Nice site!
I saw your explanation for the energy dissipated in a flash tube:
My understanding of wattage ratings for these tubes is measured in "Joules".
And according to the "ARRL Handbook", a Joule is one watt-second.
That is correct.
Watt Seconds = (Volt2) X (Capitance in Farads)/2
This is the energy in a given capacitor.
However the Capacitor in a Strobe circuit doesn't discharge to Zero Volts.
The Minimun voltage I measure is around 110 volts.
Therefore, assuming a slow flash rate and achieving a charge voltage of 340 Volts
with a capacitance of 160 Mfd, (.00016 Farads) I come up with a wattage of:
340 - 110 = 230 volts drop. 230 times 230 = 52,900
52,900 times .00016 = 8.464 8.464/2 = 4.232 Watts Dissipation
This is not correct. You need to think in energy as stored in the capacitor.
The capacitor starts out at:
340x340 x .00016 = 18.496, / 2 = 9.248 Joule
And it ends up containing:
110x110 x .00016 = 1.936, / 2 = 0.968 Joule
Thus, when the C discharges, 8.28 joules 'disappear', and not the 4.232 that
your calculation shows.
52,900 times .00016 = 8.464
8.464/2 = 4.232 Watts Dissipation
Now you could think about losses in the capacitor. That is energy that will
not reach the tube.
Another way of going about, if we think most of the losses are in the capacitor,
The voltage at which this charge is dissipated is the guessed 160 volts.
0.0368 C * 160 V = 5.888 J.
Now, 1 million flashes is quite a lot.
Good luck with the site! and a happy new year from Amsterdam,
Thomas Tonino
and inductance does not play a role. This will be a lower limit on the energy:
Let's say the voltage on the tube if it is conducting is 160 volts. The C goes from
340 to 110 volts and is 0.00016 F. That is a charge of (340-110) * 0.00016 F = 0.0368 Coulomb.
1 coulomb is the charge represented by 1 ampere during 1 second.
A capacitor of 1 F charged to 1 V holds a 1 C charge.
Not really, as you and at 110, but it must be higher that that.