SMPSU efficiency

[Michel]

Hello Folks,

While experimenting with an IN-13 bargraph tube I thought it would be
interesting to connect it up to my nixie watch HV supply and see how
far the bar would go. Since the voltage of the tube is lower than the
nixie tubes, the nixies won't light up when the bar graph tube is
wired parallel to the HV (as expected). This means all the current
from the HV supply goes through this bar graph tube and make it very
easy to measure the HV efficiency.

The drawn battery current is 130mA; battery voltage 3.37V. The average
HV supply is 112V and average tube current 3.1mA this would result in
a 79% efficiency.

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?

Michel

[Jeff Thomas]

Hi Michel, if you're using a SEPIC for HV generation, then the calculated efficiency is plausible.
A transformer design could only achieve 80%, or perhaps a little better well tuned.

Regards, Jeff

[Michel van der Meij]

Thanks for the confirmation Jeff!
Even if somewhere between 79% and 96% would still be a good result!

Michel

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[John Rehwinkel]

> 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

[Michel van der Meij]

Absolutely correct John, it would only make sense (in my case) if there was no phase shift between voltage and current. Something I haven't measured yet....

Michel

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[Michel van der Meij]

OK, I got a definite answer.

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 <jre...@mac.com> wrote: