hi,
For explanation, let the Vbty = 10V, and the Cap = 10uF and the Cap charged voltage be 350Vdc
From I * t = C * V
Say, the 't' time to charge the cap is 1sec and the discharge time is 0.01sec.
Icharg= (C * V) /t
Let the eff=100% for the step up ratio of 350/10 = 35 times, assuming 350Vdc on the cap
So, I = (10^-5 * 350) / 1 = 350µA [flowing into the cap, allowing for the step up], the charge C on the cap is 10^-5 * 350
Let cap discharge time of the flash be 0.01sec
So now,
Idis = (10^-5 * 350)/ 0.01 = 0.35A
IF it were possible to create a flash directly from the battery, the battery current would have to supply [35 * 0.35A] for 10mSec...
which for a small dry battery would be impossible.
Xenon flash tubes
require a high voltage across the electrodes to generate a 'flash', which is triggered by a HV trig pulse.
Is this any clearer.?
Flashtube - Wikipedia, the free encyclopedia