Buck Converter and Boost Converter Voltage & Duty Cycle Conversions

If you need to step down 12V to 5V at 2A using a buck converter and boost converter setup, the required buck duty cycle is exactly 41.7% with an input current of 0.93A (assuming 90% efficiency). Conversely, stepping up 5V to 12V at 1A via a boost converter requires a 58.3% duty cycle and draws 2.67A from the source. These conversions assume Continuous Conduction Mode (CCM); if your inductor is undersized and the circuit falls into Discontinuous Conduction Mode (DCM), these voltage-ratio conversions become meaningless because the output voltage will rely on load current rather than duty cycle.

Core Conversion Formulas and Boundary Assumptions

The mathematical relationship between input voltage, output voltage, and switch duty cycle is fixed by the topology. For a buck converter, the formula is D = Vout / Vin. Substituting our 12V to 5V query: D = 5 / 12 = 0.4167 (41.7%). For a boost converter, the formula is D = 1 - (Vin / Vout). Substituting our 5V to 12V query: D = 1 - (5 / 12) = 0.5833 (58.3%).

Input current conversion requires accounting for efficiency ($\eta$). The formula is Iin = (Vout × Iout) / (Vin × η). Assuming a realistic 90% efficiency for a modern synchronous IC like the TI TPS5430, the buck input current is (5 × 2) / (12 × 0.9) = 0.926A. The boost input current is (12 × 1) / (5 × 0.9) = 2.667A.

The primary assumption fixing these answers is Continuous Conduction Mode (CCM). In CCM, the inductor current never falls to zero during the switching cycle. If the load drops below the critical current threshold, the converter enters DCM, and the output voltage becomes a function of load resistance and inductance, rendering the simple duty-cycle conversions above meaningless.

Topology Shifts: From 12V DC to 230V AC Mains

The formulas above strictly assume a DC source. If your input is AC mains, the conversion math shifts dramatically based on the rectification stage. A 120V AC RMS source rectifies to a ~170V DC peak bus, while a 230V AC RMS source yields a ~325V DC peak bus. In these high-voltage scenarios, you typically abandon standard non-isolated buck converters for isolated topologies (like flyback or LLC resonant) due to safety isolation requirements and extreme duty-cycle limits.

However, if you are using a 3-phase AC supply (common in industrial 480V systems), the rectified DC bus has significantly lower ripple voltage compared to single-phase 120V/230V. This 3-phase advantage means you can drastically reduce the input bulk capacitance, but the fundamental DC-DC duty cycle on the secondary side remains fixed by the DC bus voltage, not the AC line voltage. Furthermore, if the AC-side Power Factor (PF) is unknown or uncorrected, calculating the primary AC input current becomes meaningless, as you cannot distinguish between real power (Watts) and reactive power (VAR) to size your upstream breakers.

Real-World Component Selection and Loss Margins

Theoretical conversions assume ideal components. On the bench, parasitic resistances alter the math. The DC resistance (DCR) of the inductor and the on-resistance (Rds_on) of the MOSFETs introduce voltage drops. For a boost converter operating at high duty cycles (e.g., D > 80%), the right-half-plane zero (RHPZ) in the control loop limits bandwidth, and parasitic resistances cause the actual output voltage to sag below the theoretical Vin / (1-D) calculation. According to Analog Devices application notes, pushing a boost converter beyond an 85% duty cycle often results in severe efficiency degradation and thermal runaway.

When selecting inductors, convert your target ripple current ($\Delta I_L$) using the formula: L = (Vin - Vout) × D / (f_sw × ΔI_L) for a buck converter. A standard design rule is to set $\Delta I_L$ to 30% of the maximum DC load current. For our 12V-to-5V buck at 2A, targeting a 500 kHz switching frequency and 0.6A ripple: L = (12 - 5) × 0.417 / (500,000 × 0.6) = 9.73 µH. You would select a standard 10 µH shielded ferrite inductor with a saturation current rating at least 20% above the peak current limit (2A + 0.3A = 2.3A minimum).