For a standard 12V input to -12V output inverting buck boost converter operating in Continuous Conduction Mode (CCM), the required duty cycle (D) is exactly 0.50 (50%), and the internal switch node must tolerate a minimum of 24V. The governing formula is Vout = -Vin × [D / (1 - D)]. Substituting our target values: -12V = -12V × [0.50 / (1 - 0.50)]. If you are designing this on the bench today, you need an inductor rated for at least 1.5× your maximum DC load current to prevent core saturation, and a MOSFET/diode pair with a voltage rating exceeding 30V to survive the 24V switch-node spikes plus high-frequency ringing.
The Core Conversion Formula and Duty Cycle Table
The inverting buck-boost topology is unique because it flips the polarity of the output voltage relative to the input ground. The switch node (where the inductor, diode, and internal MOSFET meet) swings aggressively between 0V and -(Vin + |Vout|). To maintain regulation when your input voltage sags or surges, the controller must adjust the duty cycle dynamically.
Below is a reference table showing how the duty cycle shifts across a ±20% tolerance range on a nominal 12V input rail, targeting a strict -12V output.
| Input Voltage (Vin) | Target Output (Vout) | Required Duty Cycle (D) | Switch Node Peak Voltage |
|---|---|---|---|
| 9.6V (-20%) | -12.0V | 0.555 (55.5%) | -21.6V |
| 10.8V (-10%) | -12.0V | 0.526 (52.6%) | -22.8V |
| 12.0V (Nominal) | -12.0V | 0.500 (50.0%) | -24.0V |
| 13.2V (+10%) | -12.0V | 0.476 (47.6%) | -25.2V |
| 14.4V (+20%) | -12.0V | 0.454 (45.4%) | -26.4V |
What Assumptions Fix This Answer (and When It Breaks)
The clean 50% duty cycle calculation above relies on two major assumptions: Continuous Conduction Mode (CCM) and ideal components. In the real world, a Schottky diode introduces a ~0.4V forward voltage drop, and the MOSFET has an RDS(on) that steals a few hundred millivolts. To compensate, the controller will nudge the duty cycle up by 1-2% to maintain exactly -12V at the load.
When the Conversion Becomes Meaningless
The CCM formula becomes entirely meaningless if your load current drops below the critical inductance threshold, pushing the converter into Discontinuous Conduction Mode (DCM). In DCM, the inductor current hits zero before the switching period ends, making the output voltage highly dependent on the exact load resistance rather than just the duty cycle.
Furthermore, if your inverting buck-boost is fed from an offline AC source rather than a DC bench supply, the AC power factor (PF) dictates front-end sizing. If the PF is unknown and uncorrected, calculating the true apparent power (VA) versus real power (W) for the upstream transformer or fuse sizing is meaningless until a Power Factor Correction (PFC) stage is verified.
How the Answer Shifts for 120V vs 230V vs 3-Phase Inputs
While the DC-DC topology only "sees" the rectified DC bus, the AC mains origin drastically shifts the required conversion ratios:
- 120V AC Mains: Rectifies to ~170V DC. To generate -12V, the duty cycle plummets to D = 0.066. Standard controllers will fail here due to minimum on-time limitations.
- 230V AC Mains: Rectifies to ~325V DC. Duty cycle drops to D = 0.035. You must use a specialized high-voltage controller with frequency foldback or skip-cycle modes.
- 3-Phase AC: Rectifies to ~540V DC. This exceeds the absolute maximum ratings of 99% of monolithic inverting buck-boost ICs, requiring a discrete gate-driver and high-voltage SiC MOSFET implementation.
How the Conversion Shifts Across Standard DC Voltage Rails
If you are staying within standard low-voltage DC distribution, the input rail choice dramatically alters your component stress:
- 5V Input to -12V Output: Duty cycle jumps to 0.705 (70.5%). This high duty cycle means the internal MOSFET is conducting 70% of the time, leading to high RMS switch currents and severe thermal stress on the IC.
- 12V Input to -12V Output: Duty cycle is a balanced 0.50. Thermal load is split evenly between the high-side switch and the low-side diode.
- 24V Input to -12V Output: Duty cycle drops to 0.333 (33.3%). The switch runs cooler, but the diode conducts for 66% of the cycle, shifting the thermal burden to the catch diode. You must upgrade the diode's current rating accordingly.
Component Selection Decision Tree
Choosing the right controller IC depends entirely on your maximum input voltage and required output current. Use this decision matrix to lock in your silicon. For a detailed topology overview, refer to the TI Buck-Boost & Inverting Regulators guide or the All About Circuits buck-boost primer.
| Input Voltage Range | Max Output Current | Recommended IC | Key Feature |
|---|---|---|---|
| 2.7V to 5.5V | ≤ 1.0A | TPS63700 (TI) | Integrated low-side switch, optimized for single-cell Li-ion. |
| 4.5V to 42V | ≤ 2.5A | LM5118 (TI) | Wide Vin, external MOSFETs required for high current. |
| 3.5V to 60V | ≤ 1.5A | TPS54160 (TI) | Integrated 60V MOSFET, 2.5A switch limit. (Default Pick) |
The Final Pick: For the vast majority of bench and embedded projects requiring a 12V to -12V conversion at 1A or less, the TPS54160 is the definitive choice. Its integrated 60V MOSFET provides massive headroom for the 24V switch-node peaks plus ringing, and its 2.5A internal current limit safely handles the ~1.5A peak inductor currents inherent to a 50% duty cycle at 1A output.
Frequently Asked Questions
Why is the output capacitor ESR so critical in an inverting topology?
Unlike a standard buck converter where inductor current smooths into the output capacitor, the inverting buck-boost delivers current to the output in discrete, high-di/dt pulses through the catch diode. If your output capacitor has high Equivalent Series Resistance (ESR), these pulses generate massive output voltage ripple. Always use low-ESR ceramic (MLCC) or polymer capacitors on the output rail.
Can I use a standard buck converter IC to build an inverting buck-boost?
Yes, but with a major caveat. You must wire the IC's ground pin to the negative output rail instead of the system ground, effectively "tricking" the IC into regulating the voltage difference between its VIN and its local GND. However, the IC's absolute maximum voltage rating must now exceed Vin + |Vout|. A standard 5V buck IC will instantly die if you try this trick to generate -5V from a 5V source, as the internal silicon sees 10V across its pins.
How do I calculate the required inductor value?
Use the formula: L = (Vin × D) / (fsw × ΔIL). For our 12V to -12V example at a 500kHz switching frequency, targeting a 30% ripple current (ΔIL = 0.3A for a 1A load), the math yields: L = (12 × 0.50) / (500,000 × 0.3) = 40μH. Select a standard 47μH shielded inductor with a saturation current rating of at least 1.5A.
