Design negative-rail DC-DC converters: duty cycle, inductor ripple current, output ripple voltage, CCM/DCM boundary, peak-current inductor rating and recommended L/C values — all in one place.
Enter your operating point and the inductor/capacitor you plan to use. The tool computes the duty cycle, ripples, the CCM/DCM boundary inductance, and the L/C values your ripple targets actually need.
All equations below are the inverting buck-boost converter continuous-conduction-mode (CCM) equations — ideal and first-order. If your inductor is smaller than Lmin, the discontinuous conduction mode warning in the calculator tells you the ideal transfer function no longer applies.
| Quantity | Equation |
|---|---|
| Duty cycle | D = |Vout| / (Vin + |Vout|) |
| Avg. inductor current | IL,avg = Iout / (1 − D) |
| Inductor ripple current | ΔIL = Vin × D / (fsw × L) |
| Output voltage ripple | ΔVout = Iout × D / (fsw × C) |
| CCM boundary inductance | Lmin = (1 − D)² × R / (2 × fsw), R = |Vout| / Iout |
| Peak inductor current | IL,peak = IL,avg + ΔIL / 2 |
| Switch / diode voltage stress | Vstress ≈ Vin + |Vout| |
| On-time | ton = D / fsw |
First-order ideal equations. Real designs add ESR/ESL, diode forward drop, MOSFET Rds(on), layout parasitics and control-loop margin — always validate with simulation and bench measurement.
12 V → −15 V at 1 A, fsw = 100 kHz, L = 47 µH, C = 100 µF.
The inverting buck-boost topology stores energy in the inductor while the switch is on, then transfers it to the output with reversed polarity when the switch turns off. The output capacitor therefore charges negative with respect to the input ground, so the ideal output is Vout = −Vin × D/(1−D).
For the ideal converter in continuous conduction mode, D = |Vout| / (Vin + |Vout|). Equivalently, |Vout|/Vin = D/(1−D). For example, 12 V in to −15 V out gives D = 15/27 = 55.56%.
At D = 50%, |Vout| = Vin, so the output is −Vin. Duty cycles below 50% step the magnitude down (buck region); duty cycles above 50% step it up (boost region).
Yes. Duty cycles above 50% produce output magnitudes larger than the input, and the converter can approach D = 100% in theory. In practice, stay well below the controller maximum (often 80–90%) because inductor current, switch stress and losses grow rapidly at extreme duty cycles.
Use ΔIL = Vin × D / (fsw × L), where fsw is the switching frequency and L the inductance. A common design target is 20–40% of the average inductor current IL,avg = Iout/(1−D).
For the inverting buck-boost in CCM, ΔVout = Iout × D / (fsw × C). Choose C so this stays within your ripple budget, typically 0.5–2% of |Vout|, then round up to a standard capacitor value with adequate voltage rating.
First size L from your ripple target: L = Vin × D / (fsw × ΔIL,target). Then check it exceeds the CCM boundary Lmin = (1−D)² × R / (2 × fsw) with R = |Vout|/Iout. Finally pick a part whose saturation current exceeds 1.3 × the peak current IL,avg + ΔIL/2.
Use C = Iout × D / (fsw × ΔVout,target) for your ripple target, round up to a standard E12 value, and choose a voltage rating of at least 1.5 × |Vout|. Low-ESR ceramic or polymer capacitors keep the real ripple close to the ideal calculation.
The converter stays in continuous conduction mode while L ≥ Lmin = (1−D)² × R / (2 × fsw). Below that the inductor current hits zero each cycle (discontinuous conduction mode), which changes the transfer function and raises peak currents. This calculator flags a DCM warning when your inductor is too small.
A buck converter only steps voltage down and keeps positive polarity (Vout = D × Vin, always below Vin). The inverting buck-boost steps up or down AND inverts polarity (Vout = −Vin × D/(1−D)). Use our buck-converter calculator for positive step-down rails and this tool for negative rails.
The classic inverting buck-boost is the simplest but its output is negative. A SEPIC steps up or down while keeping positive polarity, at the cost of two inductors (or a coupled inductor) and an extra capacitor. A 4-switch buck-boost also keeps positive polarity with higher efficiency but needs four MOSFETs and more complex control.
Both the MOSFET and the diode must block approximately Vin + |Vout| plus switching spikes. Choose parts rated at least 1.5× that sum for reliable operation.
Higher frequency shrinks the required L and C (both are inversely proportional to fsw), but increases switching losses and EMI. Typical designs use 100 kHz to 1 MHz; this calculator recomputes ripple, Lmin and the DCM check at whatever frequency you enter.
At high D the average inductor current IL,avg = Iout/(1−D) grows steeply, so conduction losses in the inductor, MOSFET and diode dominate. At very low D, gate-drive and switching losses dominate relative to the small delivered power. Staying near D = 30–70% keeps losses balanced.
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