Boost Converter Designer
Boost Converter Designer
Design the power stage of a step-up (boost) converter: duty cycle, inductance, input and peak currents, RMS currents, output capacitance and ESR for a ripple target, the right-half-plane zero and the nearest standard parts.
Boost power stage
5 V in, 12 V out at 1 A, 500 kHz, 30% ripple, 50 mV output ripple, ideal switches
Boost converter power stage (CCM)
- D
- duty cycle; with both drops at 0 it is 1 − Vin/Vout
- IL
- average inductor current = input current
- ΔIL
- inductor ripple, peak to peak = ripple % × IL
- ΔV
- output ripple target, peak to peak
- R
- load resistance at full load
Worked example
5 V in, 12 V out at 1 A, 500 kHz, 30% ripple, 50 mV output ripple, ideal switches
D = 1 − 5/12 = 58.33%; inductor (input) current IL = 1 ÷ 0.4167 = 2.40 A
ΔIL = 0.30 × 2.40 = 0.72 A
L = 5 × 0.5833 ÷ (500,000 × 0.72) = 8.102 µH → 8.2 µH (E12, rounded up); peak 2.756 A
Cmin = 1 × 0.5833 ÷ (500,000 × 0.050) = 23.33 µF → 33 µF (E6); ESR ≤ 0.050 ÷ 2.76 = 18.12 mΩ
RHP zero = (5/12)² × 12 Ω ÷ (2π × 8.2 µH) = 40.44 kHz
Designing the boost power stage
A boost converter stores energy in the inductor while the switch is on and releases it, on top of the input voltage, through the diode when the switch turns off. Volt-second balance gives the ideal duty cycle D = 1 − Vin/Vout, and charge balance on the output capacitor makes the average inductor current Iout/(1 − D): in the example, 1 A out at 12 V from 5 V draws 2.40 A through the inductor. Set the diode and switch drops to 0 for an ideal (synchronous, lossless) design; enter them to see how a Schottky or a MOSFET’s I·RDS(on) drop stretches the duty cycle.
Ripple conventions. The ripple percentage here is of the average inductor current, which is the input current; TI’s SLVA372D expresses the same 20–40% range as a fraction of Iout·Vout/Vin. On the input side the current is continuous, so the input capacitor only carries the triangular ripple. On the output side it is pulsed: while the switch is on, the capacitor alone feeds the load, which is why Cmin depends on Iout·D rather than on the inductor ripple, and why the output capacitor’s RMS current is much larger than a buck’s.
Minimum input is the hard case for current. The chart plots, with the E12 inductor, the ripple and the peak current across input voltages. The peak rises steeply as the input falls, because the same output power needs more input current; the ripple itself is largest near Vin = Vout/2. Size the inductor’s saturation current and the switch for the lowest input you expect.
The right-half-plane zero. In CCM a boost converter’s control-to-output response has a zero in the right half plane at f = (1 − D)²R/(2πL) (Erickson & Maksimović, Table 8.2; TI SLVA061): a sudden increase in duty first reduces the current reaching the output. It moves down at heavy load, at low input voltage and with a larger inductor, and a voltage-mode or current-mode loop has to cross over well below it — Ridley puts the practical limit at about a fifth of it. The figure here uses the E12 inductor at full load.
Limits of the model. Losses are left out apart from the two fixed drops; at high duty cycles the inductor’s and switch’s resistance cap the achievable ratio. Below the CCM boundary current the converter runs in DCM and these formulas no longer hold. For switch losses use the MOSFET loss calculator; to step down instead, use the buck converter designer.
Frequently asked questions
How do you calculate the inductor for a boost converter?
L = Vin × D ÷ (f × ΔIL), with D = 1 − Vin/Vout and ΔIL a ripple of 20–40% of the input current Iout/(1 − D). For 5 V to 12 V at 1 A and 500 kHz with 30% ripple, 8.102 µH; use 8.2 µH.
What is the right-half-plane zero of a boost converter?
A zero at (1 − D)² × R ÷ (2π × L), R being the load resistance. It limits how fast the feedback loop can be. In the example, 40.44 kHz at full load.
How much output capacitance does a boost converter need?
At least Iout × D ÷ (f × ΔV), because the capacitor supplies the load alone during the on-time: 23.33 µF for the example, plus an ESR no higher than ΔV divided by the peak inductor current.
Why is the input current higher than the output current?
Power in equals power out (ignoring losses), so stepping 5 V up to 12 V at 1 A takes 12 ÷ 5 = 2.4 A from the input.
Related calculators
References
- Erickson RW, Maksimović D. Fundamentals of Power Electronics, 3rd ed. Springer, 2020. Ch. 2 (inductor volt-second and capacitor charge balance, ripple), Ch. 5 (the CCM–DCM boundary), Ch. 8, Table 8.2 (right-half-plane zero: D′²R/L for the boost, D′²R/(DL) for the buck-boost).
- Hauke B. Basic Calculation of a Boost Converter’s Power Stage. Texas Instruments application report SLVA372D, revised November 2022. Ripple of 20–40% of IOUT·VOUT/VIN, COUT(min) = IOUT·D ÷ (fS·ΔVOUT) and the ESR ripple term.
- Rogers E. Understanding Boost Power Stages in Switchmode Power Supplies. Texas Instruments application report SLVA061, March 1999. The right-half-plane zero ω = (1 − D)²R/L and the critical load current.
- Ridley R. Buck-Boost Converter with Voltage-Mode Control (Design Center article 019). Ridley Engineering. Loop bandwidth limited to about a fifth of the RHP-zero frequency.
- IEC 60063:2015. Preferred number series for resistors and capacitors. The E6, E12 and E24 series used for the suggested standard parts.