To output 5A at 24V (120W) from a 12V source using a DC-DC buck-boost converter at a realistic 90% efficiency, the input current is 11.11 Amps. This conversion assumes Continuous Conduction Mode (CCM), a fixed 12V DC input, and a steady resistive load. The governing formula is I_in = (V_out × I_out) / (V_in × η). Substituting our exact query values: I_in = (24V × 5A) / (12V × 0.90) = 11.11A.
The Core Assumptions: What Fixes the Answer
The single assumption that fixes this answer is the converter efficiency (η). In a perfect, lossless world (100% efficiency), stepping 12V up to 24V at 5A would draw exactly 10A. However, real-world switching regulators lose energy to inductor core hysteresis, MOSFET switching transitions, and diode forward voltage drops.
If you are using a budget-friendly XL6009 boost module, efficiency at a 1:2 step-up ratio often drops to roughly 82%. Recalculating with η = 0.82 yields an input current of 12.20A. Conversely, a premium synchronous 4-switch buck-boost IC like the LTC3130 or TPS54360 can maintain 94-96% efficiency, dropping the input draw to roughly 10.63A. Always check your specific IC's datasheet efficiency curve at your exact input/output voltage differential.
Neighboring Values & Efficiency Derating Table
Below is a reference table showing how input current scales across a ±20% range of our 5A baseline output. This is critical for sizing your input wiring and fuses, as loads rarely sit at a perfect static draw.
| Output Current (I_out) | Output Power (24V) | Input Current @ 90% η | Input Current @ 82% η (Budget Modules) | Recommended Input Wire (AWG) |
|---|---|---|---|---|
| 4.0A (-20%) | 96W | 8.89A | 9.76A | 16 AWG |
| 4.5A (-10%) | 108W | 10.00A | 10.98A | 14 AWG |
| 5.0A (Baseline) | 120W | 11.11A | 12.20A | 14 AWG |
| 5.5A (+10%) | 132W | 12.22A | 13.41A | 12 AWG |
| 6.0A (+20%) | 144W | 13.33A | 14.63A | 12 AWG |
Note: Wire sizing assumes standard copper THHN in free air at 30°C ambient. For enclosed battery boxes or engine bays, derate by one AWG size.
AC Mains vs. DC-DC: When the Conversion Shifts
A massive point of confusion in electrical search queries is the dual use of the term "buck-boost." In DC electronics, it refers to switching regulators. In AC mains electrical work, a "buck-boost transformer" is an autotransformer used to adjust line voltage (e.g., bucking 240V down to 208V for HVAC equipment).
How the answer shifts for 120V vs 230V vs 3-Phase:
If you are calculating current for an AC buck-boost transformer, the DC formula above is invalid. You must account for AC Power Factor (PF). The formula becomes I = P / (V × PF). For a 120V single-phase circuit, a 1200W load at 0.85 PF draws 11.76A. If you shift that same load to a 230V single-phase European mains supply, the current drops to 6.14A. For a 3-phase system, the denominator expands to include the square root of 3: I = P / (√3 × V_LL × PF).
When the conversion is meaningless:
In AC systems, if the Power Factor is unknown (common with non-linear loads like cheap LED drivers or uncorrected switching power supplies), calculating exact amperage from wattage is meaningless. You must measure it with a true-RMS clamp meter. In DC-DC systems, the linear conversion formula becomes meaningless when the converter drops into Discontinuous Conduction Mode (DCM) at very light loads (typically under 10% of rated capacity). In DCM, quiescent current and switching losses dominate, and input current no longer scales linearly with output current.
For deeper topological insights, refer to the Texas Instruments Buck-Boost Topology Overview or the All About Circuits guide on DC-DC converters.
Frequently Asked Questions
Can a buck boost converter increase amps while dropping voltage?
Yes. When operating in "buck" (step-down) mode, output current increases relative to input current. For example, if you configure a buck-boost converter to step 24V down to 12V to supply a 5A load (60W output), assuming 90% efficiency, the input current draw from the 24V source will only be 2.77 Amps (60W / (24V × 0.90)). Power is conserved, so dropping voltage inherently multiplies available current.
Why does my buck boost converter draw more amps than the formula predicts?
If your bench multimeter shows higher input amps than calculated, you are likely experiencing one of three real-world losses: 1) The converter's quiescent current (the power the IC and feedback resistors consume just to stay awake, often 5-15mA). 2) High Equivalent Series Resistance (ESR) on your input capacitors, causing I²R heating losses. 3) Inductor saturation. If your load spikes and the inductor core saturates, its inductance drops to near zero, causing massive, inefficient current spikes that your multimeter might read as a higher average draw.
How do I size the input wire and breaker for a high-current buck boost converter?
Never size wire for the exact calculated continuous draw. According to NEC-style guidance (NEC 210.20(A)), continuous loads (those running for 3 hours or more) require the circuit to be rated at 125% of the load. If your calculated input is 11.11A, multiply by 1.25 to get 13.88A. Therefore, you should use 14 AWG copper wire (rated for 15A at 60°C) and protect it with a 15A fuse or DC breaker. Always place the fuse as close to the battery or power source as possible, not near the converter.
What is the difference between a 4-switch buck-boost and a SEPIC converter?
Both topologies allow the output voltage to be higher or lower than the input, but they handle energy transfer differently. A 4-switch synchronous buck-boost (like the LTC3130) uses four MOSFETs to seamlessly transition between buck and boost modes, offering peak efficiencies up to 96% but suffering from discontinuous input current during the boost phase. A SEPIC (Single-Ended Primary-Inductor Converter) uses two inductors and a coupling capacitor. Its main advantage is continuous input current, which drastically reduces input voltage ripple and EMI, making it ideal for sensitive RF or audio circuits, though it typically caps out at 85-88% efficiency due to the capacitor's ESR losses.
