LED tail strobe

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

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

anonymous coward wrote:
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


So? Stick a large capacitor and a low-value, one watt resister in series
with the circuit. Still cost less that $5(US) at the RatShack.

--
http://www.ernest.isa-geek.org/
"Ignorance is mankinds normal state,
alleviated by information and experience."
Veeduber

On Thu, 22 Apr 2004 01:14:13 +0000, Ernest Christley wrote:

anonymous coward wrote:
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


So? Stick a large capacitor and a low-value, one watt resister in series
with the circuit. Still cost less that $5(US) at the RatShack.

No reason why not, but that would have to be one whopping capacitor.

e.g. 4V voltage drop over 1/5 second @ 1A = 50,000 microFarads (OK the
battery will still be contributing some juice).

The aircraft on your website is truly neat.

AC

anonymous coward wrote:


The aircraft on your website is truly neat.

AC

That it is, nameless one.

Looks like georgous work, doesn't it.


Richard

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

(Jay) wrote in message . com...
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 white 5 watt luxeons have 6.5v forward voltage at 100 ma, rising
to 7.5 v at 1 amp. so you cant put these in series if you have a 14
volt system.

a bypass capacitor or emi filter at the circuit should eliminate emi
due to the rapid switching of the FET. i think this is a better
solution than
slowing the swtich.

i will try to post the circuit later this week.

-Jeff

(Jeff Peterson) wrote in message . com...
the white 5 watt luxeons have 6.5v forward voltage at 100 ma, rising
to 7.5 v at 1 amp. so you cant put these in series if you have a 14
volt system.

Crud, thats the worse case, oh well. Are the white leds the most
efficient? If you can use red, you might be better off. I think
those guys make white by hitting a phosphor suspended in the epoxy
encapsulant with UV which emits photons in a broader but less
energetic range. There has got to be some conversion loss.

I was talking with a buddy that makes big LED signs and we realized
that the rated life of LEDs is at the half power point. So if you
expect the full 100,000 hr life, you need to start off with an array
that exceeds the spec by 2x at time zero. But come to think of it,
that will never happen on a GA aircraft.




a bypass capacitor or emi filter at the circuit should eliminate emi
due to the rapid switching of the FET. i think this is a better
solution than
slowing the swtich.

i will try to post the circuit later this week.

-Jeff

--
Dan D.
http://www.ameritech.net/users/ddevillers/start.html


..
"Jay" wrote in message om...
(Jeff Peterson) wrote in message . com...
the white 5 watt luxeons have 6.5v forward voltage at 100 ma, rising
to 7.5 v at 1 amp. so you cant put these in series if you have a 14
volt system.

Crud, thats the worse case, oh well. Are the white leds the most
efficient? If you can use red, you might be better off. I think
those guys make white by hitting a phosphor suspended in the epoxy
encapsulant with UV which emits photons in a broader but less
energetic range. There has got to be some conversion loss.

I was talking with a buddy that makes big LED signs and we realized
that the rated life of LEDs is at the half power point. So if you
expect the full 100,000 hr life, you need to start off with an array
that exceeds the spec by 2x at time zero. But come to think of it,
that will never happen on a GA aircraft.


Oh darn, you mean I'll only get about 50,000 hours???? or...does it mean that the light output at 100,000 will be 1/2
the original level?




a bypass capacitor or emi filter at the circuit should eliminate emi
due to the rapid switching of the FET. i think this is a better
solution than
slowing the swtich.

i will try to post the circuit later this week.

-Jeff

"Blueskies" wrote in message . com...

Oh darn, you mean I'll only get about 50,000 hours???? or

Nope

...does it mean that the light output at 100,000 will be 1/2
the original level?

Ya that is right, thats how I understand it. Its the 3dB point. At
that time it won't appear to be half as bright to the human eye
because THAT sensor has a log response to give us all fantastic
dynamic range. Measured with a calibrated instrument, you'd see half
the output.

Theoretically, if you exactly met the spec at time zero, in 1 hour
you'd be under spec all other conditions being equal.