The Direct Conversion Answer (12V to -5V)
To convert a 12V DC input to a -5V DC output using an ideal Cuk converter operating in Continuous Conduction Mode (CCM), the required duty cycle (D) is exactly 0.294 (29.4%). The governing voltage conversion formula for a Cuk topology is V_out / V_in = -D / (1 - D). Substituting our target values: -5 / 12 = -D / (1 - D). Dropping the negative signs for the magnitude calculation yields 5/12 = D / (1 - D), which algebraically simplifies to D = 5 / (5 + 12), resulting in D = 5 / 17 ≈ 0.2941. This means your PWM controller must hold the main MOSFET ON for 29.4% of the switching period to maintain -5V at the output, assuming zero parasitic losses and a continuous inductor current.
Neighboring Values: Input Voltage Sweep (±20%)
In real-world bench or automotive environments, your 12V rail is rarely exactly 12.0V. The table below maps the required duty cycle shift across a ±20% input voltage variance to maintain a hard -5V output. This highlights why a closed-loop feedback controller is mandatory; an open-loop fixed-duty oscillator will fail as the input sags or spikes.
| Input Voltage (V_in) | Variance from Nominal | Required Duty Cycle (D) | ON-Time at 500kHz (ns) |
|---|---|---|---|
| 9.6V | -20% | 0.342 (34.2%) | 684 ns |
| 10.8V | -10% | 0.316 (31.6%) | 632 ns |
| 12.0V | Nominal | 0.294 (29.4%) | 588 ns |
| 13.2V | +10% | 0.275 (27.5%) | 550 ns |
| 14.4V | +20% | 0.258 (25.8%) | 516 ns |
What Fixes the Answer (and When the Math Breaks)
The 29.4% figure relies on three rigid assumptions: Continuous Conduction Mode (CCM), ideal components (0V diode forward drop, 0Ω MOSFET R_DS(on), and 0Ω inductor DCR), and steady-state equilibrium. In a physical build, a Schottky diode drop of ~0.4V will force the controller to bump the duty cycle to roughly 31% to compensate for the lost voltage. Furthermore, the Cuk topology's defining feature—the series transfer capacitor (C1)—carries the full pulsating RMS current. For a 1A output at D=0.294, C1 must handle roughly 0.76A RMS. If you use a standard X7R ceramic capacitor without accounting for DC bias derating and AC heating, it will crack or vent. Always specify a low-ESR polymer capacitor or parallel multiple MLCCs for this node.
When the Conversion Becomes Meaningless
The ideal CCM formula becomes entirely meaningless if your load current drops below the critical boundary and the converter enters Discontinuous Conduction Mode (DCM). If your 5V load drops to 10mA and your inductors are sized for a 500mA nominal load, the inductor current will hit zero before the switching cycle ends. In DCM, the output voltage is no longer dictated by the duty cycle alone; it becomes a function of the load resistance, inductor value, and switching frequency. The output will float above -5V unless your controller has a specific DCM-compensation loop or you implement a forced-PWM (F-PWM) mode to maintain continuous current via a dummy load.
How the Answer Shifts for 120V vs 230V vs 3-Phase
A Cuk converter is strictly a DC-DC stage; it cannot process AC directly. However, if your 12V DC rail is derived from an offline AC-DC front-end, the AC mains voltage dictates the front-end architecture, not the Cuk duty cycle. A 120VAC mains input yields a ~170V DC bus (requiring a step-down flyback or buck front-end), while 230VAC yields a ~325V DC bus. The Cuk converter's duty cycle remains exactly 0.294 as long as the front-end successfully regulates the 12V intermediate bus. 3-phase power is irrelevant here; 3-phase rectifiers are deployed for high-power (>5kW) motor drives or industrial DC buses, whereas Cuk converters are typically deployed in sub-500W distributed power architectures where single-phase offline supplies are standard.
Cuk Converter IC Decision Tree
Selecting the right controller prevents you from trying to hack a standard boost IC into an inverting topology. Follow this decision path to lock in your silicon.
- IF your maximum input voltage is < 40V and you need a simple, low-cost inverted output for a bench prototype → Evaluate charge pumps (e.g., ICL7660). If current > 50mA, proceed to next step.
- IF your input is < 75V, you need high efficiency, and you want a widely available IC with extensive TI reference designs → Select the Texas Instruments LM5118.
- IF your input voltage exceeds 75V (e.g., automotive load-dump or industrial 48V nominal rails) and you need robust over-voltage tolerance → Select the Analog Devices LTC3863.
Default Recommendation: For 90% of hobbyist, robotics, and bench-power applications operating from a 12V/24V bus, terminate your search at the Texas Instruments LM5118. It is a wide-voltage buck-boost controller that natively supports the Cuk topology when wired with the appropriate inverting feedback network. It handles the complex slope compensation required to keep the Cuk's right-half-plane zero stable, saving you weeks of control-loop debugging. For high-voltage industrial strings, the Analog Devices LTC3863 is the definitive 60V+ Cuk controller, offering a constant-frequency architecture that keeps EMI predictable.
Frequently Asked Questions
Why choose a Cuk converter over a standard inverting buck-boost?
The primary advantage is continuous input and output current. A standard inverting buck-boost draws pulsating (discontinuous) current from the source, which creates massive input voltage ripple and EMI. The Cuk converter uses an input inductor (L1) that smooths the source current, making it ideal for battery-powered systems or sensitive sensor rails where input ripple must be minimized. The trade-off is higher component count (two inductors, one high-stress transfer capacitor) and inverted polarity.
Can I use coupled inductors to shrink the footprint?
Yes. Because the voltage waveforms across L1 and L2 in a Cuk converter are identical in magnitude and phase, you can wind both inductors on a single toroidal or E-core. This reduces the magnetic footprint by roughly 40% and allows you to tune the coupling coefficient to steer ripple current. A 1:1 turns ratio with a coupling coefficient (k) of roughly 0.9 is the standard starting point for zero-ripple input designs.
How do I handle the negative voltage feedback to a ground-referenced IC?
This is the most common wiring error. You cannot connect the -5V output directly to the FB pin of a standard positive-voltage regulator. You must use a level-shifting resistor divider or an inverting op-amp network to translate the negative output voltage into a positive 0.8V to 1.2V reference signal that the controller's error amplifier can read relative to the system ground. The LM5118 datasheet provides a specific schematic for this level-shifting network in its inverting buck-boost/Cuk application section.
