What is a boost (step-up) converter?
A boost converter is a switched-mode DC-DC converter that produces an output voltage higher than its input. An inductor on the input side is charged from the supply through a MOSFET; when the switch opens, the inductor's stored energy is forced through a diode into the output capacitor, adding to the input voltage. Because energy is transferred by a (near) lossless inductor rather than dropped as heat, the boost converter reaches 90–98% efficiency and is the standard way to step a voltage up.
How it works — the two switching states
Over each switching period T = 1/fsw the converter alternates between two states set by the duty cycle D:
- Switch ON (0 → D·T): the MOSFET shorts the inductor to ground; the inductor current ramps up as
diL/dt = Vin/L, storing energy. The diode is reverse-biased, so the output capacitor alone supplies the load. - Switch OFF (D·T → T): the inductor current is forced through the diode into the output; because the inductor voltage adds to Vin, the output rises above the input. The current ramps down as
diL/dt = (Vin − Vo)/L.
Key equations
This simulator does not plug numbers into these formulas — it numerically integrates the real switching circuit (with your device parasitics) to steady state, then measures Vo, the ripple and the losses from the resulting waveforms and compares them to the equations above in the accuracy panel.
Continuous vs discontinuous conduction (CCM / DCM)
In CCM the inductor current never falls to zero and Vo = Vin/(1 − D) holds. At light load or with a small inductor the current hits zero each cycle — DCM — the diode stops conducting, and the output voltage rises above Vin/(1 − D). The boundary is the critical inductance Lc = D·(1 − D)²·R/(2·fsw). Try the Light load → DCM preset.
Efficiency, losses & device stress
Real converters lose power in the MOSFET (Rds(on)), the diode (Vf), the inductor (DCR) and the capacitor (ESR). Note that in a boost converter both the switch and the diode must block the full output voltage Vo, and the input inductor carries the (larger) input current Iin = Io/(1 − D) — so high step-up ratios stress the components hard. The simulator computes the conduction losses exactly from the integrated currents and shows the split in the loss-breakdown chart.
Buck vs boost vs buck-boost
| Converter | Output vs input | Ideal Vo |
|---|---|---|
| Buck | Steps down (Vo < Vin) | D·Vin |
| Boost | Steps up (Vo > Vin) | Vin/(1 − D) |
| Buck-Boost | Up or down, inverted | −D·Vin/(1 − D) |
Applications
Battery-powered devices that need a higher rail than the cell (e.g. 3.7 V → 5 V/12 V), LED-string drivers, photovoltaic MPPT stages, power-factor-correction (PFC) front-ends, and boosting to the DC-link of inverters. It is one of the three fundamental DC-DC topologies.
Frequently asked questions
What is the output voltage of a boost converter?
In CCM the ideal output is Vo = Vin/(1 − D), always higher than the input. Real device drops make it slightly lower; the simulator shows both.
What is the inductor (input) ripple current?
ΔiL = Vin·D/(L·fsw). Since the inductor is on the input, this is also the input-current ripple.
What causes discontinuous conduction (DCM)?
Light load or a small inductor lets the inductor current reach zero each cycle. Below the critical inductance Lc = D·(1−D)²·R/(2·fsw) the converter enters DCM and the output rises above Vin/(1−D).
Why is boost output ripple larger than buck?
While the switch is on, the output capacitor alone supplies the load, so ΔVo = Io·D/(C·fsw) — plus the ESR term ΔiL·ESR. Use a larger, low-ESR capacitor to reduce it.
What voltage must the MOSFET and diode withstand?
Both block the full output voltage Vo = Vin/(1 − D), so they must be rated well above it — especially at high duty cycles.