Of course Prms = Vrms * Irms makes no sense if the two signals are out of phase but also if the wave forms are not identical. In my previous measuring the phase shift was close to 0 degrees however the 2 wave forms were not the same so it was still an incorrect measurement.
So I smoothened my HV circuit with an extra capacitor and then re-calculated the efficiency by using the average voltage and current. I now come to 83% which is very close to the 85% I measured about 6 months ago. So it all makes sense again :-)
Michel
on Jan 08, 2013, John Rehwinkel <[email protected]> wrote:
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> However, you can only use the average values if either the tube--
> voltage or tube current is constant; in my case this is not so. So I
> would say I need to use the RMS values of tube voltage and tube
> current. If I measure these with a scope, I come to 113V RMS and
> 3.73mA RMS. This would then result in a 96% efficiency!!!
>
> This measuring must be correct, right?
The measuring is probably correct (subject to the definition of RMS when largish crest
factors are involved). However, the math is invalid. To compute the correct power draw
with AC waveforms, you have to multiply the instantaneous voltage by the instantaneous
current moment by moment, and sum up all the products over at least one full cycle.
Even with plain old sine waves, just multiplying RMS voltage by RMS current can be
misleading, due to phase differences. That's the real difference between VA and watts.
- John
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- [neonixie-l] SMPSU efficiency Michel
- [neonixie-l] Re: SMPSU efficiency Jeff Thomas
- Re: [neonixie-l] SMPSU efficiency Michel van der Meij
- Re: [neonixie-l] SMPSU efficiency John Rehwinkel
- Re: [neonixie-l] SMPSU efficiency Michel van der Meij
- Re: [neonixie-l] SMPSU efficiency Michel van der Meij
