Boost converter Calculator

Load current	500 mA
Inductor ripple, peak-to-peak	500 mA
Output ripple, peak-to-peak	50 mV

Load current

500 mA

Inductor ripple, peak-to-peak

500 mA

Output ripple, peak-to-peak

50 mV

Average inductor current	1 A
Peak inductor current	1.25 A
CCM boundary inductance	25 µH
Output power	5 W

Average inductor current

1 A

Peak inductor current

1.25 A

CCM boundary inductance

25 µH

Output power

5 W

How boost converter works

The switch first stores energy in L1. When it opens, the inductor sends current through the diode to the output at a voltage above the input. Output current is supplied by C1 while the switch is on. The guest starts near the calculated continuous-conduction operating point.

V_{out}=\frac{V_{in}}{1-D}

Ideal continuous-conduction, open-loop power stage with fixed PWM duty. No feedback controller, switch loss, capacitor ESR or inductor resistance is included in the equations. The guest uses near-ideal switches and diodes. Inputs outside the continuous-conduction or small-ripple range are rejected.

Change the form inputs or click a component value on the schematic. Choose Try in CircuitSim to simulate the configured circuit.

Do the math

Selected input values
V1	5 V
V2 — Switch on-time percentage	50 %
L1	100 µH
R1	20 Ω
C1	100 µF
V2 — Switching frequency	50 kHz

Duty fraction: $D=\text{duty percent}/100$
$50/100$
≈ 0.5
Ideal output: $V_{out}=\frac{V_{in}}{1-D}$
$\frac{5\,\mathrm{V}}{1-0.5}$
≈ 10 V
Inductor ripple: $\Delta I_L=\frac{V_{in}D}{L_1f}$
$\frac{2.5\,\mathrm{V}}{100\,\mathrm{\mu H}\times50\,\mathrm{kHz}}$
≈ 500 mA
Peak inductor current: $I_{pk}=I_{L,avg}+\Delta I_L/2$
$1\,\mathrm{A}+\frac{500\,\mathrm{mA}}{2}$
≈ 1.25 A
Output ripple estimate: $\Delta V=\frac{I_LD}{fC_1}$
$\frac{250\,\mathrm{mA}}{50\,\mathrm{kHz}\times100\,\mathrm{\mu F}}$
≈ 50 mV

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Schematic

Results

How boost converter works

Boost converter calculator

Configure

Schematic

Results

How boost converter works

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