LED tail strobe

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.

(Jay)
shared these priceless pearls of wisdom:

-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?

And as somebody else has pointed out, running a series string close to the
bottom limit of Vcc and hoping that a single resistor will provide constant
current to these devices will cook them when Vcc rises to the charging voltage.

No designer in his right mind would use a circuit in this manner.

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 Weir wrote:
(Jay)
shared these priceless pearls of wisdom:

-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?

And as somebody else has pointed out, running a series string close to the
bottom limit of Vcc and hoping that a single resistor will provide constant
current to these devices will cook them when Vcc rises to the charging voltage.

No designer in his right mind would use a circuit in this manner.

Jim



The criteria I had when laying out my circuit was that the lights would
operate down to 10V and still not exceed the max current limit at 15V.
I'm working from memory here, but I believe the SuperBright LEDs I used
had a forward voltage of 3.5V.

Putting 2 in series with a 220ohm, 1/4W resistor maximized the
efficiency while maintaining the range I wanted.

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

The criteria I had when laying out my circuit was that the lights would
operate down to 10V and still not exceed the max current limit at 15V.
I'm working from memory here, but I believe the SuperBright LEDs I used
had a forward voltage of 3.5V.

Putting 2 in series with a 220ohm, 1/4W resistor maximized the
efficiency while maintaining the range I wanted.


Okay, thats not too bad because your using a resistor to cover HALF the voltage
drop....of course your losing half your power in the resistor rather than
having it produce light in the LED...

My previous post was more concerned with folks both running the LED at absolute
max power ratings,,,,and trying to use the LED voltage drop to cover something
like 80 percent or more voltage range...thats when you have to be a little more
careful...

Your 1/4 watt resistor is NOT high powered enough....youll be running it close
to or past its maximum power rating for normal voltages....

Get something more like 1/2 watt or 1 watt resistors....or use 2,3,or or even 4
of the 1/4 watt resistors with appropriate resistance values in series to
spread the power load....resistors are cheap and there AINT no harm in using
ones that can handle significantly more power than expected....while the risk
to benefit ratio of trying to push their limits seems rather high....

Also make sure the resistors can cool effectively.....their power ratings dont
mean diddly if they are well insulated or packed into a tight space and can
warm up...

take care

Blll

On Wed, 28 Apr 2004 09:18:57 -0700, Jay wrote:

"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.

That wouldn't be true at threshold though. For example, if you moved the
LED away from the observer until it was only just visible then reduced
the brightness by half, it would become invisible and you'd have to move
it quite a long way back before the observer could see it again.

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.

Have you considered using a constant current regulator, instead of a
resistor? I believe there is an example circuit given in the LM337/317
datasheet showing how to build one using only the LM337 (normally used as
a voltage regulator) & one resistor.

It would need to be bolted to a heatsink, like the Luxeon Star LEDs, but
IIRC the LM337 and cousins also shut down if they overheat.

AC