Converter Efficiency & Power Loss Calculator

Efficiency and power loss from measured Vin/Iin/Vout/Iout, or from a breakdown of individual loss components.
From Vin/Iin/Vout/Iout
From Known Loss Components

Efficiency from Measured Values

η = Pout/Pin × 100%  •  Ploss = Pin−Pout
12V/2.2A in, 5V/5A out
24V/1.1A in, 12V/2A out
5V/1.05A in, 3.3V/1.5A out
V
A
V
A
Enter values and press Calculate.

Efficiency from Loss Components

Pin = Pout+ΣPloss  •  η = Pout/(Pout+ΣPloss) × 100%
25W out, cond 0.4W, sw 0.3W, other 0.1W
50W out, cond 1.2W, sw 0.8W, other 0.3W
10W out, low-power design
W
W
W
W
Enter values and press Calculate.

Converter Efficiency & Where Power Is Lost

Converter efficiency η is simply the ratio of power delivered to the load versus power drawn from the source; the difference is dissipated as heat inside the converter. That heat comes from several distinct loss mechanisms: conduction loss (I²R heating in switches, inductor windings, and PCB traces), switching loss (energy lost during each MOSFET turn-on/turn-off transition, proportional to switching frequency), and smaller contributors like gate-drive loss, core (magnetic) loss, and quiescent/control-circuit current.

QuantityFormula
Input powerPin = Vin×Iin
Output powerPout = Vout×Iout
Efficiencyη = Pout/Pin × 100%
Total power lossPloss = Pin−Pout = Σ(conduction + switching + other)

Efficiency typically varies with load current and input voltage — conduction loss grows with I², so efficiency often drops at very high load, while switching and quiescent losses are relatively fixed, so efficiency also drops at very light load (where they become a larger fraction of a small output power). Peak efficiency usually occurs somewhere in the mid-load range.

Real-World Applications & Examples

Worked examples

1. 12V→5V buck at 5A. Pin=12×2.2=26.4W, Pout=5×5=25W: η=25/26.4×100=94.7%, Ploss=1.4 W.
2. 24V→12V buck at 2A. Pin=24×1.1=26.4W, Pout=12×2=24W: η=24/26.4×100=90.9%, Ploss=2.4 W.
3. 5V→3.3V buck at 1.5A. Pin=5×1.05=5.25W, Pout=3.3×1.5=4.95W: η=4.95/5.25×100=94.3%, Ploss=0.3 W.
4. Loss breakdown, 25W output. Conduction 0.4W + switching 0.3W + other 0.1W = Ploss=0.8W: Pin=25.8W, η=25/25.8×100=96.9%.
5. Loss breakdown, 50W output. Conduction 1.2W + switching 0.8W + other 0.3W = Ploss=2.3W: Pin=52.3W, η=50/52.3×100=95.6%.
6. Light-load efficiency drop. The same 0.8W of fixed switching+quiescent loss that was negligible at 50W output becomes a much larger fraction at 5W output (η=5/5.8×100≈86.2%) — illustrating why light-load efficiency is always lower than mid-load efficiency.

Frequently Asked Questions

What is converter efficiency?

The ratio of power delivered to the load (output power) to power drawn from the source (input power), η=Pout/Pin×100%. The remaining power is lost as heat inside the converter.

What are the main sources of power loss in a DC-DC converter?

Conduction loss (I²R heating in MOSFETs, inductor, and PCB traces), switching loss (energy lost during each switch transition, scaling with frequency), plus smaller contributors like gate-drive loss, magnetic core loss, and quiescent/control current.

Why does efficiency drop at very light load?

Switching and quiescent losses are roughly constant regardless of load, so at low output power they represent a much larger fraction of the (small) input power, pulling efficiency down.

Why does efficiency drop at very heavy load too?

Conduction loss scales with the square of current (I²R), so at high load it grows disproportionately faster than output power, reducing efficiency again beyond the peak-efficiency point.

Where does peak efficiency usually occur?

Typically at 30–70% of a converter's rated maximum load, where the roughly-constant losses (switching, quiescent) and the roughly-quadratic conduction loss are both relatively small compared to output power.

How is power loss related to heat generation?

All power not delivered to the load (Ploss=Pin−Pout) is dissipated as heat within the converter's components, which is exactly what thermal/heatsink design must account for.

Does input or output voltage affect efficiency?

Yes — a larger input-to-output voltage difference (higher step-down/step-up ratio) generally increases switching and conduction losses relative to output power, typically reducing efficiency compared to a smaller conversion ratio.

How do I estimate conduction loss?

Approximately as Irms²×Rtotal, summing the on-resistance of switches (RDS(on)), inductor DCR, and any significant PCB trace/connector resistance in the current path.

How do I estimate switching loss?

Roughly as 0.5×Vsw×Isw×(trise+tfall)×fsw, using the switch voltage/current and its rise/fall times from the datasheet, multiplied by switching frequency — this is why lower switching frequency reduces switching loss but requires larger passive components.

Why do datasheets quote efficiency at multiple load points?

Because efficiency varies significantly across the load range (light-load, mid-load, full-load), a single number does not fully describe converter performance; a full efficiency-vs-load curve is far more useful for design decisions.

What is a good efficiency target for a modern DC-DC converter?

Well-designed synchronous buck converters commonly achieve 90–97% efficiency at mid-to-full load; higher ratios (like very high step-down or step-up) or very light loads typically achieve somewhat lower efficiency.

How does switching frequency trade off against efficiency?

Higher switching frequency allows smaller inductors/capacitors but increases switching loss; lower frequency reduces switching loss (improving efficiency) but requires larger, more expensive passive components for the same ripple performance.

Why is no-load/standby power important for chargers?

Energy efficiency standards (like DOE/Energy Star) specifically regulate no-load power consumption because billions of chargers remain plugged in without a connected device, so even small standby losses add up to significant global energy waste.

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