I'm making an AC to DC power supply that connects to mains. There's an AC to DC converter, buck converter and a smoothing LDO that outputs a DC controlled with pots.

I'm verifying the supply, and have been trying to get efficiency measurements. Due to limited equipment, I can only measure input Vrms, input Irms, output DC voltage, and output DC current. However, if I just do the Vdc * Idc/(Vrms * Irms) calculation, my effeciency goes above 100% at the higher loads, which is obviosuly wrong.

I know there's apparent vs real power when driving reactive loads with AC, but when I'm measuring the Vrms and Irms that goes into my circuit, that is the real power, is it not? Adding a power factor would mean my circuit becomes even more unphysically efficient.

Is there anything else that I haven't taken into account?

  • 2
    \$\begingroup\$ Vrms * Irms gives you appearant power, while Vdc * Idc is active power. You cannot tell efficiency from the quotient because appearant power also has phase shift and (for a rectifier mostly) current distortion in it. \$\endgroup\$
    – Janka
    Feb 3, 2018 at 1:39

1 Answer 1


The product of Irms and Vrms is not real power.

Imagine if the load was a perfect capacitor and input was a pure sine wave.

RMS voltage is mains voltage. RMS current is Vrms/Xc where Xc = \$\frac{1}{2 \pi f C}\$

Say 120VAC RMS and 100\$\mu\$F, so Xc is 1592\$\Omega\$ and Irms is 75.4mA, so we get 9W.

However real power is exactly zero.

To get input power you need to find the average of the instantaneous product of voltage and current. In other words, a wattmeter.

  • \$\begingroup\$ So measuring Irms and Vrms is really just measuring the apparent power? \$\endgroup\$
    – Michael E
    Feb 3, 2018 at 1:42
  • \$\begingroup\$ Yes, by definition, the product of the two is apparent power. \$\endgroup\$ Feb 3, 2018 at 1:45
  • \$\begingroup\$ So would I be able to get to the true power delivered from the data I have + an approximation of the reactive and resistive elements or would that be pretty inaccurate? \$\endgroup\$
    – Michael E
    Feb 3, 2018 at 1:51
  • \$\begingroup\$ Your input is most likely far from an ideal reactance. It's probably some spikey waveform (large crest factor). You might be able to use an oscillscope that has appropriate math functions but most are not easily used on mains voltage without danger. For cheap you could try a "Kill-a-Watt" consumer meter. \$\endgroup\$ Feb 3, 2018 at 1:58
  • \$\begingroup\$ I see, thanks! I guess the last question I have is, if my current and voltage inputs have some phase shift, that would just be a smaller number than the Irms*Vrms that I have. Then why is my efficiency still greater than 100%? If I use the data I have without accounting for power factor, it should just result in worse efficiency, right? \$\endgroup\$
    – Michael E
    Feb 3, 2018 at 2:14

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