How to use it
- Enter Vin (use the highest input voltage for the inductor), Vout and the load current.
- Enter the switching frequency in kHz from your controller’s datasheet.
- Choose the ripple current as a percentage of Iout (30 % is a typical choice).
- Enter the expected efficiency and the output capacitance.
Formulas
For continuous conduction:
- Duty cycle D = Vout ÷ (Vin × η)
- Ripple current ΔI = ripple % × Iout
- Inductance L = Vout × (Vin − Vout) ÷ (ΔI × f × Vin)
- Peak current I_peak = Iout + ΔI ÷ 2
- Output ripple ΔV = ΔI ÷ (8 × f × C)
- On-time t_on = D ÷ f
- Input power P_in = Vout × Iout ÷ η, input current P_in ÷ Vin, loss P_in − Vout × Iout
The duty cycle with η follows a common application-note approach; the inductor formula uses the ideal ratio.
Worked example
12 V to 5 V at 2 A, 500 kHz, 30 % ripple, 90 % efficiency, 22 µF:
- D = 5 ÷ (12 × 0.9) = 46.3 %
- ΔI = 0.3 × 2 = 0.6 A, peak 2.3 A
- L = 5 × 7 ÷ (0.6 × 500,000 × 12) = 9.722 µH — choose 10 µH
- ΔV = 0.6 ÷ (8 × 500,000 × 22 µF) = 6.818 mV
- t_on = 0.9259 µs; P_in = 11.111 W, input current 0.9259 A, 1.111 W lost as heat
Tips
- Check the inductor’s saturation current against the peak current, with margin.
- Light loads leave continuous mode when Iout falls below ΔI ÷ 2; the formulas above then no longer hold.
- Losses become heat. Use the loss figure with the heatsink calculator to check the switch temperature.