Boost Converter Calculator
Size the inductor and output capacitor of a step up (boost) DC to DC converter in continuous conduction mode.
- D = 1 − Vin × η ÷ Vout = 62.5 %
- IL = Iout ÷ (1 − D) = 1.333 A, ΔI = 400 mA
- L = Vin (Vout − Vin) ÷ (ΔI × f × Vout) = 14.58 µH
- C = Iout × D ÷ (f × ΔV) = 12.5 µF
Boost Converter Equations
IL = Iout ÷ (1 − D)
L = Vin × (Vout − Vin) ÷ (ΔIL × fsw × Vout)
Cout = Iout × D ÷ (fsw × ΔVout)
Worked example
5 V to 12 V at 0.5 A, 500 kHz, 30 % ripple, 50 mV ripple, 90 % efficiency: D = 62.5 %, IL = 1.33 A, L = 14.6 µH (use 15 µH), Cout ≥ 12.5 µF, peak switch current 1.53 A.
Frequently Asked Questions
What is the duty cycle of a boost converter?
Ideally 1 − Vin/Vout; for 5 V to 12 V that is 58 %, slightly more with losses.
Why is the input current higher than the output current?
Power is conserved, so stepping voltage up by a ratio divides current by the same ratio, plus losses.
Can a boost converter be turned off completely?
Not in the basic form; the input passes through the inductor and diode to the output even when the switch is off.
Learn more on the blog
Boost converter working principle
A practical, field focused guide on instrumentationblog.in that explains the theory behind this calculator and how engineers apply it on real plants.
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