's peak current are still something a real component can survive.
- Enter Input voltage (Vin) — the positive DC rail feeding the switch, such as a 12 V battery or a 24 V control-panel bus.
- Set Duty cycle (D) as a decimal between 0 and 1 — the fraction of each switching period the transistor stays closed, never a percentage.
- Read Output voltage (inverted) (Vout) — the steady-state negative rail the converter delivers once the inductor current settles.
- Watch the sign: Vout is always negative for any positive Vin and D, which is the topology's defining behaviour, not an error.
- Nudge Duty cycle past 0.5 to see the magnitude overtake Vin, or below 0.5 to see it fall short — the crossover sits exactly at D = 0.5.
Worked example — 12 V to −8 V at 40% duty cycle
Set Input voltage to 12 V — a common automotive or benchtop rail — and Duty cycle to 0.4, so the switch closes for 40 percent of every switching period. The formula gives Vout = −Vin·D ⁄ (1 − D) = −12 × 0.4 ⁄ 0.6 = −4.8 ⁄ 0.6 = −8.0 V, a clean negative rail sized for feeding the V− pin of a dual-supply op-amp from a single 12 V source.
Because 0.4 sits below the 0.5 crossover, the output's magnitude (8 V) lands below the input's (12 V) — this duty cycle range behaves like a step-down. Push Duty cycle up to 0.6 instead and the same formula gives Vout = −12 × 0.6 ⁄ 0.4 = −18.0 V, a magnitude larger than Vin, because 0.6 sits past the crossover into step-up territory.
Questions
Why is Vout negative when Vin and D are both positive?
Because polarity inversion is what this topology does by design, not a fault. The diode's cathode ties to the switch node instead of the output, so when the switch opens, the inductor dumps its current into the output capacitor from the ground side rather than the input side, pulling that node below ground. Any positive Vin and D between 0 and 1 will always return a negative Vout.
Does a duty cycle of 0.5 always make the output equal to −Vin?
Yes. Substituting D = 0.5 into Vout = −Vin·D ⁄ (1 − D) leaves −Vin·0.5 ⁄ 0.5 = −Vin exactly, regardless of what Vin is. That duty cycle is the converter's crossover point: below 0.5 the output magnitude is smaller than Vin, and above 0.5 it is larger, the same way a buck steps down and a boost steps up.
How is this different from a non-inverting buck-boost like a SEPIC?
A SEPIC or four-switch buck-boost keeps the output on the same rail polarity as the input, using a second inductor or a pair of extra switches to route the energy without flipping ground reference. This single-inductor, single-switch topology is simpler and cheaper but always inverts polarity — if a same-polarity step-up-or-down output is what a design needs, this is the wrong circuit.
What actually limits how high the duty cycle can go?
Component stress, not the formula. As D approaches 1, the diode must block a reverse voltage of Vin minus the growing negative Vout, and the inductor's peak current climbs sharply, so most designs cap duty cycle around 0.8 to 0.9. Push past that and switching losses and voltage spikes typically outrun any further gain in output magnitude.
Can this circuit generate a negative supply rail from a single positive battery?
Yes — that is its most common job. A single 12 V or 24 V rail feeding this topology at the right duty cycle produces a clean negative voltage for an op-amp's V− pin, a MOSFET gate-drive return, or an RS-232 line driver, without needing a second battery or a transformer to create the negative reference.
What happens if duty cycle is set to 1?
The calculator rejects it, and physically the circuit stops working: at D = 1 the switch never opens, so the inductor never gets a chance to discharge into the output, and the (1 − D) denominator hits zero — an undefined result. Real controllers keep duty cycle safely below 1 for exactly this reason.