I have a 48 V 65 W power supply plugged into my Kill A Watt meter.

I'm not so sure it is as efficient as the label suggests.

These are the calculations I have made:

$$S = V \cdot I$$ in VA the apparent power. $$232.6~V \cdot 0.25~A = 58.15~ VA$$

$$pf = P/S ~ [W]/[VA]$$


$$P = V\cdot I\cdot pf ~[W]$$

$$27.4~W/58.15~VA = 0.41~ [pf]$$

Does this sound like the efficiency of a power supply marked VI?

Whilst I realise pf is not efficiency, my understanding is that to be more efficient the pf has to be high (close to 1) and this looks like I have less than half?

I have seen that efficiency, of a class VI power supply, should be a minimum efficiency of 86 % and this doesn't look like its near this?

Am I wrong?

  • \$\begingroup\$ Below some power rating, which I cannot find right now, power supplies are not required to have a specified minimum power factor. \$\endgroup\$ Feb 16, 2017 at 20:00
  • \$\begingroup\$ To determine the efficiency at a given output current, apply a load (must represent a known output power- a fixed resistor will work if you measure the resistance when hot and the actual output voltage) to the power supply and use your cheapie power meter to measure the input power. Efficiency = Pout/Pin x 100%. It will be 0 at no load, will likely peak closer to the full rated DC load. \$\endgroup\$ Feb 16, 2017 at 21:17

3 Answers 3


The efficiency and the power factor are not related at all. The efficiency is the resistive input vs the output.

The power factor is most likely due to distortion in the current waveform and you have to be sure that your instruments measure RMS current and voltage and that the power meter can work with distorted waveforms.

Most quality switch-mode power supplies are power factor corrected as well and will not exhibit such a bad power factor. Even linear power supplies distort the current due to the input rectifier and capacitor.

  • \$\begingroup\$ Note that Kill A Watt is a registered trademark of an inexpensive instrument used to measure the power of plug-in appliances. I altered that and another part of your edit that appeared to change the intended meaning of the OP. \$\endgroup\$
    – user80875
    Feb 16, 2017 at 20:10
  • 1
    \$\begingroup\$ np. Watch out for cheap power meters as they only use zero crossing phase shift to determine the pf and not [W]/[VA] \$\endgroup\$
    – skvery
    Feb 16, 2017 at 20:15
  • \$\begingroup\$ My experience with the 120 V Kill A Watt P3 is that it measures real power and total pf reasonably accurately. \$\endgroup\$
    – user80875
    Feb 16, 2017 at 20:18
  • \$\begingroup\$ @user1831847 maybe you can provide a link to a meter that uses zero crossing to determine PF? Also, be very wary about altering questions like this. I see the alteration that you made and it was wrong although Charles has restored this. If a question states something like this and there is already an answer explaining the problem as originally put by the OP you must leave it alone or run the risk of getting flack. \$\endgroup\$
    – Andy aka
    Feb 16, 2017 at 21:43
  • \$\begingroup\$ @Andyaka I am sorry I lost the specific reference in my deleted emails but power-factor on Google gives about 600 000 references if you add zero-crossing and about 1 000 000 if you add RMS. I will be careful for flack. \$\endgroup\$
    – skvery
    Feb 16, 2017 at 21:50

My experience with the Kill A Watt meter is that the readings are reasonably accurate even with a distorted input current waveform. The current is total RMS including harmonics and the power factor is the total power factor as calculated from real power divided by voltage multiplied by total RMS amperes including harmonics.

To calculate efficiency, you need to measure the output DC current and calculate the output power based on rated output voltage multiplied by measured output current. Calculate efficiency by dividing the calculated output power by the input power measured with the Kill A Watt. That will provide a reasonably accurate efficiency. If the output power is less than the rated output power, you can expect the efficiency to be lower that it would be at rated output.

As explained in other answers, the power factor doesn't need to be considered since the Kill A Watt provides a real power measurement.


Power efficiency is normally measured at full load. You appear to be using a figure of 27.4 watts and this is nowhere near the 65 watts full power rating. Power efficiency has nothing to do with power factor and, to emphasise this, most electricity utilities only bill you on actual real power consumed. The actual load power figure could be 40 watts or so (given the description you have given) but I emphasise, efficiency should be determined at a full load power of 65 watts.


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