If you need 60W of output power (12V at 5A) from a 24V DC source using a standard 90% efficient buck dc converter circuit, your required input current is exactly 2.77A, and the circuit will dissipate 6.6W of heat. The governing formula is I_in = (V_out × I_out) / (V_in × η), which substitutes as (12 × 5) / (24 × 0.90) = 2.77A. This calculation assumes Continuous Conduction Mode (CCM) at a 25°C ambient temperature with a properly sized 33µH shielded inductor. If your efficiency drops to 80% due to poor inductor selection or high switching losses, input current spikes to 3.12A and heat jumps to 15W, immediately requiring a forced-air heatsink or a larger silicon die.
The Core Conversion: Output Power to Input Current
To size your input traces, select your upstream fuse, and calculate thermal dissipation, you must convert your target output requirements into input-side realities. The assumption that fixes this answer is the converter's efficiency (η), which is dictated by the switching frequency, MOSFET Rds(on), and inductor core losses.
Below is a reference table showing how a ±20% shift in your target output current (around our 5A baseline) alters the input requirements for a 24V-to-12V step-down topology at 90% efficiency.
| Target Output (12V) | Output Power | Input Current (24V) | Heat Dissipated | Minimum Input Trace (1oz Cu) |
|---|---|---|---|---|
| 4.0A (-20%) | 48W | 2.22A | 5.3W | 40 mil |
| 5.0A (Baseline) | 60W | 2.77A | 6.6W | 50 mil |
| 6.0A (+20%) | 72W | 3.33A | 8.0W | 60 mil |
How Voltage Topologies Shift the Math
A pure DC-DC converter circuit only cares about its DC input. However, in real-world builds, that DC is usually generated by an AC-DC front-end power supply (like a Mean Well LRS-150-24). When you trace the conversion all the way back to the wall, the math shifts based on AC line voltage and Power Factor (PF).
If your 60W DC load is fed by an AC-DC switching supply with 90% efficiency, the AC input current formula becomes I_ac = P_out / (V_ac × PF × η).
- 120V AC (US Standard): Assuming a typical PF of 0.85 for non-PFC supplies,
60 / (120 × 0.85 × 0.90) = 0.65A. You need a standard 1A or 2A IEC inlet fuse. - 230V AC (EU/UK Standard): PF often improves slightly to ~0.90 at higher voltages.
60 / (230 × 0.90 × 0.90) = 0.32A. The current halves, allowing thinner IEC cords. - 3-Phase AC (Industrial): Rarely used for sub-100W loads, but if feeding a larger 1kW DC bus, current per phase drops by a factor of √3.
I_phase = P_total / (V_line × √3 × PF × η).
When This Conversion Becomes Meaningless
Theoretical conversions fall apart on the bench when physical limits are hit. The math above is meaningless under three specific conditions:
- Inductor Saturation: If you use a 10µH inductor rated for 4A on a 6A load, the core saturates. The inductor stops acting like an inductor and becomes a low-value resistor. Efficiency crashes to 40%, input current spikes unpredictably, and the switching MOSFET usually detonates. The datasheet's efficiency curve assumes an unsaturated core.
- Light-Load Burst Mode: If your 60W converter is only pulling 0.1A (1.2W) at standby, it enters pulse-skipping or burst mode. Quiescent current (Iq) of the control IC dominates, and efficiency can plummet to 40-50%. The standard η=90% assumption is invalid below 10% of rated load.
- Thermal Foldback: If the IC junction temperature exceeds its limit (typically 150°C), internal protection circuitry reduces the switching frequency or skips cycles. Output voltage droops, and the input current waveform becomes erratic.
Decision Tree: Picking Your DC Converter IC
Don't guess your silicon. Use this decision matrix to terminate your design process with a specific, proven part number based on your converted current and voltage requirements.
| Condition / Requirement | If True, Go To... | If False, Go To... |
|---|---|---|
| Is V_in > V_out? (Step-Down) | Row 2 (Buck) | Row 4 (Boost) |
| Is I_out < 3A? | Pick LM2596 (Cheap, 150kHz, TO-263) | Row 3 |
| Is I_out between 3A and 5A? | Pick TPS5430 (5.5-36V in, 3A cont / 4A peak, SOIC-8) | Pick XL4015 (5A, TO-220, needs heatsink) |
| Is V_out > V_in? (Step-Up) | Pick MT3608 (Up to 2A, SOT23-6) | Pick XL6009 (Up to 4A, TO-263) |
FAQ: Real-World Bench Questions
Why does my multimeter show higher input current than the formula predicts?
Standard multimeters read average DC current, but a buck converter draws high-frequency pulsed current from the source. If your input bulk capacitor is too small (e.g., 10µF instead of 100µF), the RMS current increases due to high ripple, causing I²R losses in your wires that register as higher total power draw. Always use at least 47µF of low-ESR ceramic or polymer capacitance on the input.
Can I just use a linear regulator (LDO) instead of a switching dc converter circuit?
Only if the voltage drop and current are tiny. Dropping 24V to 12V at 5A through an LDO would dissipate (24 - 12) × 5 = 60W of pure heat. That requires a massive, actively cooled heatsink and is 50% efficient. A switching converter is mandatory for this power level.
Does the PCB layout change the efficiency calculation?
Yes. The 'hot loop' (the path from the input capacitor, through the high-side MOSFET, to the inductor and back via the low-side diode/MOSFET) must be as short and wide as possible. A poorly routed hot loop adds parasitic inductance, causing voltage ringing that forces you to slow down the switching edges, directly increasing switching losses and dropping your real-world η from 90% to 82%.
