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Feb 4, 2014
Feb 4, 2014
Hi.someone may explain me why in a 55W tube t8 with a 0,55 A rms current ,100V lamp voltage,0,775 peak current (0,55 *sqrt2) has been calculated impedance in this way with that voltage ? follow image:

sshot-3pp.jpg i dont don't understand why 310V( that is about 230V sqrt 2 that is peak value dc after bridge diodes rectification voltage).should not be calculated with this:sshot-3ujpg.jpg that is sqrt2 *230/pi = 145 volts about rms.
at the half bridge output is like i had a sine wave generator 1st harmonic square wave that is a sine wave of above value so if lamp voltage is 100 v for example i should have 145 Vrms -100v/ 2times Ip current to calculate impedance .what am i doing wrong in my reasoning?
VCC in image is bus voltage.
ip = 1/2 peak current about in image above.clear is that voltage at bus minus voltage at lamp gives voltage at inductor that divided by current gives impedance.then going ahead he calculate this:
sshot-3kk.jpg .However my question is why that voltage 310 volts is to be considered.a snap of schematic , it is a classical half bridge .sshot-3oo.jpgtVm in advance.Henry.


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