For a 1000W buck and boost converter system outputting 12V at 83.3A, the required DC input current from a 48V battery bank is 21.93A (assuming 95% efficiency). If you are sizing the upstream AC-DC power supply to feed this converter from a wall outlet, it will draw 9.75A from a 120V AC source, 5.09A from a 230V AC source, or 3.25A from a 208V 3-phase source (assuming a 0.9 Power Factor). The core DC-DC formula is I_in = P_out / (η × V_in). Substituting our values: I_in = 1000W / (0.95 × 48V) = 21.93A. For the AC mains side, we must account for Power Factor (PF) to find apparent current: I_AC = P_out / (η × V_AC × PF). Substituting for 120V: 1000 / (0.95 × 120 × 0.9) = 9.75A.
Neighboring Load Values and AC Mains Shifts
A buck and boost converter is strictly a DC-DC stage. It does not have a "power factor" or a "3-phase" input. However, in real-world bench and industrial builds, these converters are fed by an AC-DC switched-mode power supply (SMPS). Sizing the upstream AC breaker requires converting the DC power demand back to AC mains current. The table below maps a ±20% load range around our 1000W baseline, showing exactly how the current shifts across single-phase and three-phase AC topologies.
| Output Power (W) | DC Input Current @ 48V (A) | AC Input Current @ 120V (A) | AC Input Current @ 230V (A) | 3-Phase Current @ 208V (A) |
|---|---|---|---|---|
| 800 | 17.54 | 7.80 | 4.07 | 2.60 |
| 900 | 19.74 | 8.77 | 4.58 | 2.93 |
| 1000 | 21.93 | 9.75 | 5.09 | 3.25 |
| 1100 | 24.12 | 10.72 | 5.59 | 3.58 |
| 1200 | 26.32 | 11.70 | 6.10 | 3.90 |
Note: 3-phase calculations use the formula I = P / (√3 × V_L × PF). If your AC-DC supply lacks Active Power Factor Correction (PFC), drop the PF assumption to 0.6, which will nearly double your 120V/230V AC current draw.
Component Reality Check: Real-World IC Limits
The mathematical conversion above assumes the converter can actually handle the power. In practice, the conversion becomes meaningless if you push the silicon beyond its physical duty cycle or thermal limits. For example, asking a standard 500kHz buck converter to step 48V down to 47V requires a 98% duty cycle. Most ICs have a minimum off-time of ~100ns to reset the internal gate drivers, capping the maximum duty cycle around 95%. If you exceed this, the output voltage will droop regardless of your input current capacity.
When selecting the silicon for your design, use tools like the TI Power Stage Designer or Analog Devices LTpowerCAD to verify switching nodes. Below is a data-dense spec sheet of common architectures used in high-power buck/boost designs.
| IC / Controller | Topology | V_in Range | V_out Range | Max I_out | f_sw | Best Use Case |
|---|---|---|---|---|---|---|
| TI TPS5430 | Step-Down (Buck) | 5.5V – 36V | 1.22V – 31V | 3A (Int FET) | 500 kHz | 12V/24V to 5V logic rails |
| TI LM5170-Q1 | Bidirectional Buck/Boost | 2V – 85V | 2V – 85V | >10A (Ext FETs) | 50kHz – 1MHz | 48V mild-hybrid EV / Solar storage |
| ADI LT3780 | 4-Switch Buck-Boost | 4V – 38V | 0.8V – 38V | 10A (Ext FETs) | 200kHz – 700kHz | Automotive 12V to 12V/24V regulation |
| Microchip MCP1630 | Boost / SEPIC | 3.0V – 16V | V_in to 25V | 2A (Int FET) | 250kHz – 2MHz | Li-Ion to 12V LED strings |
When the Conversion Fails: Edge Cases and Dropouts
Blindly applying P = V × I to a buck and boost converter will lead to melted traces or tripped breakers if you ignore these three physical realities:
- Dropout Voltage (Buck): A buck converter cannot output a voltage equal to its input. It requires headroom (dropout voltage) to maintain regulation, typically 0.5V to 2V depending on the MOSFET R_DS(on) and inductor DCR. If your 12V input sags to 11.8V under load, a 12V output target will fail.
- Inductor Saturation (Boost): In a boost topology, the input current is significantly higher than the output current. If the inductor's saturation current rating (I_sat) is lower than the calculated peak input current, the inductor turns into a low-value resistor, spiking current and destroying the internal switching FET.
- Unknown Power Factor (AC Side): If you are sizing a branch circuit for a bank of buck/boost modules fed by cheap, uncorrected AC-DC brick supplies, the PF assumption of 0.9 is meaningless. Uncorrected supplies often run at 0.5 to 0.6 PF. You must size your AC wire and breaker for the higher apparent current, or mandate Active PFC supplies in your BOM.
Frequently Asked Questions
Does a boost converter draw more current from the source than it delivers to the load?
Yes. Because power must be conserved (minus efficiency losses), stepping up the voltage forces the input current to be higher than the output current. For a 12V to 48V boost converter delivering 2A (96W) at 90% efficiency, the input current will be 96W / (0.90 × 12V) = 8.88A. Your input wiring must be sized for the 8.88A, not the 2A output.
How do I measure the actual efficiency (η) on the bench?
Do not rely solely on the datasheet. Connect a precision shunt resistor or a calibrated Hall-effect clamp meter (like a Fluke 87V with a i17xx series clamp) on both the input and output. Measure true DC voltage and current at the exact load point. Calculate η = (V_out × I_out) / (V_in × I_in). At light loads (under 10%), efficiency often drops below 70% due to switching losses and quiescent current dominating the power budget.
Can I put a buck and boost converter directly on an AC line?
No. Buck and boost topologies rely on DC voltage differentials and unidirectional current flow through inductors and diodes/synchronous FETs. Applying AC will result in immediate catastrophic failure of the switching elements. You must use a front-end rectifier and bulk capacitor, or a dedicated AC-DC flyback/PFC stage, to create a stable DC bus first.
