Quick Answer: To convert a 5V DC input to a 12V DC output using a standard continuous conduction mode (CCM) boost converter topology, the required duty cycle (D) is exactly 58.33%. The foundational formula used is D = 1 - (Vin / Vout). Substituting our values: D = 1 - (5 / 12) = 0.5833. This means the internal MOSFET must remain ON for 58.33% of the switching period to store sufficient energy in the inductor to step up the voltage.
While the ideal formula provides a baseline, real-world designs require accounting for component tolerances and load variations. Below is a conversion table showing how the required duty cycle shifts across a ±20% range of target output voltages, assuming a fixed 5V input.
| Target Vout (±20%) | Vin (Fixed) | Ideal Duty Cycle (D) | Switch ON Time (at 1MHz) |
|---|---|---|---|
| 9.6V (-20%) | 5.0V | 47.92% | 479 ns |
| 10.8V (-10%) | 5.0V | 53.70% | 537 ns |
| 12.0V (Base) | 5.0V | 58.33% | 583 ns |
| 13.2V (+10%) | 5.0V | 62.12% | 621 ns |
| 14.4V (+20%) | 5.0V | 65.28% | 653 ns |
Real-World IC Limits and Data-Dense Specifications
The ideal formula D = 1 - (Vin / Vout) assumes 100% efficiency and ignores the forward voltage drop (Vf) of the Schottky diode and the RDS(on) of the switching MOSFET. In practice, you must select a controller IC that supports the necessary maximum duty cycle and switching frequency. If an IC hits its maximum duty cycle limit before reaching your target Vout, the converter will drop out of regulation.
According to Texas Instruments application note SLVA372, you must also derate the maximum duty cycle to allow for transient response and current limit protection. Below is a specification matrix of four widely used boost converter ICs, highlighting their hard limits.
| IC Part Number | Manufacturer | Vin Range | Max Vout | Switching Freq | Max Duty Cycle |
|---|---|---|---|---|---|
| MT3608 | Aerosemi (Generic) | 2.0V - 24V | 28V | 1.2 MHz | ~90% |
| TPS61029 | Texas Instruments | 0.5V - 5.5V | 5.5V | 720 kHz | 96% |
| LM3478 | Texas Instruments | 2.97V - 40V | 40V+ (Ext FET) | 50kHz - 1MHz | 88% |
| LT1370 | Analog Devices | 1.2V - 25V | 35V | 500 kHz | 90% |
Note: The MT3608 is ubiquitous in cheap hobbyist modules but suffers from severe thermal throttling above 1.5A output current. For reliable 2A+ designs, use the LM3478 with an external low-RDS(on) N-channel MOSFET.
Why AC Mains (120V/230V/3-Phase) Breaks the Standard Formula
When calculating duty cycles, it is critical to understand the boundary conditions of the boost converter topology. What assumption fixes the answer? The standard DC-DC calculation assumes a pure, steady DC input voltage, Continuous Conduction Mode (CCM), and ideal components.
How does the answer shift for 120V vs 230V vs 3-phase?
If your input source is AC mains (120V, 230V, or 3-phase), the standard static boost converter formula completely breaks down. You cannot simply plug 120V AC into the Vin variable. To step up AC mains efficiently and legally, you must use an active Power Factor Correction (PFC) boost stage. In a PFC boost topology, the duty cycle is not a fixed percentage; it modulates dynamically at twice the line frequency (100Hz or 120Hz) to force the input current to track the AC voltage sine wave, as detailed in All About Circuits' power electronics guides.
When is the conversion meaningless?
If you are analyzing an AC-DC front end and the Power Factor (PF) is unknown, or if you attempt to apply the static DC formula to an unrectified AC source without a bulk capacitor, the conversion is mathematically meaningless. The instantaneous input voltage is continuously swinging through zero. A static 'D' value cannot be calculated without knowing the exact phase angle, the target PF waveform, and the bulk DC bus voltage that the rectifier has already established.
Efficiency Derating and Discontinuous Mode Edge Cases
In a physical circuit, power is lost to the inductor's DC resistance (DCR), the diode's forward voltage (Vf), and switching losses. To find the real-world duty cycle, you must incorporate the estimated efficiency (η) of the converter:
Real-World Formula: D = 1 - [ (Vin × η) / Vout ]
If we return to our 5V to 12V example and assume a realistic efficiency of 85% (η = 0.85), the required duty cycle shifts significantly:
- D = 1 - [ (5 × 0.85) / 12 ]
- D = 1 - [ 4.25 / 12 ]
- D = 1 - 0.3541 = 64.58%
This 6.25% increase in ON-time is critical. If your selected IC (like the LM3478 with an 88% max duty cycle) is already operating near its limit, this efficiency drop could push the converter out of regulation under heavy load.
Furthermore, if the load current drops below the inductor's ripple current threshold, the converter enters Discontinuous Conduction Mode (DCM). In DCM, the output voltage becomes dependent on the load current and the inductance value, rendering the standard voltage-ratio formula invalid. For battery-powered IoT devices that spend most of their time in DCM sleep states, you must rely on the IC's pulse-frequency modulation (PFM) logic rather than manual duty cycle calculations.
Frequently Asked Questions
Can a boost converter topology output a lower voltage than its input?
No. A standard asynchronous boost converter relies on the diode to pass the input voltage directly to the output when the switch is off. Even at a 0% duty cycle, Vout will equal Vin minus the diode drop. If you need step-down capability, you must use a buck-boost or SEPIC topology.
Why does my 5V to 12V boost converter only output 9V under load?
This is typically caused inductor saturation. If the peak current exceeds the inductor's saturation rating (Isat), the inductance collapses, the MOSFET experiences massive current spikes, and the controller's overcurrent protection limits the duty cycle, resulting in a severe voltage drop.
