Jeff,
some of this is semantics :
Jeff commented:
> Also the energy isn't really wasted when the extinction voltage is
> reached is it? It remains in the capacitors.
my original post:
> > This is increased wasted energy, unused at end of flash.
The energy remaining stored at end of discharge is not used (my original
expression: "unused") to
produce light, so only in that sense is wasted. It is not wased as in
dissipated, but it is unused
stored energy, not available to produce light.
Jeff alsio wrote:
> How about power = V**2 / R ?
>
> The voltage across the two series tubes is equal to the voltage across
> a single tube but the resistance has doubled(?) The current flow would
> be lower for two tubes in series ... so the resistance of each would
> be also be slightly less than for a single tube?
The tubes are extremely non-linear resistances with the resistance a function
of current and
plasma temperature. So you would need to integrate I(t)^2 R(t) over discharge,
with R a fn of I.
However, it is much easier to just look at start energy in cap and end energy
in cap and know that
since load is resistive all the change in energy goes to the tubes. (halve that
energy for each
individual tube).
So original 1 tube actual energy used :
0.5C*(330V)^2-0.5C*(60V)^2
and two tubes series (sum energy to both tubes) is a bit reduced:
0.5C*(330V)^2-0.5C*(120V)^2
Where C is flash capacitance.
(@60V extinction, this is ~10% reduction of total delivered energy)
There are some other subtleties to this, which affect this a little more. The
flash circuit has a
series inductor and this applies optimal waveforem for 1 tube but less than
optimal for two tubes
and this may affect energy a bit too. Losses in switch thyristor are reduced
with two tubes. My
guess is these are second order effects.
Jeff asked:
> Do most flashes at full power dump energy into the tube until it quits
> flowing or do they cut
> off the flow?
Normally they don't cut off flow, the flash tube conducts until the tube self
extinguishes at
something like ~60V.
Tim Hughes
--- Jeff Keller <jeffreyrkeller@xxxxxxxxx> wrote:
>
> How about power = V**2 / R ?
>
> The voltage across the two series tubes is equal to the voltage across
> a single tube but the resistance has doubled(?) The current flow would
> be lower for two tubes in series ... so the resistance of each would
> be also be slightly less than for a single tube?
>
> Also the energy isn't really wasted when the extinction voltage is
> reached is it? It remains in the capacitors. Do most flashes at full
> power dump energy into the tube until it quits flowing or do they cut
> off the flow? I remember seeing something about studio strobes color
> temperature changing slightly as the voltage drops ...
>
>
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