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Greetings from Norfolk UK
The energy stored in a Capacitor is given by the
formulae
1/2 C V V (a
half C (in Farads) times Voltage squared
e.g a 500uF capacitor, charged to 300 V would
store 0.5 x 500 x 10^-6 x 300 x 300 Joules = 22.5 Joules
The same capacitor charged to 50 V would store
0.625 Joules
Thus a flash which fired with the capacitor (500
uF) initially charged to 300 V and the xeon tube discharged the capacitor to 50
V would 'use' 22.5 - 0.625 Joules = 21.875 Joules.
This is the enrergy removed from the capacitor,
electrical losses in the connections between the capacitor and the tube, and the
energy causing the tube to heat up (a certain consequence of passing the
capacitor current through the tube) must be deducted from this figure
These will necessarilly be small, but in a poorly designed unit could be worthy
of note.
Remember, immediately following the striking of the
tube the current from the capacitor can be very large, the tube has a typical
'struck' voltage of around 60 V, and the capacitor voltage will initially be (in
the example) 300 V), the resistance of the wireing is going to significantly
determine the peak current - i.e. I = ( 300 - 60)/R; here R is the
resistance of the wirening. The use of delay line techniques will
significantly modify the peak current, extending the pulse duration, and
reducing the peak current, but will also increase, in a small degree, the losses
in the wireing.
Transient effects will modify these figures, but
that is getting well beyond the area of this note.
Richard.
r.wrigley@xxxxxxxxxxxxxx "I have yet to see any problem, however
complicated, which when looked at in
the right way, did not become still more complicated" Poul Anderson |
