As seen here in the example, this particular power supply (and most others too) has higher efficiency when running on 230V. Given that computer power supplies are usually required to output a combination of 12V, 5V, and 3.3V DC, why is it that stepping down from a higher AC voltage is more efficient? It seems counter-intuitive.

Also is this a result intrinsic to the process of converting AC to DC, or is it a compromise that manufactures settle with for compatibility? In other words, if someone is to build a power supply that only works on 115V, is it more difficult to achieve the same efficiency as one built only for 230V?

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    \$\begingroup\$ Input components will handle less current at the higher input voltage, thus dissipating less heat. \$\endgroup\$
    – LShaver
    Dec 2, 2019 at 3:03
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    \$\begingroup\$ Could be something with the active PFC: it boosts unsmoothed rectified mains to >350 volts before the high frequency inverter. \$\endgroup\$
    – Oskar Skog
    Dec 2, 2019 at 6:50
  • \$\begingroup\$ Where did you get that image? \$\endgroup\$ Dec 3, 2019 at 8:31
  • \$\begingroup\$ @BruceAbbott Google. I forgot my search term though. Sorry for the bad resolution \$\endgroup\$
    – cyqsimon
    Dec 3, 2019 at 9:34
  • \$\begingroup\$ "It seems counter-intuitive" anything with higher current tends to be less efficient because of the induced currents and heating effects. Science isn't intuitive, so don't try to use it. \$\endgroup\$
    – UKMonkey
    Dec 3, 2019 at 11:50

3 Answers 3


As the law \$ P = U * I\$ , to acheive the same power at lower voltage, you need to increase the current.

In resistive components, like wires, pcb traces, transformer wire (green), losses increase to the square of the current, as \$P(loss) = R * I ^ 2\$.

In switching components and other diodes/rectifiers, (Green) the losses equal to \$ P(loss) = V(bandgap) * I \$. V is bound to the component regardless of the voltage input, like ~1V for a rectifier.

Eddy currents losses (Red) will also increase in any core as the current (and thus electromagnetic field) increases.

enter image description here

Losses related to capacitor leakage are negligible.

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    \$\begingroup\$ I remember learning this from high school physics. But are you sure this really applies? Power lines benefit from high voltage because it's essentially a very long resistor, whereas in the case of power supplies all the traces aren't that long so the resistance of the wires shouldn't be that significant. I'm doubtful that this is the correct answer. At least I don't think it tells the full story. \$\endgroup\$
    – cyqsimon
    Dec 2, 2019 at 7:39
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    \$\begingroup\$ The "full answer" is more complex than that and supplies can be optimized depending on the input but as a general simplification without going into a thesis in supply design, it is true. R here does not stand for resistance but impedance in general, the same rule applies for any bridge rectifier, transformer, the losses will increase with the increase of current. \$\endgroup\$
    – Damien
    Dec 2, 2019 at 7:44
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    \$\begingroup\$ @cyqsimon To convert AC to DC, you typically need diodes, diodes have a voltage drop of about 0.5, 0.5V of 120V is twice as much as 0.5V of 240V (Simplified example),there might be other non-linear components in a power supply \$\endgroup\$
    – Ferrybig
    Dec 2, 2019 at 13:48
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    \$\begingroup\$ @Ferrybig I'd take your statement as being more likely the culprit than this answer. \$\endgroup\$
    – horta
    Dec 2, 2019 at 23:37
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    \$\begingroup\$ I don't really understand this answer because couldnt you say: P = V^2 / R therefore "Losses increase to the square of the voltage"? \$\endgroup\$
    – Loocid
    Dec 4, 2019 at 6:35

As Oskar Skog proposed, the power factor corrector (PFC) is the main suspect.

The PFC is usually a precisely controlled boost converter that converts the pulsating rectified mains to something like 350-400 V. A boost converter's efficiency depends on the difference between the input and the output voltage - the more the input, the less it has to convert.

if someone is to build a power supply that only works on 115V, is it more difficult to achieve the same efficiency as one built only for 230V?

Generally, making a PSU that accepts a wider range of input is harder and leads to more compromises with other parameters (say, efficiency, weight, price).

To a lesser extent, using modern components and in the power range of the computer PSUs, making 230 V-only input is marginally easier and a bit more efficient than 115 V-only input.

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    \$\begingroup\$ "Boost converter's efficiency depends on the difference between the input and the output voltage" That is also an I^2 R effect though. (Assuming the PWM frequency of the converter is constant, which may not be a valid assumption) \$\endgroup\$
    – abb
    Dec 3, 2019 at 0:13
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    \$\begingroup\$ even before PFC: a full bridge rectifier will drop about 1.4V on the input. The lower the AC input voltage, the more will that drop reduce the efficiency. \$\endgroup\$ Dec 3, 2019 at 4:37
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    \$\begingroup\$ @Florian The rectifier drop could account for some 0.5% difference in the efficiency - and I am not sure how the Shottky diodes perform these days at these voltages. I know they are available. \$\endgroup\$
    – fraxinus
    Dec 3, 2019 at 7:45
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    \$\begingroup\$ @abb given that the whole input current passes the boost converter inductor, and the modern element base, probably I2R is the main component of the loses. Then again, you have switching loses, hysteresis loses in the core, dielectric loses in the capacitors - and depending on the design, in some marginal cases I2R may not dominate the heat output. \$\endgroup\$
    – fraxinus
    Dec 3, 2019 at 7:55
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    \$\begingroup\$ @Damien Well, you are right. I tried to calculate the dependency from the input voltage using the basic principles, but yes, at the end almost all loses in the PFC get I2R multiplier. \$\endgroup\$
    – fraxinus
    Dec 6, 2019 at 8:56

I'm surprised nobody has yet mentioned the forward voltage drop across a semiconductor junction. For a Silicon p-n junction, it's about 0.65 volts.

I'm not up to date with exactly how switched-mode power supplies work these days. They used to start with a bridge rectifier to convert AC mains to a high DC voltage. On 110V, 1.3 volts is lost in such a bridge rectifier (0.65V across each conducting diode), or just over 1%. On 230V, just over 0.5%.

There will be further losses in subsequent components. Power FETs are preferred where possible to bipolar transistors, because the irreducible voltage drop across them is less.


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