LED tail strobe [Archive] - AviationBanter

Jay

April 17th 04, 06:15 AM

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 here:
> >
> > 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.

Jim Weir

April 17th 04, 07:00 AM

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

Blueskies

April 17th 04, 11:38 AM

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

GeorgeB

April 17th 04, 01:34 PM

On 16 Apr 2004 22:15:44 -0700, (Jay) wrote:

>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).

Your concept is reasonable, but there are some significant problems.
The forward voltage varies with temperature, and the power supply
LIKELY varies over time. As you have it, if the fwd voltage dropped
0.1v, and the supply did not change, you would have 0.7 vs 0.3 across
the resistor, for over 2x the current ... maybe a real problem.

Now, let's have the alternator charging the battery, and have it at
15.5V or so ... now I have 3.8 volts across that current determining
resistor ... 12 times the "design". OUCH.

Now let's have the alternator fail, the battery voltage drop to 10.5V.
Your series string will draw no current and give no light ... and you
are in an emerency situation that is exactly when someone needs to see
you.

What is the solution ...

There are "constant current" devices. I have used them, and they
work allowing operating this string with probably 3 LEDs over the
range at visually constant brightness.

You can put 2 in a string allowing the current determining resistor to
allow a power range varying over about 2:1, 10V to 15V.

You can put 3 in a string like above, power will vary over about 4:1.

You can design a pulse system turning the LED on for (maybe) 0.1ms
then off for maybe 5ms and PROBALBY not overdrive (into damage) the
LED and put "as many" as you want in parallel. The driver will likely
be a FET.

>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.

I was taught to allow about half the voltage for the resistor, half
for the LED string unless I had current control. In the "old days",
for current control we used an emitter resistor in a common emitter
circuit, 2 or 3 diodes to set bias (single vs darlington), and the
LEDs between collector and V+. There are other (better) ways, but
everyone understood this one.

>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.

I disagree that there will be damage with any in backwards. The
reverse voltage will almost certainly be higher than the forward
voltage, so there won't be any current drawn. If there is, you still
would have less than correctly wired.

>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 here:
>> >
>> > 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.

Jay

April 18th 04, 07:15 AM

Read comments below...
GeorgeB > wrote in message
>...
> Your concept is reasonable, but there are some significant problems.
> The forward voltage varies with temperature,

Indeed it does, diodes are often used as the sense element for
temperature sensing, but it's a very small change, and you need to
amplify it when you do. Also, the self heating of the device will
swamp ambient temp effects I should guess.

> and the power supply
> LIKELY varies over time. As you have it, if the fwd voltage dropped
> 0.1v, and the supply did not change, you would have 0.7 vs 0.3 across
> the resistor, for over 2x the current ... maybe a real problem.

Thats what makes it a self regulating circuit, as the current comes
up, the foward drop of the diodes go up as well, thus reducing the
drop across the resistor.

> Now, let's have the alternator charging the battery, and have it at
> 15.5V or so ... now I have 3.8 volts across that current determining
> resistor ... 12 times the "design". OUCH.

Same thing as before...

> Now let's have the alternator fail, the battery voltage drop to 10.5V.
> Your series string will draw no current and give no light ... and you
> are in an emerency situation that is exactly when someone needs to see
> you.

You can handle this case by dropping one diode off the string and
recalculating the resistor as before. Cuts your efficiency a little
but hey, some poeple drop over half the power delivered as heat into
their "current limiting device" for 28V applications.

> What is the solution ...
>
> There are "constant current" devices. I have used them, and they
> work allowing operating this string with probably 3 LEDs over the
> range at visually constant brightness.

You can do that, but for driving LEDs, since they form a nice self
regulating circuit with a single resistor I didn't feel it was
neccesary. Please share with the group which part you've had success
with as a "constant current" device.

> You can design a pulse system turning the LED on for (maybe) 0.1ms
> then off for maybe 5ms and PROBALBY not overdrive (into damage) the
> LED and put "as many" as you want in parallel. The driver will likely
> be a FET.

You could do this also but each LED would need its own current
limiting resistor in series because the forward drop of the LEDs vary
from part to part and with temperature as you've mentioned and the one
with the lowest drop would eat the most power without those resistors.
But again, with a pulse width modulation circuit, why so complex?

> I was taught to allow about half the voltage for the resistor, half
> for the LED string unless I had current control. In the "old days",
> for current control we used an emitter resistor in a common emitter
> circuit, 2 or 3 diodes to set bias (single vs darlington), and the
> LEDs between collector and V+. There are other (better) ways, but
> everyone understood this one.

So you're building a constant current supply from each group of 2 or 3
LEDS, thats pretty complex if a single resistor will work. What
you're suggesting is too complex for the average guy and I see no
practical benefit. Don't light bulbs vary in brightness with supply
voltage? Sure they do, and they vary more than the single resistor
method I've sketched out.

> >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.
>
> I disagree that there will be damage with any in backwards. The
> reverse voltage will almost certainly be higher than the forward
> voltage, so there won't be any current drawn. If there is, you still
> would have less than correctly wired.

The reverse drop on the LEDS will be the supply divided by the number
of diodes. 12/4=3V. Last data sheet I looked at said the reverse
voltage limit was 5V. Thats why I also said that if you put one
backwards it will cook. It will see the full 12V.

What I've outlined is a simple method to build LEDs lights. Yes, you
could build a constant current supply, and the LEDS would see the
exact same current from 10V to 15V but your light bulbs will vary in
brightness (acnd color) over that range anyway more that my suggest
circuit due to the self limiting nature of a diode(s) in series with a
resistor.

Jay

April 18th 04, 07:25 AM

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

TaxSrv

April 19th 04, 03:24 AM

"GeorgeB" > wrote in message

> Now, let's have the alternator charging the battery, and have it at
> 15.5V or so ... now I have 3.8 volts across that current determining
> resistor ... 12 times the "design". OUCH.
>
15.5V is not a normal charging voltage, and if sustained will ruin the
battery. It would only be a fault condition with a regulator gone
bad. I think you can safely calculate dropping resistors based on
14.5V max.

Fred F.

Blueskies

April 20th 04, 10:08 PM

That is a good point...bad regulator putting out high voltage. How high can the alternator go with a runaway regulator?

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

..
"TaxSrv" > wrote in message ...
> "GeorgeB" > wrote in message
>
> > Now, let's have the alternator charging the battery, and have it at
> > 15.5V or so ... now I have 3.8 volts across that current determining
> > resistor ... 12 times the "design". OUCH.
> >
> 15.5V is not a normal charging voltage, and if sustained will ruin the
> battery. It would only be a fault condition with a regulator gone
> bad. I think you can safely calculate dropping resistors based on
> 14.5V max.
>
> Fred F.
>
>
>

anonymous coward

April 21st 04, 10:24 PM

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

Ernest Christley

April 22nd 04, 02:14 AM

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

John

April 22nd 04, 03:50 AM

At full field and normal cruise RPM with low load the alternator will
put out about 90 volts DC. This is enough to fry all of your
electronic goodies!
This is how you convert your 12 volt auto alternator to power your 110
volt electric drill with the flip of a switch.

On Tue, 20 Apr 2004 21:08:07 GMT, "Blueskies" > wrote:

>That is a good point...bad regulator putting out high voltage. How high can the alternator go with a runaway regulator?

anonymous coward

April 23rd 04, 12:53 AM

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

Jay

April 23rd 04, 01:53 AM

Hopefully breakers and/or fuses pop before your electronics do.

(John) wrote in message >...
> At full field and normal cruise RPM with low load the alternator will
> put out about 90 volts DC. This is enough to fry all of your
> electronic goodies!
> This is how you convert your 12 volt auto alternator to power your 110
> volt electric drill with the flip of a switch.
>
> On Tue, 20 Apr 2004 21:08:07 GMT, "Blueskies" > wrote:
>
> >That is a good point...bad regulator putting out high voltage. How high can the alternator go with a runaway regulator?

Richard Lamb

April 23rd 04, 02:30 AM

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

Jay

April 25th 04, 03:14 AM

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 >...
> 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

Jeff Peterson

April 26th 04, 12:08 AM

(Jay) wrote in message >...
> 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

Jay

April 26th 04, 05:32 PM

(Jeff Peterson) wrote in message >...
> 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

Blueskies

April 27th 04, 02:16 AM

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

..
"Jay" > wrote in message om...
> (Jeff Peterson) wrote in message >...
> > 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

Jay

April 28th 04, 05:18 PM

"Blueskies" > wrote in message >...

> 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.

Jim Weir

May 1st 04, 12:18 AM

(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

Ernest Christley

May 1st 04, 02:53 PM

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

BllFs6

May 1st 04, 04:04 PM

>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

anonymous coward

May 5th 04, 01:18 AM

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

> "Blueskies" > wrote in message >...
>
>> 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.

anonymous coward

May 5th 04, 01:23 AM

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

Ernest Christley

May 5th 04, 03:46 AM

anonymous coward wrote:
> 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
>

Yes. I considered it. I opted for the simplicity of a single current
limiting resistor and the constant voltage regulator that is already
there. Number one rule of fault management. If it ain't there, there's
no way to break it. Regulators not only add an additional active
component with its list of failure modes, it also adds severl solder
connections and more heat, making it even more difficult to rig the
system into a 1/4" piece of plexiglass.

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

Jay

May 6th 04, 05:17 AM

Jim Weir > wrote in message
> 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 more the supply varies the less effective the technique is. I
picked the simple resistor design with certain assumptions. When it
comes down to it, the voltage doesn't vary all that much on a properly
operating electrical system. That big 'ole battery is like an anchor
on the supply bus. Over voltage condition blows the breaker.

What if the alternator quits? Well how far down do you want to go on
the discharge? You could get a range of zero to 14 volts if you
consider a discharged battery, thats a pretty tough range to design
for isn't it? I can think of a lot of systems that will malfunction
when supplied from a discharged battery. Even normal lights will go
out of spec on a discharged battery.

anonymous coward

May 11th 04, 09:01 AM

On Wed, 05 May 2004 02:46:06 +0000, Ernest Christley wrote:

> anonymous coward wrote:
>> 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
>>
>
> Yes. I considered it. I opted for the simplicity of a single current
> limiting resistor and the constant voltage regulator that is already
> there. Number one rule of fault management. If it ain't there, there's
> no way to break it. Regulators not only add an additional active
> component with its list of failure modes, it also adds severl solder
> connections and more heat,

A regulator will produce no more heat than a resistor passing the same
current with the same voltage drop. The ones I'm thinking of are 'in-line'
devices, so their (negligible) supply current does not have to be factored
in as an extra source of heat.

I buy what you're saying about complexity, but you would probably only
need a single current regulator + one resistor for each parallel bank of
series LEDs (eugh, but I can't think how better to put it).

BTW, do you need to use zener diodes / transorbs with aircraft power
systems as you do in cars, to avoid problems with voltage spikes?

AC

> making it even more difficult to rig the system into a 1/4" piece of plexiglass.

Ernest Christley

May 11th 04, 06:04 PM

anonymous coward wrote:

> BTW, do you need to use zener diodes / transorbs with aircraft power
> systems as you do in cars, to avoid problems with voltage spikes?
>
> AC
>

In one form or another, I'd say the answer is a qualified yes. Again, I
didn't add any sort of regulation to the LED array. As I remember it,
and it has been quite a while since I looked at it, the larger LEDs can
absorb rather large transients themselves. Their construction is not
that far removed from zeners or transorbs after all.

As I understand it, the limiting factor of the LED's ability to suck
down transients is their ability to dump the internal heat is the reason
why 'overdriving' them with higher but pulsating current works. You can
drive more current through them, just not for very long. In my opinion
(vs me sitting down and running actual numbers which will only happen
when I'm ready to build my own prodution models), if you derate the LED
to about 80%, it'll be able to take whatever a barely functioning
regulator will ever throw at it. Anything more is overdesign, adds
complexity, and one more solder joint is just one more chance for me to
screw something up.

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

Jeff Peterson

May 11th 04, 09:27 PM

I posted new photos of the LED tail strobe, now with the circuit
completed, at
http://w1.lancair.net/pix/led-strobe

i also posted the schematic.

the particular component choices give a 1 second period and
50% duty cycle. tinker with values to get your favorite flash pattern

i used monolithic ceramic capacitors because they are small, but
they do have a bit of variattion of C with temp. might be
better to use something more stable. as the circuit warms it flashes
faster.

The circuit works quite well, but after some soul-searching I have
tenatively decided NOT to put the system on the LNC 360.

reasons:
-its heavy...235 grams. Installing the strobe will require more than 3
times its mass added to the rudder counterweight. perhaps a kilogram
total. all this at the maximum rearward moment arm. yuck.

-it gets quite warm. after a minute or two the temp is maybe 40C at
the
base of the LEDs and thats with a fan pushing air past at about 10
MPH. in still air it overheats. if i forget to turn it off on
landing,
I could possibly slag my rudder.

-the system satisfys the FAA Regs, but with no margin...I prefer a
safety factor 2.

-its a complicated system. lots of machining. tight fits. odd angles.
no finesse.

it just doesnt seem like a bulletproof system to me, so i may just
chalk
this one up to experience and develop plan B.

Plan B: get a bunch of 5 mm diameter superbright white leds and
mill the hemispherical ends off the plastic cases. this will give
them a
lambertian pattern. no need to polish the ends, might even sandblast
them
to make the pattern even broader.

glue these into holes drilled in the curved trailing
edge of the lancair wingtip. i will need many more of these LEDS than
with
luxeons, but that will help distribute the thermal dissipation.

the weight might actually be lower this way than with luxeons since
the
aluminum heat sink will no longer be needed. and the weight will be
near the center of lift, and will not require a counterweight.

or maybe i will come up with plan C...

-Jeff

John

May 12th 04, 02:11 AM

A linear regulator will produce the SAME amount of heat as a resistor
at the same current. W=VI If the volts and the amps are the same
the power is the same. However since the car mfg's are installing
LED's in the center brake light they have paid for the development of
simple constant current switching power supply IC's to drive the
LED's. The power dissipated in the LED is the same however the power
that was dissipated in the current limit ballast resistor has been
greatly reduced so that the TOTAL system power and heat generated is
about 50% of what it was to get the same light out of the LED using
resistors.
You have to add a small inductor and since it is a switcher you may
have radio interference if your layout and EMI filter is not good.
John

On Tue, 11 May 2004 09:01:23 +0100, anonymous coward
> wrote:

>On Wed, 05 May 2004 02:46:06 +0000, Ernest Christley wrote:
>
>> anonymous coward wrote:
>>> 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
>>>
>>
>> Yes. I considered it. I opted for the simplicity of a single current
>> limiting resistor and the constant voltage regulator that is already
>> there. Number one rule of fault management. If it ain't there, there's
>> no way to break it. Regulators not only add an additional active
>> component with its list of failure modes, it also adds severl solder
>> connections and more heat,
>
>A regulator will produce no more heat than a resistor passing the same
>current with the same voltage drop. The ones I'm thinking of are 'in-line'
>devices, so their (negligible) supply current does not have to be factored
>in as an extra source of heat.
>
>I buy what you're saying about complexity, but you would probably only
>need a single current regulator + one resistor for each parallel bank of
>series LEDs (eugh, but I can't think how better to put it).
>
>BTW, do you need to use zener diodes / transorbs with aircraft power
>systems as you do in cars, to avoid problems with voltage spikes?
>
>AC
>
>> making it even more difficult to rig the system into a 1/4" piece of plexiglass.

anonymous coward

May 12th 04, 08:00 AM

On Tue, 11 May 2004 17:04:22 +0000, Ernest Christley wrote:

> anonymous coward wrote:
>
>> BTW, do you need to use zener diodes / transorbs with aircraft power
>> systems as you do in cars, to avoid problems with voltage spikes?
>>
>> AC
>>
>
> In one form or another, I'd say the answer is a qualified yes. Again, I
> didn't add any sort of regulation to the LED array. As I remember it,
> and it has been quite a while since I looked at it, the larger LEDs can
> absorb rather large transients themselves. Their construction is not
> that far removed from zeners or transorbs after all.

Do you know where I might find any links to material about this? I'm
building a computer controlled LED flasher device (not for a tail strobe
- something unrelated) and naturally I would like to make the flashes as
bright as possible. Luxeon reckon that when pulse-width modulating their
LEDs, the current should never exceed 500-550 mA (for the 1W versions).
Given that their normal current is only 350 mA this isn't much of an
increase. Also, this is meant to be at pulse-width-modulation frequencies
of 100Hz or more. The pulse duration I need is > 1/10 of a second.

On the other hand, your experience echoes my experience. My Luxeons aren't
heat-sunk, yet due to software faults I've unintentionally passed 1.5A
through some of them for several seconds. I won't be surprised if their
life expectancy is greatly reduced but they have lasted the 'development'
phase of my device surprisingly well.

> As I understand it, the limiting factor of the LED's ability to suck
> down transients is their ability to dump the internal heat is the reason
> why 'overdriving' them with higher but pulsating current works.

What worries me is how quickly the die can dump heat to the aluminium
casing.

AC

Ernest Christley

May 12th 04, 04:24 PM

anonymous coward wrote:
> On Tue, 11 May 2004 17:04:22 +0000, Ernest Christley wrote:

>>In one form or another, I'd say the answer is a qualified yes. Again, I
>>didn't add any sort of regulation to the LED array. As I remember it,
>>and it has been quite a while since I looked at it, the larger LEDs can
>>absorb rather large transients themselves. Their construction is not
>>that far removed from zeners or transorbs after all.
>
>
> Do you know where I might find any links to material about this? I'm
> building a computer controlled LED flasher device (not for a tail strobe
> - something unrelated) and naturally I would like to make the flashes as
> bright as possible. Luxeon reckon that when pulse-width modulating their
> LEDs, the current should never exceed 500-550 mA (for the 1W versions).
> Given that their normal current is only 350 mA this isn't much of an
> increase. Also, this is meant to be at pulse-width-modulation frequencies
> of 100Hz or more. The pulse duration I need is > 1/10 of a second.
>
> On the other hand, your experience echoes my experience. My Luxeons aren't
> heat-sunk, yet due to software faults I've unintentionally passed 1.5A
> through some of them for several seconds. I won't be surprised if their
> life expectancy is greatly reduced but they have lasted the 'development'
> phase of my device surprisingly well.
>

Superbrightled.com (I think) has some of this info. The rest is just
general knowledge of semi-conductors. If Luxeon gave you a Imax for
pulsed current, they should have also given you a pulse duration and
duty cycle. That is, how long and how often the current flows. Again,
the limiting factor is how quickly can you get the heat from the CENTER
of the device. The surface temperature is really only a side effect.
It's the temperature of that little piece of doped glass in the center
of that large chunk of plastic that is critical, and it can melt while
the exterior is still cool. So you can briefly drive a lot of current,
but then you have to stop and let the casing suck the heat out.

>
>>As I understand it, the limiting factor of the LED's ability to suck
>>down transients is their ability to dump the internal heat is the reason
>>why 'overdriving' them with higher but pulsating current works.
>
>
> What worries me is how quickly the die can dump heat to the aluminium
> casing.
>
> AC

You are well informed to be worried. All of these PC overclockers
thinking they can crank up the juice if they just add a bigger heat sink
are fooling themselves. The heat energy has a somewhat tortuous path to
traverse through the IC packaging before it can even be transferred to
the heatsink. To be most effective, a heat sink has to be as 'close' to
the heat source as possible. Close being defined as the least amount of
insulation between the two, ie. a 1/2" gap filled with copper plate is
probably better than a 1/10" air gap. The base of the LED is NOT the
heat source. (Though the leads are some nice metal heat conductors
connected directly to the glass in the middle 8*)

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

Jay

May 12th 04, 08:48 PM

From what I've read, the bond wire that connects the semiconductor die
to the leadframe (the part you solder) can be the limiting factor on
pulse current. It has a lower thermal time constant that the chip
itself. You can smoke that little gold wire if you try to shove too
many e-'s through it at once.

I'm not sure what you're doing with your alternative application, but
in general, you pulse them to multiplex them for moving signs or to
lower the observed brightness (via PWM) for brightness controls.

anonymous coward > wrote in message >...
> On Tue, 11 May 2004 17:04:22 +0000, Ernest Christley wrote:
>
> > anonymous coward wrote:
> >
> >> BTW, do you need to use zener diodes / transorbs with aircraft power
> >> systems as you do in cars, to avoid problems with voltage spikes?
> >>
> >> AC
> >>
> >
> > In one form or another, I'd say the answer is a qualified yes. Again, I
> > didn't add any sort of regulation to the LED array. As I remember it,
> > and it has been quite a while since I looked at it, the larger LEDs can
> > absorb rather large transients themselves. Their construction is not
> > that far removed from zeners or transorbs after all.
>
> Do you know where I might find any links to material about this? I'm
> building a computer controlled LED flasher device (not for a tail strobe
> - something unrelated) and naturally I would like to make the flashes as
> bright as possible. Luxeon reckon that when pulse-width modulating their
> LEDs, the current should never exceed 500-550 mA (for the 1W versions).
> Given that their normal current is only 350 mA this isn't much of an
> increase. Also, this is meant to be at pulse-width-modulation frequencies
> of 100Hz or more. The pulse duration I need is > 1/10 of a second.
>
> On the other hand, your experience echoes my experience. My Luxeons aren't
> heat-sunk, yet due to software faults I've unintentionally passed 1.5A
> through some of them for several seconds. I won't be surprised if their
> life expectancy is greatly reduced but they have lasted the 'development'
> phase of my device surprisingly well.
>
> > As I understand it, the limiting factor of the LED's ability to suck
> > down transients is their ability to dump the internal heat is the reason
> > why 'overdriving' them with higher but pulsating current works.
>
> What worries me is how quickly the die can dump heat to the aluminium
> casing.
>
> AC

Blueskies

May 13th 04, 12:54 AM

Luxeon has a patented mechanical design that allows them to dissipate the power better, thus getting more lumens. It is
not like the superbrights...

See: http://www.lumileds.com/pdfs/protected/AB23.PDF

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

..
"Ernest Christley" > wrote in message om...
> anonymous coward wrote:
> > On Tue, 11 May 2004 17:04:22 +0000, Ernest Christley wrote:
>
> >>In one form or another, I'd say the answer is a qualified yes. Again, I
> >>didn't add any sort of regulation to the LED array. As I remember it,
> >>and it has been quite a while since I looked at it, the larger LEDs can
> >>absorb rather large transients themselves. Their construction is not
> >>that far removed from zeners or transorbs after all.
> >
> >
> > Do you know where I might find any links to material about this? I'm
> > building a computer controlled LED flasher device (not for a tail strobe
> > - something unrelated) and naturally I would like to make the flashes as
> > bright as possible. Luxeon reckon that when pulse-width modulating their
> > LEDs, the current should never exceed 500-550 mA (for the 1W versions).
> > Given that their normal current is only 350 mA this isn't much of an
> > increase. Also, this is meant to be at pulse-width-modulation frequencies
> > of 100Hz or more. The pulse duration I need is > 1/10 of a second.
> >
> > On the other hand, your experience echoes my experience. My Luxeons aren't
> > heat-sunk, yet due to software faults I've unintentionally passed 1.5A
> > through some of them for several seconds. I won't be surprised if their
> > life expectancy is greatly reduced but they have lasted the 'development'
> > phase of my device surprisingly well.
> >
>
> Superbrightled.com (I think) has some of this info. The rest is just
> general knowledge of semi-conductors. If Luxeon gave you a Imax for
> pulsed current, they should have also given you a pulse duration and
> duty cycle. That is, how long and how often the current flows. Again,
> the limiting factor is how quickly can you get the heat from the CENTER
> of the device. The surface temperature is really only a side effect.
> It's the temperature of that little piece of doped glass in the center
> of that large chunk of plastic that is critical, and it can melt while
> the exterior is still cool. So you can briefly drive a lot of current,
> but then you have to stop and let the casing suck the heat out.
>
> >
> >>As I understand it, the limiting factor of the LED's ability to suck
> >>down transients is their ability to dump the internal heat is the reason
> >>why 'overdriving' them with higher but pulsating current works.
> >
> >
> > What worries me is how quickly the die can dump heat to the aluminium
> > casing.
> >
> > AC
>
> You are well informed to be worried. All of these PC overclockers
> thinking they can crank up the juice if they just add a bigger heat sink
> are fooling themselves. The heat energy has a somewhat tortuous path to
> traverse through the IC packaging before it can even be transferred to
> the heatsink. To be most effective, a heat sink has to be as 'close' to
> the heat source as possible. Close being defined as the least amount of
> insulation between the two, ie. a 1/2" gap filled with copper plate is
> probably better than a 1/10" air gap. The base of the LED is NOT the
> heat source. (Though the leads are some nice metal heat conductors
> connected directly to the glass in the middle 8*)
>
> --
> http://www.ernest.isa-geek.org/
> "Ignorance is mankinds normal state,
> alleviated by information and experience."
> Veeduber

Ernest Christley

May 15th 04, 04:52 AM

Jay wrote:
> From what I've read, the bond wire that connects the semiconductor die
> to the leadframe (the part you solder) can be the limiting factor on
> pulse current. It has a lower thermal time constant that the chip
> itself. You can smoke that little gold wire if you try to shove too
> many e-'s through it at once.

Yeah. What he said.

>
> I'm not sure what you're doing with your alternative application, but
> in general, you pulse them to multiplex them for moving signs or to
> lower the observed brightness (via PWM) for brightness controls.
>
>

You can also overrate them and pulse higher current for short time
periods. As you noted above, this technique definitely has its limit.
The human eye is a peak detection device for time constants greater than
about 50ms (it'll average anything shown less time than that). So, you
can drive more current, then turn it off at the right rate to get a
brighter LED without burn it up. Again, this technique has its limits.

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

anonymous coward

May 15th 04, 03:13 PM

On Sat, 15 May 2004 03:52:20 +0000, Ernest Christley wrote:

> Jay wrote:
>> From what I've read, the bond wire that connects the semiconductor die
>> to the leadframe (the part you solder) can be the limiting factor on
>> pulse current. It has a lower thermal time constant that the chip
>> itself. You can smoke that little gold wire if you try to shove too
>> many e-'s through it at once.
>
> Yeah. What he said.
>
>>
>> I'm not sure what you're doing with your alternative application, but
>> in general, you pulse them to multiplex them for moving signs or to
>> lower the observed brightness (via PWM) for brightness controls.
>>
> You can also overrate them and pulse higher current for short time
> periods. As you noted above, this technique definitely has its limit.
> The human eye is a peak detection device for time constants greater than
> about 50ms (it'll average anything shown less time than that). So, you
> can drive more current, then turn it off at the right rate to get a
> brighter LED without burn it up. Again, this technique has its limits.

FWIW I heard that in the olden days, LEDs worked most efficiently at
currents that were high enough to burn them out if these currents
were applied continuously. But it meant that if you drove an LED at 1A
for 1/20 duty cycle, you could get more light out than if you put 1/20 amp
in continuously (hypothetical numbers). I gather that newer LEDs tend to
work most efficiently with a steady drive current.

I assume that for a strobe application it's important to pack the
maximum power into a pulse that's equal to or shorter than the eye's
integration time. I have to say I don't know what that integration time is
- you can certainly see 20hz flicker at photopic levels, but at scotopic
(night-vision) levels the frequency can be much lower.

On the other hand, given that we can see that Mars is red (only photopic
vision lets us see colour) at night when adapted to scotopic light
levels I assume that we use photopic vision to sense small point sources
such as stars or strobe lights - perhaps someone out there actually knows
what the optimal pulse duration is?

AC