How this instrument works
A boost converter is a switch-mode circuit that raises a DC voltage without a transformer, using only a transistor, a diode, an inductor and a capacitor. The relationship this instrument solves, Vout = Vin ⁄ (1 − D), links three quantities: the source voltage Vin, the fraction D of each switching cycle the transistor spends closed, and the resulting output Vout. D is a pure ratio — the transistor's on-time divided by the total period — so it carries no unit of its own, while Vin and Vout share whatever voltage unit you choose.
The shape of the formula comes from volt-second balance across the inductor, the same principle used to analyse every switch-mode topology. While the transistor is closed for a duration D·T, the source drives current through the inductor and it stores energy, accumulating Vin·D·T volt-seconds. While the transistor is open for (1 − D)·T, that stored energy is forced into the output through the diode, and the inductor now sees (Vout − Vin) across it for (1 − D)·T volt-seconds. In steady state the two must balance exactly — Vin·D = (Vout − Vin)·(1 − D) — and rearranging that single equation produces Vout = Vin ⁄ (1 − D).
The formula assumes an ideal converter running in continuous conduction, where inductor current never reaches zero between switching cycles, and it ignores the diode's forward drop and the resistance in the switch, inductor and wiring — all of which shave a few percent off a real circuit's output. It also breaks down as duty cycle approaches 1: the denominator 1 − D shrinks toward zero, so the predicted output climbs toward infinity, but real boost converters rarely run past about 80-90 percent duty cycle because parasitic losses grow faster than the theoretical gain and efficiency collapses.
- Enter the source voltage in the Input voltage field — the battery, solar cell or rail you are stepping up, in V or mV.
- Set Duty cycle to the transistor's on-time fraction, a decimal from 0 up to just under 1 (never exactly 1).
- Read the boosted value in Output voltage — it updates immediately as you adjust either input.
- To hit a target rail, work backward: raise Duty cycle to push Output voltage higher, keeping it below about 0.9 for a realistic design margin.
Worked example — 5 V to 10 V at 50% duty cycle
A guitar-pedal power supply runs from a 5 V USB power bank, but the pedal's op-amp stage needs roughly 10 V of headroom to avoid clipping loud signals, so the board carries a small boost converter between the USB jack and the analogue stage. Setting Duty cycle to 0.5 — the transistor closed for exactly half of every switching period — gives Vout = 5 ⁄ (1 − 0.5) = 5 ⁄ 0.5 = 10 V, doubling the input exactly.
This result also demonstrates the converter's built-in floor: because D can only range from 0 up to just under 1, the factor 1 ⁄ (1 − D) can never fall below 1, so Vout can never dip under Vin no matter what duty cycle is dialled in. Nudge Duty cycle down to 0 and the instrument returns Vout = 5 V — the transistor never closes, and the inductor and diode simply pass the source straight through to the output capacitor.
Questions
What does 'duty cycle' mean in a boost converter?
It is the fraction of each switching period that the transistor is held closed, conducting current through the inductor. A duty cycle of 0.5 means the switch is on for half of every cycle and off for the other half — entered here as a decimal from 0 up to just under 1, never as a full 1.
Why does the formula divide by (1 − D) instead of multiplying by D?
Because the boost happens during the off-time, not the on-time. Volt-second balance across the inductor requires Vin·D = (Vout − Vin)·(1 − D) over one switching period; solving that equation for Vout produces Vin ⁄ (1 − D). As the on-time D grows, (1 − D) shrinks, and dividing by a smaller number is what drives the output higher.
What happens at a duty cycle of 1?
The instrument rejects it — a duty cycle of 1 means the transistor never opens, so the inductor never gets a chance to release its stored energy into the output, and Vout is undefined (division by zero). Real converters cap the usable duty cycle well below 1, typically around 0.8 to 0.9, because parasitic resistance dominates as (1 − D) shrinks.
Can a boost converter step voltage down?
No — that is a buck converter's job. Because D always sits between 0 and 1, the factor 1 ⁄ (1 − D) is always at least 1, so Vout can only equal or exceed Vin. A circuit that needs to both raise and lower voltage from the same input uses a buck-boost topology, which rearranges the transistor and diode of this same basic circuit.
Does this formula hold for a real, lossy converter?
It is the ideal, lossless approximation that design guides start from. A real converter loses a little voltage to the diode's forward drop, the switch's on-resistance and the inductor's winding resistance, so a measured Vout typically runs a few percent below this calculation, with the gap widening as duty cycle nears its practical ceiling.
What operating assumption does this calculation depend on?
Continuous conduction mode: the inductor current must never fall to zero between switching cycles. At light loads or low switching frequencies a converter can slip into discontinuous conduction, where the current does hit zero, and this formula then overstates the true output voltage.