For a standard 12V to 5V, 2A buck converter design, the ideal duty cycle is 41.67% and the required inductance for a 30% peak-to-peak ripple current at a 500 kHz switching frequency is 9.72 µH. These numbers assume Continuous Conduction Mode (CCM) and ideal components. In a real-world bench build using a controller like the TI TPS5420 or LM2596, you must derate for ~90% efficiency, the internal MOSFET voltage drop, and the catch diode forward voltage. This shifts your actual duty cycle closer to 44.5% and requires selecting the next standard off-the-shelf inductor value, typically 10 µH.
Core Buck Converter Equations and Substituted Values
To size the power stage of a step-down DC-DC regulator, we rely on two foundational formulas. The calculations below use our baseline parameters: $V_{in} = 12V$, $V_{out} = 5V$, $I_{out} = 2A$, $f_{sw} = 500\text{ kHz}$, and a ripple coefficient ($K_{ind}$) of 0.3 (30% of load current, meaning $\Delta I_L = 0.6A$).
1. Ideal Duty Cycle ($D$):
$D = \frac{V_{out}}{V_{in}}$
$D = \frac{5}{12} = 0.4167 \text{ (or 41.67%)}$
2. Inductor Value ($L$):
$L = \frac{V_{out} \times (1 - D)}{f_{sw} \times \Delta I_L}$
$L = \frac{5 \times (1 - 0.4167)}{500,000 \times 0.6} = \frac{2.9165}{300,000} = 9.72 \text{ µH}$
Because 9.72 µH is not a standard off-the-shelf component value, you would round up to 10 µH. Rounding up slightly reduces your peak-to-peak ripple current, which lowers output voltage ripple and improves transient response, at the cost of a physically larger and slightly more expensive inductor.
Neighboring Values: ±20% Input Voltage Range
In automotive or unregulated solar applications, your input voltage is rarely a perfect 12.0V. The table below shows how the required duty cycle and theoretical inductance shift across a ±20% input tolerance band, assuming the output remains locked at 5V / 2A.
| Input Voltage ($V_{in}$) | Variance | Ideal Duty Cycle ($D$) | Calculated Inductor ($L$) | Standard Inductor Choice |
|---|---|---|---|---|
| 9.6V | -20% | 52.08% | 7.98 µH | 8.2 µH |
| 10.8V | -10% | 46.30% | 8.95 µH | 10 µH |
| 12.0V | Nominal | 41.67% | 9.72 µH | 10 µH |
| 13.2V | +10% | 37.88% | 10.35 µH | 12 µH |
| 14.4V | +20% | 34.72% | 10.88 µH | 12 µH |
Note: If your design must operate across this entire 9.6V–14.4V range, you must select the 12 µH inductor to prevent the converter from saturating or falling out of regulation at the high end, while verifying that the controller can achieve the 52.08% duty cycle required at the low end without dropping out of regulation.
Assumptions, Voltage Shifts, and Meaningless Conversions
The equations above are highly reliable, but only because they are anchored to specific assumptions. If those assumptions break, the math breaks with them.
What Assumptions Fix These Answers?
The primary assumption fixing these buck converter equations is Continuous Conduction Mode (CCM). CCM assumes the inductor current never falls to zero during the switching cycle. We also assume a fixed 30% ripple ratio ($K_{ind} = 0.3$), which is the industry-standard sweet spot balancing inductor size against output ripple. Finally, we assume a steady-state DC load; highly dynamic loads (like an ESP32 waking up and transmitting on WiFi) require control-loop bandwidth analysis, not just steady-state inductor sizing.
How the Math Shifts for 120V vs 230V vs 3-Phase
A common point of confusion for beginners is applying DC-DC buck equations to AC mains voltages. Buck converters are strictly DC-DC topologies. If you are rectifying 120V AC (yielding a ~160V DC bus) or 230V AC (yielding a ~320V DC bus) to feed a buck stage, the duty cycle collapses. For example, stepping 160V DC down to 5V yields a duty cycle of just 3.1%.
At this extreme ratio, standard non-isolated buck converter equations become practically useless. The required MOSFET on-time ($t_{on}$) will violate the minimum on-time limits of almost all commercial PWM controllers. Furthermore, the inductor would need to be massive to prevent the circuit from falling into Discontinuous Conduction Mode. For 120V/230V AC mains step-down, or 3-phase industrial rectification, you must abandon the buck topology entirely in favor of isolated topologies like flyback, forward, or LLC resonant converters, which utilize transformer turns ratios rather than pure duty cycle to achieve massive step-down ratios safely.
When the Conversion is Meaningless
The standard $D = V_{out}/V_{in}$ equation becomes completely meaningless when the load current drops below the critical boundary, pushing the circuit into Discontinuous Conduction Mode (DCM). This happens when $I_{out} < \frac{\Delta I_L}{2}$. In DCM, the inductor current hits zero before the next switching cycle begins. The duty cycle is no longer a simple linear ratio of voltages; it becomes a complex square-root function dependent on load current, inductance, and switching frequency. If your load is highly variable and dips into light-load or sleep states, your control IC will automatically shift its internal PWM mathematics to handle DCM or burst mode—you cannot rely on basic CCM equations to predict behavior at 10mA loads.
Frequently Asked Questions
How do buck converter equations change for discontinuous conduction mode (DCM)?
In DCM, the duty cycle equation changes to $D = \frac{V_{out}}{V_{in}} \times \sqrt{\frac{2 \times L \times f_{sw} \times I_{out}}{V_{in} - V_{out}}}$. Notice that load current ($I_{out}$) and inductance ($L$) are now inside the duty cycle calculation. This is why DCM control loops are harder to stabilize; the plant transfer function changes fundamentally based on how much current the load is drawing.
What is the formula for calculating buck converter output voltage ripple?
Output voltage ripple ($\Delta V_{out}$) is dominated by two factors: the capacitive ripple and the Equivalent Series Resistance (ESR) ripple. The combined equation is:
$\Delta V_{out} = (\Delta I_L \times ESR) + \frac{\Delta I_L}{8 \times f_{sw} \times C_{out}}$
In modern designs using MLCC (ceramic) capacitors, the ESR is so low that the capacitive term dominates. However, if you use older electrolytic or tantalum capacitors, the $\Delta I_L \times ESR$ term will completely overwhelm the math, resulting in massive voltage spikes.
Why does my calculated buck converter inductor value not match the evaluation board?
Evaluation boards from manufacturers like Texas Instruments or Analog Devices often use inductors that are 20% to 50% larger than the bare minimum CCM calculation suggests. Engineers designing eval boards optimize for transient response and thermal derating, not just steady-state ripple. A larger inductor stores more energy, which helps the control loop recover faster when a microcontroller suddenly wakes up and draws 500mA in a few nanoseconds. Furthermore, inductors lose inductance as they heat up and approach their saturation current ($I_{sat}$); oversizing the part ensures the 10 µH part actually acts like a 10 µH part at 85°C.
How do you calculate the input capacitor RMS current for a buck converter?
The input capacitor in a buck converter takes a massive beating because it supplies the pulsing current drawn by the high-side MOSFET. The RMS current equation is:
$I_{cin(rms)} = I_{out} \times \sqrt{D \times (1 - D)}$
For our 12V to 5V, 2A example, this equates to $2 \times \sqrt{0.4167 \times 0.5833} = 0.98A$. You must select an input capacitor (or parallel bank of MLCCs) rated to handle at least 1A of RMS ripple current at your specific switching frequency, otherwise the capacitor will overheat, crack, and fail short.
References:
1. Texas Instruments, Basic Calculation of a Buck Converter's Power Stage (SLVA477B).
2. All About Circuits, Buck Converter Topology and Operating Principles.
