> >I discovered that a camera flash inverter will power
> >some length of EL-wire if you intercept it at the
> >HVAC part. While drawing more than an Amp from the
> >1.5V cell.  That's NOT efficient enough :-)
>
> Yeah, these converters are optimised to charge up a capacitor in a short
> time :)

Several people have commented similarly.
I know nothing about El light efficiency but, if the inverter is 50%
efficient end to end (which is quite a challenge at 1.5 volts or less at
that current level) then there is far less than a watt available for light.
If the EL wire is no more efficient than many LEDs or incandescent bulbs,
then say 3/4 of a watt is quite believable.

Consider a 3v LED at 20 mA = 60 mW.
750 mW/60 mW = about 12 LEDs. if these are modern high efficiency LEDS rated
at say 3000 - 6000 milli-Candella each then you'd get 'lots'* of light. Use
slightly older (or less efficient cheaper) ones and you could well get less
light than you see from "a length" of EL wire. Note that LED light is
focussed and wire may put out a substantially higher light energy level than
is obvious.

Given all the above, I suspect that an efficient supply may need less power
than the flash does BUT not as little as one may have expected. Perhaps 100
mW - 500 mW range ?


        Russell McMahon


* arcane engineering optical flux unit

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