thanks medved, but this system drav 160W!

I know about ballast losses, but i'm bit scared about tube loading.

A questions:

- How did you calculate the power?

- What instrument did you used to measure the arc voltage? Was it the true-rms meter?

Meter problem:

Low cost meters measure the "rectified average" and then multiply the result by a correction factor so, the reading of the sine wave voltage match the rms value:

Reading = average(abd(Vin))*1.11

But with non-sinewave shape the reading of this meter is incorrect, with the rectangular shape of the discharge arc voltage it mean the "1.11" factor, so the real arc voltage would be 130/1.11=117V.

Power calculation:

When calculating power, the only correct way is multiply the current and voltage waveforms (multiply each time point separately, so you get the waveform of the power transfer) and average the result.

This method is rather impractical for everyday use with resistive circuits with only in phase sinewaves of the base frequency, so the voltage and current single number values were defined (= the rms values) to allow simple calculation of "P=Vrms*Arms".

But as this simplification is designed only for sinewave in phase, it give the correct result only with sinewaves in phase (well, the result remain correct with any shape, when the voltage and current waveform shape exactly match and are exactly in phase), so only for pure linear, noninercial resistive loads.

But the arc is not a linear load: Current fed by the ballast is nearly sinewave, but the voltage is nearly rectangular. So when you use the "P=Vrms*Irms" equation, the result would be higher by the factor of 1.11 (you may do the integration math; for exact sine and an exact rectangle it would be 2*sqrt(2)/Pi).

This shape mismatch correction factor (better say it's inverted value) is then called "power factor" and it is unity only when both shape and phase match (not the case with nonlinear component like discharge, not the case for inertial components like L or C).

So with the discharge fed by sinewave current the real power would be (I use your case):

Plamp = Irms * Vrms * 0.9 = 1.25 * 117 * 0.9 = 131W

And that is not as much off the rated power...