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