To step down 24V DC to 12V DC at a 5A load using a standard buck DC-DC converter operating at 90% efficiency, the input current draw is exactly 2.78A. The formula used to find this is Iin = (Vout × Iout) / (Vin × η). Substituting the exact query values: Iin = (12V × 5A) / (24V × 0.90) = 60W / 21.6V = 2.78A. The primary assumption that fixes this answer is a steady-state DC load with a known, fixed converter efficiency (η). Unlike AC circuits, pure DC conversions do not require complex power factor adjustments, making the math straightforward once efficiency is locked in.
The Core Conversion Formula and Neighboring Values
When sizing upstream wiring, fuses, or battery banks for DC-DC converters, you must calculate the input current based on the output power demand divided by the converter's efficiency. Conservation of energy dictates that input power must always exceed output power. The missing energy is dissipated as heat.
The governing equations are:
- Output Power: Pout = Vout × Iout
- Input Power: Pin = Pout / η
- Input Current: Iin = Pin / Vin
Below is a reference table showing how the input current shifts across a ±20% range of output current for our 24V-to-12V scenario, assuming a constant 90% efficiency. This is critical for sizing input traces and fuses to handle transient load spikes without nuisance tripping.
| Output Current (Iout) | Output Power (Pout) | Input Power @ 90% eff (Pin) | Input Current at 24V (Iin) |
|---|---|---|---|
| 4.0A (-20%) | 48W | 53.3W | 2.22A |
| 4.5A (-10%) | 54W | 60.0W | 2.50A |
| 5.0A (Nominal) | 60W | 66.7W | 2.78A |
| 5.5A (+10%) | 66W | 73.3W | 3.07A |
| 6.0A (+20%) | 72W | 80.0W | 3.33A |
Why Power Factor and AC Phase Shifts Are Meaningless Here
A common point of confusion for makers and junior engineers is attempting to apply AC power formulas to DC-DC converters. For a pure DC-DC converter—whether it is a buck, boost, or buck-boost topology—power factor (PF) and AC phase angles are strictly meaningless. Power factor only exists when converting alternating current, where voltage and current waveforms can fall out of phase due to reactive components (inductance/capacitance).
However, if you are analyzing the upstream AC-DC power supply that feeds your DC-DC stage, the assumptions shift dramatically. Let's look at how the answer changes if we are pulling that 66.7W of input power from an AC wall outlet instead of a 24V battery:
- At 120V AC (Single Phase): Assuming a typical active PFC (Power Factor Correction) supply with a PF of 0.95, the AC current draw is I = 66.7W / (120V × 0.95) = 0.58A. If the supply lacks PFC and has a poor PF of 0.65, the current jumps to 0.85A, even though the real power consumed remains 66.7W.
- At 230V AC (Single Phase): The higher voltage halves the current requirement. At 0.95 PF, the draw drops to 0.30A.
- In a 3-Phase System: The load is divided across three legs. The line current calculation shifts to I = P / (√3 × Vline × PF), drastically reducing the current per conductor.
When is the conversion meaningless? If you are trying to calculate the AC input current of a front-end power supply and the power factor is unknown, the conversion is meaningless. You cannot calculate apparent current (VA) from real power (W) without the PF. In these cases, theoretical math fails; you must use a true-RMS clamp meter with a power analysis function to measure the actual draw, as detailed in All About Circuits' guide on DC-DC converter design.
Real-World Efficiency Derating and Thermal Limits
Theoretical calculations assume a fixed η (efficiency), but in reality, efficiency is a moving target that shifts with load, input voltage, and temperature. When sourcing DC-DC converters, the part number dictates your thermal reality.
Take the ubiquitous LM2596 step-down modules found on Amazon for $2 each. While they claim 3A output, their efficiency at a 24V-to-12V conversion often drops to 75% or lower. At 60W output, 75% efficiency means 20W of waste heat. Without a massive heatsink and active airflow, the silicon will hit its thermal shutdown threshold (typically 150°C) and throttle or fail.
For professional or high-reliability bench builds, upgrade to modern synchronous buck controllers. A Texas Instruments TPS5430 or LMR336 operates at 90-95% efficiency, reducing that 20W heat penalty down to roughly 3.5W. For point-of-load 5V or 12V rails, drop-in switching replacements like the Murata OKI-78SR series offer 95% efficiency in a TO-220 footprint, completely eliminating the need for heatsinks at loads up to 1.5A.
Pro-Tip: Always check the datasheet's efficiency curve at your specific input voltage. A converter might be 92% efficient at 12V-in, but drop to 84% at 36V-in due to increased switching losses and gate-charge requirements.
DC-DC Converters Frequently Asked Questions
How do I size an input fuse for DC-DC converters?
Never size a fuse based purely on the steady-state calculated input current. DC-DC converters draw massive inrush currents to charge their input bulk capacitors upon startup. Size your input fuse for 1.5× to 2.0× the maximum continuous input current, and use a slow-blow (time-delay) fuse. For our 2.78A nominal input, a 5A slow-blow automotive blade fuse or a 4A ceramic tube fuse is appropriate. If you use a fast-acting fuse, it will likely blow the moment you flip the switch.
Can I wire two DC-DC converters in series to double the voltage?
No, you cannot wire the outputs of two standard non-isolated DC-DC converters in series. Most cheap buck/boost modules share a common ground between the input and output. Wiring them in series creates a dead short through the ground path of the lower converter. If you need to double voltage (e.g., 12V to 24V), use a dedicated boost converter, or use an isolated DC-DC converter where the input and output grounds are galvanically separated by a high-frequency transformer.
Why does my DC-DC converter output voltage drop under heavy load?
Output voltage sag under load is almost always caused by one of three physical bottlenecks: inadequate input decoupling, inductor saturation, or trace resistance. First, measure the voltage directly at the converter's input pins under load; if it drops below the minimum dropout voltage, your input wires are too thin. Second, if using a cheap module, the power inductor may be saturating (losing inductance) past 2A, causing the control loop to lose regulation. Finally, ensure your output wiring is sized correctly—a 5A load on 22 AWG wire will drop over 0.2V per foot before it even reaches your load.
