To step up a 3.7V Li-ion cell to a 12V, 1A output (12W) using a standard 1.2 MHz switching frequency boost converter circuit, you need a 74% duty cycle and a 2.2µH shielded inductor rated for at least 5.0A saturation current. This direct answer assumes 85% overall efficiency and Continuous Conduction Mode (CCM). If you are using a ubiquitous module like the MT3608 or a modern silicon equivalent like the TI TPS61230, these exact values will keep your ripple current under 30% and prevent the inductor core from saturating and melting at peak load.
The Core Conversion Formulas (Duty Cycle & Inductance)
The math behind stepping up DC voltage relies on two fixed assumptions: your target efficiency ($\eta$) and your switching frequency ($f_{sw}$). According to standard power electronics theory documented by Texas Instruments, the duty cycle ($D$) dictates the energy transfer ratio, while the inductor ($L$) smooths the current ripple.
1. Duty Cycle ($D$):
Formula: $D = 1 - \frac{V_{in} \times \eta}{V_{out}}$
Substituted: $D = 1 - \frac{3.7V \times 0.85}{12V} = 1 - 0.262 = 0.738$ (74%)
2. Average Input Current ($I_{in}$):
Formula: $I_{in} = \frac{P_{out}}{V_{in} \times \eta}$
Substituted: $I_{in} = \frac{12W}{3.7V \times 0.85} = \mathbf{3.82A}$
3. Inductor Value ($L$):
Assuming a 30% ripple current ($\Delta I_L = 0.30 \times 3.82A = 1.146A$) and a 1.2 MHz switching frequency ($1,200,000$ Hz):
Formula: $L = \frac{V_{in} \times D}{f_{sw} \times \Delta I_L}$
Substituted: $L = \frac{3.7 \times 0.74}{1,200,000 \times 1.146} = 1.99\mu H$
Nearest standard E12 value: 2.2µH
4. Peak Saturation Current ($I_{sat}$):
$I_{peak} = I_{in} + \frac{\Delta I_L}{2} = 3.82A + 0.573A = 4.39A$. Adding a 15% bench safety margin for transient spikes yields a required saturation rating of 5.0A.
Neighboring Values: Input Voltage Variance (±20%)
A Li-ion cell is rarely exactly 3.7V. It swings from 3.0V (depleted) to 4.2V (fully charged). Designing for the nominal voltage without checking the extremes is how you end up with an inductor that saturates and shorts the input to ground. Here is how the required parameters shift across a ±20% input variance for a fixed 12V/1A output:
| Input Voltage ($V_{in}$) | Avg Input Current | Duty Cycle ($D$) | Peak Inductor Current | Risk Factor |
|---|---|---|---|---|
| 3.0V (Depleted) | 4.70A | 82% | 5.50A | High: Inductor saturation if rated < 6A |
| 3.7V (Nominal) | 3.82A | 74% | 4.39A | Normal operating range |
| 4.2V (Full Charge) | 3.36A | 71% | 3.86A | Low stress, highest efficiency |
| 4.4V (Overcharge/Hot) | 3.21A | 70% | 3.69A | IC max voltage limit check required |
Takeaway: Your inductor must be sized for the 3.0V depleted state, not the 3.7V nominal state. Always specify a 6.0A+ saturation current rating for a 3.7V-to-12V design to survive dead-battery conditions.
Decision Path: Picking the Right Boost IC and Inductor
Choosing the right silicon and magnetics depends entirely on your step-up ratio and output current. Use this decision tree to terminate your part selection process with a concrete bill of materials.
| Condition / Profile | Recommended Boost IC | Recommended Inductor Type |
|---|---|---|
| IF $V_{in}$ is 3-5V, $V_{out}$ < 12V, $I_{out}$ < 2A | TI TPS61230 or MT3608 | 2.2µH - 4.7µH Shielded Ferrite (e.g., Wurth 744774) |
| IF $V_{in}$ is 12V, $V_{out}$ is 24V, $I_{out}$ is 3-5A | TI TPS55340 | 10µH - 15µH High-Current Molded (e.g., Coilcraft XEL) |
| IF $V_{in}$ is 3.7V, $V_{out}$ > 24V (High Step-Up) | LT3757 (Flyback/Boost) | Coupled Inductor / Transformer (e.g., Coilcraft MSD) |
| IF Ultra-low noise / RF application required | TI TPS61070 (with LDO post-reg) | Shielded + external RC snubber network |
How Assumptions Shift Across Voltage Profiles
The 3.7V to 12V calculation above is anchored to Continuous Conduction Mode (CCM) and a reasonable step-up ratio (1:3.2). When you change the voltage profile, the underlying physics shift, and blindly applying the standard formula becomes dangerous.
- Low Step-Up (5V to 6V): The duty cycle drops below 20%. The inductor value must increase significantly (often to 10µH+) to maintain CCM and prevent the output voltage from sagging under transient loads. Here, synchronous rectification (using an internal MOSFET instead of a Schottky diode) becomes mandatory to maintain efficiency.
- Medium Step-Up (12V to 24V): The input current is lower, but the switch node voltage spikes are brutal. The assumption that fixes the answer here shifts from inductor saturation to MOSFET drain-source breakdown voltage ($V_{DS}$). You must select an IC with a minimum 35V internal switch rating to survive the 24V output plus the diode forward voltage and ringing spikes.
- When the Conversion is Meaningless: If your step-up ratio exceeds 1:6 (e.g., 3.7V to 24V), standard single-inductor boost math breaks down. Parasitic resistances (inductor DCR, PCB trace resistance, and switch $R_{DS(on)}$) dominate the duty cycle equation. The theoretical duty cycle approaches 100%, but the physical circuit will never reach the target voltage. For ratios > 1:6, you must abandon the single inductor and use a coupled inductor (flyback topology) or a multi-stage charge pump.
FAQ: Boost Converter Sizing Edge Cases
What happens if I use a 4.7µH inductor instead of the calculated 2.2µH?
Your ripple current ($\Delta I_L$) will drop, which is generally good for efficiency and reduces output voltage ripple. However, the physical size of the inductor increases, and its transient response time slows down. If your load switches rapidly (like an ESP32 transmitting on WiFi), a 4.7µH inductor might cause the output voltage to dip during sudden 300mA current spikes because the inductor resists rapid changes in current.
Do I need a Schottky diode or can I use a standard 1N4007?
You must use a Schottky diode (like the SS34 or MBRS140) or a synchronous MOSFET. A standard 1N4007 has a reverse recovery time ($t_{rr}$) measured in microseconds. A boost converter switching at 1.2 MHz has a period of 0.83 microseconds. The 1N4007 will remain forward-biased while the internal MOSFET turns on, creating a dead short from input to ground that will instantly destroy the IC.
How does inductor ESR affect the calculation?
The formulas above assume an ideal inductor. In reality, a cheap unshielded inductor might have a DC Resistance (DCR) of 150mΩ. At 4A of input current, that wastes $I^2R = 16 \times 0.15 = 2.4W$ purely as heat inside the inductor, dropping your assumed 85% efficiency down to 65%. Always check the datasheet for DCR and specify parts with < 50mΩ for high-current boost applications. For deeper topology analysis, refer to All About Circuits' guide on boost converter losses.
