Buck Converters Input-to-Output Current Conversion: Formulas & FAQ

To get 3.0A at 5.0V (15W) out of a standard buck converter with 90% efficiency from a 12V DC source, you must supply exactly 1.39A of input current. The foundational formula used here is I_in = (V_out × I_out) / (V_in × η). Substituting our exact query values: I_in = (5V × 3A) / (12V × 0.90) = 15W / 10.8W = 1.388A. This conversion assumes continuous conduction mode (CCM) and a fixed 90% efficiency (η), which is typical for modern synchronous controllers like the TI TPS5430 at moderate loads.

The Core Conversion Formula and Fixing Assumptions

Unlike a linear regulator (like an LM7805) where input current roughly equals output current, a buck converter is a switching power supply that conserves power, not current. It trades higher input voltage for higher output current. The universal power balance equation is P_in = P_out / η. Therefore, the input current formula is:

I_in = (V_out × I_out) / (V_in × η)

The primary assumption that fixes this answer is efficiency (η). If you do not know the efficiency of your specific buck converter at your specific load point, the conversion is merely an estimate. Efficiency is not a static number; it forms a bell curve. A chip like the MP2315 might hit 95% efficiency at 1A, but drop to 75% at a 10mA light load due to switching losses and quiescent current (I_q). Furthermore, this formula assumes the converter is operating in Continuous Conduction Mode (CCM), where the inductor current never drops to zero during the switching cycle.

Neighboring Values: Input Current vs. Output Load

When designing a power tree, you rarely size for just one exact load. Below is a reference table showing how input current scales across a ±20% range around our 3.0A baseline target, assuming a 12V input, 5V output, and a fixed 90% efficiency.

Output Current (Iout)	Output Power (Pout)	Required Input Power (Pin)	Input Current (Iin) @ 12V
2.4A (-20%)	12.0W	13.33W	1.11A
2.7A (-10%)	13.5W	15.00W	1.25A
3.0A (Baseline)	15.0W	16.67W	1.39A
3.3A (+10%)	16.5W	18.33W	1.53A
3.6A (+20%)	18.0W	20.00W	1.67A

Note: Always add a 20% safety margin to your input trace width and upstream fuse sizing to account for transient spikes and efficiency degradation at thermal limits.

How the Conversion Shifts Across Input Voltages (and AC Mains)

A common point of confusion in forums is how this math shifts for 120V AC, 230V AC, or 3-phase power. The direct answer is: a buck converter cannot process AC mains or 3-phase power directly. Buck converters are strictly DC-DC step-down devices. If you apply 120V AC to a standard DC-DC buck module, you will destroy the input capacitors and the switching MOSFETs instantly. To use 120V or 230V AC, you must first use an AC-DC rectifier and filter stage to create a high-voltage DC bus (yielding ~170V DC or ~325V DC), or utilize a specialized isolated AC-DC switching supply.

However, if we look at how the conversion shifts across standard DC bus voltages—12V (automotive), 24V (industrial), and 48V (telecom/PoE)—the input current drops proportionally as voltage rises. For our 15W (5V @ 3A) output at 90% efficiency:

• At 12V input: I_in = 1.39A
• At 24V input: I_in = 0.69A
• At 48V input: I_in = 0.35A

While higher input voltages drastically reduce input current (allowing for thinner input wiring), they introduce a new penalty: increased switching losses. Stepping 48V down to 5V requires a much lower duty cycle and MOSFETs with higher V_DS ratings, which typically have higher R_DS(on). As detailed in TI Application Report SLVA372, this high step-down ratio can push overall efficiency down to 82-85%, slightly increasing the actual input current required compared to the theoretical 90% calculation.

When This Conversion Becomes Meaningless

There are three specific scenarios where applying the I_in = P_out / (V_in × η) formula will give you dangerously incorrect design data:

1. Dropout Voltage Violation: Every buck converter has a minimum dropout voltage (the minimum difference between V_in and V_out). If you try to calculate the input current for a 5.5V input to a 5.0V output using a chip with a 1.0V dropout, the converter will simply pass the input voltage through (or drop out of regulation). The math becomes meaningless because the switch is operating at 100% duty cycle.
2. Discontinuous Conduction Mode (DCM): At very light loads (e.g., drawing 10mA from a 3A rated converter), the inductor current drops to zero before the next switching cycle. In DCM, the efficiency curve collapses, and quiescent current (I_q) dominates. The standard formula will under-calculate the input current by a massive margin.
3. Unknown or Unstated Efficiency: If a cheap, unbranded module claims '95% efficiency' on the box but uses a high-R_DS(on) asynchronous diode topology, real-world efficiency at full load might be 70%. Using 95% in your formula will result in undersized input traces that overheat.