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