Power efficiency measurements

you must remember power factor aswell.......i presume this is a buck design..............

the easiest way to do it is to get the average value of your DC bus....(after the rectifier)......then you get the average value of the FET current........then you multiply these and that gives you your input power........you then calculate output power = V * I..........efficiency is then Pout/Pin *100%

It is a buck converter, input voltage 230VAC, output voltage 5VDC 100mA; MDCM operation. The input voltage is rectified by a half wave rectifier as in the example circuit in the LNK304 data-sheet. I measured the input DC current (before the rectifier) through an analog ammeter getting a reading of 1mA (without output load).
The input current flows only in one direction (due to the half-wave rectifier).
Let's suppose a power factor of 1, in such conditions I could compute the input power as I * V/2 where:
I = input DC pulsating current (before the rectifier)
V = input RMS voltage

P = 0.001 * 230/2 = 0.11W

I divided the input RMS voltage by 2 as the rectifier only allows current flowing during one half of the sine wave.
Is this correct?

…..ok, I think I must re-do my answer to you……………one quite cheap and quick way to get the efficiency measurement is to buy one of those cheap domestic power meters which cost some 20 dollars…they also show your power factor…

…but if you want to calculate it as well….we can get a good scope with a sample and hold…………….you must get the input current waveform up on the scope….the quickest way to do this is to just put a series resistor in the “hot” line and look at the voltage across it…then work out the corresponding current waveform by OHM’s law……………….use the scope to do a sample-and-hold of the input current waveform……………….Now when you have a few cycles of the current waveform “frozen” on the scope (dont forget a complete cycle or period of the current waveform is 20ms long) get the vertical cursors 20ms apart and place them at the start and finish of one whole current cycle………………..then push the button labelled “FFT” and get the fundamental sine of this waveform…..its period will be 20ms and you need to get its amplitude………………this is the only bit of the input current that contributes to real power………find the rms of this fundamental sine with rms = 0.707 * Peak…….then you have the RMS of the input current that you can use in the REAL input power calculation……-but first you must find the phase of this fundamental current with respect to the input mains voltage………call this phase “theta”……get the input voltage rms by 0.707 * peak voltage……….then multiply Vrms * I rms * cos(theta) to get the REAL input power…………find output power as I said before…..-then [P(out)/ P(in)] * 100% = efficiency…………………….hopefully you will be able to find the phase of the fundamental input current by hoping that your scope will agree to display the fundamental sine current along with the input voltage sine…then you can read off the phase difference…………………………………..however, I would be doubtful if such a scope exists……………..though I could be wrong here ?………………………………………………….This method is rather complex, and so therefore I would be tempted to just use my previous method, -even though it does not take into account the losses in the half bridge rectifier………..nor does it take into account the power loss in the input “hot” line thermistor………………nor any losses in the EMC filter……………………at light load therefore, my method will be very inaccurate……………….in fact the lower the power output of your converter , the more inaccurate my simple (previous post) method will be……………your “divide by two” method I think will give you a bad result……with rms values you cant just divide by two just because its half-wave and only conducting for half of the mains cycle……………………………………doing an accurate calculation of the efficiency of an SMPS is not trivial……………….in fact it’s a closely guarded industrial secret and company design engineers have their methods squirreled away secretly on their hard drives, so nobody (-especially ‘n00b’ junior engineers) can find it !!…….

There is a pretty good video on measuring efficiency on the Apps TV section. Visit
http://www.powerint.com/en/design-support/appstv
Click the Apps lab tip tab (on the right of the video display window. Then select the measuring Efficiency video.

Thanks Treez, I understood that the simplest way is to place a digital power meter to the input of my regulator... it will do all the dirty computing on input waveform and give me a clear reading on power consumption and power factor.
However your discussion about the problem has been very stimulating and interesting. Thank you.

I found some interesting readings,

power measuring for non-linear loads:
http://www.enetsystems.com/~lorenzo/pmeter/PFCHarmonics.pdf

power in nonsinusoidal systems:
http://ecee.colorado.edu/~pwrelect/book/slides/Ch15slide.pdf