To step down a 24V nominal DC source to 12V DC at 5A (60W output) using a buck converter, you need an input current of 2.78A and a PWM duty cycle of 42.8%, assuming a realistic 90% efficiency ($\eta$). The exact formulas with values substituted are:
- Input Current: $I_{in} = \frac{V_{out} \times I_{out}}{V_{in} \times \eta} = \frac{12 \times 5}{24 \times 0.90} = 2.78A$
- Duty Cycle: $D = \frac{V_{out}}{V_{in(actual)}} = \frac{12}{28} \approx 42.8\%$ (using 28V, the actual charging voltage of a 24V nominal battery system).
The Core Conversion: Input Current and Duty Cycle
A buck converter does not create power; it converts higher voltage and lower current into lower voltage and higher current, minus switching and conduction losses. The calculation above assumes Continuous Conduction Mode (CCM), where the inductor current never drops to zero during the switching cycle.
Below is a reference table showing how the input current and duty cycle shift across a ±20% range of output current, holding the 28V actual input and 90% efficiency constant.
| Output Current ($I_{out}$) | Output Power | Input Current ($I_{in}$) | Duty Cycle ($D$) | Inductor Ripple (Est. 30%) |
|---|---|---|---|---|
| 4.0A (-20%) | 48W | 2.22A | 42.8% | 1.2A peak-to-peak |
| 5.0A (Nominal) | 60W | 2.78A | 42.8% | 1.5A peak-to-peak |
| 6.0A (+20%) | 72W | 3.33A | 42.8% | 1.8A peak-to-peak |
System Voltage Shifts: AC Mains vs. DC Batteries
A frequent and dangerous point of confusion is asking how a buck converter conversion shifts for 120V AC, 230V AC, or 3-phase power. It doesn't. A raw buck converter is strictly a non-isolated DC-DC topology. If you apply 120V RMS AC (170V peak) or 230V RMS AC (325V peak) directly to a buck IC, the internal MOSFET will avalanche, short circuit, and likely catch fire. To step down mains AC, you must first use a bridge rectifier and bulk capacitor to create a high-voltage DC bus, followed by an isolated flyback or forward converter for safety. Never use a non-isolated buck topology directly off rectified mains unless the entire downstream circuit is treated as lethal line voltage.
Within the DC realm, shifting your baseline system voltage drastically alters the component requirements:
- 12V Nominal (14.4V actual): Stepping down to 5V requires a duty cycle of ~34%. Standard 20V-rated MOSFETs and ICs are sufficient.
- 24V Nominal (28.8V actual): Stepping down to 12V requires a duty cycle of ~41%. You must upgrade to ICs with at least a 40V absolute maximum input rating.
- 48V Nominal (57.6V actual): Stepping down to 12V drops the duty cycle to ~20%. This requires a 60V or 80V rated IC, and the low duty cycle forces either a lower switching frequency or a smaller inductor to maintain CCM.
When is this conversion math meaningless? First, if your converter drops into Discontinuous Conduction Mode (DCM) at light loads (typically below 10-20% of max current). In DCM, the duty cycle is no longer a simple $V_{out}/V_{in}$ ratio; it becomes a complex function of the inductor value, switching frequency, and load current. Second, the math is useless if you guess the efficiency ($\eta$). A poorly designed board with high-ESR capacitors and long, thin traces might run at 70% efficiency instead of 90%, pushing your actual input current up to 3.57A and causing thermal failure.
Decision Tree: Picking the Right Buck IC
Do not default to the cheap, blue LM2596 modules found on Amazon for serious 5A loads; their TO-263 packages and aging silicon struggle with thermal management above 3A without massive heatsinks. Use this decision path to select a modern, reliable silicon part:
| Condition / Requirement | Path | Recommended Action |
|---|---|---|
| Input voltage > 36V (e.g., 48V systems) | Requires high $V_{IN}$ tolerance | Select a 60V+ rated IC (e.g., TI LMR36015) |
| Input ≤ 36V, Output 12V at 5A continuous | Standard 24V battery/solar system | Concrete Pick: TI TPS54560BDDAR |
| Need > 95% efficiency, tight PCB space | Requires integrated FETs, high $F_{sw}$ | Select a QFN-package sync buck (e.g., MPS MP8765) |
| Budget is under $1.00, efficiency is secondary | Hobbyist / non-critical load | Use an asynchronous controller with external N-FET |
The Default Recommendation: For a 24V-to-12V, 5A conversion, buy the Texas Instruments TPS54560BDDAR. It supports up to 60V input, handles 5A continuous output, integrates a low-resistance high-side MOSFET, and costs roughly $3.50 on Mouser. It requires an external Schottky diode (like the CSD17571Q5A or a 60V/5A diode like the SS56) but offers robust thermal performance in an SOIC-8 package with an exposed thermal pad.
Component Reality Check: Inductor and Diode Sizing
The IC is only half the battle. The physical "conversion" happens in the inductor. For the TPS54560 running at 400kHz, targeting a 12V output and 30% ripple current (1.5A), the required inductance is:
$L = \frac{V_{out} \times (1 - D)}{f_{sw} \times \Delta I_L} = \frac{12 \times (1 - 0.428)}{400,000 \times 1.5} \approx 11.4\mu H$
Use a standard 10μH or 12μH shielded ferrite inductor rated for at least 7A saturation current ($I_{sat}$). If the inductor saturates, its inductance drops to near zero, the current spikes uncontrollably, and the IC's internal MOSFET will destroy itself. Always check the inductor datasheet for the $I_{sat}$ curve at 100°C, not just the 25°C rating.
Frequently Asked Questions
Can I put a buck converter in reverse to boost 12V to 24V?
No. A buck converter topology is strictly unidirectional. The internal high-side switch and low-side diode/synchronous FET are arranged to pull current from the high-voltage input and push it to the low-voltage output. Forcing 12V into the $V_{out}$ pin and drawing from the $V_{in}$ pin will not boost the voltage; it will simply forward-bias the internal body diodes, effectively shorting your source or yielding an unregulated diode-drop voltage. To step 12V up to 24V, you must use a boost converter topology.
Why does my 24V-to-12V buck converter output 24V when the load is disconnected?
If your output reads exactly the input voltage under no-load conditions, your low-side switch (or external catch diode) has failed short, or the high-side MOSFET is stuck in the "on" position due to a blown bootstrap capacitor or gate driver failure. In a non-isolated buck converter, a shorted high-side switch directly connects $V_{in}$ to $V_{out}$. This is why critical 12V loads on a 24V system should always be protected by a downstream crowbar circuit or a fast-blow fuse sized just above the maximum load current.
