How it works
All three converters move energy through one inductor. While the switch is on, the inductor sees a fixed voltage and its current ramps up; while the rectifier conducts it ramps down. In steady state the two slopes balance over a period, which fixes the duty cycle, and the ramp height ΔI_L follows from the on-time voltage and the inductance.
The ripple ratio r = ΔI_L/I_L,avg ties the inductor to the load. For a buck the inductor carries the output current and the ripple grows with input voltage, so the highest V_in sets L. For a boost and a buck-boost the inductor current is I_out/(1 − D). In a boost the ripple-to-current ratio peaks near D = 1/3; in an inverting buck-boost it keeps rising as D falls, so the highest V_in governs. The inductance is found by scanning the whole input range, and the tool reports which V_in set it.
When the load drops so far that the current valley touches zero, the converter enters discontinuous conduction. That boundary load is ΔI_L/2 for a buck and (ΔI_L/2)(1 − D) for the other two. Below it the duty cycle starts to depend on the load, and the waveform chart switches to the DCM shape.
The output capacitor sees the inductor ripple in a buck but the full pulsed diode current in a boost and a buck-boost, which is why those two need far more output capacitance and RMS rating for the same ripple. The ESR term is subtracted from the ripple budget first, and the capacitance has to cover what is left.
Design procedure
- 1Enter the input range, output voltage and current, and the switching frequency. For an inverting buck-boost, enter the magnitude of the negative output.
- 2Pick a ripple ratio, usually 0.2 to 0.4, and read the minimum inductance and the input voltage that sets it. Choose a standard part at or above that value with a saturation current above the peak shown.
- 3Check the boundary load: below it the converter runs in DCM. If CCM is needed down to a lighter load, raise the inductance.
- 4Set the output ripple target and the capacitor ESR, read the capacitance, and confirm the capacitor's RMS rating against the table. In a boost or buck-boost the RMS current is usually what decides the capacitor.
- 5Use the switch and diode RMS and average currents and the voltage stress to select the devices, adding margin for ringing on the switch node.
Formulas
- Duty cycle with losses
- Buck: D = (V_out + V_f)/(V_in − I_L·R_ds + V_f); Boost: D = 1 − (V_in − I_L·R_ds)/(V_out + V_f); Buck-boost: D = (V_out + V_f)/(V_in − I_L·R_ds + V_out + V_f)
- Inductor average current
- I_L,avg = I_out (buck), I_out/(1 − D) (boost, buck-boost)
- Ripple and required inductance
- ΔI_L = V_on·D/(f_s·L), V_on = V_in − V_out (buck), V_in (boost, buck-boost); L = max over V_in of V_on·D/(f_s·r·I_L,avg)
- CCM/DCM boundary load
- I_out,crit = ΔI_L/2 (buck), (ΔI_L/2)·(1 − D) (boost, buck-boost)
- Output capacitance
- Buck: C ≥ ΔI_L/(8·f_s·ΔV_C), ΔV_ESR = ΔI_L·ESR; Boost, buck-boost: C ≥ I_out·D/(f_s·ΔV_C), ΔV_ESR = I_L,peak·ESR; ΔV_C = ΔV_out − ΔV_ESR
- RMS currents
- I_rms = √(δ·(I_L² + ΔI_L²/12)) with δ = D (switch), 1 − D (diode); I_Cin,buck ≈ I_out·√(D(1 − D)), I_Cout,boost ≈ I_out·√(D/(1 − D))
Worked example
The default buck converts 10.8–13.2 V (12 V nominal) to 5 V at 3 A, switching at 500 kHz with synchronous rectification and 20 mΩ switches. At 12 V the duty cycle is 0.419. A 35 % ripple ratio needs at least 5.90 µH, set at 13.2 V, so the next E12 value of 6.8 µH is used: the ripple is then 0.855 A at 12 V and 0.911 A at 13.2 V, and the inductor peaks at 3.46 A. The converter stays in CCM down to about 0.46 A of load. With 5 mΩ ESR, the ESR takes 4.6 mV of a 50 mV ripple budget, so the capacitance needed is 5.0 µF and capacitance dominates. The input capacitor carries up to 1.50 A RMS, and switch and diode see 13.2 V.
Common mistakes
- Sizing a boost inductor only at the minimum input voltage. The ripple ratio can be higher elsewhere in the range, so scan the range or check the input voltage closest to 2/3·V_out.
- Choosing the inductor by inductance alone. The saturation current must exceed the peak current at the worst input with margin, or the inductance collapses exactly where it is needed.
- Rating a boost output capacitor for the ripple voltage only. Its RMS current, about I_out·√(D/(1 − D)), often requires several capacitors in parallel.
- Using the nominal ceramic capacitance. DC bias can remove half or more of a small MLCC's capacitance at its working voltage.
Frequently asked questions
Why does the boost inductance come from the middle of the range rather than the minimum input?
The absolute ripple of a boost is largest at D = 0.5, but the ripple ratio compares it with the inductor current, which grows as 1/(1 − D). The ratio peaks near D = 1/3, that is V_in ≈ 2/3·V_out. If that point lies outside the input range, the end of the range closest to it governs.
What ripple ratio should I choose?
Between 0.2 and 0.4 is the usual compromise. Lower ripple means a larger inductor and slower transient response; higher ripple means more core loss, a higher peak current, more output ripple and an earlier move into DCM at light load.
Is it a problem if the converter enters DCM at light load?
Not by itself. Most controllers handle it, and many skip pulses at very light load. It does change the small-signal behavior: the double pole of the LC filter disappears in DCM, so a loop tuned for CCM may respond differently there.
Why do V_f and R_ds(on) change the duty cycle?
The switch drop reduces the voltage applied to the inductor, and the diode drop adds to the voltage it must overcome during the off-time. Both push the duty cycle up compared with the ideal ratio, most noticeably at low output voltage and at high current.
Results are engineering estimates. Verify your design with datasheets, simulation and measurement.