To convert a 12V DC input to a -5V output using a classic inverting buck boost converter circuit, the required duty cycle (D) is 0.294 (29.4%). Using the continuous conduction mode (CCM) formula D = |Vout| / (Vin + |Vout|), we substitute the exact values: D = 5 / (12 + 5) = 0.294. For a 1A output at a 500kHz switching frequency, this converts to a minimum 15µH inductor. The concrete, default IC pick for this exact conversion is the Texas Instruments TPS54360B, configured in the inverting buck-boost topology.
The Core Conversion Formulas and Fixing Assumptions
The mathematical conversion from voltage requirements to component values relies on three fixing assumptions. If any of these shift on the bench, your calculated values will fail to regulate.
- Continuous Conduction Mode (CCM): The inductor current never falls to zero during the switching cycle. If your load drops below the critical current threshold, the converter enters Discontinuous Conduction Mode (DCM), and the duty cycle becomes load-dependent rather than purely voltage-dependent.
- Ideal Components: The formula
D = |Vout| / (Vin + |Vout|)assumes 0V drop across the switching MOSFET (Rds_on) and the catch diode (Vf). - Steady-State DC Input: The input is a pure DC source. Ripple on the input rail will modulate the duty cycle cycle-by-cycle if the controller uses voltage-mode control.
To find the inductor value, we convert the duty cycle into a physical component requirement using the ripple current formula. Assuming we want a 30% ripple current relative to the maximum output current (1A), the target ripple (ΔIL) is 0.3A. However, in an inverting topology, the average inductor current is Iout / (1 - D). Therefore, the actual average inductor current is 1A / (1 - 0.294) = 1.41A. Setting ripple to 30% of 1.41A gives ΔIL = 0.42A.
L = (Vin × D) / (fsw × ΔIL)
L = (12 × 0.294) / (500,000 × 0.42) = 16.8µH
We round down to the nearest standard E12 value: 15µH, which slightly increases ripple but improves transient response.
Neighboring Values: ±20% Input Variance Table
In real-world applications, a "12V" source is rarely exactly 12V. An automotive or sealed lead-acid (SLA) battery will sag to 9.6V under heavy load or spike to 14.4V during alternator charging. Here is how the duty cycle conversion shifts across a ±20% variance band.
| Input Voltage (Vin) | Variance | Duty Cycle (D) | Avg Inductor Current | Ripple Current (at 15µH, 500kHz) |
|---|---|---|---|---|
| 9.6V | -20% (Cranking/Sag) | 0.342 | 1.52A | 0.44A |
| 10.8V | -10% | 0.316 | 1.46A | 0.46A |
| 12.0V | Nominal | 0.294 | 1.41A | 0.47A |
| 13.2V | +10% | 0.275 | 1.38A | 0.48A |
| 14.4V | +20% (Charging) | 0.258 | 1.35A | 0.50A |
Topology Shifts: 12V vs 24V vs AC Mains
The conversion math above is strictly for DC-to-DC non-isolated topologies. Applying this specific duty cycle conversion to other voltage domains requires fundamental architectural shifts.
- 24V and 48V DC Systems: If you attempt to convert 24V to -5V, the duty cycle plummets to
D = 5 / (24 + 5) = 0.172. At 500kHz, this demands an on-time of just 344 nanoseconds. Most standard controllers cannot guarantee this minimum on-time, leading to pulse-skipping and massive output voltage ripple. For 24V/48V inputs, you must either lower the switching frequency or switch to an isolated flyback topology. - 120V / 230V AC Mains: Using this DC-DC buck-boost conversion for AC mains is meaningless. A buck-boost converter requires a unipolar DC bus. To step down 120VAC or 230VAC, you must first pass the AC through a bridge rectifier and bulk capacitor (yielding ~170VDC or ~325VDC). At those DC bus voltages, a non-isolated inverting buck-boost poses a lethal shock hazard and exceeds the voltage ratings of standard MOSFETs. You must use an isolated topology (like a flyback or LLC resonant converter) with a high-frequency transformer to provide galvanic isolation.
- 3-Phase AC: Similarly meaningless for this topology. 3-phase requires a 6-diode rectifier front-end and a 3-phase PFC (Power Factor Correction) stage before any DC-DC conversion can occur.
Decision Tree: Selecting Your Converter IC
Do not waste time breadboarding discrete MOSFETs and 555 timers for a buck boost converter circuit. Use a dedicated controller. Follow this decision path to lock in your part number.
| Condition / Requirement | If True... | If False... |
|---|---|---|
| Is the output current > 500mA? | Require integrated MOSFET or external gate driver. | Use a low-power charge pump (e.g., ICL7660). |
| Is the input voltage > 36V? | Select a high-Vin controller (e.g., LTC3891). | Standard 5V-36V input range is sufficient. |
| Do you need negative (inverting) output? | Ensure IC supports inverting buck-boost config. | Use a standard 4-switch non-inverting buck-boost. |
| Final Concrete Pick | Texas Instruments TPS54360B (Supports up to 60V Vin, easily configured for inverting -5V at 1A+ using the INV pin topology). | |
When the Mathematical Conversion Fails (Edge Cases)
Theoretical formulas break down on the bench under three specific conditions. If you hit these, your converter will fail to regulate, overheat, or destroy the switching node.
- Extreme Duty Cycles (D > 0.85 or D < 0.15): If your input sags so low that D exceeds 0.85, the inductor spends almost all its time charging. The parasitic resistances (DCR of the inductor, Rds_on of the MOSFET) dominate the circuit, and the output voltage will collapse under load regardless of what the math says. Conversely, if D drops below 0.15, you hit the controller's minimum on-time limit.
- Right-Half-Plane Zero (RHPZ): The classic inverting buck-boost topology suffers from a RHPZ in its control loop. This means that when the load suddenly increases, the output voltage initially dips further before recovering. If you do not compensate your error amplifier with a Type II or Type III network specifically calculated for the RHPZ frequency, the circuit will oscillate and destroy your load.
- Inductor Saturation: The math gives you an inductance value (15µH), but not a current rating. The peak inductor current is
Iavg + (ΔIL / 2). For our 12V-to-5V example, peak current is1.41A + 0.235A = 1.645A. If you select a 15µH inductor with a 1.5A saturation current rating (Isat), the core will saturate, inductance will drop to near zero, and the IC's internal MOSFET will instantly vaporize from overcurrent. Always specify an inductor with an Isat rating at least 30% above your calculated peak current (e.g., a 2.2A Isat part).
Frequently Asked Questions
Can I use a standard buck converter IC to make a negative voltage?
Yes, but only if the IC's maximum voltage rating (Vin to SW pin) exceeds the sum of your positive input and absolute negative output. For a 12V input and -5V output, the switch pin sees 17V when off. A standard 5V buck IC will explode; a 60V rated IC like the TPS54360B will handle it easily.
Why does my output voltage have a massive spike at startup?
Inverting buck-boost circuits have a direct path from input to output through the inductor and catch diode before the switch even turns on. You will always see a startup spike equal to Vin - Vf(diode). To mitigate this, add a soft-start capacitor to the controller's SS pin to slowly ramp the duty cycle.
