The advantage of putting your emmiters in series (rather than
parallel) is that you pull less current. The mark to space ratio
people have been discussing isn't that small. I don't think it would
be a problem, but if you wanted to reduce the surge you could charge
and discharge the gate of that big FET more slowly thus having a
slower turn-on speed. You could do that by putting a high value
resistor in series with the 555 output.

The high voltage switching supplies that power those flash tubes can
draw progressively more current as they degrade over time.

anonymous coward wrote in message e...
Something I've been pondering...

4 Luxeon star 1W emmitters in parallel will draw 1A or so. Your 555 based
timer circuit will draw large(ish) currents for short periods of time,
because of the small mark:space ratio. Might this interfere with other
systems powered by the battery?

The switching regulators for standard Xenon flash tubes draw a lower
but much more constant current (though goodness knows they can generate
radio noise if they're not shielded right).

AC




On Sat, 17 Apr 2004 23:25:24 -0700, Jay wrote:

I think someone may have already pointed this out, and maybe I didn't
make it as clear as I should have... I stacked the forward drop of
MULTIPLE LEDs up until I got somewhere near the bottom end of the
supply voltage. So for the example I gave, I got to 4 LEDS in series.
Why waste all that power as long IR (heat) off a big resistor when we
want red and green light right?

Regarding 2.8V- The forward drop of these devices now-a-days is all
over the place. The new chemistries seem to be making higher forward
drops, plus the trend is to package multiple die into one larger
device and this can effect the forward drop of the composite device.

By the way, anyone building my circuit should try one instance of it
(4 LEDS and resistor) on your bench supply before you go fly at night
cross country.



Jim Weir wrote in message . ..
Before everybody in the Western Hemisphere blows a bucket full of light emitting
diodes, would you care to calculate the resistor one more time? And perhaps
post a retraction?


(Jay)
shared these priceless pearls of wisdom:

-You can find examples on how to power the LEDs on the manufacturer web
-site.
-
-Having said that...


So lets say the recommended current for
-the LED is 20mA. Ohms law is R=E/I, so that gives you a resistor
-value of .3V/.02A=15 ohms.


Um, no. Suppose the diode has a forward voltage drop of 2.8 volts (that's not a
common value, but I'll give it to you for argument.

Now the power supply (battery) is a 12 volt supply, but 14.2 volts at full
charge with the alternator going, so the drop across the series resistor is
going to be

14.2 minus 2.8 equals 11.4 volts, which is the voltage across the resistor.
This current limiting resistor is going to have 20 mA flowing through it, so Ohm
tells us that resistance equals voltage divided by current. In this case, 11.4
volts divided by 20 mA gives us a resistor of 570 ohms (560 is the nearest
standard value).

You put your calculated 15 ohm resistor in series with this diode and I
guarantee you that the SNAP you hear is the gallium aluminum arsenide
semiconductor of the diode being sacrificed on Ohm's altar.

I'm serious. You owe the newsgroup a correction before somebody takes your
error and blows up a whole bunch of LEDs.

Jim


Jim Weir (A&P/IA, CFI, & other good alphabet soup)
VP Eng RST Pres. Cyberchapter EAA Tech. Counselor
http://www.rst-engr.com