On Sun, 5 Jul 1998 19:15:06, -0500, JNVS44B@xxxxxxxxxxx (MR THOMAS N
CURLEE) wrote:
>A simple way to drop the 6V+ of the lead acid battery is to put 2
>power diodes in series with the battery. Each diode will give
>approximately .6V drop. The diodes cost less than $1.00 each US
>worst case.
>
>Tom
This works allright if you are not too demanding about regulating the
voltage. It is cheap ( my guess would be $ 0.20 per diode...).
Adding a 3-position switch would give you the choice between full
battery voltage, f.b.v minus 0.6, and f.b.v minus 1.2.
Make sure that the diodes are capable of handling a 3 amps current,
because the modeling lights of a T-28 twin set will take about that
(my battery is a 3.4 Ah Panasonic, which lasts for about one hour when
used continuously)
Since the full battery voltage changes with time you will never know
what exact voltages result from your diode system. If you want that
precision, you will have to use a low drop regulated stabilizer
instead.
BTW, the first real white LEDS (color temperature 8000K) have come on
the market. Rather expensive yet, but in the near future suited for
the perfect modeling lights, I guess. Maybe you can even use them for
shooting, once they get powerful enough....
Frank van Lindert.
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