LED tail strobe

I am building an LED tail light for my Lancair 360. It uses 6 ea 5
watt luxeon LEDs. I can strobe 1 amp through each of these which
should give enough light to satisfy the FAA regs. Runs surprisingly
cool. I have posted photos he

http://w1.lancair.net/pix/album01\

Its not quite finished, but I thought you might like to see my
progress.

cheers,

Jeff
N273CK stilll building.

ps thanks to Eric Jones for his posts on the Lancair Mail List, his
web pages on the subject and many emails and phone calls.

here is a more specific url...
http://w1.lancair.net/pix/led-strobe

-Jeff


(Jeff Peterson) wrote in message . com...
I am building an LED tail light for my Lancair 360. It uses 6 ea 5
watt luxeon LEDs. I can strobe 1 amp through each of these which
should give enough light to satisfy the FAA regs. Runs surprisingly
cool. I have posted photos he

http://w1.lancair.net/pix/album01\

Its not quite finished, but I thought you might like to see my
progress.

cheers,

Jeff
N273CK stilll building.

ps thanks to Eric Jones for his posts on the Lancair Mail List, his
web pages on the subject and many emails and phone calls.

try: http://www1.lancair.net/pix/led-strobe

--
Dan D.



..
"Jeff Peterson" wrote in message om...
here is a more specific url...
http://w1.lancair.net/pix/led-strobe

-Jeff


(Jeff Peterson) wrote in message . com...
I am building an LED tail light for my Lancair 360. It uses 6 ea 5
watt luxeon LEDs. I can strobe 1 amp through each of these which
should give enough light to satisfy the FAA regs. Runs surprisingly
cool. I have posted photos he

http://w1.lancair.net/pix/album01\

Its not quite finished, but I thought you might like to see my
progress.

cheers,

Jeff
N273CK stilll building.

ps thanks to Eric Jones for his posts on the Lancair Mail List, his
web pages on the subject and many emails and phone calls.

Jeff,
Would you consider sharing your design for the driver electonics?
Dean
Cozy MK4
BKV FL

"Jeff Peterson" wrote in message
om...
I am building an LED tail light for my Lancair 360. It uses 6 ea 5
watt luxeon LEDs. I can strobe 1 amp through each of these which
should give enough light to satisfy the FAA regs. Runs surprisingly
cool. I have posted photos he

http://w1.lancair.net/pix/album01\

Its not quite finished, but I thought you might like to see my
progress.

cheers,

Jeff
N273CK stilll building.

ps thanks to Eric Jones for his posts on the Lancair Mail List, his
web pages on the subject and many emails and phone calls.

Neither link will work for me????


"Dean Head" wrote in message
...
Jeff,
Would you consider sharing your design for the driver electonics?
Dean
Cozy MK4
BKV FL

"Jeff Peterson" wrote in message
om...
I am building an LED tail light for my Lancair 360. It uses 6 ea 5
watt luxeon LEDs. I can strobe 1 amp through each of these which
should give enough light to satisfy the FAA regs. Runs surprisingly
cool. I have posted photos he

http://w1.lancair.net/pix/album01\

Its not quite finished, but I thought you might like to see my
progress.

cheers,

Jeff
N273CK stilll building.

ps thanks to Eric Jones for his posts on the Lancair Mail List, his
web pages on the subject and many emails and phone calls.

sure...I haven't designed that yet, but when I do I will post it...Jeff

"Dean Head" wrote in message .. .
Jeff,
Would you consider sharing your design for the driver electonics?
Dean
Cozy MK4
BKV FL

You can find examples on how to power the LEDs on the manufacturer web
site.

Having said that...

What is typically done is the LEDs are just put in series with a
current limiting resistor. This forms a circuit akin to a kind of
voltage regulator called a "zener regulator". LEDs have a fixed
forward voltage for the recommended drive current, say for example its
2.8 volts for a green LED. Divide the power supply (e.g. 12V) by the
forward voltage of the LEDs and drop any fraction (12/2.8=4.3 make
that 4 even) Put those in series with a current limiting resistor that
will drop the fraction (.3V). 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. Check power to make sure it won't over
heat (P=IV) so thats .3V*.02A=.006W so a typicial 1/4 watt resistor is
fine. You must of course have some kind of resistor in series to limit
the current. If you math works out that you need no resistor, put one
less LED in series and then recalculate the limiting resistor. Put
the LEDs and resistor in series (in any sequence) observing the proper
polarity of the LEDS. If you hook up your entire string backwards, no
harm will be done, but if you happen to solder one LED backwards, it
will likely be toasted on power up.

Need more than 4 LEDS? Replicate this circuit in parallel as many
times as you need to get the luminous flux you need.

Of course the numbers (Vf, If) used here are for the older style
single chip LEDS. The parts that are getting everybody excited these
days are the multi-chip variety whose forward voltage and current will
vary alot from my example.

"Dean Head" wrote in message .. .
Jeff,
Would you consider sharing your design for the driver electonics?
Dean
Cozy MK4
BKV FL

"Jeff Peterson" wrote in message
om...
I am building an LED tail light for my Lancair 360. It uses 6 ea 5
watt luxeon LEDs. I can strobe 1 amp through each of these which
should give enough light to satisfy the FAA regs. Runs surprisingly
cool. I have posted photos he

http://w1.lancair.net/pix/album01\

Its not quite finished, but I thought you might like to see my
progress.

cheers,

Jeff
N273CK stilll building.

ps thanks to Eric Jones for his posts on the Lancair Mail List, his
web pages on the subject and many emails and phone calls.

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

Jim, he put 4 LEDs in series, each with it's '2.8' volt drop, and had only the remaining voltage to drop across. It does
look incorrect, however, in the voltage to drop across. 4 x 2.8 is 11.2. While-running voltage in 12 volt system is 14.2
(to use your number) leaves 3 volts to drop across....

--
Dan D.



..
"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

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